Physics 741 – Graduate Quantum Mechanics 1
You may use (1) class notes, (2) former homeworks and solutions (available online), (3) online routines, such as Clebsch, provided by me, or (4) any math references, such as integral tables, Maple, etc. If you cannot do an integral, or if you cannot solve an equation, try to go on, as if you knew the answer. Feel free to contact me with questions.
1. [20 points] A particle of mass m in one dimensions has wave function
  

Nxe
0
  x x x

0

0
(a) [5] Determine the correct normalization constant N. one, so
The normalization condition is that the square of the amplitude must integrate to
N
1
 

0
Nxe
  x
 
3 2
2 .
2 dx

N
2

0

2 x e

2
 x dx

N
2
  
3
2

N
2

3
4 ,
(b) [5] Find the probability that if you measure the position x, you find x
  
1
.
This is a straightforward but boring integral:

  
1






1

4

3
2
 x
2
Nx e dx
 
4

3

  
1 x
2    
2 x
   
3
 e

2
 x

 
1
2
1

3

2
1

3

4
1

3 e

2 
5 e

2 
0.6767
(c) [5] Find the Fourier transform
 
  of this wave function.
The Fourier transform is simply given by
  
1
2



 e
 ikx
 
2


3


0
 ikx
  x xe e dx

2


3
1
   ik

2
.

(d) [5] Find the probability that if you measure the momentum p, you find p
  
.
The probability is simply given by
  
 



 
 
  2 k dk

2


3


 dk
   ik

2
2

2


3





2 dk
 k
2

2

2

3
 
 arctan

 x
2
 x
 
2




   
2 4 2
1

 
4 2
1


9.08%

1




2
 lim x
 x
2
 x
 
2

4 2


2
2


2. [25 points] A particle of mass m lies in a one dimensional potential given by
     
V
0
  
V
0
V ( x )
where
 
is the Dirac delta function,
  
is the x
Heaviside function (equals 1 for x > 0 and 0 for x < 0).
In other words, the potential is zero to the left, + V
0
to the right, and negative infinity at the origin, as sketched at right. We are going to search for bound states with energy E < 0.
(a) [6] Solve Schrödinger’s time-independent equation in the regions x < 0 and x > 0 for an arbitrary energy E

0 sure you pick solutions that have appropriate behavior at

!
. Make
E
    
 2 d
2
  
2 m dx
2
Keeping in mind that E is negative, the solutions of this equation are of the form
   
Ae
  x
, where E
   2

2
However, we want solutions that will go to zero at negative infinity, so we choose the solution
   
Ae
  x
.
For V
0
, so Schrödinger’s equation is
E
    
 2 d 2

2 m dx 2
 

V
0
  
The solution to this equation is identical to before, but with E
 
0
, so
   
Be
  x
, where

0
   2

2

This time, however, we prefer the solution
   
Be
  x
, so that it vanishes at infinity.
(b) [6] What boundary conditions are appropriate for the wave function and its derivative at the boundary x = 0?
Let’s start by examining Schrödinger’s equation right at the boundary, so we have
E
    
 2 d
2
2 m dx
2
         
 2 d
2
2 m dx
2
   
V
0
          
.
Integrate this equation across a small region near the origin. We find
E


 
  
 2

 d
2 m
  dx
2
2
   
0


0
 

     
 

.
The first term on the left is tiny, because the integral is finite and taken over a small region. The same can be said of the middle term on the right. The remaining terms can be integrated, the first on the right via the fundamental theorem of calculus, the last by using the fact that integrating delta functions is easy. The result is
0
 2
 
2 m
 
 
 
   

2 m


 2

The finite discontinuity in the derivative implies the wave function itself must be continuous, so

