Physics 741 – Graduate Quantum Mechanics 1
Solutions to Final Exam, Fall 2011
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. Some possibly useful integrals can be found at the end of
the test. 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.
Work: 758-4994
Home: 724-2008
Cell:
407-6528
1. [15 points] A particle of mass m lies in the one-dimensional potential
V  x   12  x 4 . Let the uncertainty in the position be x  a .
(a) [7] Find an inequality for the uncertainty in the momentum. Assuming this
inequality is saturated (that means it’s an equality), and assuming that we
can replace x  x and p  p , estimate the energy in terms of a.
The uncertainty principle states that xp  12  . Changing this inequality to an
equality, and setting x  a , we have
p 



2x 2a
We then substitute these expressions into the Hamiltonian, which is
p2
p2 1 4
2
a4
V  x 
 x 

.
H
2m
2m 2
8ma 2
2
(b) [8] Estimate the ground state energy by minimizing the function you found in
part (a).
This function goes to infinity at either a = 0 or a   . The minimum, therefore,
must lie between these two values. To find this minimum, we take the derivative and set
it to zero to yield
d
2
H 
 2 a 3 ,
da
4ma 3
8m a 6   2 ,
0
1/3
 2 
a 

 8m 
2
1/3
1  2 
 
 .
2  m 
We then substitute this back into the expression to estimate the energy:
2 2  m 
 1  2 
E



 
8m   2 
2 4  m 
1/3
2/3
3 4/3 1/3
.

8m 2/3
2. [20 points] A particle in one dimension has wave function
 N xe  x
0

x0,
  x  
x  0.
.
(a) [4] What is the normalization constant N?
To find the normalization constant, we demand that the integral of the square of
the wave function equal 1. So we have

1
   x
2

dx  N
2

 xe

2 x
0
1
N2
 x 2 x

dx  N  
e
,
 2 e 2 x  
2
4
 2
 0 4
2
N  2 .
(b) [6] What is the probability that if the particle’s position is measured, it will
give a value x  1  ?
We simply repeat the previous integral, changing the limits appropriately and
substituting our value for N.
P  x  a 


  x  dx  N 2
2
1


x 2 x
1 2 x 
2 x
2 
1  xe dx  4   2 e  4 2 e 1 
1 
 1
 4 2  2  2  e 2  3e 2  0.406 .
4 
 2
(c) [10] If you were to measure the momentum of this particle, what is the
probability that it will give a value p   ?
We must calculate the Fourier transform, which is given by
  k  



dx
2
  x  e ikx 
2
2


xdxe  ikx  x 

2 

3/2
.
  ik  
3
2 2
2   ik 
Since the momentum is given by p  k , demanding that p   is the same as
demanding that k   . The probability of this is given by


P  k        k  dk  

2


 2 dk
4   ik    ik 
3
3


 2 dk
4  2  k 2 
3
This equation can then be evaluated by making the trigonometric substitution
k   tan  , and we obtain
.
P k    
 /2
 2 d  tan  
 4
4  2   2 tan 2  

 /2
 12 sin   /4 
1
2
 /2
3
1   
2
2
 /2
1
sec 2  d
1
 
  cos  d
3/2
2  4 1  tan 2  
2 4
1
2
 14 2  0.146 .
3. [25 points] A particle of mass m is in a one-dimensional harmonic oscillator
potential V  x   12 m 2 x 2 .
(a) [6] Work out the first two eigenstates, n = 0 and n = 1, of the harmonic
oscillator (equation 5.13). What are the energies of these states?
Equation (5.13) tell us that the first two wave functions of the Harmonic oscillator
are
 m x 2 
 m 
 0  x  
,
 exp  
2 
  

1/4
 m 
1  x  

  
1/4
 m 


  
1/4
 m 


  
1/4
 m
 m x 2 
 d 

x
 exp  

2
2m dx 
2 


 m
 m x 2 
 m x 2  
 m
exp  
x exp  
x


2
2 
2m 
2  



 m x 2 
2m
x exp  
.
2 


The energies are En    n  12  , or E0  12  and E1  32  .
(b) [8] The wave function at t = 0 is given by
 m x 2 
 m x 2 
 m x 2 
A
exp
Bx
exp
  x, t  0    A  Bx  exp  







