Name _________________
Solutions to Test 3
November 10, 2014
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
Reflection off a step:
 E  E  V
0

R   E  E  V0


1

2

 if E  V0


if E  V0
Barrier penetration:
E
E
T  16 1   e 2 L
V0  V0 

2m V0  E 

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)
Hydrogen
En
13.6 eV  Z 2


n2
1D square well:
 2 2 n2
En 
2mL2
2
  nx 
sin 
 n  x 

L
 L 
n  1, 2,3,
1. Which of the following must be true of the wave function   x, t 
A)
B)
C)
D)
E)
It must be normalized (only)
It must vanish in any region where the potential is bigger than the energy (only)
It must be continuous (only)
Both A and B are true, but not C
Both A and C are true, but not B
2. Suppose   x  is a solution of Schrödinger’s time-independent equation with energy E.
What is the corresponding solution   x, t  of Schrödinger’s time-dependent equation?
A)   x  e iE t B)   x  e iE  t C)   x  e iEt  D)   x  e iEt 
E) None of these
3. For the infinite square well with allowed region from x = 0 to L, we included solutions that
went like sin  nx L  but rejected those that went like cos  nx L  . Why?
A)
B)
C)
D)
E)
The function must vanish at the boundaries, so sine works and cosine doesn’t
You can normalize sine, but not cosine
Cosine has negative energy, which is impossible
Sine satisfied the differential equation in the interior, cosine did not
We didn’t have a reason, I just hate cosine.
4. The smallest energy a harmonic oscillator with angular frequency  can have is
A) 0
B) 12 
C) 
D) 32 
E) None of these
5. According to quantum mechanics, if you hit a step potential with less energy than the height
of the potential, E < V0, what happens in the classically forbidden region?
A) The wave continues with the same amplitude as before
B) The wave continues as a wave, but with a much smaller amplitude
C) The wave grows exponentially in the forbidden region
, eventually
becoming infinite
D) The wave damps exponentially in the forbidden region, falling towards zero
E) The wave is automatically zero in the forbidden region
6. Which of the following is the lowest energy state in a typical atom?
A) 3p
B) 4p
C) 4f
D) 3d
E) 4d
7. Which of the following corresponds to the momentum as operator, pop?
dx
 
p2
A) mv
B) m
C)
D) k
E)
V
2m
i x
dt
8. If you measured the total spin-squared of an electron, S2, the result you would get would be
A) 0
B) 14  2
C) 12  2
D) 34  2
E)  2
9. Which of the following is true when you do quantum mechanics in three dimensions?
A) The momentum operator gets replaced by a single radial momentum operator
B) The wave function becomes a function of all three positions: x, y, and z
C) There are three wave functions, one for each of the dimensions
D) The energy is generally the product of the energies in each of the three dimensions
2
E) The probability density is no longer represented by the magnitude squared  of the
wave function
10. If we calculate the expectation value of the Hamiltonian, H , the result tells you what you
would get, on average, if you measure the
A) Position B) Momentum C) Angular momentum D) Mass
E) Energy
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 we were doing reflection from a step barrier for V0 < E, we found solutions in the
“transmitted” region x > 0 that looked like   x   Ceikx  De  ikx , but we set D = 0.
Why? Later, when we redid it for V0 > E, we found solutions that looked like
  x   e x , but we threw out e  x . Why?
The D term is multiplied by e ikx , which represents a wave moving to the left. But since
we are in the region x > 0, this would represent an incoming wave from the right. The problem
we were dealing with had a particle coming in from the left, so this isn’t a term we want. For V0
> E, the solution that goes like e  x grows to infinity, and therefore will not be normalizable.
12. Explain qualitatively how a scanning tunneling electron microscope works.
In a scanning tunneling microscope, a very sharp needle with a tip with just one atom is
held over some conducting material. The electrons tunnel across the tiny gap between the needle
and the material, as allowed by the tunneling. The tip is then scanned across the material, and
the needle’s position is adjusted to keep the current constant. The height of the needle is then
used to map the height of the surface, with atomic scale precision.
13. According to the Pauli exclusion principle, you can only have one electron for any given
set of quantum numbers. For the 1s level of hydrogen (or anything), you have quantum
numbers n = 1, l = 0, and m = 0, but you can still fit two electrons into this level. How is
this possible?
In addition to the listed quantum numbers, an electron also has spin, which is described
for an electron by the quantum numbers ms   12 . Thus you can put two electrons into this
state, one with each value of ms.
Part III: Calculation: [60 points] Choose three of the following four questions and perform the
indicated calculations (20 points each).
14. An electron of mass m  5.11105 eV / c 2 is trapped in a one-dimensional trap of size
L  3.00 109 m
(a) What is the energy of this electron, in eV if it has its minimum possible energy?
The lowest energy corresponds to n = 1, and so this energy is
E1 
 22
2mL2

 2  6.582 1016 eV  s   2.998 108 m/s 
2
 2  2c 2

2  5.11105 eV  L2
2  5.11105 eV  3.00 109 m 
2
 0.04178 eV
(b) What energy of photon would be required to knock it up to the level n = 5?
The amount of energy required is the difference between the ground state and the new
level, so
E  E5  E1 
 2  2 52

