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)