Name _________________
Solutions to Test 3
November 9, 2015
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 J  s  4.136 1015 eV  s
  1.055 1034 J  s  6.582 1016 eV  s
1 eV  1.602 1019 J
Reflection off a step:
 E  E  V
0

R   E  E  V0


1

Barrier penetration:
E
E
T  16 1   e 2 L
V0  V0 

2

 if E  V0


if E  V0
Hydrogen
En
2m V0  E 
13.6 eV  Z 2


n2
1D square well:
 2 2 n2
En 
2mL2
2
  nx 
 n  x 
sin 

L
 L 
n  1, 2,3,

Harmonic Osc.
En    n  12 
n  0,1, 2,
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. Which of the following functions might represent a plane wave   x, t  in quantum
mechanics?
A) cos  kx  t  B) sin  kx  t 
C) eikx it
D) Ne  Ax
2
2
E)
2 L sin  nx L 
2. In the harmonic oscillator, the difference in energy between the n = 100th state and the n =
101st state is
A) 0
B) 12 
C) 
D) 100.5 E) 101.5
3. In experiments where you take a single photon and use half-silvered mirrors to split it in half,
which observation indicates that the single photon apparently goes both ways?
A) Occasionally you can actually observe two photons created from the single photon
B) The two pieces can be recombined, and the resulting interference demonstrates the
effect
C) The momentum transferred to subsequent collisions can be measured, demonstrating that
half of the momentum went each way
D) Detectors can see the quantum wave passing by on each of the two paths
E) The probability of the photon coming out on each path is 50%, proving it went both ways
4. Suppose you have a barrier potential with a finite but large width L but height V0 > E, the
energy. On the other side of the barrier, what does the wave look like?
A) There is no wave at all, because the barrier is too high to get across
B) The wave always penetrates the barrier, so it has the same magnitude as it had before
C) The wave continues as a wave, but the magnitude is slightly reduced
D) The wave damps exponentially, falling off to zero eventually
E) The wave has a very small chance of penetrating, coming out as a much smaller
amplitude wave
5. When calculating the reflection of a wave off a step potential, which of the following must be
done?
A) Match the wave function at the boundary (only)
B) Match the derivative of the wave function at the boundary (only)
C) Eliminate the portion of the wave that is non-physical, as it represents a wave coming in
from the wrong direction (only)
D) All of the above
E) None of the above
6. If a hydrogen atom has an electron in n = 4, l = 2, and m = 1, which of the following is the
equation for the wave function?
A) R42Y21
B) R41Y21
C) R42Y41
D) R21Y41
E) R21Y42
7. Which of the following is impossible for the orbital angular momentum Lz of an electron in
an atom?
A) 0
B) 
C) 
D) 2
E) 12 
8. A particle is in a bound state if its energy is
A) Larger than the potential at infinity
B) Smaller than the potential at infinity
C) Equal to the potential at infinity
D) Larger than the potential at zero
E) Smaller than the potential at zero
9. Which of the following is the equivalent of the momentum operator in one dimension?
 

p2
A)
B) i
C) x
D)
E) None of these
V  x
i x
2m
t
10. The graviton is a hypothetical particle with spin s = 2. The operator S2 then has value
A) 2 2
B) 4 2
C) 5 2
D) 6 2
E) none of these
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. When solving the infinite square well, the solutions we found for an allowed region
0  x  L are given by  n  x   2 L sin  nx L  . Explain (i) why sin is used instead of
cos, (ii) Why we have sin  nx L  rather than some other value sin  kx  is there, and
(iii) why the factor of
2 L is there.
The infinite square well has infinite potential energy outside of the allowed region, and
therefore the wave function must vanish out there. To make it continuous, it must vanish at the
origin, and hence we must use sine rather than cosine. To make it vanish at the other boundary,
x  L , it must take the form sin  nx L  rather than some other value sin  kx  . Finally, the
factor of
2 L is just there to make sure it is normalized, so that   dx  1 .
2
12. Explain qualitatively what the Hamiltonian is (give a formula), and what use the
expectation value of the Hamiltonian H has for a general wave function; that is, what
measurement it would correspond to.
The Hamiltonian is the formula for the energy in terms of the position and momentum.
1 2
We normally use the formula H 
Pop  V  x  for the Hamiltonian. Its expectation value for
2m
a given wave function is the average energy that you would find if you measured the energy of
the particle.
13. For Manganese (Z = 25), give the complete electron
configuration (1s2 …). Assume the conventional rules work.
We fill up the quantum state according to the standard rules, as
sketched at right. The electron configuration is:
1s2 2s2 2p6 3s2 3p6 4s2 3d5 .
1s2
2s2
3s2
4s2
5s2
2p6
3p6 3d10
4p6 4d10 4f14
5p6 5d10 5f14
Part III: Calculation: [60 points] Choose three of the following four questions and perform the
indicated calculations (20 points each).
14. A proton consists of three quarks each of mass m = 5.58  10–28 kg (or m = 313 MeV/c2),
which can be thought of as trapped inside a proton, (treat as a 1D infinite square well)
which has a size of about L= 1.754 fm (1.754  10–15 m).
(a) What is the energy of one quark in this infinite square well in the ground state in
MeV? What is the energy of all three quarks? What fraction is this of rest energy
of the proton, mpc2 = 938 MeV?
We can work in either eV units of SI units. Let’s use SI units. The energy of a single
quark is:
E1 
 2  212
2
2mL

