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Physics 6210/Spring 2007/Lecture 24
Lecture 24
Relevant sections in text:
§
3.5, 3.6
Angular momentum eigenvalues and eigenvectors (cont.)
Next we show that the eigenvalues of
J
2
are nonnegative and bound the magnitude
of the eigenvalues of
J
z
. One way to see this arises by studying the relation
J
2

J
2
z
=
1
2
(
J
+
J

+
J

J
+
) =
1
2
(
J
†

J

+
J
†
+
J
+
)
.
Now, for any operator
A
and vector

ψ
i
we have that (exercise)
h
ψ

A
†
A

ψ
i ≥
0
,
so that for any vector

ψ
i
(in the domain of the squared angular momentum operators)
(exercise)
h
ψ

J
2

J
2
z

ψ
i ≥
0
.
Assuming the eigenvectors

a, b
i
are not of the “generalized” type,
i.e.,
are normalizable,
we have
0
≤ h
a, b

J
2

J
2
z

a, b
i
=
a

b
2
,
and hence
a
≥
0
,

√
a
≤
b
≤
√
a.
The ladder operators increase/decrease the
b
value of the eigenvector with out changing
a
. Thus by repeated application of these operators we can violate the inequality above
unless there is a maximum and minimum value for
b
such that application of
J
+
and
J

,
respectively, will result in the zero vector. Moreover, if we start with an eigenvector with
a minimum (maximum) value for
b
, then by successively applying
J
+
(
J

) we must hit
the maximum (minimum) value. As shown in your text, these requirements lead to the
following results. The eigenvalues
a
can only be of the form
a
=
j
(
j
+ 1)¯
h
2
,
where
j
≥
0 can be a nonnegative integer or a half integer only:
j
= 0
,
1
/
2
,
1
,
3
/
2
, . . . .
For an eigenvector with a given value of
j
, the eigenvalues
b
are given by
b
=
m
¯
h,
1
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View Full Document Physics 6210/Spring 2007/Lecture 24
where
m
=

j,

j
+ 1
, . . . , j

1
, j.
Note that if
j
is an integer then so is
m
, and if
j
is a halfinteger, then so is
m
. Note also
that for a ﬁxed value of
j
there are 2
j
+ 1 possible values for
m
. The usual notational
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