This test consists of three parts. Please note that in parts II and III, you can skip one question of those offered. Some possibly useful formulas can be found below. h

6.626 10

34    
15 
 
1.055 10

34    
16 
Harmonic Osc.
E n
 n

   n

1
2
0,1, 2,


E n
Hydrogen
 

13.6 eV n
2

Z
2
 n
1D square well:
E n


2  2 n
2
2 mL
2

2 sin

L n

1, 2, 3,
 nx
L
T
Barrier penetration:

E
16 1
V
0

 
E
V
0

 e

2

L
,
 
2

0

E


Reflection off a step:
R



E
E




0
0


2
if E

V
1 if E

V
0
0
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. For what values of the energy E will a particle in a potential V ( x ) be a bound state?
A) E

V
 
B) E

V
 
C)

V D) E

V E) E

V
2. In order, when the quantum number l takes on the values 0, 1, 2, 3, we denote these by the latters
A) s, d, p, f B) s, p, f, d C) s, f, p, d D) s, d, p, f E) s, p, d, f
3. For which types of problems does it make sense to divide the wave function into a product of a radial and angular part,
       
?
A) Problems where the boundaries are easily specified in terms of x , y , and z , like the
3D box
B) Problems where the forces fall off quickly at large distances
C) Problems where the potential depends on time
D) Problems where the potential depends only on distance from the origin,
    
E) Problems involving electrons
4. Which of the following is not generally a restriction you put on the wave function
  
that solves Schrodinger’s time independent equation?

A) It must be normalized
B) It must be continuous
C) Its derivative must be continuous
D) It must satisfy Schrodinger’s time independent equation
E) It must be real
5. The solutions of the 1D infinite square well are given above as
 n
  
2 L sin
  nx L

. Why is that factor of 2 L there for?
A) To normalize it correctly
B) To make sure it vanishes at the boundaries x = 0 and x = L
C) To make sure it satisfies Schrodinger’s equation
D) To make the energy formula come out correctly
E) To make it continuous at the boundaries
6. Suppose I have found a solution of Schrodinger’s time independent equation with energy E and wave function
  
. What would the solution
   of
Schrodinger’s time-dependent equation look like?
A)
   i Et 
B)
    i Et 
C)
   i t E
   i t E
E)
    i t E
7. If a single photon is passed through a half-silvered mirror, and two perfect detectors
X and Y are then placed along both subsequent paths, what happens, according to the standard interpretation of quantum mechanics?
A) The photon will always go to X, or always go to Y, depending on the exact position on the mirror it strikes
B) The photon will go to X 50% of the time, and Y 50% of the time, but these are uncorrelated, so 25% of the time it goes to both
C) The photon will go to X 50% of the time, and Y 50% of the time, but never both and never neither
D) A half energy photon will go to each detector, every time
E) A full energy photon will go to each detector, every time
8. Written in terms of the Hamiltonian H , Schrodinger’s time-dependent equation is simply
A) i


 t

B)
C)
 i

 

 t



 i H t


D)
E) i
   i
 t
H


 t

H

9. In classical physics, you normally specify the initial conditions by stating a particle’s position and velocity. In addition to the initial wave function
 
,

0

, what else must be specified in quantum mechanics?
A) The initial position (only)
B) The initial momentum (only)
C) The initial position AND the initial momentum
D) The time derivative of the wave function,

,

0
E) Nothing;
 
,

0

is sufficient

10. Which of the following is true about electrons if you have multiple electrons?
A) There is one wave function for each electron
B) Because of the repulsion between electrons, you can’t usually make more than one electron go into an atom
C) Two electrons cannot go into the same quantum state
D) You can only have one electron for each value of n in the atom
E) Different electrons have different values of s , the total spin quantum number

Part II: Short answer [20 points]
Choose two of the following three questions and give a short answer (2-4 sentences) (10 points each).
11. Classically, if a particle with energy E reaches a barrier of height V
0
, it will continue through (100% of the time) if E > V
0
, and it will be reflected (100% of the time) if E < V
0
. What happens qualitatively quantum mechanically in these two cases?
Quantum mechanically, if the incoming energy is sufficient to make it over the barrier, E > V
0
, then there is a finite probability it will penetrate and a finite probability it will be reflected. If the energy is insufficient, E < V
0
, then it will always be reflected.
Note: Because of the ambiguous wording of the problem, and choice of the word
“barrier,” many different answers were accepted for this question.
12. When we first solved the infinite square well for a particle restricted to the region 0 x L , we initially tried functions like sin
 
and cos
 
, but eventually settled on sin
 
and also restricted k. How did we narrow it down like this?
The wave function had to be continuous, and therefore it must vanish at the ends so it will match the zero value outside the allowed region. To vanish at the origin, we had to pick sin
 
since it didn’t vanish otherwise. To vanish at the other boundary, we had to make sin
  vanish, which happens at multiples of

, so kl
 n

.
13. An electron in hydrogen is described by four numbers, n, l, m, and m s
. Which of these does the radial function R(r) depend on? Which of these does the angular function (spherical harmonics) Y (

,

) depend on?
The wave function for hydrogen is given by
 nlm
 r
 
R nl
   
.
Hence the radial function depends on n and l , while the spherical harmonics depend on l and m .

Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated calculations (20 points each).
14. A harmonic oscillator has an electron in it. It is found that when the electron transitions from the n = 5 to the n = 4 state, a photon is emitted with energy E =
0.356 eV.
(a) What is the angular frequency

for this Harmonic oscillator?
The energy of a particle in the harmonic oscillator is given by E
    n

1
2

.
Therefore the energies of the two states named are 9  
and
2
 
E
5

E
4
  
. Hence the angular frequency

is just
11
2
 
, for a difference of
 


E

0.356 eV
6.582 10

16 
  14

1
(b) What is the minimum energy (zero point energy) that this electron can have?
The minimum energy corresponds to n = 0, which has energy
E
0

1
2
  
1
2

0.356 eV
 
0.178 eV .
(c) Visible light photons have energy ranging from about 1.65–3.26 eV. If this harmonic oscillator starts in the lowest energy state and absorbs one visible light photon, what are the possible states n that it could end up with (I want a list)?
The initial energy is difference in energy is then
E
0

1
2
 
, and the final energy is E n
 
 
E n

E
0
 n
 
. We therefore have n

1
2

 
. The
1.65 eV
1.65 eV
 n

 n
  
3.26 eV ,
0.356 eV
 
3.26 eV ,
4.63
9.16 .
Since n must be an integer, the only possibilities are n = 5, 6, 7, 8, 9.

15. A group of electrons with mass m
  5
5.11 10 eV / c
2 have kinetic energy E =
15.0 eV approach a barrier of height V
0
= 20.0 eV and thickness L = 0.530 nm.
(a) What is the probability that a given electron makes it through the barrier?
We first need to find the damping coefficient

, which is given by
 

2

0

E
  c
2
  


6.582 10

16 

 
 
16 

 8
6.582 10 eV s 2.998 10 m/s



 10

1 
11.46 nm

1
We now substitute this into the transmission coefficient formula, which yields
T

E
16 1
V
0

 
E
V
0

 e

2

L
 
5

16

15 eV
20 eV

1

15 eV
20 eV exp 



1
 
0.530 nm

(b) The thickness L is now increased to a new length, and now it is found that only 1.00×10 –5 of the electrons make it through the barrier. What is L now?
This time we solve for L . So we have
 
5 
E
1.00 10 16 1
V
0



E
V
0

6

2

L

 e

2

L

,

6


 e

2

L 
16

15 eV

15 eV
1
20 eV 20 eV
 
12.61 ,
L

12.61
2


12.61


1


0.550 nm .
e

2

L 
3 e

2

L
,

16. A particle has wave function given by
  

Nxe
 x a
0
 for x

0 .
Some possibly useful integrals appear below.
(a) What is the correct normalization factor N?
To get the normalization correct, we simply demand that the normalization integral yields one, so we have
N
1

 

   2 dx

N
2

0
  xe
 x a

2 dx

N
2

0

2 x e

2 x a dx

N
2
  
3
2!

1
4
2 3
N a ,

4

2
.
a
3 a
3
(b) What are x and
2 x for this wave function?
To calculate these, we simply insert a factor of x or x
2
into the previous integral, yielding x
2 x
 

 x
   2 dx

N
2

0
 
 x a

2 dx

4 a
3

0

3

2 x a x e dx

4 2

4
3!
3 a a
 a
4
4 6
16 a 3

3
2 a ,
 

 x
2
   2 dx

N
2

0
 x
2
 xe
 x a

2 dx

4 a
3

0

4

2 x a x e dx

4 2

5
3 a a
4!
 a
5
4 24
32 a
3
 a
2
3 .
(c) What is the uncertainty
 x for this wave function?
The uncertainty can be found with the expression
 
2  x
2  x
2 
3 a
2   
2 
3 a
2 
9
4 a
2 
2
3 a .
4 a
2 
3
4 a
2
,
Useful integral:

0
 n
  x x e dx
  n n ! , where n
    n .

17. Sometimes it is useful to discuss the total angular momentum of an electron, which is the sum of the orbital angular momentum, and the spin angular momentum. For example, the z-component of the total angular momentum is given by
J z

L z

S z
(a) Write a simple formula for the z-component of the total angular momentum
J z
in terms of some combination of two of the standard quantum numbers, m, and m s
.
We know that L z
  m and S z
  m s
. Combining these, we have
J z

L z

S z
  m
  m s
.
(b) Suppose that an electron in hydrogen has n = 5 and J
 
5

. List all z 2 possible values for m and m s
that would make this possible.
The only allowed values for m are s m
 
1 . Clearly we have s 2 m
  m s

5
2
. Hence the only two solutions are m
 m s
 
5
2
, so
 m
  m s
 
2
 and
 m
  m
 
2

(c) For each of the values m and m s
that you found in part (b), work out explicitly all possible values of s and l, subject to the restrictions n = 5, and it is an electron.
Electrons always have s

1 , so there really isn’t any choice in the matter. As for
2 l , it must be less than n = 5. We also know that m takes on values from –l to + l , which tells us that l must be at least m in size. Hence the possibilities are m
 
2 : m
 
3 : m s
 
1
2
: l

2, 3, 4; l

3, 4; s

1
2
.
(d) For each of the values of s and l you found in part (c), work out the corresponding values of
2
L and
2
S .
The formula for these quantities is L
2   2
 l
2  l

and S
2   2
 s
2  s

. Hence l

2 : L
2   2
6 ; l

3 : s

1
2
:
L
2   2
12 ;
S
2 
3
4
 2
.
l

4 : L
2   2
20 ;