1
PHYS4221 Suggested Solution of Homework 5
Question 1
(a) The Hamiltonian
H
=
p
2
x
+
p
2
y
2
m
0
+
m
0
ω
2
0
2
(
x
2
+
y
2
)
+
λxy
can be easily diagonized by a change of variable. We let
x
±
=
x
±
y
√
2
,
p
±
=
p
x
±
p
y
√
2
.
This change of variables corresponds to a
unitary transformation
, because the new variables preserve the commutation
relations
[
x
+
, p
+
] = [
x

, p

] =
i
¯
h
[
x
+
, p

] = [
x

, p
+
] = 0
[
x

, x
+
] = [
p

, p
+
] = 0
.
As an example, using [
x, p
x
] = [
y, p
y
] =
i
¯
h
and [
x, p
y
] = [
y, p
x
] = 0,
[
x
+
, p
+
] =
1
2
([
x
+
y, p
x
+
p
y
]) =
1
2
([
x, p
x
] + [
y, p
y
]) =
1
2
(
i
¯
h
+
i
¯
h
) =
i
¯
h
[
x
+
, p

] =
1
2
([
x
+
y, p
x

p
y
]) =
1
2
([
x, p
x
]

[
y, p
y
]) = 0
.
Furthermore, we can write
H
as
H
=
p
2
+
+
p
2

2
m
0
+
m
0
ω
2
0
2
(
x
2
+
+
x
2

)
+
λ
2
(
x
2
+

x
2

)
=
p
2
+
2
m
0
+
1
2
m
0
ω
2
+
x
2
+
+
p
2

2
m
0
+
1
2
m
0
ω
2

x
2

≡
H
+
+
H

where
H
±
=
p
2
±
2
m
0
+
1
2
m
0
ω
2
±
x
2
±
,
ω
±
=
r
ω
2
0
±
λ
m
0
.
To summarize, in normalmode coordinates
{
x
+
, x

}
, the Hamiltonian describes two
decoupled
harmonic oscillators
with frequencies
ω
+
and
ω

respectively. Hence the eigenstates of
H
are given by
{
n
+
, n

i}
, with eigenvalues
H

n
+
, n

i
=
¯
hω
+
n
+
+
1
2
+ ¯
hω

n

+
1
2

n
+
, n

i
,
n
±
= 0
,
1
,
2
,
. . . .
(i) In coordinate space (
x
+
, x

), the eigenfunction of
H
is just the product of the eigenfunctions of
H
+
and
H

:
Ψ
n
+
,n

(
x
+
, x

)
≡ h
x
+
, x


n
+
, n

i
=
ψ
(+)
n
+
(
x
+
)
*
ψ
(

)
n

(
x

)
,
where
ψ
(
±
)
n
(
x
) =
1
q
2
n
n
!
√
πx
±
0
H
n
x
x
±
0
exp
(

1
2
x
x
±
0
2
)
,
x
±
0
=
s
¯
h
m
0
ω
±
.
(c.f. lecture notes no.3, eq.(42)). Rewriting the required eigenfunction in terms of (
x, y
),
Ψ
n
+
,n

(
x, y
) =
ψ
(+)
n
+
x
+
y
√
2
ψ
(

)
n

x

y
√
2
.