Physics 334 Modern Physics Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on the.
Download ReportTranscript Physics 334 Modern Physics Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on the.
Slide 1
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 2
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 3
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 4
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 5
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 6
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 7
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 8
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 9
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 10
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 11
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 12
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 13
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 14
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 15
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 16
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 17
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 18
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 19
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 20
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 21
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 22
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 23
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 24
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 25
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 26
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 27
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 28
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 29
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 30
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 31
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 32
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 33
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 34
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 35
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 36
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 37
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 38
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 39
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 40
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 41
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 42
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 43
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 44
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 45
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 2
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 3
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 4
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 5
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 6
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 7
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 8
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 9
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 10
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 11
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 12
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 13
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 14
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 15
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 16
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 17
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 18
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 19
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 20
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 21
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 22
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 23
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 24
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 25
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 26
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 27
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 28
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 29
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 30
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 31
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 32
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 33
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 34
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 35
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 36
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 37
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 38
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 39
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 40
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 41
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 42
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 43
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 44
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is
Slide 45
Physics 334
Modern Physics
Credits: Material for this PowerPoint was adopted from Rick Trebino’s lectures from Georgia Tech which were based on
the textbook “Modern Physics” by Thornton and Rex. Many of the images have been used also from “Modern Physics” by
Tipler and Llewellyn, others from a variety of sources (PowerPoint clip art, Wikipedia encyclopedia etc), and contributions
are noted wherever possible in the PowerPoint file. The PDF handouts are intended for my Modern Physics class, as a
study aid only.
CHAPTER 5
The Wavelike Properties of Particles
The de Broglie Hypothesis
Measurement of Particle wavelength
Wave Motion and Wave Packets
The Probabilistic Interpretation of the
Wave Function
The Uncertainty Principle
Some Consequences of the
Uncertainty Principle
Louis de Broglie
(1892-1987)
I thus arrived at the overall concept which guided my studies: for both matter
and radiations, light in particular, it is necessary to introduce the corpuscle
concept and the wave concept at the same time.
- Louis de Broglie, 1929
De Broglie Waves
In his thesis in 1923, Prince Louis V.
de Broglie suggested that mass
particles should have wave properties
similar to electromagnetic radiation.
The energy can be written as:
If a light-wave could also act
like a particle, why shouldn’t
matter-particles also act like
waves?
hf = pc = plf
Thus the wavelength of a matter
wave is called the de Broglie
wavelength:
Louis V. de Broglie
(1892-1987)
Bohr’s Quantization Condition revisited
One of Bohr’s assumptions in his hydrogen atom model was that the
angular momentum of the electron in a stationary state is nħ.
This turns out to be equivalent to
saying that the electron’s orbit
consists of an integral number
of electron de Broglie
wavelengths:
electron
de Broglie
wavelength
Circumference
Bohr’s Quantization Condition revisited
Exercise1: What is the de Broglie wavelength for a very small but
microscopic object of mass 10-9 g moving with speed 3x10-8 m/s?
Exercise2: What is the de Broglie wavelength of a 10-eV electron?
Measurement of Particle Wavelength
First measurement of the wavelength of electrons were made by
Davisson and Germer in 1927
Electrons scattered at an angle φ from a nickel crystal are detected in
an ionization chamber
12.26
l
1.67 A
54
Measurement of Particle Wavelength
Exercise 3: Calculate the wavelength of scattered electrons by using
Bragg diffraction condition.
Another Demonstration
G. P. Thomson and son J. J. Thomson share the Nobel prize in 1937
with Davisson.