,
2 
2 
2 



where A and B are real constants. What is the wave function   x, t  at all
times?
We see that the two terms are proportional to eigenstates of the Hamiltonian.
Eigenstates simply evolve as n  x  e iEnt  , so the A term will simply get multiplied by
e it 2 and B by e 3it 2 . So we have
 m x 2 
  x, t    Ae it 2  Bxe i 3t 2  exp  
.
2 

(c) [11] Find the probability density   x, t  and probability current j  x, t  at
all times and all locations, and check that probability is locally conserved.
The probability density is given by
  *  x, t    x, t    Aeit 2  Bxei 3t 2  Ae it 2  Bxe i 3t 2  exp   m x 2  
  A2  B 2 x 2  ABxe  it  ABxe  it  exp   m x 2  
  A2  B 2 x 2  2 ABx cos t   exp   m x 2   .
The probability current (normally a vector, but we are in one dimension) is given by
j




Im  *  x, t    x, t  
x
m 


 m x 2   
 m x 2   


 it 2
i 3t 2
Im  Aeit 2  Bxei 3t 2  exp  
Ae
Bxe
exp



 

 
m 
2  x 
2   




 m x 2   i 3t 2 m
 m x 2  

 i t 2
i 3 t 2 


Im  Aeit 2  Bxei 3t 2  exp  
Be
x
Ae
Bxe
exp

   2 


m 
2  


 

 m x 2 

it 2
exp  
 Bxei 3t 2  Be i 3t 2  Axm e  it 2   Bx 2 m e  it 2  
 Im  Ae
m




 m x 2 

 it
it
2
2
exp  
 Im AB 1  x m   e  AB  x m   e
m
 


 m x 2 
 m x 2 

AB
2
2
exp  
sin
1
exp
AB

t
x
m

x
m














 sin t 
m
m
 
 






Conservation of probability says that we should have
 m x 2  AB  2m x 
 m x 2 
 j
  2 AB x sin t  exp  


exp


 sin t   0 .


t x
  m 
 
 


4. [15 points] Consider the three operators
A  X 2  Y 2 , B  2 XY , Lz  XPy  YPx ,
Where the third of these is simply the standard angular momentum operator.
(a) [6] Work out the three commutators of these operators with each other. Note
that the commutators of Lz with various operators were worked out in
problem 3.4, part (a).
It is clear that A and B will commute with each other. The remaining
commutators need to be worked out:
 Lz , A   Lz , X 2  Y 2   X  Lz , X    Lz , X  X  Y  Lz , Y    Lz , Y Y
 iXY  iYX   iYX    iXY   4iXY  2iB ,
 Lz , B   2  Lz , XY   2 X  Lz , Y   2  Lz , X Y  2iXX  2iYY  2iA .
(b) [5] Based on the results of part (a), deduce two non-trivial inequalities
relating the uncertainties of these operators and their expectation values.
The product of the uncertainties of two operators can always be related to the
expectation value of the commutator. This isn’t helpful for the trivial commutator, but
for the other two we obtain the relation
Lz A 
1
2
i  Lz , A 
1
2
2B   B ,
Lz B 
1
2
i  Lz , B  
1
2
2 A   A .
(c) [4] Show that if you are in an eigenstate of Lz then A  B  0 .
As we proved in a previous homework, if you are in an eigenstate of any operator,
then the uncertainty in that operator is zero. As a quick review, if the eigenvalue of Lz is
m , then
 Lz 
2
  L2z    Lz 
From our previous part, this implies 0  B
2
  m    m   0 .
2
and 0  A
2
But an absolute value can
only be non-positive if its argument is zero. So A  B  0 .
5. [25 points] A particle is about to be measured by one of the two operators
1 0 0
0 1 0




A  0 0 1 , B  1 0 0 .
0 1 0
0 0 1




(a) [8] For each of these operators, find the eigenvalues and orthonormal
eigenvectors. For each operator, if a measurement of an arbitrary state
occurs, what are the possible outcomes?
I have drawn in dashed lines which demonstrate that each of these operators is
block diagonal. This immediately gives us one eigenstate with eigenvalue 1. The
remaining 2  2 matrix is familiar enough that we can also know its eigenvalues and
eigenvectors. In fact, this sub-matrix is the same sub-matrix which we called H2 on page
51 of the notes, and whose eigenvectors are given by (3.61). In summary, the three
eigenvectors for A and the three for B work out to
1
 
