224 CHAPTER14. THEADDITIONOFANGULARMOMENTUM
andweget
Φ (^32) − 12 =
√
2
3
Y 10 χ−+
√
1
3
Y 1 − 1 χ+ (14.17)
Theremainingstatewithj=^32 hastobethelowestweightstate(14.8).Butjustas
acheckofthealgebra,weobtainitbyapplyingtheladderoperatoronemoretime
J−Φ (^32) − 12 = (L−+S−)
√
2
3
Y 10 χ−+
√
1
3
Y 1 − 1 χ+
̄h
√
3 Φ (^32) − 32 = ̄h
√
1
3
(√
2 (L−Y 10 )χ−+(L−Y 1 − 1 )χ++√
2 Y 10 (S−χ−)+Y 1 − 1 (S−χ+))= ̄h√
1
3(2Y 1 − 1 χ−+ 0 + 0 +Y 1 − 1 χ−)= ̄h√
3 Y 1 − 1 χ− (14.18)andtherefore
Φ (^32) − 32 =Y 1 − 1 χ− (14.19)
asitshould.
ButisthisalltheeigenstatesofJ^2 ,Jz thatwecan formfromthel=1,s=^12
states?Sofarwehaveconstructedfourorthonormalstates
{Φ^32 jz, jz=
3
2
,
1
2
,−
1
2
,−
3
2
} (14.20)
whichareeigenstatesofJ^2 ,Jz,outofsixorthonormalstates
Y 11 χ+ Y 10 χ+ Y 1 − 1 χ+
Y 11 χ− Y 10 χ− Y 1 − 1 χ−(14.21)
Sooutofthesesixstates, weoughtto beabletobuildtwomorestateswhichare
orthogonaltothefour{Φ (^32) jz}.Nowtheseextratwostatescan’thavej= 3 /2,because
wehavealreadyconstructedallofthosestates,andtheycan’tbej> 3 /2,basically
becauseastatewithjz>^32 isn’tavailable. Soletstrytofindstateswithj= 1 /2.
Thereareexactlytwosuchstates
Φ 1212 and Φ (^12) − 12 (14.22)
sothiswouldbringthetotaltosix.Buthowdowegetthem?Thetrickistofirstfind
Φ 1212 fromthefactthatithastobeorthogonaltoΦ 3212 ,andalsonormalized. Then
wecanusetheladderoperatortogettheotherstate.
WebeginfromthefactthatΦ 1212 isaneigenstateofJz,withjz=^12. Thereare
twostatesinthesetofsix(14.21)whichhavethisvalueofjz,namely
Y 11 χ− and Y 10 χ+ (14.23)