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PHY4604
Final Exam Solutions
Department of Physics
Page 1 of 8
PHY 4604 Final Exam Solutions
Monday December 5, 2005
(Total Points = 100)
Problem 1 (10 points):
Circle true or false for following (1 point each).
(a) (True or False)
If the expectation value of the operator O
op
is a real number then O
op
is
hermitian.
(b) (True or False)
If 
ψ
> is a (normalized) “ket” vector, then the operator O
op
= 
ψ
><
ψ

has the
property that (O
op
)
2
= 1.
(c) (True or False) If
the operator U
op
is unitary then, its inverse is equal it its hermitian
conjugate.
(d) (True or False)
If P
op
is the parity operator and if
)
1
(
2
1
op
op
P
O
±
=
±
, then
±
±
=
op
op
O
O
2
)
(
.
(e) (True or False)
One can simultaneously know the precise values of commuting observables.
(f) (True or False)
The overall wave function (space × spin) for two identical spin 3/2 particles is
always symmetric under the interchange of the two particles.
(g) (True or False)
A
µ

particle has the same charge as an electron but its mass is 207 times
greater, hence the ground state energy of a muonic atom (
µ

and a proton) is lower than the
ground state of hydrogen.
(h) (True or False)
In SU(2), 3 × 3 = 4 + 3 + 2.
(i) (True or False)
If one neglects the electronelectron interactions then the ground state of
helium would have a degeneracy of 2.
(j) (True or False)
The maximum number of electrons that can be in the n
th
energy level of an
atom is n
2
.
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View Full DocumentPHY4604
Final Exam Solutions
Department of Physics
Page 2 of 8
Problem 2 (30 points):
Consider a bead of mass
m
that slides frictionlessly
around a circular wire ring of radius
R
.
The bead is constrained to lie on the
circle.
This is similar to the free particle with Hamiltonian
2
2
2
2
2
2
2
2
2
2
2
θ
d
d
mR
ds
d
m
m
p
H
h
h
−
=
−
=
=
.
where
s = R
θ
is the distance around the circle. We look for stationary state
solutions of the form
h
/
)
(
)
,
(
iEt
e
t
−
=
Ψ
ψ
with the timeindependent
Schrödinger equation given by
E
d
d
mR
=
−
2
2
2
2
2
h
.
Consider separately the solutions corresponding to clockwise,
+
, and counterclockwise,
−
,
movement around the wire. The probability density
2


)
(
±
±
=
ρ
is the probability of finding
the particle between
θ
and
θ
+d
θ
and is normalized as follows:
1
)
(
2
0
=
∫
±
π
d
.
(a) (20 points)
Find the stationary states
±
(with the correct normalization) and the
corresponding allowed energies. (Hint: require
)
(
)
2
(
=
+
since
s
and
s+2
π
R
correspond
to the same point on the wire)
Answer:
in
n
e
m
2
1
)
(
=
±
,
2
2
2
2
mR
n
E
n
h
=
±
(n = 0, 1, 2, …)
Solution:
We see that
α
2
2
2
−
=
d
d
with
2
2
2
2
h
E
mR
=
and
2
2
2
2
mR
E
h
=
with solution
t
i
t
i
Be
Ae
−
+
+
=
)
(
For
α
> 0 the first term corresponds to counterclockwise movement and the second term
corresponds to clockwise movement.
We also know that
αθ
i
i
i
i
Be
Ae
Be
Ae
−
+
+
−
+
+
+
=
=
+
=
+
)
(
)
2
(
)
2
(
)
2
(
which implies that
0
)
1
(
)
1
(
2
2
=
−
+
−
−
−
+
+
i
i
i
i
e
Be
e
Ae
.
This equation implies that
1
2
=
±
i
e
which implies that
α
= n = 0, ±1, ±2, …and hence
in
n
e
−
+
=
2
1
)
(
and
2
2
2
2
mR
n
E
n
h
=
+
(n = 0, 1, 2, …)
in
n
e
+
−
=
2
1
)
(
and
2
2
2
2
mR
n
E
n
h
=
−
(n = 0, 1, 2, …).
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