a1   0  ,
 
0
0
1  
a2 
1 ,
2  
1
1
1  
b1 
1 ,
2  
0
0
1  
a3 
1 ,
2  
 1
1
0
1  
 
1 , b3   0  .
b2 

2 
1
0
 
For A¸ the first two have eigenvalue +1 and the third has eigenvalue -1. For B, the first
and third have eigenvalue 1 and the middle one has eigenvalue -1. Whichever operator
you measure, the only possible outcomes are 1 .
(b) [5] Operator A is measured first, and the resulting value is negative. What is
the quantum state immediately thereafter?
After measuring an operator, the system will always be in an eigenstate with the
corresponding eigenvalue. Therefore it must be in the state a3 .
(c) [5] Operator B is then measured. What is the probability for each possible
outcome?
The probabilities are the square of the overlap with the corresponding
eigenvectors, so
2
P  1  b1 a3
2
 b3 a3
2
2
0
0
1
1
1 1 3
 
 1 1 0   1  
 0 0 1  1     ,
2
4 2 4
2
 1
 1 
 
 
2
P  1  b2 a3
2
0
1
1
 
 1 1 0   1   .
2
4
 1
 
(d) [7] Assuming the outcome in part (c) was positive, what is the quantum state
after the measurement?
If the result is positive, the state vector is now supposed to be
 
1
P  1
b
1
b1 a3  b3 b3 a3


1
 0  0
 0 
4  1  1
    1

1  1 1 0   1    0 
 0 0 1  1  


3 2 2
    2
 
0
 1  1 
 1 

 2 1 2   0   
 
4  1  


0
  
  
3  2 2   1   

 0   2    


1
.
6

2 
3 
1
6
6. [25 points] A particle of mass m is in a one-dimensional harmonic oscillator with
angular frequency  is in the quantum state


  N 0 i 1  2 2 .
(a) [3] What is the normalization constant N?
We simply use the normalization condition:

1    N2 0 i 1  2 2
N  12 .
 0  i 1 

2 2  N 2 1  1  2   4 N 2 ,
(b) [17] Find the expectation values X , X 2 , P , and P 2 .
This is most easily done by using raising and lowering operators. We have
X 

X 

1

 a  a†  0  i 1  2 2
2 2m




1


i 0 2 1  1 i 2 2  6 3 
i 0  1 i 2 2  6 3
2 2m
8m
1

2 8m
 0 i 1 
2 2
i  i  2i

0,
2
8m
 1  1  2  6 
i 0  1 i 2 2  6 3 
8m
i 0  1  i




2 2  6 3 


i 0  1  i 2 2  6 3
8m
5

,
4m
i m †
m
P 
a  a 0  i 1  2 2 
i 1  2 2  i 6 3  0  2i 1 ,

2
2
8
X2 

P 
1 m
2
8

 0 i 1 
2 2

 0  3i 1 
1 m
m
,
1  3  2   3
2
8
8
m

0  3i 1  2 2  i 6 3
8
m
9m

.
1  9  2  6  
8
4

2 2 i 6 3


P2

 0  3i 1 
2 2 i 6 3

(c) [5] Find the uncertainties X and P and check that they satisfy the
uncertainty relation.
We have
X 
X2  X
P 
P2  P
X P 
2
2
5
,
4m


9
4
m  89 m 
9
8
m ,
5 9m
45

 1.186 .
4m
8
32
The uncertainty relation says the product must exceed 0.5 , which it certainly does.
7. [15 points] A particle of mass m in three dimensions has Hamiltonian
H
1
Px2  Py2  Pz2   12   X 4  Y 4    X 2Y 2   cos  kZ 

2m
(a) [4] Show the Hamiltonian is invariant under rotations around the z-axis by
90 degrees, R    zˆ , 12    for an appropriate angle  .
The kinetic term is always invariant under an arbitrary rotation of any angle. The
potential terms must be checked after rotation by seeing if V  X , Y , Z    V  X , Y , Z  ,
where
 X , Y , Z     Y , X , Z  .
We easily see that
V  Y , X , Z   12   Y   X 4     Y  X 2  sin  kZ   V  X , Y , Z  .


