QMGreensite_merged

223

FromthisoneeigenstateofJ^2 (thehighestweightstate),itissimpletoconstruct
alltheothereigenstatesofJ^2 andJz,bysuccessivelyapplyingtheladderoperator

J−=L−+S− (14.11)

andtherelations

J−|jjz> =

√

j(j+1)−jz(jz−1) ̄h|j,jz− 1 >

L−|lm> =

√

l(l+1)−m(m−1)) ̄h|l,m− 1 >

S−|ssz> =

√

s(s+1)−sz(sz−1) ̄h|s,sz− 1 > (14.12)

Toseehowitworks,letspickaparticularexampletoworkout,withl=1. Inthis
case,thehighestweightstateis

Φ 3232 =Y 11 χ+ (14.13)

NowapplytheJ−operatoronbothsidesofthisequation,anduseeq.(14.12)

J−Φ 3

2

3

2

= (L−+S−)Y 11 χ+

̄h

√

3

2

(

3

2

+1)−

3

2

(

3

2

−1)Φ 3212 = (L−Y 11 )χ++Y 11 (S−χ+)

̄h

√

3 Φ 3212 = ̄h





√

1(1+1)− 0 Y 10 χ++Y 11 ·

√

1

2

(

1

2

+1)−

1

2

(

1

2

−1)χ−





= ̄h

(√

2 Y 10 χ++Y 11 χ−

)

(14.14)

Dividingbothsidesby

√

3 ̄h,wehavefoundtheeigenstateofJ^2 ,Jzwithj=^32 and
jz=^12

Φ 3212 =

√

2

3

Y 10 χ++

√

1

3

Y 11 χ− (14.15)

Togetthestatewithjz=−^12 ,wejustapplyJ−again,tobothsidesof(14.15)

J−Φ 3212 = (L−+S−)





√

2

3

Y 10 χ++

√

1

3

Y 11 χ−





̄h

√

3

2

(

3

2

+1)−

1

2

(

1

2

−1)Φ 3

2 −

1

2

= ̄h

√

1

3

(√

2(L−Y 10 )χ++(L−Y 11 )χ−

+

√

2 Y 10 (S−χ+)+Y 11 (S−χ−)

)

2 ̄hΦ (^32) −^12 = ̄h
√
1
3
(
2 Y 1 − 1 χ++

√

2 Y 10 χ−

+

√

2 Y 10 χ−+ 0

)

(14.16)