2mL2
 22

2mL2
24 2  2
 24  0.04178 eV   1.003 eV
2mL2
(c) Infrared photons have energy less than 1.65 eV. To which levels can this electron be
elevated if it absorbs a single infrared photon?
In a manner similar to before, the energy must be given by
E  En  E1 
 2 2 n2
2
2mL

 22
2
2mL
n

2
 1  2  2
2
2mL
  n 2  1  0.04178 eV 
We want this to be in the infrared range, so E  1.65 eV . We therefore have
n
2
 1  0.04178 eV   1.65 eV
1.65 eV
 39.43
0.04178 eV
n 2  39.43  1  40.43
n2  1 
n  40.43  6.36
Hence it can lift things up to the level n = 6, so n = 1, 2, 3, 4, or 5.
15. A group of 106 electrons with kinetic energy E = 18.00 eV approach a step with positive
potential V0 = +16.00 eV.
(a) How many of the electrons are reflected?
The reflection probability is
 E  E  V0
R
 E  E V
0

2
  18.00 eV  2.00 eV 
2
  0.5000   0.2500
  



  18.00 eV  2.00 eV 
2
If we multiply this by the number of electrons, we get
N R   0.25  106   250, 000 .
(b) The potential is now changed to a negative but unknown value. It is nonetheless
found that the same number of electrons are reflected. What is the new potential?
The reflection probability must be unchanged, so we must still have
 E  E  V0

 E  E V
0

2

  0.2500


Now, if we simply take the positive square root, it will still have the same value we had before.
Hence we must take the negative square root. We therefore have
E  E  V0
E  E  V0
 0.500 ,
E  E  V0  0.500 E  0.500 E  V0 ,
1.500 E  0.500 E  V0 .
We now square both sides, yielding
2.2500 E  0.2500  E  V0   0.2500 E  0.2500V0 ,
2.000 E  0.2500V0 ,
V0  
2.000
E  8.00 E  144.0 eV .
0.2500
2a 3 
. Some possibly
x2  a2
useful integrals appear below. This wave function has p 2   2  2a 2 
16. A particle has normalized wave function given by   x  
(a) What are x and x 2 for this wave function?
We simply set up the appropriate integrals in each case, and then perform the integral.
We have
1
1
2a 3
x     x  x  x  dx 
x
dx 

  x 2  a 2 x 2  a 2



*
2a 3
x 2    *  x  x 2  x  dx 


2a 3

x

1
1
2a 3
2

x
dx
  x 2  a 2 x 2  a 2




xdx
2
 a2 


x
2
0,
x 2 dx
2
 a2 
2

2a 3 

 a2.
 2a
(b) What is the uncertainty x for this wave function?
We simply use the formula
 x 
2
 x2  x
2
 a 2  02 ,
x  a .
(c) What is p ? What is the uncertainty p for this wave function?
Since the wave function is real, we conclude with no work that p  0 . The uncertainty
is then
 x 
2
2
2
2
 p  p  2 0  2
2a
2a

p 
.
a 2
2
2
(d) Check that it satisfies the uncertainty relation.
We simply multiply our previous results, and have
 x  p   a
So it satisfies the uncertainty relation.

a 2


 0.7071  12  .
2
17. An electron in hydrogen is in the state n = 3, l = 1, m = –1.
(a) If we were to measure the energy E, the angular momentum squared L2, and the zcomponent of angular momentum Lz, what values would we get?
There are formulas for each of these quantities, so we find
13.6 eV  Z 2

E
13.6 eV
 1.51 eV ,
32
n
L2   l 2  l   2  12  1  2  2 2 ,
2

Lz  m   .
(b) Write down the explicit form of the wave function  nlm  r ,  ,   . Several radial and
angular wave functions are given below.
We use the formula
 n ,l ,m  r   Rn ,l  r  Yl ,m  ,   ,
 3,1,1  r   R3,1  r  Y1,1  ,   

4 2r 
r 
3
1   e  r 3a
sin  e  i
5 
8
27 3a  6a 
r 

1   e  r 3a sin  e  i

27  a 5  6a 
2r
(c) For what locations can we be certain the particle cannot be found? You can give
your answers as values of r, , or  as needed.
We have several factors multiplied. If any of them vanish, the product vanishes, and the
particle cannot be at those points. The factor e  i can never vanish, and neither can e  r 3a . The
three factors that can vanish are r, 1  r  6a  , and sin  . For r, the second option is r = 6a. To
solve sin   0 , we recall that 0     , and the places sine vanishes in this region are 0 and .
So the places where it vanishes are:
r  0, 6a and   0, 
The two values of  correspond to the positive and negative z-axis, so the z-axis (which includes
r = 0) and a sphere of radius 6a.
R1,0 
2
a
Y1,0 
3
e  r a , R3,0 
 2r 2r 2   r 3 a
4 2r
e
, R3,1 
1  
2 
3
3 3a  3a 27 a 
27 3a 5
2
r  r 3a

1   e
 6a 
3
3
21
cos  , Y1,1 
sin  e  i , Y3,1 
sin   5cos 2   1 e  i
4
8
64