M
2 , M
2 , z
Â
Ã
Â
à ¼ 2i M
2 , M y x
Â
à À M
2 , M x y
Â
à þ i M
2 , z
Â
Ã
È
É
¼ 2i M y M
2 , x
Â
à À M x M
2 , y
Â
à þ i M
2 , z
Â
Ã
È
É
¼ 2i 2iM y yM z À M y z
À
Á À 2iM x M x z À xM z
ð
Þþi M
2 , z
Â
Ã
È
É
¼ À2 2 M x x þ M y y þ M z z
À
Á
M z À 2 M x
2
þ M y
2
þ M z
2
À
Á
z
È
þM
2 z À z M
2
g ¼ 2 M
2 z þ z M
2
À
Á :
ð4:75Þ
In the above calculations, (i) we used [M
2 , M y ] ¼ [M
2 , M x ] ¼ 0 (with the second
equality); (ii) we used (4.74) (the third equality); (iii) RHS was modified so that we
can use the relation M ⊥ x from the definition of the angular momentum operator,
i.e., M x x + M y y + M z z ¼ 0 (the second last equality). We used [M z , z] ¼ 0 as well.
Similar results are obtained with x and y. That is, we have
M
2 , M
2 , x
Â
Ã
Â
à ¼ 2 M
2 x þ xM
2
À
Á
,
ð4:76Þ
M
2 , M
2 , y
Â
Ã
Â
à ¼ 2 M
2 y þ yM
2
À
Á :
ð4:77Þ
Rewriting, e.g., (4.75), we have
M
4 z À 2M
2 z M
2
þ z M
4
¼ 2 M
2 z þ z M
2
À
Á :
ð4:78Þ
Using the relation (4.78) and taking inner products of both sides, we get, e.g.,
l
0
jM
4 z À 2M
2 z M
2
þ z M
4
jl
¼ l
0
j2 M
2 z þ z M
2
À
Á jl
:
ð4:79Þ
That is,
l
0
jM
4 z À 2M
2 z M
2
þ z M
4
jl
À l
0
j2 M
2 z þ z M
2
À
Á jl
¼ 0:
Considering that both terms of LHS contain a factor hl
0
j zj li in common, we have
l
0 2 l
0
þ 1
ð
Þ
2 À 2l
0 l l
0
þ 1
ð
Þ l þ 1
ð
Þþl
2 l þ 1
ð
Þ
2 À 2l
0 l
0
þ 1
ð
ÞÀ2l l þ 1
ð
Þ
h
i
 l
0
jzjl
h
i ¼ 0,
ð4:80Þ
where the quantum state j li is identical to j l, mi in (3.151) with m omitted.
To factorize the first factor of LHS of (4.80), we view it as a quartic equation with
respect to l
0 . Replacing l
0 with Àl, we find that the first factor vanishes, and so the first
factor should have a factor (l
0 + l). Then, we factorize the first factor of LHS of (4.80)
such that
4.4 Selection Rules
145
2 , M
2 , z
Â
Ã
Â
à ¼ 2i M
2 , M y x
Â
à À M
2 , M x y
Â
à þ i M
2 , z
Â
Ã
È
É
¼ 2i M y M
2 , x
Â
à À M x M
2 , y
Â
à þ i M
2 , z
Â
Ã
È
É
¼ 2i 2iM y yM z À M y z
À
Á À 2iM x M x z À xM z
ð
Þþi M
2 , z
Â
Ã
È
É
¼ À2 2 M x x þ M y y þ M z z
À
Á
M z À 2 M x
2
þ M y
2
þ M z
2
À
Á
z
È
þM
2 z À z M
2
g ¼ 2 M
2 z þ z M
2
À
Á :
ð4:75Þ
In the above calculations, (i) we used [M
2 , M y ] ¼ [M
2 , M x ] ¼ 0 (with the second
equality); (ii) we used (4.74) (the third equality); (iii) RHS was modified so that we
can use the relation M ⊥ x from the definition of the angular momentum operator,
i.e., M x x + M y y + M z z ¼ 0 (the second last equality). We used [M z , z] ¼ 0 as well.
Similar results are obtained with x and y. That is, we have
M
2 , M
2 , x
Â
Ã
Â
à ¼ 2 M
2 x þ xM
2
À
Á
,
ð4:76Þ
M
2 , M
2 , y
Â
Ã
Â
à ¼ 2 M
2 y þ yM
2
À
Á :
ð4:77Þ
Rewriting, e.g., (4.75), we have
M
4 z À 2M
2 z M
2
þ z M
4
¼ 2 M
2 z þ z M
2
À
Á :
ð4:78Þ
Using the relation (4.78) and taking inner products of both sides, we get, e.g.,
l
0
jM
4 z À 2M
2 z M
2
þ z M
4
jl
¼ l
0
j2 M
2 z þ z M
2
À
Á jl
:
ð4:79Þ
That is,
l
0
jM
4 z À 2M
2 z M
2
þ z M
4
jl
À l
0
j2 M
2 z þ z M
2
À
Á jl
¼ 0:
Considering that both terms of LHS contain a factor hl
0
j zj li in common, we have
l
0 2 l
0
þ 1
ð
Þ
2 À 2l
0 l l
0
þ 1
ð
Þ l þ 1
ð
Þþl
2 l þ 1
ð
Þ
2 À 2l
0 l
0
þ 1
ð
ÞÀ2l l þ 1
ð
Þ
h
i
 l
0
jzjl
h
i ¼ 0,
ð4:80Þ
where the quantum state j li is identical to j l, mi in (3.151) with m omitted.
To factorize the first factor of LHS of (4.80), we view it as a quartic equation with
respect to l
0 . Replacing l
0 with Àl, we find that the first factor vanishes, and so the first
factor should have a factor (l
0 + l). Then, we factorize the first factor of LHS of (4.80)
such that
4.4 Selection Rules
145