218 3 Quantum Mechanics – IINowLzg(φ)=−i∂φ∂g=mg(φ)
Thus the z-component of angular momentum is quantized with eigen
value.
3.81 One can do similar calculations forLxandLyas in Problem 3.80 and obtain
Lx
i=sinφ∂
∂θ+cotθcosφ∂
∂φ
Ly
i=−cosφ∂
∂θ+cotθsinφ∂
∂φ
3.82 UsingL^2 =L^2 x+L^2 y+L^2 z, the commutator with total angular momentum
squared can be evaluated[L^2 ,Lz]=[
L^2 x+L^2 y+L^2 z,Lz]
=
[
L^2 x+L^2 y,Lz]
=Lx[Lx,Lz]+[Lx,Lz]Lx+Ly[
Ly,Lz]
+
[
Ly,Lz]
Ly (1)
=−iLxLy−iLyLx+iLyLx+iLxLy= 0
Similarly[
L^2 ,Lx]
=[L^2 ,Ly]=[L^2 ,L]= 03.83 L^2 =L^2 x+L^2 y+L^2 z
Using the expressions forLx,LyandLzfrom problem (3.81)
L^2
(i)^2=
(
sinφ∂
∂θ+cotθcosφ∂
∂φ)(
sinφ∂
∂θ+cotθcosφ∂
∂φ)
+
(
−cosφ∂
∂θ+cotθsinφ∂
∂φ)(
−cosφ∂
∂θ+cotθsinφ∂
∂φ)
+
(
−
∂
∂φ)(
−
∂
∂φ)
=sin^2 φ∂^2
∂θ^2+cot^2 θcos^2 φ∂^2
∂φ^2−sinφcosφcosec^2 θ∂
∂φ+cos^2 φcotθ∂
∂θ+cos^2 φ∂^2
∂θ^2+cot^2 θsin^2 φ∂^2
∂φ^2+sinφcosφcosec^2 θ∂
∂φ+sin^2 φcotθ∂
∂θ+
∂^2
∂φ^2
The cross terms get cancelled and the expression is reduced to∂^2
∂θ^2+
1
sin^2 θ∂^2
∂φ^2+cotθ∂
∂θ∇^2 ψ=1
r^2∂
∂r(
r^2∂ψ
∂r)
+
1
r^2 sinθ∂
∂θ(
sinθ∂ψ
∂θ)
+
1
r^2 sin^2 θ∂^2 ψ
∂φ^2