226 CHAPTER14. THEADDITIONOFANGULARMOMENTUM
Theset of allsix eigenstatesof J^2 ,Jz thatcan be formedfromthe sixstates
Y 1 mχ±, togetherwiththe operationusedto construct each, aredisplayedin Fig.
[14.1].Butthereisnothingspecialaboutl=1,ofcourse,andtheprocedureshown
aboveworksforarbitraryl, asshowninFig. [14-2]. Notethat,since−l≤m≤l
thereare(2l+1)valueform,and 2 possiblevaluesofsz=±^12 .Thismeansthereare
2(2l+1)orthogonalYlmχ±statesforeachfixedl,s=^12 .Fromthese,oneconstructs
asetofeigenstatesofJ^2 withjmax=l+^12 ,andjmin=l−^12. Thetotalnumberof
J^2 eigenstatesisthesameasthenumberofYlmχ±states,i.e.
N = (2jmax+1)+(2jmin+1)
= (2(l+
1
2
)+1)+(2(l−
1
2
)+1)
= 2(2l+1) (14.32)
NowletsgobacktotheHydrogenatom.LetH 0 betheHydrogenatomHamilto-
nianwithoutthespin-orbitcontribution,i.e.
H 0 =−
̄h^2
2 m
∇^2 −
e^2
r
(14.33)
Includingelectronspin,theeigenstatesofH 0 ,L^2 ,Lz,S^2 ,Szarethesetofstates
{|nlmssz>} ⇔ {Rnl(r)Ylm(θ,φ)χsz, n= 1 , 2 ,...; l= 0 , 1 ..,n−1;
m=−l,...,l, sz=
1
2
,−
1
2
} (14.34)
What has been done in this chapter is to show how to construct eigenstates of
H 0 ,J^2 ,Jz,L^2 ,S^2
{|njjzls>} ⇔ {Rnl(r)Φjjz, n= 1 , 2 ,...; l= 0 , 1 ,..,n− 1 ,
j=l+
1
2
,l−
1
2
; jz=−j,...,j} (14.35)
Now,takingintoaccountthespin-orbitterm,theHamiltonianoftheHydrogenatom
isreally
H=H 0 +H′ (14.36)
where
H′=
e^2
4 M^2 c^2
1
r^3
(J^2 −L^2 −S^2 ) (14.37)
SinceHstillcommuteswithL^2 ,S^2 ,J^2 ,Jz,theeigenfunctionsofHwillstillhavethe
form
R′nlj(r)Φjjz (14.38)
butnowR′(r)willbealittledifferentfromRnl(r),andalsothenewenergyeigenvalues
E′njlwillbealittledifferentfromtheBohrenergiesEn. Sincespin-orbitcouplingis