Quantum Mechanics
Solutions to Graduate Exam
Summer, 2014
Each problem is worth 25 points. The points for individual parts are marked in square
brackets. Some useful formulas are given at the end of the exam. To ensure full
credit, show your work. Do any four (4) of the following five (5) problems. If you
attempt all 5 problems you must clearly state which 4 problems you want to have graded.
1. A two-state quantum system has Hamiltonian given by
3 4 
H   
,
 4 3 
where  is a constant.
(a) [8 points] Find the eigenstates and normalized eigenvectors of this
Hamiltonian.
We start by pulling out the common factor of  , and then try to find the
eigenvalues of the remaining matrix (which must be multiplied by  ) and
corresponding eigenvectors (which do not). We subtract  times the identity matrix and
set the determinant to zero, to yield
3 
0  det 
 4
   .
4 
2
2
   3    3     4  4  9  3  3    16    25 ,
3   
The eigenvectors can then be found by solving the equation
 3 4   
 

    5   ,
 4 3    
 
3  4   5 and 4  3  5 ,
 3  5   4
and 4 =  3  5   .
We now need to work out each of the cases. If we take the upper sign, the equations
become 2  4  and 4  8 , which are identical equations, so   2  . Getting
the normalization right demands that 1   2   2  5 2 , which then gives you your first
normalized eigenstate. If we take the lower sign, then equations become 8  4  and
4  2 , so   2 . Normalizing again demands 1   2   2  5 2 . So we have
our two eigenvalues and eigenvectors:
 
1  2
 
5 1
and
 
1 1
  .
5  2 
These states form a complete orthonormal set. The eigenvalues are 5 .
1
(b) [8 points] The system starts in the state   t  0   1    . Find   t  .
0
To do so, we first have to write   t  0  in terms of our eigenstates. This is
pretty straightforward, since we can use the fact that  is an orthonormal set:
 t  0     t  0     t  0 
1
5
2 
  .
Each eigenstate can then be evolved as a function of time, so that it is multiplied by
e iEt  , so we have
 t  
1
5
 2  e5it   e5it  
1 1   2  5it  1  5it 

  e 
2   e
5 5  1
 2 

5it
 e5it 
1  4e
 
.
5  2e 5it  2e5it 
(c) [5 points] At time t what is the probability that it is in state 1 ?
The probability is simply
1  t 
2
 4e5it  e5it  4e5it  e5it 
1 16  4e 10it  4e10it  1  17  8 cos 10t .
 25

 25 25  

1
25
4e 5it  e5it
2

1
25
(d) [4 points] At time t, the energy is measured and found to be positive. What is
the state vector after this measurement?
Because the energy is positive, it must be in the positive energy state, or  .
Technically, we should extract that portion of the state vector that has the correct energy
and then renormalize, so that
 after 
   before
P 
The phase is, however, irrelevant.

5
4

2 e 5it
5
  e5it 
1  2  5it
.
 e
5 1
2. A hydrogen atom has its electron in a superposition of various states:


  N i 1, 0, 0,  12  2 2, 0, 0,  12  2,1, 1,  12 ,
in the notation where the quantum numbers are n, l , ml , ms where n is the
principal quantum number, l is the quantum number associated with the
operator L2, ml is the quantum number associated with Lz, and ms is the
quantum number associated with Sz.
(a) [2 points] What is the normalization N?
1    N
 N
2
2
i 1, 0, 0,  12  2 2, 0, 0,  12  2,1, 1,  12 
  i 1, 0, 0,  12  2 2, 0, 0,  12  2,1, 1,  12 
1  2  1  4 N 2 ,
N  12 .
The phase is, of course, unknown and irrelevant.
(b) [8 points] For each of the following operators, if we measured any one of
them, what would be the possible outcomes and probabilities: H, Lz, Sz, L2, S2?
All of these states are manifestly eigenstates of these. For the Hamiltonian, the
energy is given by the formula for En, given at the end of the test. For Lz and Sz, the

eigenvalues are ml and ms respectively. For L2 and S2, the eigenvalues are  l 2  l


and  s 2  s , but s 
1
2

for the electron, so there is never any uncertainty in this
quantity. For all the others, the probabilities are just the (sum of the) amplitudes of the
relevant parts (after substituting N  12 ), so we have
H : P  13.6 eV   14 , P  3.4 eV   34 ,
Lz : P  0   34 , P      14 ,