Diffraction of x-rays to produce Laue patterns
a) Diffraction
from Al target
b) Diffraction
pattern for xrays of Al
c) Diffraction
pattern for
electron of Al.
fig. a and b
show
similarities
θ satisfies the Bragg condition. 2θ is the scattered beam
Diffraction of Other Particles
O. Stern and I. Estermann 1930 demonstrated diffraction of beams of
helium atoms and hydrogen molecules from lithium fluoride crystals
a) Diffraction of
He atom from
LiF crystal
b) Experimental
setup
c) Intensity as a
function of
detector angle
Diffraction of Other Particles
Top left: Diffraction pattern
produced by neutrons
Top right: Laue pattern of NaCl
produced by neutrons
Bottom right: Diffraction pattern
produced by protons from
oxygen nuclei
Wave Motion and Wave Packets
In Ocean it is the water that “waves”, for sound it is the air molecules,
for light it is the E and the B field. So for matter “What is waving?”
De Broglie matter waves should be described in the same manner as
light waves (no ether). The matter wave should be a solution to a wave
equation like the one for electromagnetic waves:
In order to understand matter waves we need to review properties of
classical waves first
Exercise 4: Consider a stretched string displaced from its normal
position, derive the wave equation and explore the possible solutions
to this wave equation.
What is a wave?
A wave is anything that moves. A
disturbance or a pulse.
To displace any function f(x) to the
right, just change its argument from
x to x-a, where a is a positive
number.
If we let a = v t, where v is positive
and t is time, then the displacement
will increase with time.
f(x)
f(x-2)
f(x-1)
f(x-3)
So f(x - v t) represents a rightward,
or forward, propagating wave.
Similarly, f(x + v t) represents a
leftward, or backward, propagating
wave.
v will be the velocity of the wave.
0
1
2
3
x
The one-dimensional wave equation
In the previous exercise we derived the wave equation. Here it is in
its one-dimensional form for scalar (i.e., non-vector) functions, f:
2f
2
x
1 2f
2
2
v t
0
Matter waves will be a solution to this equation. And v will be the
velocity of de Broglie waves.
The solution to the one-dimensional
wave equation
The wave equation has the simple solution:
f ( x, t ) f ( x vt )
where f (u) can be any twice-differentiable function.
Solution of a Wave equation
Exercise 5: Show that the 1-D wave equation has a simple solution
of the form f(x,t)=f(x±t), where f(u) can be any twice differentiable
function
Proof that f (x ± vt) solves the wave equation
Write f (x ± vt) as f (u), where u = x ± vt. So u 1 and
x
f f u
x u x
Now, use the chain rule:
2
f f
f 2f
So
2
x u
x
u2
f f u
t u t
2
2f
f
f
f
2
v
v
and
t
u
t2
u2
Substituting into the wave equation:
2f 1 2f
2
2
x
v t2
u
v
t
2f
1 22f
2 v
0
2
2
u
v u
Definitions: Amplitude and Absolute phase
y(x,t) = A cos[(k x – wt ) – q]
A = Amplitude
q = Absolute phase (or initial phase)
p
Definitions
Spatial quantities:
Wave number=
number of radians
Corresponding t a
wave train 1m long
Temporal quantities:
ν=f
The Phase Velocity
How fast is the wave traveling?
Velocity is a reference distance
divided by a reference time.
The phase velocity is the wavelength / period: v = l / t
Since f = 1/t:
v = lf
In terms of the k-vector, k = 2p/ l, and
the angular frequency, w = 2p/ t, this is:
v =w/k
Human wave
A typical human wave has a phase velocity of about 20 seats per
second.
Complex numbers
Consider a point,
P = (x,y), on a 2D
Cartesian grid.
Let the x-coordinate be the real part
and the y-coordinate the imaginary part
of a complex number.
So, instead of using an ordered pair, (x,y), we write:
P = x+iy
= A cos(j) + i A sin(j)
where i = (-1)1/2
Euler's Formula
exp (ij) = cos(j) + i sin(j)
so the point, P = A cos(j) + i A sin(j), can be written:
P = A exp(ij)
where
A = Amplitude
j
= Phase
Proof of Euler's Formula exp(ij) = cos(j) + i sin(j)
Use Taylor Series:
x
x2
x3
f ( x) f (0) f '(0)
f ''(0)
f '''(0) ...
1!
2!
3!
x x 2 x3 x 4
exp( x) 1 ...
