178 CHAPTER11. ANGULARMOMENTUM
Now,we mustbeableto reachthehighest stateφabmax byacting successively on
φabminwiththeraisingoperator
φabmax∝(L ̃+)nφabmin (11.41)whichimplies,sincetheraisingoperatorraisesbtob+1,that
bmax=bmin+n n= apositiveintegeror 0 (11.42)or,equivalently,
bmax = bav+n
2
bmin = bav−n
2
bav =1
2
(bmax+bmin) (11.43)Next,usingtheexpressionsforL ̃^2 in(11.29)L ̃^2 φabmax = (L ̃−L ̃++L ̃^2 z+ ̄hL ̃z)φabmax
̄h^2 a^2 φabmax = h ̄^2 bmax(bmax+1)φabmax (11.44)sothat
a^2 = bmax(bmax+1)= (bav+1
2
n)(bav+1
2
n+1) (11.45)Likewise,
L ̃^2 φabmin = (L ̃+L ̃−+L ̃^2 z− ̄hL ̃z)φabmin
̄h^2 a^2 φabmin = h ̄^2 bmin(bmin−1)φabmin (11.46)sothat
a^2 = bmin(bmin−1)= (bav−1
2
n)(bav−1
2
n−1)= (−bav+1
2
n)(−bav+1
2
n+1) (11.47)Equatingtheright-handsidesof(11.45)and(11.47)
(bav+1
2
n)(bav+1
2
n+1)=(−bav+1
2
n)(−bav+1
2
n+1) (11.48)