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PHY4604
Exam 1 Solutions
Department of Physics
Page 1 of 8
PHY 4604 Exam 1 Solutions
Monday October 3, 2005
(Total Points = 100)
Problem 1 (10 points):
Circle true or false for following (1 point each).
(a) (True or False)
One of the “breakthroughs” that lead to quantum mechanics was the idea of
associating differential operators with the dynamical variables.
(b) (True or False)
The wave function
Ψ
(x,t)
must vanish in a region of infinite potential.
(c) (True or False) It is possible for a free particle to have a definite energy.
(d) (True or False)
In quantum mechanics particles can enter the “classically forbidden” region
where V
0
> E (
i.e.
KE < 0).
(e) (True or False)
The operator
(A
op
A
↑
op
)
is hermitian.
Note is we let
O = A
op
A
↑
op
then
O
↑
= A
↑
op
 A
op
=  (A
op
A
↑
op
)= O
.
(f) (True or False)
If
A
op
and
B
op
are hermitian then
A
op
B
op
is also hermitian.
Note is we let
O = A
op
B
op
then
O
↑
= A
↑
op
B
↑
op
= A
op
B
op
= O
.
(g) (True or False)
If
P
op
is the parity operator,
P
op
ψ
(x) =
ψ
(x)
, then
P
op
2
= 1
.
(h) (True or False)
Solutions of Schrödinger’s equation of the form
)
(
)
(
)
,
(
t
x
t
x
φ
ψ
=
Ψ
correspond to states with definite energy E.
(i) (True or False)
Solutions of Schrödinger’s equation of the form
)
(
)
(
)
,
(
t
x
t
x
=
Ψ
correspond to states in which the probability density
2

)
,
(

)
,
(
t
x
t
x
Ψ
=
ρ
is independent of time.
(j) (True or False)
Schrödinger’s equation is valid for all velocities even when v
→
c.
Problem 2 (30 points):
Consider an infinite square well defined by
V(x) = 0
for
–L/2 < x < +L/2
,
V(x) = +
∞
otherwise.
We look for stationary states of the form
h
/
)
(
)
,
(
t
iE
n
n
n
e
x
t
x
−
=
Ψ
.
Parity is a
good quantum number in this problem since V(x) = V(x) and hence the
stationary state solutions are either even or odd under parity as follows:
)
(
)
(
)
(
)
(
)
(
)
(
x
x
x
P
n
n
n
+
+
+
+
+
+
=
−
=
)
(
)
(
)
(
)
(
)
(
)
(
x
x
x
P
n
n
n
−
−
−
−
−
−
−
=
−
=
(a) (5 points).
Calculate the (normalized) even parity wave functions
)
(
)
(
x
n
+
+
and their
corresponding energies,
+
n
E
.
V = +infinity
V = +infinity
Infinite Square Well
L/2
+L/2
x
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View Full DocumentPHY4604
Exam 1 Solutions
Department of Physics
Page 2 of 8
Answer:
)
/
)
1
2
cos((
2
)
(
)
(
L
x
n
L
x
n
π
ψ
−
=
+
+
+
,
2
2
2
2
2
2
)
1
2
(
2
2
−
=
=
+
+
+
n
mL
m
k
E
n
n
h
h
, where n
+
= 1, 2, 3, …
Solution:
We look for solutions of the timeindependent Schrödinger equation
)
(
)
(
)
(
)
(
2
2
2
2
x
E
x
x
V
dx
x
d
m
=
+
−
h
or
)
(
))
(
(
2
)
(
2
2
2
x
x
V
E
m
dx
x
d
−
−
=
h
with
h
/
)
(
)
,
(
iEt
e
x
t
x
−
=
Ψ
. In the region –L/2 < x < L/2 for E > 0 and V(x) = 0 we have
)
(
)
(
2
)
(
2
2
2
2
x
k
x
mE
dx
x
d
−
=
−
=
h
with
2
2
h
mE
k
=
and
m
k
E
2
2
2
h
=
The most general solution is
ikx
ikx
Be
Ae
x
−
+
+
=
)
(
.
The even parity solution have
ψ
(x) =
ψ
(x) and hence
A = B
and hence
)
cos(
2
)
(
)
(
)
(
kx
A
e
e
A
x
ikx
ikx
=
+
=
−
+
+
The boundary condition at x = L/2 is
)
2
/
cos(
2
0
)
2
/
(
)
(
kL
A
L
=
=
+
,
which implies that kL/2 = (n
+
1/2)
π
,
where n
+
= 1, 2, 3, …
Thus,
)
/
)
1
2
cos((
2
)
(
)
(
L
x
n
A
x
n
−
=
+
+
+
.
The normalization is determined by requiring that
1
2
4
)
2
sin(
2
)
1
2
(
4
)
(
cos
)
1
2
(
4
)
/
)
1
2
((
cos
4

)
(

2
)
(
)
(
2
)
(
)
(
2
2
2
/
2
/
2
2
2
/
2
/
2
)
(
2
1
2
1
2
1
2
1
=
=
+
−
=
−
=
−
=
−
−
−
+
−
−
−
+
−
+
−
+
+
+
+
+
+
∫
∫
∫
L
A
y
y
n
L
A
dy
y
n
L
A
dx
L
x
n
A
dx
x
n
n
n
n
L
L
L
L
n
.
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 Spring '07
 Field
 mechanics

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