Solutions to Schrodinger`s Equation
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Transcript Solutions to Schrodinger`s Equation
Back To Solutions of
Schrödinger's equa.
Particle in a Box
Particle in a Box
E4
Energy
E3=9E1
E2=4E1
E1
E=0
Particle in a Box (Quantization
of Momentum)
Using the momentum operator we can
determine the avg momentum.
Particle in a Box (Quantization
of Momentum)
Using the momentum operator we can
determine the avg momentum.
P Pdx n (
)n dx
i x
0
L
*
*
Particle in a Box (Quantization
of Momentum)
Using the momentum operator we can
determine the avg momentum.
P Pdx n (
) n dx
i x
0
L
*
*
L
0
2
L
sin
nx
L
d
i dx
2
L
sin
nx
L
dx
Particle in a Box (Quantization
of Momentum)
2n
sin
2
iL 0
L
nx
L
cos nLx dx
Particle in a Box (Quantization
of Momentum)
2n
sin
2
iL 0
L
nx
L
cos nLx dx
Integrating by parts we get
2n L
2
P
sin
2
iL 2n
L
nx
L
0
Particle in a Box (Quantization
of Momentum)
2n
sin
2
iL 0
L
nx
L
cos nLx dx
Integrating by parts we get
2n L
2
P
sin
2
iL 2n
L
0
nx
L
0
Particle in a Box (Quantization
of Momentum)
2n
sin
2
iL 0
L
nx
L
cos nLx dx
Integrating by parts we get
2n L
2
P
sin
2
iL 2n
L
0
nx
L
Hence avg. momentum =0
0
Particle in a Box (Finite Square
Well)
The infinite potential is an
oversimplification which can never be
realised.
Particle in a Box (Finite Square
Well)
The infinite potential is an
oversimplification which can never be
realised.
A more realistic finite square well is
shown
U
E
0
L
Particle in a Box (Finite Square
Well)
The infinite potential is an
oversimplification which can never be
realised.
A more realistic finite square well is
shown
U
Regions of interest
are shown (1-3)
E
1
2
0
3
L
Particle in a Box (Finite Square
Well)
Practical, given enough energy a particle
can escape any well.
Particle in a Box (Finite Square
Well)
Practically, given enough energy a
particle can escape any well.
A classical particle with E>U can enter
the well where it moves freely with a
reduced energy (E-U).
Particle in a Box (Finite Square
Well)
Practically, given enough energy a
particle can escape any well.
A classical particle with E>U can enter
the well where it moves freely with a
reduced energy (E-U).
A classical particle with E<U is trapped
within the well (can not escape).
Particle in a Box (Finite Square
Well)
Practically, given enough energy a
particle can escape any well.
A classical particle with E>U can enter
the well where it moves freely with a
reduced energy (E-U).
A classical particle with E<U is trapped
within the well (can not escape).
Particle in a Box (Finite Square
Well)
In QM there is a probability of it being
outside the well.
Particle in a Box (Finite Square
Well)
In QM there is a probability of it being
outside the well. That is the waveform is
nonzero outside the region 0<x<L.
Particle in a Box (Finite Square
Well)
In QM there is a probability of it being
outside the well. That is the waveform is
nonzero outside the region 0<x<L.
For stationary states (x) is found from
the time independent Schrödinger’s
equation.
Particle in a Box (Finite Square
Well)
In QM there is a probability of it being
outside the well. That is the waveform is
nonzero outside the region 0<x<L.
For stationary states (x) is found from
the time independent Schrödinger’s
equation.
2 d 2
U E
2
2m dx
Particle in a Box (Finite Square
Well)
In QM there is a probability of it being
outside the well. That is the waveform is
nonzero outside the region 0<x<L.
For stationary states (x) is found from
the time independent Schrödinger’s
equation.
2 d 2
d 2
U E
2
2
2m dx
dx
2 m U E
2
0
Particle in a Box (Finite Square
Well)
Therefore outside the well where
2
d
U(x)=U, 2 2 0
dx
Particle in a Box (Finite Square
Well)
Therefore outside the well where
2
d
U(x)=U, 2 2 0
dx
2mU E
where
2
2
Particle in a Box (Finite Square
Well)
Therefore outside the well where
2
d
U(x)=U, 2 2 0
dx
2mU E
where
2
2
Since U>E, this term is positive.