   
.
(c) [13] Find an equation for the energy of the bound state in terms of m,

, and
V
0
.
Our two boundary conditions work out to

 
0
  
0
 
2 m


 2

A

B
 
A
 
B

2 m

 2
A
Combining these, we see that
2 m
  2
We also know that
E
 
 2

2
2 m
Subtracting these equations, we see that and

0
 
 2

2
2 m

V
0

 2

2
2 m

 2

2
2 m

 2
2 m


2  
2


 2
2 m
   
 2
2 m
2 m

 2
      
,
V
0

By subtracting this equation from 2 m
  2
we can get a formula for

, namely
  m
 2


V
0
2

But recall we need
 
0 , so this actually only makes sense if

2   2
V m
0
2 . You can then substitute this in to get the energy, which is
E
 
 2

2
2 m
 2
 
2 m m

 2

V
0
2

 2
  m

2
2
 2

V
0

 2 2
2 8 m
V

0
2

, ,
† a a a a
1 2 1
†
2

be the usual raising and lowering operators of the
2d harmonic oscillator. Define the three operators
J x

1
2


†  † a a a a
1 2 2 1

, J y

1
2


†  † i a a a a
2 1 1 2

, J z

1
2


†  † a a a a
1 1 2 2

Show that these three operators J have angular momentum-like commutation relations.
We simply work them out, using the known commutator rules for each of the operators. We have x y

 i
2
1
4 i
4 i
 2
1
†
2
 †
2 1
,
†
2 1

1
† a a
2
 2
 2


†
,
† a a a a
1 2 2 1
†
,
†
1 2 1 2 a †
1 a a †
2 2 a
1
 a †
2 a a a
1
,
1 2
†
,
†
2 1 2 1

 i
2
 2

†
,
†
2 1 1 2 a a
1 1
 a a
2 2

 
 z
,
 i
2
 2
J , J y z
 z x
 



1
4
1
4
1
4
1
4
 2 †  †
,
†
1 2 1 1
 † i a a a a a a a a
2 1 2 2 i
 2
 2


†
,
†
2 1 1 1
†
,
1 1
† 
†
,
†
2 1 2 2
† †
,

†
,
†
1 2 1 1
† †
,
†
,
†
1 2 2 2
 †
,
†
 i a a a a a a a a a a a a a a a a
2 1 2 2 2 1 1 1 1 2 1 2 2 2 i
 2

† a a
2 1
 † a a
2 1
 † a a
1 2
 † a a
1 2

 i

J x
,




1
4
1
4
1
4
1
4
 2
 2
 2
 2



†  †
,
†
2 2 1 2
 † a a a a a a a a
1 1 2 1
†
,
†
1 1 1 2
†
,
†
1 1 2 1
†
,
†
2 2 1 2 y
.
†
,
†
2 2 2 1

†
,
1 1
†  † †
,
 † †
,
 †
,
† a a a a a a a a a a a a a a a a
1 2 2 1 1 1 1 2 2 2 2 2 2 1

† a a
1 2
 † a a
2 1
 † a a
1 2
 a
† a
2 1

 
†
,
†
1 2 2 1

The other commutators are all trivially zero or related to these.
4. [25 points] The spin operators for a spin one particle are
S z
 


1 0 0

0 0 0
 and
 0 0

1 
S x




0 1 0
1 0 1


2
 0 1 0 
(a) [10] For a general spin, what are the possible outcomes if S z
is measured?
What if S x
is measured? Find all three properly normalized eigenvectors of
S z
. Also find the normalized eigenvector corresponding to the zero eigenvalue for S x
.
S z
The possible outcomes are the eigenvalues. The eigenvalues and eigenvectors of
are trivial, because it is already diagonalized. The eigenvalues are
 
, 0,
 
and the eigenvectors are
 
, z

1
  0,
  z

  and
 
 
, z

0 .
 
For S x
we must do some work. We first ignore the factor of

2 and find the eigenvalues of the remaining matrix using the standard determinant trick:
0
 det
 





0 1 0
1
0
0
1
1
0
 


 


1 0 0
 






0 1 0
 det 1

0 0 0   

1 0
0 1


1




  

2 
2

The solutions of this equation are
 
0 and
  
2 , which after multiplying by the factor of

2 , implies the eigenvalues of the original matrix are
  , 0,
 
, the same as for S z
. To find the eigenvector with zero eigenvalue, we solve the equation


0 1 0
0

1 0 1
0 1 0



 
 


These equations imply
 
0 and
   
, so, normalizing, we find
0, x

1
2
1
0 .
 