4
2
(b) [6] Show that the Hamiltonian is invariant under translations along the z-axis
by an amount a, T  azˆ  . What is the smallest such translation possible?
We similarly wish to have V  X , Y , Z  a   V  X , Y , Z  . Clearly, this demands
sin  kZ  ka   sin  kZ  . Since sine is 2 -periodic, ka must be a multiple of 2 , and
the smallest value of a for which this will work is
a
2
.
k
(c) [5] Suppose states are found that are eigenstates of these two operators, so
R    zˆ , 12    R , T  R R , T
and T  azˆ  R , T  T R , T .
What restrictions, if any, are there on the eigenvalues R and T ?
Symmetry operators are always unitary, and therefore must have magnitude 1.
However, if we perform the rotation four times, we end up back where we started, so we
can conclude that  R    zˆ , 12      1 . It follows that
4
R , T   R    zˆ , 12     R , T  R4 R , T ,
R4  1 ,
4
R2  1 ,
R  1, 1, i or  i .
For the other eigenvalue, we only demand that T* T  1 .
8. [20 points] The   mesons are a pair of spin 1 (s = 1) particles with charge  e
and identical mass m  775 MeV/c 2 .Although it wouldn’t work in practice, one
could imagine binding them together to make a hydrogen-like system. Assume
only electric interactions are relevant for the resulting system.
(a) [5] What would be the binding energy of this state in the n = 1 state?
We first need to calculate the reduced mass, which is given by
1


1
1
1 1 2

   ,
m1 m2 m m m
  12 m  12  775 MeV/c 2   387.5 MeV/c 2 .
The energy is then given by
En  
 2c2
2n 2
2
1 1 
 
  387.5 MeV   0.0103 MeV  10.3 keV .
2  137 
(b) [7] If one were to measure the total spin squared S 2 for this system in the n =
1 state, what are the possible outcomes? For each of these possible outcomes,
what are the possible outcomes for Sz?
Because it is in n = 1, and l is always less than n, there is no orbital angular
momentum. Hence the total angular momentum of this “atom” will come just from the
spins of the two mesons. Since these each have spin s1  s2  1 , the total spin must be in
the range from s1  s2  0 to s1  s2  2 , so the total spin can
2
be either s = 0, 1 or 2. The corresponding eigenvalue of S is
given by  2  s 2  s  , which is either 0, 2 2 or 6 2 . The
eigenvalues of Sz are then given by ms , where ms runs from
 s to  s . A complete table of possibilities is given at right.
S2
0
2 2
6 2
Sz
0
, 0, 
2, , 0, , 2
(c) [8] The value of S 2 and Sz are actually measured. S 2 yields the maximum
possible value, and Sz yields 0. Write the spin state of the two mesons in form
s, ms and explicitly in terms of the individual spin states ms1 , ms 2 . You
may use the online Clebsch routine. If the spin S1z of just the   were
measured, what would be the possible outcomes, and what are their
probabilities?
The largest values of S 2 corresponds to s = 2, and obviously Sz = 0 corresponds to
m = 0. So the state is s, ms  2, 0 . We now want to write this in terms of the
underlying spins, that is, we want to write
2, 0 

m1s , m2 s m1s , m2 s 2, 0 .
m1 s , m2 s
This is the purpose of Clebsch-Gordan coefficients. The coefficients we want are
1,1; m1s , m2 s 2, 0 . This can only be non-vanishing if m1s  m2 s  0 , so there are only
three coefficients that need to be calculated. We will use the online routine.
> for m1 from -1 to 1 do clebsch(1,1,m1,-m1,2,0);end do;
We find:
s, ms  2, 0 
1
6
 1, 1  2 0, 0
 1,1  .
We now can straightforwardly calculate the probability that the spin of the first spin is in
each of these states. We find:
P  Sz    
 
1
6

 16 , P  S z  0   2
2
1
6

2
 23 , P  S z    
 
1
6
2
 16 .
9. [20 points] An electron of mass  lies in a region with electric and magnetic
fields
E
02
e
 xxˆ  yyˆ  ,
B  Bzˆ
(a) [5] Find a suitable scalar potential U  r  . Demonstrate that the vector
potential A  r   12 B  xyˆ  yxˆ  is one way to describe this magnetic field.
Since the vector potential has no time-dependence, it does not contributes only to
the magnetic field, for which we have
 A Ay
B    A  xˆ  z 
z
 y

 Ax Az

  yˆ 
x
 z

  Ay Ax

  zˆ 
y
  x

1
1
  0  0  zˆ  2 B  2 B   Bzˆ .

