PHY4604 Fall 2006
Final Exam Solutions
Department of Physics
Page 1 of 25
PHY 4604 Final Exam Solutions
Wednesday December 6, 2006
(Total Points = 100)
Problem 1 (25 points):
Consider the following onedimensional potential:
∞
+
−
=
0
0
)
(
V
x
V
L
x
L
x
L
x
≥
<
≤
<
0
Where
V
0
is a positive (real) constant.
(A)
Suppose that particles with mass m and energy
E > 0
enter from the left in
region 1 and travel to the right and encounter the potential
V(x)
.
Let
2
2
2
2
ε
mL
E
h
=
and
E
V
r
0
1
+
=
.
(1) (5 points)
Calculate the probability that particles entering from the left in region 1 will be
reflected back (
i.e.
calculate the ratio of the probability flux in region 1 traveling to the left to
the incident flux in region 1 traveling to the right,
1
1
/
j
j
P
R
r
s
=
).
Express your answer in terms of
ε
and r.
Answer:
P
R
= 1
.
Solution:
We look for solutions of the timeindependent Schrödinger equation
)
(
)
(
)
(
)
(
2
2
2
2
x
E
x
x
V
dx
x
d
m
ψ
ψ
ψ
=
+
−
h
or
)
(
))
(
(
2
)
(
2
2
2
x
x
V
E
m
dx
x
d
ψ
ψ
−
−
=
h
with
h
/
)
(
)
,
(
iEt
e
x
t
x
−
=
Ψ
ψ
. In the region x < 0 (region 1) for E > 0 and V(x) = 0 we have
)
(
)
(
2
)
(
2
2
2
2
x
k
x
mE
dx
x
d
ψ
ψ
ψ
−
=
−
=
h
with
L
mE
k
ε
=
=
2
2
h
and
2
2
2
2
2
2
2
ε
mL
m
k
E
h
h
=
=
The most general solution is
ikx
ikx
Be
Ae
x
−
+
+
=
)
(
1
ψ
,
In the region 0 < x < L (region 2) we have
iqx
iqx
De
Ce
x
−
+
+
=
)
(
2
ψ
,
with
rk
V
E
m
q
=
+
=
2
0
)
(
2
h
, where
E
V
k
q
r
0
1
+
=
=
.
In the region x > L (region 3) we have
0
)
(
3
=
x
ψ
,
The boundary conditions at x = L implies that
0
)
(
2
=
L
ψ
and hence
0
=
+
−
+
iqL
iqL
De
Ce
or
iqL
Ce
D
2
+
−
=
.
Thus,
)
(
)
(
2
2
iqx
iqL
iqx
e
e
e
C
x
−
+
+
−
=
ψ
,
At x = 0 the wave function must be continuous with a continuous derivative which yields
x
V(x)
V
0
V = +infinity
L
Region 1
Region 2
Region 3
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PHY4604 Fall 2006
Final Exam Solutions
Department of Physics
Page 2 of 25
)
0
(
)
0
(
2
1
ψ
ψ
=
and
)
1
(
2
iqL
e
C
B
A
+
−
=
+
0
2
0
1
)
(
)
(
=
=
=
x
x
dx
x
d
dx
x
d
ψ
ψ
and
)
1
(
2
iqL
e
Ciq
ikB
ikA
+
+
=
−
Thus,
)
1
(
2
iqL
e
C
B
A
+
−
=
+
and
)
1
(
2
iqL
e
Cr
B
A
+
+
=
−
And
)
)
1
(
)
1
((
2
2
iqL
e
r
r
C
A
+
−
+
+
=
and
)
)
1
(
)
1
((
2
2
iqL
e
r
r
C
B
+
+
+
−
−
=
Thus,
iqL
e
r
r
A
C
2
)
1
(
)
1
(
2
+
−
+
+
=
and
A
e
r
r
e
r
r
B
iqL
iqL
2
2
)
1
(
)
1
(
)
1
(
)
1
(
+
+
−
+
+
+
+
−
−
=
.
The relfection probability is given by
1
)
1
)(
1
(
)
1
)(
1
(
)
1
(
)
1
(
)
1
)(
1
(
)
1
)(
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(




2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
=
+
−
+
+
−
+
−
+
+
−
+
+
−
+
+
+
+
−
=
−
+
+
+
+
−
−
+
+
+
+
−
=
−
+
+
+
+
−
=
=
−
+
−
+
−
−
+
+
+
+
iqL
iqL
iqL
iqL
iqL
iqL
iqL
iqL
iqL
iqL
m
k
m
k
R
e
r
r
e
r
r
r
r
e
r
r
e
r
r
r
r
e
r
r
e
r
r
e
r
r
e
r
r
e
r
r
e
r
r
A
B
P
h
h
(2) (5 points)
Calculate the probability that particles traveling to the right in region 2 will be
reflected back (
i.e.
calculate the ratio of the probability flux in region 2 traveling to the left to
flux in region 2 traveling to the right,
2
2
/
j
j
R
r
s
=
).
Express your answer in terms of
ε
and r.
Answer:
R = 1
.
Solution:
We see that
1




2
2
2
2
=
=
=
+
iqL
m
q
m
q
e
C
D
R
h
h
.
(3) (5 points)
Calculate the ratio of the probability flux in region 2 traveling to the right to the
incident flux in region 1 traveling to the right,
1
2
/
j
j
P
r
r
=
).
Express your answer in terms of
ε
and r. What are the maximum and minimum values of P(
ε
) (express your answer in terms of r)?
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