(b) [2] The z-component of the spin S z
is measured, yielding a value of
 
. What must the state vector be?
It must be in an eigenstates of S z
with this eigenvalue. Since this eigenvalue is non-degenerate, it must be in the state
 
, z .
(c) [5] After the measurement in part (b) is performed, the spin is measured in the x-direction, S x
. What is the probability that it will yield the result zero?
What will the state vector be afterwards, assuming zero is the result?
The probability of obtaining the value zero is just
 x

0
 
0, x
 
, z
2 
1
2

1 0
 
1
 
2

1
2
2

1
2
Afterwards, the state vector must be in this eigenstates of S x
, so
 
0, x .
(d) [8] After the measurement in part (c) is completed, and the result zero was obtained, S z
is measured again. What is the probability of each possible outcome now?
The probability of each outcome now is
 z
 m

  m

, z 0, x
2
, in other words


 z z

0

 
0,

, z 0, x z 0, x
2 
2 
1
2

1 0 0

1
0
 
2

1
2
2

1
2
,
1
2

0 1 0

1
0
 
2

0
2 
0, z


, z 0, x
2 
1
2

0 0 1

1
0
 
2
 
1
2
2

1
.
2

5. [15 points] A particle of mass m lies in a one-dimensional harmonic oscillator with angular frequency

. At t = 0 it is in the state
 t 0
 
N
 cos

0
 sin

2


are constants.
(a) [2] What is the normalization constant N?
It must be normalized, so
1
   
N
2
 cos

0
 sin

2
 cos

0
 sin

2


N
2
 cos
2
  sin
2



N
2
Hence, N = 1.
(b) [4] What is the state vector
  
at all times t? and 5
2
  states 2 are eigenstates of the Hamiltonian, with eigenvalues 1
2
. It follows that 0 e
 
2 and 2 e

5

2 are solutions of the time-dependant
 
Schrödinger equation. So also will be linear combinations of these, so the solution is
    cos

0 e
 
2  sin

2 e

5 2
(c) [9] Find the expectation value of the position X , the position squared and the uncertainty

X as functions of time.
2
X ,
 
2 m

 a
 a
†

, we see that
X
   


2 m


2 m
 cos

 a
 a
†

0 e
 
2  sin

 a
 a
†

2 e

5

2 cos

1 e
 
2  sin

2 1 e

5

2  sin

3 3 e

5 2
We immediately see that between X
  
and
X
   
X
   
0 , since there are no states in common
  
. However, for X
2
we have
  
X
2    
X
   2


2 m



2 m
 cos

1 e
 i t 2  sin

2 1 e

5 2  sin

3 3






 cos
 e

2 sin

3 e


5

2 sin
2

 e

5

2 sin

3 e 5
 cos
2

 e
2  e

5

2
2
2 sin
 e
5 2









2 m


2 m

 cos
2
 
5sin
2
 
 cos
2
 
5sin
2
 
 
 
 e
2
The uncertainty will simply be the square root of this factor.
 e

2
  t



2 m
 cos 2
 
5sin 2
      t

6. [15 points] A particle of mass m in three dimensions has Hamiltonian
H

1
2 m

P x
2 
P y
2 
P z
2 
X 4 
Y 4

 
X Y 2
(a) [5] Argue that there are two independent symmetries that this Hamiltonian satisfies. One of them will be an appropriate sized rotation.
We first note that the Hamiltonian is independent of Z . This implies it is invariant under translation in the z -direction, which is obvious.
It is also obvious that the x and y -directions are treated in a symmetric manor.
This suggests our other symmetry could be some sort of rotation symmetry that changes x into y and vice versa. We see, for example, from equation (6.27) in the notes that a ninety degree rotation around the z -axis turns x into – y and y into x . It is obvious our potential is invariant under this change, so our two symmetries are
 
for any a and


1
2
 
.
(b) [5] Based on part (a), there should be two operators that can be simultaneously diagonalized with the Hamiltonian. What are they? What restrictions are there on their eigenvalues, if any?
Because we have translation symmetry in the z -direction, our Hamiltonian should commute with P z
, and indeed it does. It also commutes with 
 z ˆ, 1
2
 
, and these two commute with each other, so we can commute them simultaneously. If call our eigenvalue of P z
by the name
 k real. If we call our eigenvalue of
, then k is unrestricted other than demanding that it be


1
2
 
by the name

, then we know that rotating four times brings us back to where we start, so notes, are
    i .

