SJP QM 3220
3D 1
To prove, need two very useful identities:
[
]
[
]
[
]
[
]
[
]
[
]
B
C
A
C
B
A
C
AB
C
B
C
A
C
B
A
,
,
,
,
,
,
+
=
+
=
+
Proof:
[
]
[
]
=


=
z
x
y
z
y
x
xp
zp
zp
yp
L
L
,
,
z
x
y
L
i
yp
xp
i
=

+
=
)
(
(Have used
[
]
[
]
[
]
[
]
etc.
0
0
x,
0
,
,
,
,
,
,
,
=
=
=
=
y
x
y
x
p
p
p
y
x
i
p
x
I'm dropping the ˆ over operators when no danger of confusion.
Since [L
x
, L
y
] ≠ 0, cannot have simultaneous eigenstates of
.
and
y
x
L
L
ˆ
ˆ
However,
2
2
2
2
z
y
x
L
L
L
L
L
L
+
+
=
⋅
=
does
commute with L
z
.
Claim:
[
]
0
2
=
z
L
L
,
, i = x, y, or z
Proof:
[
]
[
]
[
]
[
]
0
2
2
2
2
z
z
z
y
z
x
z
L
L
L
L
L
L
L
L
,
,
,
,
+
+
=
[
]
[
]
[
]
[
]
y
x
z
y
x
z
y
y
x
y
z
x
y
z
x
x
L
L
i
L
L
L
i
L
L
L
L
L
i
L
L
L
i
L
L
L
+
+
+
+

+

=
,
,
,
,
= 0
(Note cancellations)
[L
2
, L
z
] = 0 => can have simultaneous eigenstates of
i)
any
(or
2
2
i
z
L
L
L
L
ˆ
,
ˆ
ˆ
,
ˆ
Page H3
M. Dubson, (typeset by J. Anderson) Mods by S. Pollock
Fall 2008
[
]
[
]
[
]
[
]
[
]
[
]
=
+
+