S z : P  12   14 , P  12   43 ,
 
L2 : P  0   34 , P 2 2  14 ,
S2 : P
 34 2   1.
(c) [8 points] Define the total angular momentum J  L  S . If we measure Jz,
what would be the possible outcomes and corresponding probabilities? If we
measure J 2 , what would be the possible outcomes and corresponding
probabilities?
It is obvious that each of these states is an eigenstate of Jz. The first two have
eigenvalue  12  , while the last one has eigenvalue  32  .
Trickier is realizing that these states are also eigenvalues of J2. For the first two
terms, you are adding l = 0 to s  12 , which can only result in j  12 . For the last term,
you are adding l = 1 to s  12 , which generally can lead to j 
1
2
or j  32 . But since
m j  32 , and we must have m j  j , then we must have j  32 . Hence they are all


eigenstates of J2 as well, with eigenvalues  2 j 2  j . So we have




J z : P  12   34 , P  32   14 ,
J2 : P
 34 2   34 , P  154 2   14 .
(d) [7 points] Simplify, as much as possible, the expectation value of the three
position operators,  R  , for this state.
To work this out, we must, in principle, write out
 R  
1
4
i 1, 0, 0,  12 


2 2, 0, 0,  12  2,1, 1,  12 R
 i 1, 0, 0,  12  2 2, 0, 0,  12  2,1, 1,  12

Now, what terms might contribute? Because the position operators commute with spin,
they can’t change the spin, and hence only those terms that leave the spin intact might
contribute:
 R  
1
4
i 1, 0, 0,  12 
 
2 2, 0, 0,  12 R i 1, 0, 0,  12  2 2, 0, 0,  12

 14 2,1, 1,  12 R 2,1, 1,  12
However, it is also known that dipole moments like these are non-zero only if l changes
by one. Hence all the terms on the right vanish, and we conclude  R   0 .
3. Lame Larry is attempting to find the energy of the ground state of the 3D
harmonic oscillator with Hamiltonian H  21m P 2  12 m 2 R 2 . Unaware that it
has an exact solution, he attempts to estimate the ground state energy using the
variational principle with trial wave function   r   exp  12  r . Find the


resulting estimate of the ground state energy. Compare your answer to the true
ground state energy.
To calculate this, we have to put three pieces together, which we now calculate:

    d 3r e r
   d 0 r 2dr e r  4  23  83 ,
2 2

 P 2    2  d 3r e r

 4  2  dr
0


 R 2    d 3r r 2 e r
2

 2  e r 2   2  d 0 r 2dr e r 2 4
2



e
r 
 r 2
 2 2   2  2
,
  r e  r  4  2  3  2  
 4





 4
24 96
  d   r dr e  r  4 5  5 .
1  2r 2
4
2 2



0

We then assemble them to get the energy as a function of  :
E   

 H
3  1
1


 P 2   m 2  R 2  


8  2m
2

 3  2  2 m 2 96   2  2 6m 2


.


8  2m
2  5 
8m
2
We now minimize this function by first taking the derivative with respect to  :
d
  2 12m 2
0
E   
,

d
4m
3
4 
48m 2 2
,
2
4 3m
2 
.

We then substitute it back into the expression to get an estimate of the ground state:
E0 
 3
3
 2  4 3m 
 
2
  


  6m 
  3  1.732 .

8m   
2 
 4 3m 
 2
The exact answer is E0  1.5 , which is slightly less than the approximation, as it must
be.
4. Consider the two-dimensional harmonic oscillator, both dimensions having
classical angular frequency  , and neglect spin.
(a) [4 points] For a single particle, what are the energy eigenvalues of this
Hamiltonian?
The resulting Hamiltonian is just H  H x  H y , and eigenstates of these
Hamiltonians can be found separately. Therefore the states can be labeled nx , n y , and
the energies will just be sums, so






Enx ny   nx  12   n y  12   nx  n y  1 .
(b) [6 points] Consider two non-interacting distinguishable particles in this
oscillator. Write down the general two-particle wave function and
corresponding energies in terms of the single particle wave functions
 nx ny  r1  and  nx ny  r2  .
In general, the wave functions will just be products, and the energies sums, so we
have
n
n n ny 2
x1 y 1 x 2
 r1, r2    n n  r1  n
x1 y 1
x2
ny 2
 r2  .
The corresponding energy will be


Enx1ny1nx 2 ny 2  Enx1ny1  Enx 2 ny 2   nx1  n y1  nx 2  n y 2  2 .
(c) [6 points] Same as (b), but assume identical bosons and write down all
degenerate wave functions for the first excited state. What is the degeneracy
and the corresponding energy?

The wave functions are the same, but they must be symmetrized. Now, if
nx1, n y1  nx 2 , n y 2 , then there is no need to symmetrize, but otherwise we have
 
n

n n ny 2
x1 y 1 x 2
 r1, r2  
1
2
 n n  r1  n n  r2    n n  r1  n n  r2   .
x2 y2
x2 y2
x1 y 1
 x1 y 1