4 
1 . The solutions, as described in the

P z

(c) [5] The eigenstates of this Hamiltonian can be written as functions of x, y, and z. Write explicitly the dependence of the wave function on one of these variables.
Since our solutions are eigenstates of P z
with eigenvalue  k
  k

, so
, we must have
  i
 z



, ,
   k
 
, ,

,

, ,
  e ikz
The proportionality constant will be a function of x and y , however, so we are a long way from finding the eigenfunctions.
7. [20 points] Bottomonium describes a series of particles which are bound states of a bottom quark and a bottom anti-quark. One can crudely approximate the
Hamiltonian as
H
P
1
2 P
2
2
  
2 m 2 m a
R R
1

2
 b R R
1

2 but are not identical particles (and ignore spin). Our goal is to learn as much as we can about the energy eigenvalues of this Hamiltonian.
(a) [4] Explain, giving relevant equations as needed, how you can reduce this two-body problem to a one-body problem. What is the reduced mass

for this one-body problem?
As explained in chapter 7 section E, you can simplify this problem a great deal by changing to a new set of relative and center of mass operators given by
P cm

P
1

P
2
P
 m P
1
 m P
2 m
 m

1
2

P
1

P
2

R cm
 m R

R
1 m m
 m
2

1
2

R
1

R
2

R

R
1

R
2 where in this case m
1
 m
2
 m In terms of these, as shown in the homework, the
Hamiltonian becomes
H tot

P
2
2 cm
M

P
2

2
 a
R
 b R
The reduced mass

is given by
  m m
1 2
 m
1
 m
2
  m
2
2 m
 m
M
 m
1
 m
2

2 m .
2 . The total mass is

(b) [4] The resulting wave function can be written in spherical coordinates as
  r
  
. Explain how you can factor this in a sensible way, and give the form of one of the factors.
Because the potential is spherically symmetric, the wave function can be factored into a radial and angular part, like this:
  r
   
R nl
   
The angular part Y l m
 
will simply be the spherical harmonics, worked out in detail in section 7D.
(c) [4] Find a differential equation (not a partial differential equation) for the other factor. You are not expected to solve this equation. so
This is straightforward. The radial wave function now satisfies equation (7.27),
 
 2
2



1 d r dr
2
2

   l
2  l r
2 

 a r


 
Obviously the solution will depend on l . There will also tend to be several solutions for any given l , which we label by n , which is why our solutions are labeled R nl
 
.
(d) [4] The angular momentum L z
is measured, and it is discovered that it has the value

2

. If L 2 were measured, what would be the minimum value that might be measured?
 l momentunm m

, where m runs from
to l . It follows that l must be at least two. Since L 2   2
 l 2  l

, we must have
L
2 
6
 2
.
(e) [4] L 2 is also measured, and it is found to have the value 12  2 . What does this tell you about the wave function?
L
2   2
 l
2  l

, l
2 l 12 , so l

3 . Our wave function is therefore
  r
   
R n 3
   
, where Y
3

2
  
105 e

2 i

4 2
 sin
2
 cos

The explicit form was simply cut and pasted from the class notes.

8. [15 points] A sodium atom is in an excited state with total angular momentum j

3 , initially in the state
2
  
3
2
, m

1
2
. It is going to decay by emission of a photon to the state

, j

1
2
, m . It turns out the rate for this to occur depends on the matrix elements

1
2 m j
R
 
3 1
2 2
. Let us define the unknown double-bar matrix element as
A
 
1
2
R
 
3
2
(a) [5] Find all non-zero components of the matrix elements

1
2 m R q
 
3 1 in
2 2 terms of A, where R q
are the three components of the spherical tensor corresponding to the vector R . Feel free to use the Clebsch-Gordan routine, or some other source for the coefficients.
According to the Wigner-Eckart theorem, the matrix elements are given by

