Transcript 1AMQ, Part II Quantum Mechanics

1AMQ, Part II Quantum Mechanics
4 Lectures
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Wave Particle Duality, the Heisenberg
Uncertainty Principle, de Broglie
(wavelength) hypothesis
Schrodinger Wave Equation.
Wavefunctions and the Free Particle.
Electrons in a box, energy quantization.
•K.Krane, Modern Physics, Chapters 4 and 5
• Eisberg and Resnick, Quantum Physics,
Chapters 3 (and 5 & 6)
1AMQ P.H. Regan
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Wave-Particle Duality.
Light behaves like: (a) a wave (diffraction,
interference) and/or (b) a stream of massless
particles or photons, (black body spectrum,
photoelectric effect, Compton scattering).
Consider diffraction
from a slit, one photon at
a time.
The experiment shows
that individual particles
of light gradually build
up the diffraction pattern
predicted by the classical
wave theory.
The wave pattern
describes the probability
of detecting a photon at
that point.
The wave pattern measures the probability
detecting a photon at that point.
1AMQ P.H. Regan
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Individual photons unpredictable in detail,
BUT their average behaviour is predictable!
At any point P in the interference pattern, the
intensity gives the probability: ie
I(P) is prop. to prob. that a photon hits pos. P.
For waves, the intensity = (amplitude)2 and thus
the wave and particle pictures are connected,
Probability
a (amplitude of wave)2
This is called the Born Approximation (Max
Born, Nobel Prize, 1954).
The question then arises, if waves can act like
particles, is the reverse true ?
The de Broglie Hypothesis
(Nobel Prize, 1929)
Louis de Broglie (for his PhD thesis!) postulated
that in some phenomena, particles can act like
waves!
1AMQ P.H. Regan
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He guess that for particles, the momentum, p,
could be related to a wavelength, l , by
p = h/l where l is the de Broglie wavelength.
Note this is consistent with the momentum for
photons, ie p = E/c = hn/nl = h/l
This predicts, for example, that electrons should
show diffraction effects (like photons), if they
pass through a slit with width ~ ldB.
Calculating the de Broglie Wavelength.
Example.
e-s in a TV Ke-=20 keV, me=9.1x10-31kg, gives
ldB=0.009nm. (Note that the typical atomic
spacing is ~0.05nm).
ldB=0.05nm corresponds to Ke-=601 eV for
electrons. [from Ke=p2/2me=(1/2m) . (h/l)2]
ie. e-s accelerates across 600 eV should be
diffracted exactly like x-rays with l=0.05nm.
1AMQ P.H. Regan
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X-ray diffraction
pattern
Electron diffraction
pattern
1AMQ P.H. Regan
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Note how (2 slit)
e- interference
pattern builds up
with increasing
number of e-s.
Electron and neutron diffraction
are widely used in crystallography
studies.
1AMQ P.H. Regan
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The y-position for any transmitted photon is
specified to an accuracy (uncertainty) D y=a.
Making the slit smaller (ie. reducing D y)
increases the uncertainty in the angle, Dq.
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Thus, DyDpy~ h (for single slit diffraction).
This relates to the physical limits we can put
on determining either the position or
momentum of the individual photons as they
pass through the slit. These are a subset of the
Heisenberg Uncertainty Principle.
In general, if p = momentum, t = time and
E= energy, h = Planck’s constant ,then,
DxDp x  h
DxDp x  h
2
, DxDp x  h
2
, DEDt  h
2
2
The division by 2 comes about due to the
use of wave number (k) rather than
wavelength (l=2/k) in the definition of
momentum (see Krane p114).
1AMQ P.H. Regan
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The Schrodinger Equation.
For particles, forces change momentum (p=mv)
and each force has an associated potential energy.
Newton’s 2nd law related potential to the
change in momentum, via,
dp
d 2x
dV
F=
= ma = m 2 and for 1 - D, F = 
dt
dt
dx
dV
d 2 x ie. given V(x), solve for
thus, 
=m 2
dx
dt x(t) or V(x,t).
For matter waves, p=h/l. Thus, we need a
differential equation to calculate how forces
(described by a potential energy) affect l or p.
We will solve this problem for a wave,
which has an amplitude (y ) for each
value of position (x) and time (t).
1AMQ P.H. Regan
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Mathematical Forms of Waves.
Y( x,t)=Yo cos(kx-wt) , k=2/l and w=2n
We can re-write this formula in terms of e, as
Y(x,t)=Yoei(kxw t) + Yoei(kxwt)
2
Recalling eix = cosx + isinx, the simplest form
of a wave is Y(x,t)=Yoei(kxw t)
Formulation of the Schrodinger Equation
To relate p and V we use,
Total Energy (E) = Kinetic (K) + Potential (V)
1AMQ P.H. Regan
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p2
h
h
E = K +V =
+ V , where p = = k , ( =
)
2m
l
2
Now, if
Y = Y0 e
i ( kx wt )
dY
= ik Y
, then, if t = const ,
dx
d 2Y d
2
Y
k