 2 1.055 1034 J  s 
2
2  5.58 1028 kg 1.754 1015 m 
2
 3.200 1011 J 
eV
1.602 1019 J
 1.997 108 eV  199.7 MeV
For three quarks, we would simply add these numbers together to get
Etot  3E1  3 199.7 MeV   599 MeV .
As a fraction of the proton energy, this is
fp 
Etot
599 MeV

 63.9% .
2
mpc
938 MeV
So a substantial part of the proton’s energy may just come from confining the quarks into a small
space.
(b) Suppose one of the three quarks is excited to the first excited state of this infinite
square well, while the others are still in the ground state. What is the energy now?
What fraction is this of the excited nucleon energy mN c 2  1440 MeV ?
The energy of the excited quark only will be E2 
 2  2 22
2mL2
 4 E1 .
The total energy, therefore, will be
  2 E1  E2  2 E1  4 E1  6 E1  1198 MeV .
Etot
As a fraction of the total energy, this is
fN 
Etot
1198 MeV

 83.2% .
2
mN c 1440 MeV
Again, the majority of the energy of this excited nucleon is just the kinetic energy of the trapped
quarks.
15. The harmonic oscillator for a particle of mass m has potential V  x   12 m 2 x 2
(a) Write down the one-dimensional time-independent Schrödinger equation for this
potential.
This is straightforward:
E  x   
2 d 2
2 d 2
1
x
V
x
x





  x   m 2 x 2  x 






2
2
2m dx
2m dx
2
(b) Show explicitly that the wave function   x   e  Ax
2
2
satisfies this equation if
A  m  , and that the energy is the right value for the ground state.
We start by taking two derivatives of the wave function:
2
2
d  Ax2 2
d
 e  Ax 2   12 Ax 2    Axe  Ax 2 ,
e
dx
dx
2
2
d
d
d  Ax2 2
 Axe  Ax 2    Ax  e  Ax 2  Ax
e
dx
dx
dx
2
2
2
2
2
d
  Ae  Ax 2  Axe  Ax 2   12 Ax 2    Ae  Ax 2  A2 x 2 e  Ax 2   A2 x 2  A  e  Ax 2 .
dx
d
  x 
dx
d2
  x 
dx 2




We now substitute this into Schrödinger’s equation, taking advantage of A  m  :
Ee  Ax
2
2
2
2
2
1
A2 x 2  A  e  Ax 2  m 2 x 2 e Ax 2

2m
2
2
2 2
2
  m  2 m   Ax2 2 1

 m 2 x 2 e  Ax 2
 2 x 
e
2m  
2
 

  12   12 m 2 x 2  e  Ax
2
2
 12 m 2 x 2 e  Ax
2
2
 12  e  Ax
2
2
.
E  12  .
The energy came out to E  12  , which matches the formula given for the n = 0 or ground state
of the harmonic oscillator.
16. A particle has normalized wave function given by
 30 a 5  ax  x 2  for 0  x  a ,
  x  
. This wave function is already normalized.
0
otherwise.

Some possibly useful integrals appear below. It has expectation values x 2  72 a 2 and
p 2  10 2 a 2 .
(a) Find the expectation values x and p for this wave function.

We calculate each of these using     *  x    x  dx , and use the integral tables

below.

x    *  x  x  x  dx 


p    *  x  pop  x  dx 

a
30
30 2a 6
a
2
2
ax
x
x
ax
x
dx




 ,



5 
5
a 0
a 456 2
a
a
30
30
 d
ax  x 2 
ax  x 2  dx  5   ax  x 2   a  2 x  dx

5 
a 0
i dx
ai 0
30  a 3
a4 
 5 a 
 2
0,
3  4 
a i  23
We could have more quickly determined p  0 because the wave function is real.
(b) What are the uncertainties x and p for this wave function?
The uncertainties are given by:
2
x 
x2  x
p 
p2  p
2
2

2a 2  a 
2 1
a
,
   a
 
7 2
7 4
28

10 2
 10
2
 0 
.
2
a
a
(c) Check that it satisfies the uncertainty relation.
The product of these is
 x  p  
This is larger than
1
2
a
Useful integrals:
 , as it must be.
2
n
 x  ax  x  dx 
0
a  10
5


  0.598
14
a
28
a n 3
,
 n  2  n  3
a
2
n
 x  ax  x  dx 
0
2
2a n  5
.
 n  3 n  4  n  5
17. An electron in hydrogen has l = 2.
(a) What is the value of L2?
The formula is L2   l 2  l   2   22  2   2  6 2 .
(b) What are the possible values of m, and the corresponding possible values for Lz?
The quantum number m ranges from –l to +l in steps of one, so m  2, 1, 0,1, 2 . The
formula for Lz is Lz  m , so the corresponding values are Lz  2,  , 0, , 2 .
(c) What are the possible values for ms, and the corresponding possible values for Sz?
The possible values for ms are always ms   12 , and the corresponding value is
S z  ms    12  .
(d) What are the possible values for the principal quantum number n? What is the
lowest (most negative) energy possible for this electron?
The principal quantum number n must be an integer that exceeds l, so n  3, 4,5, .
The smaller n is, the lower the energy, so the lowest energy is
E  E3  
13.6 eV
 1.51 eV .
32
(e) For a single value of n, and l = 2, how many total quantum states are possible?
Since m takes on five values, and ms takes on two values, the number of pairs  m, ms  is
5  2  10 .