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)