Particle in a Box (Finite Square
Well)
Therefore outside the well where
2
d
U(x)=U, 2 2 0
dx
2mU E
where
2
2
Since U>E, this term is positive.
x
ns
Independent sol to the differential are e ,
e
x
Particle in a Box (Finite Square
Well)
To keep the waveform finite as x
and x
Particle in a Box (Finite Square
Well)
To keep the waveform finite as x
and x
Therefore the exterior wave takes the
form
( x) Aex , x 0
( x) Be x , x L
Particle in a Box (Finite Square
Well)
To keep the waveform finite as x
and x
Therefore the exterior wave takes the
form
( x) Aex , x 0
( x) Be x , x L
The internal wave is given as before by
( x) C cos kx D sin kx
Particle in a Box (Finite Square
Well)
The coefficients are determined by
matching the exterior wave smoothly
onto the wavefunction for the well
interior.
Particle in a Box (Finite Square
Well)
The coefficients are determined by
matching the exterior wave smoothly
onto the wavefunction for the well
interior.
That is, (x) and d dx are continuous at
the boundaries.
Particle in a Box (Finite Square
Well)
The coefficients are determined by
matching the exterior wave smoothly
onto the wavefunction for the well
interior.
That is, (x) and d dx are continuous at
the boundaries.
This is obtained for certain values of E.
Particle in a Box (Finite Square
Well)
Because is nonzero at the boundaries,
the de Broglie wavelength is increase
and hence lowers the energy and
momentum of the particle.
Particle in a Box (Finite Square
Well)
The solution for the finite well is
n 2 2 2
En
2
2mL 2
Particle in a Box (Finite Square
Well)
The solution for the finite well is
n 2 2 2
En
2
2mL 2
As long as is small compared to L.
Particle in a Box (Finite Square
Well)
The solution for the finite well is
n 2 2 2
En
2
2mL 2
As long as is small compared to L.
1
where
2mU E
Particle in a Box (Finite Square
Well)
The solution for the finite well is
n 2 2 2
En
2
2mL 2
As long as is small compared to L.
1
where
2mU E
The approximation only works for bounds
states. And is best for the lowest lying states.
Particle in a Box (Finite Square
Well)
The Quantum Oscillator
We now look at the final potential well for
which exact results can be obtained.
The Quantum Oscillator
We now look at the final potential well for
which exact results can be obtained.
We consider a particle acted on by a
linear restoring force F kx .
The Quantum Oscillator
We now look at the final potential well for
which exact results can be obtained.
We consider a particle acted on by a
linear restoring force F kx .
-A
X=0
A
The Quantum Oscillator
We now look at the final potential well for
which exact results can be obtained.
We consider a particle acted on by a
linear restoring force F kx .
x rep. its displacement from stable
equilibrium(x=0).
The Quantum Oscillator
We now look at the final potential well for
which exact results can be obtained.
We consider a particle acted on by a linear
restoring force F kx .
x rep. its displacement from stable
equilibrium(x=0).
Strictly F kx applies to any object limited
to small excursions about equilibrium.
The Quantum Oscillator
The motion of a classical oscillator with
mass m is SHM at frequency k m
The Quantum Oscillator
The motion of a classical oscillator with
mass m is SHM at frequency k m
If the particle is displaced so that it
oscillated between x=A and x=-A, with
2
1
E
kA
total energy
the particle can be
2
given any (nonnegative) energy
including zero.
The Quantum Oscillator
The quantum oscillator is described by
introducing the potential energy U ( x) 12 kx2
into Schrödinger's equation.
The Quantum Oscillator
The quantum oscillator is described by
introducing the potential energy U ( x) 12 kx2
into Schrödinger's equation.
2
2
d
1 2
So that,
2 kx E
2
2m dx
The Quantum Oscillator
The quantum oscillator is described by
introducing the potential energy U ( x) 12 kx2
into Schrödinger's equation.
2
2
d
1 2
So that,
2 kx E
2
2m dx
d 2 2m
2
2
dx
1
2
kx2 E 0
The Quantum Oscillator
The quantum oscillator is described by
introducing the potential energy U ( x) 12 kx2
into Schrödinger's equation.
2
2
d
1 2
So that,
2 kx E
2
2m dx
d 2 2m
2
2
dx
d 2 2m
2
2
dx
1
2
1
2
kx2 E 0
m 2 x 2 E 0
The Quantum Oscillator
We can consider properties which our
wavefunction must and can’t have.
The Quantum Oscillator
1.
We can consider properties which our
wavefunction must and can’t have.
Exponential and trig forms will not work
2
because of the x in the equation.
The Quantum Oscillator
1.
2.
We can consider properties which our
wavefunction must and can’t have.
Exponential and trig forms will not work
2
because of the x in the equation.
should be symmetric about x 0 .
The Quantum Oscillator
1.
2.
3.
We can consider properties which our
wavefunction must and can’t have.
Exponential and trig forms will not work
2
because of the x in the equation.
should be symmetric about x 0 .
should be nodeless, but approach
zero for x large.
The Quantum Oscillator
Condition (2) requires that the
2
waveform be a function of x .
