3.3 Solutions 225
3.90 With reference to the Table 3.2, the functionψ(r,θ,φ)isthe2pfunction for
the hydrogen atom.
(a)Lz=−i∂φ∂
ApplyingLzto the wavefunctionLzψ(r,θ,φ)=−i∂ψ(r,θ,φ)
∂φ
=(−i)(−i)ψ(r,θ,φ)
=−ψ(r,θ,φ)
Therefore, the value ofLzis−
(b)Asitisap-state,l=1 and the parity is (−1)l=(−1)l=−1, that is an
odd parity.3.91 Lx=i
(
sinφ∂
∂θ+cotθcosφ∂
∂φ)
(1)
Ly=i(
−cosφ∂
∂θ+cotθ sinφ∂
∂φ)
(2)
L+=Lx+iLy=i(
sinφ∂
∂θ+cotθcosφ∂
∂φ)
−
(
−cosφ∂
∂θ+cotθsinφ∂
∂φ)
(3)
Apply (3) to them=+1 state which is proportional to sinθeiφ
L+(sinθeiφ)=i(sinφcosθ+icotθcosφ)eiφ
−(−cosφcosθi+cotθsinφsinθ)eiφ
=i(sinφcosθ−cotθsinφsinθ)eiφ
−(cosθcosφ−cosφcosθ)eiφ= 0
Thus the state withm=2 does not exist. Similarly by applyingL−to the
state withm=−1, it can be shown that them=−2 state does not exist.3.92 Particles with even spin (0, 2 , 4 ...) obey Bose statistics and those with odd
spin (1/ 2 , 3 / 2 , 5 / 2 ...) obey Fermi – Dirac statistics.
Consider a diatomic molecule with identical nuclei. The total wave function
may be written as
ψ=ψelecζvibρrotσnuc
Letpbe an operator which exchanges the space and spin coordinates.
Nowpψelec=±ψelec
It is known from molecular spectroscopy, that for the ground state it is pos-
itive. Furthermore,Pζvib=+ζvib, becauseζvibdepends only on the distance
of separation of nuclei.
Nowρ=Plm(cosθ)eimφ, whereθis the polar angle andφthe azimuth
angle;ρis represented by the associated Legendre function.