1st Midterm Exam Solutions

1 st Midterm Exam Solutions
1. (20%)
(a)
Consider special relativity, the relationship between energy and momentum
reads
E
E m c
c
Since the de Broglie wavelength of a particle is inversely proportional to its
momentum, i.e.
h
(1%)
p
Therefore, the de Broglie wavelength of a particle is never shorter than a
photon with the same energy. (2%)
(b)
Since the particle’s kinetic energy is never greater than its total energy, i.e.
2
2
E mc K K E mc (3%)
So, the particle’s kinetic energy is never greater than the photon’s energy,
since the photon has the same energy as particle’s total energy. (2%)
(c)
From (a), since the total energy of particle is much greater than its rest energy,
the momentum of the particle reads
1 2 2 4 E
p E m c (3%)
c
c
which is the same as photon’s momentum. Since the particle and photon have
approximately the same momentum, they have almost the same de Broglie
wavelength. (2%)
(d)
From (a), since the particle’s kinetic energy is much less than its rest energy,
we have
1 2 2 4 1
2
2 1
1 E
p E m c E
mc E
mc K2E
K 2KE
(3%)
c
c
c
c c
Therefore, the de Broglie wavelength of particle is much longer than
photon’s wavelength. (2%)
2 2 2 4 1 2 2 4
p c m c p (2%)
2. (5%)
From Schrodinger equation, the curvature of the wavefunction is negative
(positive) if the wavefunction is positive (negative), which lead to oscillating

ehavior.
3. (25%)
(a)
2
p
The Hamiltonian for a free particle is Hˆ
. Its eigen-value corresponds to
2ˆ
m
the total energy of the particle. Therefore, its eigen-value is allowed to be any
positive real number. (5%)
(b)
For a free particle, satisfies the Schrodinger equation
2 2
d
i x, t
x,
t
2
t
2m
dx
Apply the separation of variables and let the eigen-value be E , we have the
following solution for Schrodinger equation
x, t exp i t
k x (3%)
E 2mE
Where , k . (2%)
(c)
The most general form of wavefunction for free particle is the superposition of
all possible modes which reads
Ak
x
t
dkAk
exp
it
k x
, (5%)
Where is arbitrary function satisfies normalization condition.
(d)
Phase velocity describes the propagating velocity of phase, which means if an
observer travels with the phase velocity of a travelling wave, the observed phase
keep unchanged. It is not the actual propagating velocity of wave when the
dispersion relation is not linear. For matter waves, since the dispersion relation is
quadratic, its phase velocity reads
E p v
v p
(5%)
k p 2m
2
Which is obviously not the propagating velocity of matter wave.
(e)
The group velocity of matter wave is
v g
2
p p
k
E v
k p p
m
(3%)
2 m
Which is the actual propagating velocity of matter wave. This is only valid

for wave-packet, which means the wave-vector must be localized in k -space
as well as its position in real space. (2%)
4. (35%)
(a)
The energy eigen-value of given eigen-states are
2 2
2 2 2
2 2 2
d n
n
En
n
E
2 n
2 n
n
(5%)
2
2m
dx 2ma
2ma
(b)
The Heisenberg uncertainty principle tells
x p
(2%)
2
We may simply estimate x and p by
x a , p
2 mEg
(1% each)
a
x p
(1%)
2
Which satisfies Heisenberg uncertainty principle.
(c)
From (a), if the size of box is halved, the eigen-energies become four times
of original. Similarly, they become one fourth of original eigen-energy if the
size of box is doubled. (5%)
(d)(5%)
From Fourier transformation (uncertainty principle), the larger spread in
position space lead to smaller spread in momentum space, and vice versa
(e)
(f)
(g)
1
5
i 2 3 x
4 i
2x
3
4e
sin e sin
1 2
(5%)
a 5 a 5 a
(The phase factor can be ignored here.)
2
2 2
2 2
p a
*
d 9 16 73
E dx
E1
E2
0
2
2m
2m
dx
(5%)
25 25 50 ma
W E1
2
hen we measure the energy of the system, only and E are possible to
be obtained for the given state. If we do the same measurement on the same
system with the same given state many times, the average value of the result
is given by (f), although all these measurement only give E
1
and E 2
each

time. (5%)
5. (15%)
(a) (5%)
The top figure has highest oscillation frequency,
which means it has highest kinetic energy. Therefore,
it corresponds to E
A
. The bottom figure shows that
the total energy is lower than V 0
. The exponential
decay of probability density at right hand side of
bottom figure is because of tunneling effect. Then, we
may deduce that the bottom figure corresponds to E
C
.
The middle figure shows that its total energy is higher
than V
0
, but lower than top figure, so it corresponds to
E
B
.
(b) (5%)
For the case that the particle has total energy higher
than the potential energy at region 0
, a
as shown in
top and middle figure, the particle has higher kinetic
energy in region
a,0
. Which means it passes
through this region requires less time so that it is harder
to be found in this region. For the bottom figure, since
the particle’s total energy is not enough to encounter the
potential barrier in region 0 , a, it can only move back and forth in region
a,0. However, it is still possible to find the particle in region 0,
a
because
of tunneling effect.
(c) (5%)
E
C
is not the ground state because its probability density has several nodes.
The probability density of ground state is expected as shown in the following
figure. Single peak appears in left hand side, with very small magnitude of
“leaking” to right hand side which is caused by tunneling effect.