1
2 m R q
 
3 1
2 2

1
2 m 3
2
1; 1
2
2
 
2
1 q

1
2
R
 
3
2

A 1 m
2
3
2
2
1; 1
2 q
These matrix elements are only non-zero if restricted to m
 
. Because the final values of m are
2 q m
 
1 , there will only be two non-vanishing matrix elements. I got the
2
Clebsch’s using my routine:
> clebsch(3/2,1,1/2,0,1/2,1/2);
> clebsch(3/2,1,1/2,-1,1/2,-1/2);

1 1
2 2

1
2
,

1
2
R
0
 
3 1

2 2
A
R

1
 
3 1
2 2

A
1 1 3
2 2 2
1; 0 2
2
 
A
1
2
,

1 3
2 2
1; , 1 2
2
 
A
6 ,
12 ,
R

1

(b) [5] Find all non-zero components of the matrix elements

1
2 m j
1
2
The spherical tensors are related to the vector operators by


X
 iY

. From these it is pretty easy to see that
X

Y

1
1
2
2

R

1

R

1

,


1

R

1

.
R
 
3 1
2 2
R
0

Z and
.
We now realize that the only non-zero matrix elements of R will therefore be

1 1
2 2

1
2
,

1
2
Z
X

1
2
,

1
2
Y
 
3 1
 
2 2
A
 
3 1
2 2
 
3 1
2 2


6 ,
1
2
1
2 i


1
2
,

1
2
1
2
,

1
2

R

1

R

1


R

1

R

1

 
3 1
2 2
 
3 1
2 2

A 24 ,
2
 iA 24

(c) [5] The probability for it to decay spontaneously
 
3 1

2 2

1
2 m can be shown to be proportional to

1
2 m R
 
3 1
2 2
2
. Based on this fact, make a prediction about what fraction of the time it ends up in the state m
 
1 and
2 what fraction it ends up in the state m
 
1 .
2
We calculate the two matrix elements squared:

1 1
2 2
R
 
3 1
2 2

1
2
,

1
2
R
 
3 1
2 2
2 
2 

1 1
2 2
Z

1
2
,

1
2
 
3 1
2 2
X
 
2 
1
6
A
2
,
3 1
2 2
2  
1
2
,

1
2
Y
 
3 1
2 2
2 
1
24
A
2 
1
24
A
2 
1
12
A
2
.
So the probability in the first case is twice as big as the second case. We would expect the first decay to occur 2
3
of the time, and the second to occur 1
3
.
9. [30 points] An electron of mass m lies in a region of constant crossed electric and magnetic fields, with electric field E

E ˆ B

B ˆ
(a) [6] Find an electrostatic potential U and a vector potential A that can account for these two fields. I recommend choosing them to depend only on the variable x if possible.
The magnetic field is given by B
 
A . To get a magnetic field in the z direction, we need A to point in a direction perpendicular to B and have a space dependence perpendicular to both B and A . For example, we could choose it to be oriented in the y -direction and depend on x . A little trial and error will convince you that
A

Bx ˆ will work.
The electric field is given by E easy to see we can get this to work if we choose U
, but A has no time dependence. It is
 
Ex . So, in summary,
U
 
Ex and A

Bx ˆ .
This form will prove useful, since everything is independent of both y and z , which will make the problem easier to solve.
(b) [7] Write the Hamiltonian explicitly. What two momentum operators of the three P ’s does it commute with? Which one of the three operators S does it commute with?
The Hamiltonian is given by (9.21):
H

1
2 m

P
 e A

2  eU
 ge
2 m
1
2 m


P x
2 

P y
 eBX

2 
P z
2

 eEX
 geB
S z
2 m
This Hamiltonian commutes with P y
, P z
, and S z
, and these all commute with each other, so it should be possible to simultaneously diagonalize all of these.