For the electric field, we want E  U  r  , which suggests we need
02
U
x and

e
x
02
U
y.

e
y
These equations are easy to integrate to yield
U 
02
2e
x
2
 y2  .
(b) [6] Write the Hamiltonian explicitly.
The Hamiltonian is given by
1
ge
2
 P  eA   eU  r   S  B
2
2
2
1 
ge
2
1
1




 Pz2   12 02  X 2  Y 2  
P
eBY
P
eBX
BS z




x
y
2
2


2 
2
H

1
ge
 Px2  Py2  Pz2  eB  XPy  YPx   14 e2 B 2  X 2  Y 2    12 02  X 2  Y 2  
BS z

2 
2
1
1  2 e2 B 2  2
eB
2
2
2

Px  Py  Pz     0 
X Y2  
 Lz  gS z  .

2 
2
2 
4 
2
(c) [9] Demonstrate that this Hamiltonian commutes with one of the three
momentum operators P, one of the three angular momentum operators L,
and one of the three spin operators S. Which one, in each case? Do they
commute with each other? Call the three eigenvalues under the operators k ,
m , and m s respectively. Are there any restrictions on these eigenvalues?
Every term manifestly commutes with Pz and Sz. Also, it is easy to see that Lz
commutes with the combination X 2  Y 2 , since this is just the distance squared from the
z-axis, and invariant under rotation about this axis. So the three operators that commute
with the Hamiltonian are Pz, Lz, and Sz. Since they also commute with each other, they
can all be simultaneously diagonalized, and we can assume our states are eigenvectors
under these three operators. We will label these eigenvalues as appropriate, so
Pz   k  , Lz   m  , S z   ms  .
As always, for a spin one-half particle, ms   12 , and the angular momentum quantum
number m must be an integer, but there are no restrictions on k.
10. [20 points] Two non-interacting spinless particles are in the n = 1 and n = 2 state
of an infinite square well with allowed region 0  x  a .
(a) [8] Write explicitly the wave function   x1 , x2  if the particles are (i) nonidentical, with the first particle in n = 1, (ii) identical bosons, and (iii)
identical fermions
The wave function for a non-interacting single particle in an infinite square well is
just  n  x   2 a sin  nx a  . For two particles, we simply take the product, and have
2
a
  x1   2 x2 
 sin 
.
 a   a 
  x1 , x2   sin 
However, boson states must be symmetrized, and fermion states anti-symmetrized.
Taking this into account, we have
 B  x1 , x2  
2    x1   2 x2 
 2 x1    x2  
sin 
 sin 
  sin 
 sin 
 ,

a   a   a 
 a   a 
 F  x1 , x2  
2    x1   2 x2 
 2 x1    x2  
sin 
 sin 
  sin 
 sin 
 .

a   a   a 
 a   a 
(b) [12] Find the probability that if the position of the two particles is measured,
they will both lie in the region 0  x1,2  12 a if the particles are (i) nonidentical, (ii) bosons (iii) fermions.
The probability in each case is simply given by
P   dx1  dx2   x1 , x2  .
1a
2
0
1a
2
2
0
For non-identical particles, we have
4
P 2
a
1a
2
1a
2
 dx  dx
1
0
2
0
x 
 2 x2 
sin 2  1  sin 2 

 a 
 a 
1a
1a
2
2
4 1
4
a
a
 2 x1    1
 4 x1  
 2  x1 

sin 
sin
x
  2

  2
8
16
a 2
a
 a 0  2
 a  0
a a 1
 4   4   4 .
For the fermions and bosons, the computation is much more complicated, but it is clearly
wise to do them together to save time. We have
2
P  2
a
1a
2
1a
2
   x   2 x 
 2 x    x  
0 dx1 0 dx2 sin  a 1  sin  a 2   sin  a 1  sin  a 2 
2
 2   x1  2  2 x2 
2   x1 
2   x2  
sin  a  sin  a   sin  a  sin  a  
2








 2  dx1  dx2 

a 0
  x1   2 x1   2 x2    x2  
0
 2sin 
 
 sin 
 sin 
 sin 
 a   a   a   a  

1a
2
2
 2
a
1a
2
 a  2  a  2  4
       2
 4   4   a
1a
2
1a
2
  x   2 x 
 2 x    x 
0 sin  a 1  sin  a 1  dx1 0 sin  a 2  sin  a 2  dx2
2
a
2
2
  a
x  a
 3 x    1 4  2a  1 16
sin 
  sin 

    2     2 .
4 9
 a  6
 a   0  4 a  3 
  2
1
1 4
  2
4 a
The probabilities work out to PB  0.4301 and PF  0.0699 . We see the strong tendency
for bosons to tend to be near each other, while fermions avoid each other.