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    Pdx   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    Pdx   n (
) n dx
i x

0
L
*
*
L

0
2
L
sin
nx
L
 d 


 i dx 
2
L
sin nLx dx
Particle in a Box (Quantization
of Momentum)
2 n

sin
2

iL 0
L
n x
L
cos
n x
L
dx
Particle in a Box (Quantization
of Momentum)
2 n

sin
2

iL 0
L

n x
L
cos
n x
L
dx
Integrating by parts we get
2n L
2
P 
sin
2
iL 2n
L
nx
L
0
Particle in a Box (Quantization
of Momentum)
2 n

sin
2

iL 0
L

n x
L
cos
n x
L
dx
Integrating by parts we get
2n L
2
P 
sin
2
iL 2n
L
0
nx
L
0
Particle in a Box (Quantization
of Momentum)
2 n

sin
2

iL 0
L

n x
L
cos
n x
L
dx
Integrating by parts we get
2n L
2
P 
sin
2
iL 2n

L
0
nx
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
2mU  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
2mU  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
2mU  E 
 where  
2
2


Since U>E, this term is positive.
x
ns
Independent sol to the differential are e ,
x
e
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)  Aex , 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)  Aex , x  0
 ( x)  Bex , x  L

The internal wave is given as before by
 ( x)  C coskx  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
2mL  2 
Particle in a Box (Finite Square
Well)

The solution for the finite well is
n 2 2 2
En 
2
2mL  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
2mL  2 


As long as  is small compared to L.
1

where    
2mU  E 
Particle in a Box (Finite Square
Well)

The solution for the finite well is
n 2 2 2
En 
2
2mL  2 



As long as  is small compared to L.
1

where   

2mU  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 d 2 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 d 2 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 d 2 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 ofxthe 
Gaussian form  ( x)  C0e
2
So that
d
x 2
 2xC 0 e
dx
The Quantum Oscillator

The simplest solution is of the 
x

(
x
)

C
e
Gaussian form
0
2

So
d
x 2
 2xC 0 e
that dx
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
 2xC 0 e
that dx
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
 and  2  2  
The Quantum Oscillator

Comparing coefficients we have that,
2m 1
m
2
4  2 m   
 2
2
2

and 2mE
2

m
 E  12 
 2 

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


*
2

(
x
)

(
x
)
dx

C
 0 0
 0e


 m 2
x

dx  1
The Quantum Oscillator

Normalising,

*

 0 ( x) 0 ( x)dx  1


So
that
we
have


*
2

(
x
)

(
x
)
dx

C
 0 0
 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)