(c) [6] Give appropriate names to the eigenvalues of the three operators you found in part (b). Replace the corresponding terms in the Hamiltonian with their eigenvalues.
We’ll let the eigenvalues of P y
, P z
, and S z
be  k y
,  k z
, and  m s
respectively.
The last one has eigenvalues m
 
1 , but the other two are unrestricted. s 2
If we were to attempt to find the eigenstates of this Hamiltonian, we would know that the eigenstates , , y z
, s
would simplify a good deal when acted on by this
Hamiltonian. We would have
, , y z
, s

1
 2 m


P x
2 

 k y
 eBX

2   2 2 k x

 eEX
 geB
2 m
 m s 
, , y z
, s
We see this is simply a shifted harmonic oscillator.
(d) [7] The remaining terms in the Hamiltonian should be identifiable as a
shifted harmonic oscillator. Define a shifted operator X
 and rewrite the
Hamiltonian in terms of X. If your notation and work looks like mine, the shift will be
X

X
 
 k y  mE eB eB
2
We make the suggested substitution and find
H

1
2 m


P x
2   k y
 eB X
   k y
 mE
 2
B
  2 2 k z

 
 eE X

 
 k y  mE eB eB
2





1
2 m


2  2 2
P e B X x

2 
2 emEX
  m E 2
B
2
2
1 m
P x
2  2 2 e B X

2 

2
2 2 k z m


Ek y
B

 mE
2
2 B
2
 2 2 k z


 eEX

2
 m s
 
.

Ek y
B
 mE 2
B
2

 s
2 m
 s
2 m
The leading term is a Harmonic oscillator. If we identify the cyclotron frequency

B
 eB m , the first part of the Hamiltonian will be P m m X x
2
2

1
2
 2
B

2
, which is identical with a Harmonic oscillator.
(e) [4] Find the energies of the Hamiltonian
The eigenstates of P m m X x
2
2

1
2

2
B

2
have energies
 
B
 n

1
2

, so in summary, the energy of the state , , y z
, s
will be
E

 eB  n
2
 m
 2 k 2 z
2 m


Ek y
B
 mE 2
2 B
2
 geB m s .
2 m

10. [20 points] Two particles lie in identical infinite square wells of width a, with
Hamiltonian
H

1
2 m

P
1
2 
P
2
2

     
,
  0
 if 0 x a otherwise
(a) [5] If the particles are non-identical spinless particles, find the three lowest energies, give me a list of all quantum states having those energies, and tell me the degeneracy (number of states with that energy) for each.
The eigenstates for single particles are simply labeled
 n
for positive integers n , with energy E n
 
2  2 n
2
2 ma
2
. The states for two non-identical particles, the state are
  n m
with energy

2  2
E ma
2 state(s) #
 
1 1
1
E nm


2  2
2 ma
2
 n
2  m
2

5
 2  2
4

2  2
2 ma
2 ma
2
 
1 2
,
 
2 1
 
2
1
The ground state will have n m 1 , the first excited state(s) will have one of these increased to two, and the second excited state will have both of these increased to two.
The lowest and highest states are non-degenerate, but the middle state is two-fold degenerate. The answers appear in the table above.
(b) [5] Repeat part (a) if the two particles are spinless bosons.
We must symmetrize our state vectors.
The answers will be the same as for the previous states, except the waves get symmetrized. The result is that the intermediate state is now also 5

2  2

2  2
E ma
2
2 ma
2 1
2 non-degenerate.
4

2  2 ma
2
(c) [10] Repeat part (a) if the two particles are spin ½ fermions.
 state(s) #
 
1 1
1
 
1 2
  
2 1

1
 
2 2
1
This time there is the added complication that there is spin. The energies of the states will once again be the same. For fermions, we want the total wave function to be anti-symmetric.
For the ground state and second excited state, we have no choice but to make the space part symmetric, so the spin part is forced to be symmetric. For the

2  2
E ma
2
5

2  2
4

2  2
2 ma
2 ma
2
1
2
1
2


  
1 state(s) #
2

    

1
 
1 2
  
2 1

,
1
2

 
1 2
  
2 1

  
,
4
 
1 2
  
2 1

,
1
2
 

2 2
    

1
2


  
    

,
1 first excited state, we can make the space part symmetric or anti-symmetric, and therefore the spin part can be correspondingly anti-symmetric or symmetric. In all, there are four different quantum states for this energy.