1! 2! 3! 4!
x 2 x 4 x 6 x8
cos( x) 1 ...
2! 4! 6! 8!
x x3 x5 x 7 x9
sin( x) ...
1! 3! 5! 7! 9!
If we substitute x = ij
into exp(x), then:
ij j 2 ij 3 j 4
exp(ij ) 1
...
1! 2! 3! 4!
j2 j4
j j 3
1
... i ...
2! 4!
1! 3!
cos(j ) i sin(j )
Complex number theorems
If exp(ij ) cos(j ) i sin(j )
exp(ip ) 1
exp(ip / 2) i
exp(-ij ) cos(j ) i sin(j )
1
cos(j ) exp(ij ) exp( ij )
2
1
sin(j ) exp(ij ) exp( ij )
2i
A1exp(ij1 ) A2 exp(ij 2 ) A1 A2 exp i (j1 j 2 )
A1exp(ij1 ) / A2 exp(ij 2 ) A1 / A2 exp i (j1 j 2 )
More complex number theorems
Any complex number, z, can be written:
z = Re{ z } + i Im{ z }
So
Re{ z } = 1/2 ( z + z* )
and
Im{ z } = 1/2i ( z – z* )
where z* is the complex conjugate of z ( i –i )
The "magnitude," | z |, of a complex number is:
| z |2 = z z* = Re{ z }2 + Im{ z }2
To convert z into polar form, A exp(ij):
A2 = Re{ z }2 + Im{ z }2
tan(j) = Im{ z } / Re{ z }
We can also differentiate exp(ikx) as if
the argument were real.
d
exp(ikx) ik exp(ikx)
dx
Proof :
d
cos(kx) i sin(kx) k sin(kx) ik cos( kx)
dx
1
ik sin( kx) cos(kx)
i
But 1/ i i, so :
ik i sin( kx) cos( kx)
De Broglie Velocity
Exercise 6: What is the phase velocity of de Broglie waves? Justify your
result.
Let vp=fλ be the de Broglie phase velocity, then the de Broglie wave
function can be written as
0
y( x, t ) y cos(wt kx)
where k is a propagating constant
The amplitude of the de Broglie wave
that correspond to a moving body
reflects the probability that it may be
found at a particular place at a
particular time.
The above equation is an indefinite
series with same amplitude yo
We expect that matter waves travel
with different amplitudes in a wave
packet or a wave group
De Broglie Group Velocity
Example of waves that can travel in a group is Beats
Beats are two waves of same amplitude, but different frequencies that
are produces simultaneously.
Tuning forks of 440 Hz and 442 Hz sounded together produces two
loudness peaks per second.
How to find de Broglie group velocity?
dv p
dw
vg
v p k
dk
dk
Classical Uncertainty Relations
Figure b - ∆x~1/12
Figure c - ∆k~4π~12
k x 1
w t 1
Classical Uncertainty Relations
Example: Standing in the middle of a 20-m long pier, you notice that at
any given instant there are 15 wave crest between the two ends of the
pier. Estimate the minimum uncertainty in the wavelength that could be
computed from this information.
Particle Wave Packets
Exercise 7: Show that for an electron the phase velocity of the wave is
half its particle velocity and that the group velocity is the same as its
particle velocity.
Probabilistic Interpretation of the Wave
Function
C. Jönsson of Tübingen,
Germany, succeeded in 1961
in showing double-slit
interference effects for
electrons by constructing very
narrow slits and using
relatively large distances
between the slits and the
observation screen.
This experiment demonstrated
that precisely the same
behavior occurs for both light
(waves) and electrons
(particles).
Probabilistic Interpretation of the Wave
Function
Probability, Wave Functions, and the
Copenhagen Interpretation
Okay, if particles are also waves, what’s waving?
Probability
The wave function determines the likelihood (or probability) of
finding a particle at a particular position in space at a given time:
P( x) ( x)
2
The probability of the
particle being between
x1 and x2 is given by:
x2
( x) dx
2
x1
( x) dx 1
2
The total probability of finding the particle
is 1. Forcing this condition on the wave
function is called normalization.
