Physics 512
Winter 2003
Final – 4:00–7:00 (3 hours), April 23, 2003 – Solutions
[30 pts]
1. A particle of mass
m
and charge
q
sits in an isotropic threedimensional harmonic
oscillator potential, and is described by the Hamiltonian
H
0
=
p
2
2
m
+
1
2
mω
2
r
2
Note that eigenstates of
H
0
,
L
2
and
L
z
may be labeled by

n, , m
where
n
= 0
,
1
,
2
, . . .
and
E
n
= (
n
+
3
2
)¯
hω
. At time
t
=
−∞
the oscillator is in its ground state,

0
,
0
,
0 . It
is then acted upon by a spatially uniform but time dependent electric field
E
(
t
) =
E
0
e
−
(
t/τ
)
2
ˆ
z
(where
E
0
and
τ
are constant).
[5]
a
) Show that, to first order in the perturbation, the only possible excited state the
oscillator could end up in is the

1
,
1
,
0
state.
The transition amplitude is given by first order time dependent perturbation the
ory:
A
k
←
s
=
−
i
¯
h
∞
−∞
e
iω
ks
t
k

V
(
t
)

s dt
Since, for a constant electric field in the
ˆ
z
direction, the perturbation interaction
is given by
V
(
t
) =
−
q
E
·
d
=
−
q
E
0
ze
−
(
t/τ
)
2
we find
A
k
←
s
=
i
¯
h
q
E
0
∞
−∞
e
iω
ks
t
−
t
2
/τ
2
k

z

s dt
(1)
The possible excited states that the oscillator could end up in (to first order) are
determined by selection rules on the matrix element
k

z

s
.
First of all, it is easy to see that
O
=
z
(which is a dipole operator) is the
q
= 0
component of a spherical vector
T
k
=1
q
=0
. The WignerEckart theorem immediately
yields the selection rules
δ
=
±
1
,
δm
= 0
. As a result, from the ground state

0
,
0
,
0
we can only end up in an angular momentum one state

n,
1
,
0
. Note,
however, that the angular momentum selection rule does not provide a constraint
on the principle quantum number
n
.
To show that the only possible final state has
n
= 1
, we recall that the eigen
states of the threedimensional harmonic oscillator can also be given in a basis of
individual
x
,
y
and
z
oscillators. Then, since
z
=
¯
h
2
mω
(
a
z
+
a
†
z
)
(2)
1
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the action of
z
can only increase the
z
oscillator level,
n
z
, by one.
Since
n
=
n
x
+
n
y
+
n
z
, this indicates that we can only end up in the
n
= 1
state

1
,
1
,
0
.
[15]
b
) What is the probability for the oscillator to be found in this excited state at time
t
=
∞
? [Note that
∞
−∞
e
−
x
2
dx
=
√
π
]
We may perform the time integral in
(1)
by completing the square
A
k
←
s
=
i
¯
h
q
E
0
1
,
1
,
0

z

0
,
0
,
0
∞
−∞
e
−
(
t
−
iω
ks
τ
2
/
2)
2
/τ
2
−
ω
2
ks
τ
2
/
4
dt
=
iq
E
0
¯
h
τ
√
πe
−
τ
2
ω
2
ks
/
4
1
,
1
,
0

z

0
,
0
,
0
Note that we have assumed it to be valid to shift the contour of integration away
from the real axis. Now, the matrix element may be evaluated using
(2)
and noting
that the properly normalized

1
,
1
,
0
state is identical to the state with a single
excitation in the
z
oscillator,
n
z
= 1
. Since
1
,
1
,
0

z

0
,
0
,
0 =
¯
h
2
mω
and since
ω
ks
=
ω
, we obtain
A
k
←
s
=
iq
E
0
√
2
m
¯
hω
τ
√
πe
−
τ
2
ω
2
/
4
The probability is
P
=
A
k
←
s

2
=
q
2
E
2
0
π
2
m
¯
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 Winter '03
 Unknow
 Physics, Electron, Charge, Mass, Fundamental physics concepts, matrix element

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