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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 265
(e) Angular nodes occur when these wavefunctions pass through zero, and
so for ψa occur when cos ϕ = 0, that is when ϕ = π/2, 3π/2 : this corre-
sponds to the yz plane . For ψb they occur when sin ϕ = 0, that is when
ϕ = 0, π , which is the xz plane .
P7F.10 �e Hamiltonian for a hydrogen atom is
Ĥ = − ħ2
2mr
∂2
∂r2
r + ħ2 l̂ 2
2mr2
− e2
4πε0r
�e operators l̂z and l̂ 2 only operate on the angles θ and ϕ, and so commute
with the operators that a�ect r only, such that
[ l̂z , Ĥ] = ħ2
2mr2
[ l̂z , l̂ 2] [ l̂ 2 , Ĥ] = ħ2
2mr2
[ l̂ 2 , l̂ 2]
For the latter, for an arbitrary function f , [ l̂ 2 , l̂ 2] f = l̂ 2 l̂ 2 f − l̂ 2 l̂ 2 f = 0. Hence,
[ l̂ 2 , l̂ 2] = 0, and so [ l̂ 2 , Ĥ] = 0, meaning that the total angular momentum
operator commutes with the Hamiltonian.
For the former, [ l̂z , l̂ 2] f = [ l̂z , l̂ 2x + l̂ 2y + l̂ 2z ] f = [ l̂z , l̂ 2x ] f + [ l̂z , l̂ 2y] f + [ l̂z , l̂ 2z ] f .
Using the relation given in the question, [ l̂z , l̂ 2x ] = [ l̂z , l̂x] l̂x+ l̂x[ l̂z , l̂x] = iħ l̂y l̂x+
iħ l̂x l̂y , and [ l̂z , l̂ 2y] = [ l̂z , l̂y] l̂y + l̂y[ l̂z , l̂y] = −[ l̂y , l̂z] l̂y − l̂y[ l̂y , l̂z] = −iħ l̂y l̂x −
iħ l̂x l̂y . Finally, [ l̂z , l̂ 2z ] = l̂z l̂ 2z − l̂ 2z l̂z = l̂ 3z − l̂ 3z = 0. Hence, [ l̂z , l̂ 2] = iħ l̂y l̂x +
iħ l̂x l̂y − iħ l̂y l̂x − iħ l̂x l̂y + 0 = 0, meaning that the z component of the angular
momentum operator commutes with the Hamiltonian.
�is is important as it implies that all functions that are eigenfunctions of the
Hamiltonian of a H atom are also eigenfunctions of l̂z and l̂ 2.
P7F.12 (a) Write the solutions ψ(r, θ , ϕ) = R(r)Y(θ , ϕ), and then the Schrödinger
equation, Ĥψ = Eψ becomes
− ħ2
2mr
∂2(rRY)
∂r2
− ħ2
2mr2
Λ̂2(RY) = ERY
Note that the derivatives in r only a�ects R, and Λ̂2 only operates on Y ,
such that
−ħ
2Y
2mr
∂2(rR)
∂r2
− ħ2R
2mr2
Λ̂2Y = ERY
Dividing this equation by the function RY then gives
− ħ2
2mrR
∂2(rR)
∂r2
− ħ2
2mr2Y
Λ̂2Y = E