QMGreensite_merged

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)