QMGreensite_merged

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