QMGreensite_merged

11.2. THEEIGENVALUESOFANGULARMOMENTUM 177

Therefore,
L ̃+φab=Cab+φa,b+1 (11.33)

whereCab+ isa constant. This establishesthe fact that L+is araising operator,
analogousto thea†operatorusedinsolving theharmonic oscillator. Inasimilar
way,wecanshowthatL−isaloweringoperator.Define

φ”=L ̃−φab (11.34)

Then

L ̃zφ” = L ̃zL ̃−φab

= (L ̃−L ̃z− ̄hL−)φab

= ( ̄hb− ̄h)L ̃−φab

= ̄h(b−1)φ” (11.35)

sothatφ”=L ̃−φabisalsoaneigenstateofL ̃z,withtheeigenvalue ̄h(b−1).Itisalso
aneigenstateofL ̃^2 :

L ̃^2 φ” = L ̃^2 L ̃−φab

= L ̃−L ̃^2 φab

= ̄h^2 a^2 L ̃−φab

= ̄h^2 a^2 φ” (11.36)

ThereforeL ̃−isaloweringoperator

L ̃−φab=Cab−φa,b− 1 (11.37)

whereCab−isaconstant.
InthecaseoftheHarmonicoscillator,wededucedfromthefactthat≥ 0
thattheremustbeastateoflowestenergyφ 0 ,andthereforealowestenergyeigenvalue
E 0. SincetheloweringoperatoracannotproduceastateoflowerenergythanE 0 ,it
mustannihilatethegroundstate
aφ 0 = 0 (11.38)

Theargumentforangularmomentumisquitesimilar. Wehavealreadyfoundthat
theeigenvaluesofL ̃zhavebothalowerandandupperbound,forfixed ̄h^2 a^2 ,since

−a≤b≤a (11.39)

Soletusdenotedtheminimumvalueofbasbmin andthemaximumvalueasbmax,
where|bmin|,|bmax| ≤a.Thecorrespondingeigenstatesmusthavethepropertythat

L ̃+φabmax = 0

L ̃−φabmin = 0 (11.40)