225
butthesearenoteigenstatesofJ^2 .Sowelookforalinearcombination
Φ 1212 =aY 11 χ−+bY 10 χ+ (14.24)
whichhastheorthonormalityproperties
<Φ 3
2
1
2
|Φ 1
2
1
2
> = 0
<Φ^1212 |Φ^1212 > = 1 (14.25)
Substituting(14.15)and(14.24)intothefirstequationor(14.25)
0 =
√
1
3
<Y 11 χ−|+
√
2
3
<Y 10 χ+|
[a|Y 11 χ−>+b|Y 10 χ+>]
=
√
1
3
[a+
√
2 b] (14.26)
Therefore
Φ 1212 =a
Y 11 χ−−
√
1
2
Y 10 χ+
(14.27)
Imposethenormalizationconditiononthisstate
1 = a^2
<Y 11 χ−|−
√
1
2
<Y 10 χ+|
|Y 11 χ−>−
√
1
2
|Y 10 χ+>
= a^2 (1+
1
2
) ⇒ a=
√
2
3
(14.28)
So
Φ 1212 =
√
2
3
Y 11 χ−−
√
1
3
Y 10 χ+ (14.29)
Thelastthingtodoistoapplytheladderoperatortobothsidesofthisequationto
getthelastofthesixstates
J−Φ 1212 = (L−+S−)
√
2
3
Y 11 χ−−
√
1
3
Y 10 χ+
̄hΦ (^12) − 12 =
√
2
3
(L−Y 11 χ−−
√
1
3
(L−Y 10 )χ++
√
2
3
Y 11 (S−χ−)−
√
1
3
Y 10 (S−χ+)
=
̄h
√
3
[
2 Y 10 χ−
√
2 Y 1 − 1 χ++ 0 −Y 10 χ−
]
(14.30)
andfinally
Φ^12 −^12 =
√
1
3
Y 10 χ−−
√
2
3
Y 1 − 1 χ+ (14.31)