14.1. THEGENERALMETHOD 229
or,inketnotation
{|j 1 ,m 1 ,j 2 ,m 2 >} (14.46)
witheigenvalues
J 12 ψj^11 m 1 ψj^22 m 2 = j 1 (j 1 +1) ̄h^2 ψ^1 j 1 m 1 ψj^22 m 2
J 1 zψj^11 m 1 ψj^22 m 2 = m 1 ̄hψj^11 m 1 ψ^2 j 2 m 2
J 22 ψj^11 m 1 ψj^22 m 2 = j 2 (j 2 +1) ̄h^2 ψ^1 j 1 m 1 ψj^22 m 2
J 2 zψj^11 m 1 ψj^22 m 2 = m 2 ̄hψj^11 m 1 ψ^2 j 2 m 2 (14.47)If,forexample,J 1 andJ 2 werebothorbitalangularmomenta,then
ψ^1 j 1 m 1 ψ^2 j 2 m 2 =Yj 1 m 1 (θ 1 ,φ 1 )Yj 2 m 2 (θ 2 ,φ 2 ) (14.48)Denotethesumofthetwoangularmomentaby
J%=J% 1 +J% 2 (14.49)GiventheeigenstatesofJ 12 ,J 1 z,J 22 ,J 2 z,wewanttofindtheeigenstatesofΦjmofthe
totalangularmomentumJ^2 ,Jz,J 12 ,J 22.
Onceagain,theideaistostartfromthehighest-weightstate,namely
Φjj=φ^1 j 1 j 1 φ^2 j 2 j 2 (j=j 1 +j 2 ) (14.50)andgetalltheotherstateswithj=j 1 +j 2 byapplyingtheladderoperators. Then
weconstructstateswiththenextlowestvalueofangularmomentum,j=j 1 +j 2 −1,
andthenstateswithj=j 1 +j 2 −2,andsoon.Thefirstquestionis: wheredoesthis
procedureend?Ifj=j 1 +j 2 isthehighestpossiblevalueoftotalangularmomentum,
whatisthelowestpossiblevalue?
Thisquestioncanbeansweredbycountingstates. Foragivenj 1 andj 2 thereare
atotalof 2 j 1 + 1 possiblevaluesofm 1 ,and 2 j 2 + 1 possiblevaluesofm 2 ,sothetotal
numberoforthonormalstatesψ^1 j 1 m 1 ψj^22 m 2 is
N=(2j 1 +1)(2j 2 +1) (14.51)Denotebyjmin the lowest possiblevalueof totalangularmomentumthat can be
constructedfromstatesofgivenj 1 andj 2. Thenthetotalnumberofeigenstatesof
totalangularmomentumwillbe
N′=
j (^1) ∑+j 2
j=jmin
(2j+1) (14.52)
andjminisdeterminedfromtherequiringN=N′. Thesolutiontothisconditionis
that
jmin=|j 1 −j 2 | (14.53)