The Quantum Oscillator
Condition (2) requires that the
2
waveform be a function of x .
Further, condition (3) dictates that
function have no zeros (other than at
infinity).
The Quantum Oscillator
The simplest solution is of the
x
(
x
)
C
e
Gaussian form
0
2
The Quantum Oscillator
The simplest solution is ofxthe
Gaussian form ( x) C0e
2
So that
d
x 2
2xC0 e
dx
The Quantum Oscillator
The simplest solution is of the
x
(
x
)
C
e
Gaussian form
0
2
So
d
x 2
that dx 2xC0 e
d 2
2 2
4
x 2
2
and dx
The Quantum Oscillator
The simplest solution is of the
x
(
x
)
C
e
Gaussian form
0
2
So
d
x 2
that dx 2xC0 e
d 2
2 2
4
x 2
2
and dx
2m
Therefore 4 x 2 2
2
2
1
2
m x 2 E
The Quantum Oscillator
Comparing coefficients we have that,
2m 1
4 2 m 2
2
2
The Quantum Oscillator
Comparing coefficients we have that,
2m 1
m
2
4 2 m
2
2
2
The Quantum Oscillator
Comparing coefficients we have that,
2m 1
m
2
4 2 m
2
2
2
2mE
m
2
and 2
The Quantum Oscillator
Comparing coefficients we have that,
2m 1
m
2
4 2 m
2
2
2
m
and 2mE
2
2
E 12
The Quantum Oscillator
Therefore we find that the ground state
for the harmonic oscillator has a
m x 2
(
x
)
C
e
0
waveform 0
2
The Quantum Oscillator
Therefore we find that the ground state
for the harmonic oscillator has a
waveform 0 ( x) C0e m x 2
2
C0is found after normalization.
The Quantum Oscillator
Normalising,
*
0 ( x) 0 ( x)dx 1
The Quantum Oscillator
Normalising,
*
0 ( x) 0 ( x)dx 1
So that we have
*
0 ( x) 0 ( x)dx
2
C
0e
m 2
x
dx 1
The Quantum Oscillator
Normalising,
*
0 ( x) 0 ( x)dx 1
So
that
we
have
*
0 ( x) 0 ( x)dx
2
C
0e
dx 1
m 2
x
We can show that e
>0.
ax 2
dx
a
for a
The Quantum Oscillator
Which can use to integral our integral.
Hence
2
C
0e
m 2
x
dx 1
The Quantum Oscillator
Which can use to integral our integral.
Hence
2
C
0e
m 2
x
dx 1
C
2
0
1
m
The Quantum Oscillator
Which can use to integral our integral.
Hence
2
C
0e
m 2
x
dx 1
C
2
0
1
m
m
C0
1
4
The Quantum Oscillator
In summary we determined for the
oscillator ground state
E0 12
and
0 ( x) C0e
m x 2 2
The Quantum Oscillator
To determine the excited states a similar
procedure is followed.
The Quantum Oscillator(First
Excited State)
The first excited state must have the
following properties:
The Quantum Oscillator(First
Excited State)
1.
The first excited state must have the
following properties:
Should be antisymmetric about x=0.
The Quantum Oscillator(First
Excited State)
1.
2.
The first excited state must have the
following properties:
Should be antisymmetric about x=0.
Should be exactly one node.
The Quantum Oscillator(First
Excited State)
1.
2.
The first excited state must have the
following properties:
Should be antisymmetric about x=0.
Should be exactly one node.
Because of antisymmetry the node
must occur at x=0.
The Quantum Oscillator(First
Excited State)
1.
2.
The first excited state must have the
following properties:
Should be antisymmetric about x=0.
Should be exactly one node.
Because of antisymmetry the node
must occur at x=0.
x 2
A suitable trial solution is ( x) xe
The Quantum Oscillator(First
Excited State)
x 2
Substituting ( x) xe
d 2 2m
2
2
dx
1
2
m 2 x 2 E 0
into
The Quantum Oscillator(First
Excited State)
x 2
Substituting ( x) xe
d 2 2m
2
2
dx
1
2
into
m 2 x 2 E 0
3
E
Gives the same and 1 2
The Quantum Oscillator(First
Excited State)
x 2
Substituting ( x) xe
d 2 2m
2
2
dx
1
2
into
m 2 x 2 E 0
3
E
Gives the same and 1 2
Continuing in this manner gives higher
energy levels.
The Quantum Oscillator (Energy
levels)
However using a power series
expansion we find that the allowed
energies are defined by
En n 12
The Quantum Oscillator (Energy
levels)
However using a power series
expansion we find that the allowed
energies are defined by
En n 12
NB: there is uniform spacing of energy
levels ( E ).
The Quantum Oscillator (Energy
levels)