Uncertainty Principle: Energy
Uncertainty
The energy uncertainty of a wave
packet is:
w
E hf h
w
2p
Combined with the angular
frequency relation we derived
earlier:
wt
Energy-Time Uncertainty
Principle:
.
E
1
t
2
The factor ½ accounts for
the fact that in Gaussian
distribution
the product of the
uncertainty are 1/2
Momentum Uncertainty Principle
The same mathematics relates x and k:
k x ≥ ½
So it’s also impossible to measure simultaneously the precise values
of k and x for a wave.
Now the momentum can be written in terms of k:
p
h
l
h
(h / 2p )k
2p / k
So the uncertainty in momentum is:
But multiplying k x ≥ ½ by ħ:
p k
k x
2
And we have Heisenberg’s Uncertainty Principle:
p k
How to think about Uncertainty
The act of making one measurement perturbs the other.
Precisely measuring the time disturbs the energy.
Precisely measuring the position disturbs the momentum.
The Heisenbergmobile. The problem was that when you looked at
the speedometer you got lost.
Consequences of the Uncertainty Principle
Kinetic Energy Minimum: Since we’re always uncertain as to the exact
position, x , of a particle, for example, an electron somewhere
inside an atom, the particle can’t have zero kinetic energy:
p
x
The average of a positive quantity must always exceed its
uncertainty:
pave p
x
so:
K ave
2
2
pave
(p) 2
2m
2m
2m
Particle in a Box
A particle (wave) of mass m is in a one-dimensional
box of width ℓ.
The box puts boundary conditions on the wave. The
wave function must be zero at the walls of the box
and on the outside.
In order for the probability to vanish at the walls, we
must have an integral number of half wavelengths in
the box:
The energy:
2
2
p
h
E K .E. 12 mv 2
2m 2ml 2
The possible wavelengths
are quantized and hence
so are the energies:
Width of Spectral Lines
Exercise 8: Use uncertainty principle to find the spread in spectral
lines of an electron jumping to the ground state in an atom.
Wave-particle duality
It’s somewhat disturbing that everything is both a particle and a wave.
The wave-particle duality is a little less disturbing if we think in terms
of:
Bohr’s Principle of Complementarity: It’s not possible to describe
physical observables simultaneously in terms of both particles and
waves.
When we’re making a measurement, use the particle description, but
when we’re not, use the wave description.
When we’re looking, fundamental quantities are particles; when
we’re not, they’re waves.
Waves or Particles?
Dimming the light in
Young’s two-slit experiment
results in single photons at
the screen. Since photons
are particles, each can only
go through one slit, so, at
such low intensities, their
distribution should become
the single-slit pattern.
Each photon
actually goes
through both
slits!
Can you tell which slit the photon went
through in Young’s double-slit exp’t?
When you block one slit, the one-slit pattern returns.
One-slit
pattern
Two-slit
pattern
At low intensities, Young’s two-slit experiment shows that light
propagates as a wave and is detected as a particle.
Which slit does the
electron go through?
Shine light on the double slit and observe with a microscope. After the
electron passes through one of the slits, light bounces off it; observing
the reflected light, we determine which slit the electron went through.
The photon momentum is:
The electron momentum is:
Need lph < d to
distinguish the slits.
Diffraction is significant only
when the aperture is ~ the
wavelength of the wave.
The momentum of the photons used to determine which slit the electron
went through is enough to strongly modify the momentum of the electron
itself—changing the direction of the electron! The attempt to identify
which slit the electron passes through will in itself change the diffraction
pattern! Electrons also propagate as waves and are detected as
particles.
Probability of the particle vs. position
Note that E0 = 0 is not a
possible energy level.
The concept of energy
levels, as first discussed in
the Bohr model, has
surfaced in a natural way
by using waves.
The probability of
observing the particle
between x and x + dx in
each state is