PHYS851 Quantum Mechanics I, Fall 2009
HOMEWORK ASSIGNMENT 12
Topics Covered:
Motion in a central potential, spherical harmonic oscillator, hydrogen atom, orbital
electric and magnetic dipole moments
1. [20 pts] A particle of mass
M
and charge
q
is constrained to move in a circle of radius
r
0
in the
x

y
plane.
(a) If no forces other than the forces of constraint act on the particle, what are the energy levels
and corresponding wavefunctions?
If the particle is forced to remain in the xy plane, then it can only have angular momen
tum along the zaxis, so that
L
=
L
z
e
z
and
L
2
=
L
2
z
.
The kinetic energy can be found two ways:
Method 1:
Using our knowledge of angular momentum.
We start by choosing
φ
as our co
ordinate
H
=
L
2
2
I
=
L
2
z
2
Mr
2
0
(1)
so that the eigenstates are eigenstates of
L
z
→ 
i
partial
φ
, from which we see know that the energy levels are then
E
m
=
2
m
2
2
Mr
2
0
, where
m
=
0
,
±
1
,
±
2
,
±
3
. . . ...
, and the wavefunctions are
φ

m
=
1
√
2
π
e
imφ
.
Method 2:
Solution from first principles.
We start by choosing
s
as our coordinate, where
s
is the distance measured along the circle. The classical Lagrangian is then
L
=
M
˙
s
2
2
(2)
the canonical momentum is
p
s
=
∂
s
L
=
M
˙
s
. The Hamiltonian is then
H
=
p
˙
s
 L
=
p
2
s
2
M
(3)
promoting
s
and
p
s
to operators, we must have [
S, P
s
] =
i
, so that in coordinate representation,
we can take
S
→
s
, and
P
s
→ 
i ∂
s
, which gives
H
=

2
2
M
∂
2
s
(4)
the energy eigenvalue equation is then

2
2
M
∂
2
s
ψ
(
s
) =
Eψ
(
s
)
(5)
or equivalently
∂
2
s
ψ
(
s
) =

2
ME
2
ψ
(
s
)
(6)
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This has solutions of the form:
ψ
(
s
)
∝
e
±
i
√
2
ME
s
(7)
singlevaluedness requires
ψ
(
s
+ 2
πr
0
) =
ψ
(
s
)
(8)
which means
√
2
ME
2
πr
0
= 2
πm
(9)
where
m
is any integer. This gives
E
=
2
m
2
2
Mr
2
0
(10)
so that
ψ
m
(
s
) =
e
ims/r
0
√
2
πr
0
(11)
Both methods agree because
s
=
r
0
φ
.
(b) A uniform, weak magnetic field of amplitude
B
0
is applied along the
z
axis. What are the new
energy eigenvalues and corresponding wavefunctions?
Using the angular momentum method, we now need to add the term

qB
0
2
M
L
z
to the Hamil
tonian to account for the orbital magnetic dipole moment, which gives
H
=
L
2
z
2
Mr
2
0

qB
0
2
M
L
z
(12)
so that the eigenstates are still
L
z
eigenstates,
ψ
m
(
φ
) =
e
imφ
√
2
π
, where
m
= 0
,
±
1
,
±
2
, . . .
, but the
degeneracy is lifted so that
E
m
=
2
m
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 Fall '08
 Moore,M
 mechanics, Charge, Energy, Kinetic Energy, Mass, Work, Magnetic moment

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