=
Y
ik
.
ik
=
)
Y
ik
(
=
and
2
dx
dx
2
2
2
p
 d
p = (k ) , so
Y=
Y
2
2m
2m dx
2
2
dY
= iw Y, and if E = hn = w , then,
dt
d
+ i Y = w Y = E Y
dt
Thus, p 2
d
(
+ V )Y = i Y can be written as
2m
dt
2 d 2
d
(+ V )Y = i Y
2
2m dx
dt
This is the Schrodinger equation.
1AMQ P.H. Regan
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Using the Schrodinger Equation.
(i) specify V=V(x,t) (ie details of force)
(ii) specify initial conditions (ie. at t=0) and
(iii) specify boundary conditions (x=fixed values)
1AMQ P.H. Regan
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Time Independent Schrodinger Equation
In many cases, (eg. Coulomb potential in H
atom), V=V(x), ie. not time dependent, then
Y( x, t ) = y ( x)ei w t
where y(x) satisifies
2 d 2
[
+ V ( x)]y = Ey or, Hy = Ey
2
2m dx
Properties of the T.I.S.E.
•Applies when V=V(x), ie. V is independent of
time.Eg. an e- wave acted on by a fixed nucleus,
but NOT an atom in an oscillating magnetic field.
•Applies only in the non-relativistic limit
(ie. assumes K=(1/2)mv2=p2/2m).
•Assumes p=h/l and E=hn.
•Is linear in y, ie no terms like y2, y3 etc. Laws
of superposition are valid. If ya and yb are
two possible solutions, then y=caya+cbyb
is a solution for all constants ca and cb.
1AMQ P.H. Regan
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Wavefunctions
We have 2 forms of Schrodinger equation,
(a) general, time-dependent solve for Y (x,t),
when V(x,t) is given, and
(b) TISE, for V=V(x) only. Solve for y(x) when
V(x) is given. Solution is a standing wave,
ie. Y (x,t)=y(x) e-iw t
Wavefunctions are solutions to the
Schrodinger wave equation. The w.function,
Y (x,t) describes physical state of the
particle, such as its momentum, energy etc.
and also where the particle (probability
wise) is.
1AMQ P.H. Regan
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The Born Approximation
As in the earlier electron and photon diffraction
examples, the wave intensity is intepreted as
giving the probability of finding the particle in a
particular state. Recalling that the
Intensity =|Amplitude|2,
we interpret |Y (x,t)|2dx =probability the particle
will be found between x and x+dx at time t.
Solving the Schrodinger equation specifies
Y (x,t) completely, except for a constant,
ie. if Y is a solution, so is A xY .
From the Born approx. the normalisation
condition of the wavefunction is that

 | Y ( x, t ) |

2
dx = 1
ie. The probability
that particle is
somewhere = 1
Note that probability is a real number.
AlthoughY is complex, |Y 2| is real.
1AMQ P.H. Regan
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Solution of the Wave Equation
for a Free Particle
`Free particle’, means V=0 everywhere, such as
an electron in the absence of external forces.
Using the TISE,
 2 d 2y
d 2y
2mE
Hy = 
= Ey , ie. 2 =  2 y
2
2m dx
dx