The ground state,  0000 , is already symmetrized, but the first excited state is not. So this
state is
 0001  r1, r2  
1
2
 00  r1  01  r2   01  r1  00  r2   ,
 0010  r1 , r2  
1
2
 00  r1  10  r2   10  r1  00  r2   .
These states are degenerate and have energy 3 . There is a degeneracy of two states.
(d) [9 points] Same as (b), but assume identical fermions and write down all
degenerate wave functions for the ground state. What is the degeneracy and
the corresponding energy? What is the degeneracy and energy of the first
excited state?
This time we need to anti-symmetrize the two states, so we have
n
n n ny 2
x1 y 1 x 2

 r1, r2  
1
2
 
 n n  r1  n n  r2   n n  r1  n n  r2   .
x2 y2
x2 y2
x1 y 1
 x1 y 1


However, if nx1, n y1  nx 2 , n y 2 , then the state is not already anti-symmetrized, and
indeed, attempting to do so yields zero. Hence states like  0000 simply do not exist.
Hence our ground state is degenerate, and given by
 0001  r1, r2  
1
2
 00  r1  01  r2   01  r1  00  r2   ,
 0010  r1 , r2  
1
2
 00  r1  10  r2   10  r1  00  r2   .
Of course, these still have energy 3 .
For the first excited state, we can’t go to  0101 or  1010 , but we can go to  0002 ,
 0011 ,  0020 or  0110 . These states have energy 4 , and, of course, a degeneracy of
four.
5. A particle of mass m is confined in a one-dimensional infinite square well, with
allowed region 0  x  L
(a) [4 points] What are the energy eigenvalues and properly normalized
eigenstate wave functions of the particle?
The states must be eigenstates of the Hamiltonian, which in the allowed region is
simply H  P 2 2m , so we have
E  H 
 2 d 2
1 2
P  
2m
2m dx 2
Hence the function must be proportional to its own second derivative, which implies a
function of the form   sin  kx  or   cos  kx  . But because the potential is infinite at
x  0 and x  L , the wave function must vanish, so we must use sin  kx  , and we must
have sin  kL   0 , so k   n L . This implies the wave function must be of the form
 n  x   N sin  nx L  . The normalization is fixed by demanding that
1 
L
0
2   nx 
L
2
x
L
L
 2 nx  
 LN
2 L
N sin 
dx

N

cos

N

1

1

.





 2 4 n 
2
 L 
 L 0

 2 4 n
2
2
It is then an easy matter to determine the energy by direct substitution. So our
normalized wave functions and energy are
 n  x 
2
 22n2
  nx 
sin 
,
E
.

n

L
2mL2
 L 
(b) [6 points] The particle is in its lowest possible energy state when the
separation between the walls is slowly (adiabatically) increased to 2L. How
does the expectation value of the energy change?
The process is describe as the adiabatic approximation. The quantum state
therefore, since it started in the lowest energy state, will still be in the lowest energy state,
and hence is in the ground state. The initial energy was E1   2  2 2mL2 , but because L
has doubled, the energy drops by a factor of 4. Since it is in an eigenstate of the energy
So the change in energy is
E  E1  E1 
 22
2m  2 L 
2

 22
2mL2

3 2  2
8mL2
.
(c) [6 points] Again, let the particle start out in its lowest energy state, trapped
between the walls separated by L. Now let the right-hand wall move
instantaneously to the right, increasing the distance to 2L. How much has
the expectation value of the energy changed immediately afterwards?
In the sudden approximation, the Hamiltonian changes so quickly that the
quantum state cannot change. But the Hamiltonian in the allowed region is still
H  P 2 2m , and hence it will still have the same expectation value of the energy. Hence
the change in energy is E  0 .
(d) [9 points] As in part (c), when the wall is moved to the right, what is the
probability that it will end up in the ground state?
The probability of it remaining in the ground state is the square of the overlap of
the ground state vectors before and after, so that
P 1  1    1  1
2

L
0
2 2
x  x 
sin 
 sin 
 dx
L 2L
 L   2L 
L
2   1 2 L
  x  1 2L
 3 x   
 2 
sin 
sin 

 
 2 L  2 3
 2 L   0 
L   2 
2
2
2
2
2
2 L
2 L L 
2  4L 
32
  L
 3  
sin     2     2 
 2  sin   
  2.
L 
L   3 
L  3 
9
 2  3
 2 
Possibly Useful Formulas:
Hydrogen atom: E  
13.6 eV
.
n2
Derivative in spherical coordinates:
  rˆ
 2 
 1 ˆ 
1 ˆ 
 


,
r r  r sin  
 2 2 
1
 


 2
 sin 
2
r
r r r sin   

1
 2


.
 2 2
2
 r sin  
Integrals: For all integrals below, n is assumed to be a positive integer, and  and 
are assumed to be positive, and    .
 n  x
0
x e
dx 
n!
 n1
,
x 1
2
 cos  x  dx  2  4 sin  2 x  ,
x 1
2
 sin  x  dx  2  4 sin  2 x  ,
sin     x 

sin     x 

 cos  x  cos   x  dx 
 sin  x  sin   x  dx 
2    
 sin  x  cos   x  dx  
2    
cos     x 
2    
sin     x 
sin     x 

,
2    
2    
,
cos     x 
2    
.