Try a solution of the form, y=A eax,
2
d ( Ae ax )
d
y
2mE
ax
2
= a ( Ae ) ,
=ay = 2 y
2
dx
dx

2mE
2mE
ie. a = i
, thus if, k = 
,
2


y ( x) = Aeikx , which is a solution t o the TISE
The energy is given by Hy = Ey, thus
(k )
E=
2m
2
Since k (the wavenumber) is
real, we have shown that any
positive value of energy is
allowed for a free particle.
1AMQ P.H. Regan
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Particles (electrons) in a Box: Solving
the Wavefunction for Bound Electrons.
When electrons are acted on by a force, this
acts to restrict the allowed energies (and hence
other quantities eg. momentum, spin..) This
force is specified by the potential, V(x,t).
Example: Electrons in a 1-D Box
Suppose an e- (in 1-d) is placed
in a `box’ with impenetrable
walls at x=0 and x=L. The
electron can move freely (in x
direction) in the box, such that:
V=0
V = 0 0  x  L and
V   x  0 and  L
x=0
x=L
Note if V tends to infinity, the prob. of finding
the e- there must tend to zero.ie y>0 outside box.
There are also boundary conditions that
y(x) = 0 for x=0 and x=L, for all times, t
1AMQ P.H. Regan
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Outside the box, y=0, now we solve for inside.
Since inside the box,
 2 d 2y
= Ey
V=0 for all t, we can Hy = 
2
2m dx
use the T.I.S.E, ie.
For a free particle, we saw that Aei( k x) was
a solution for any value of k and that
(k ) 2
E=
2m
For a particle in a box, we consider the most
general solution for a given energy, and then
apply the boundary conditions.
General solution is of the form
y ( x) = A1e + A2 e
ikx
ikx
, with k =
2mE

Thus we can write,
y ( x) = ( A1 + A2 ) cos kx + i ( A1  A2 ) sin kx
= A sin kx + B cos kx, where A, B are consts.
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One of the boundary conditions is y(0)=0, thus,
B=0 and hence, y(x)=Asin(kx).
We also have y(L)=0, so 0=Asin(kl), hence,
either, A=0 or sin(kL)=0.
If A=0, the wavefunction =0 everywhere! (ie.
there is no probability of the a particle having
this wavefunction!(solution can be disregarded).
nx
y ( x) = A sin(
)
L
Hence, kL = n,
where n=1,2,3….,
ie k=n/L and thus
Determine A using the normalisation condition,

nx
2 L
|y ( x) | dx = 1 =| A | 0 sin ( L )dx =| A | 2
L
2
Thus
y ( x) =
2
2
nx
sin(
) and
L
L
2
nx iwt
Y ( x, t ) =
sin(
)e
L
L
2
BCs mean that only
certain modes are
allowed for Y(x,t).
A plot of the real
(or imag.) part
of Y(x,t) looks like
standing waves.
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Energy Quantization
For particles (eg, electrons) in box, the
boundary conditions mean that only certain
modes are allowed for the wavefunction, ie.
2
nx iwt
Y( x, t ) =
sin(
)e
L
L

which arises from
kL
=
n
from the restriction,
But the particle energy is given by,
2
2
p
p
(k )
E=
+V =
=
2m
2m
2m
2
The restriction on k, means only certain energies
are allowed, ie. the energy is quantized.
Thus, by simple substitution, the energy levels
are given by,
n 2
(
)
2
2 2
(k )
h
n
L
En =
=
=
, with n = 1,2,3...
2
2m
2m
8mL
1AMQ P.H. Regan
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Note the ZPO is consistent with the
uncertainty principle, ie. particles with zero
momentum can not be localised.
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