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262 7 QUANTUM THEORY
(ii) For ψ = e−2iϕ
T̂ψ = −(ħ2/2I)d2(e−2iϕ)/dϕ2 = −(ħ2/2I)(−2i)2e−2iϕ = (4ħ2/2I)e−2iϕ
Hence ψ is an eigenfunction with eigenvalue 4ħ2/2I .
(iii) For ψ = cos ϕ
T̂ψ = −(ħ2/2I)d2(cos ϕ)/dϕ2 = (ħ2/2I) cos ϕ
Hence ψ is an eigenfunction with eigenvalue ħ2/2I .
(iv) For ψ = cos χeiϕ + sin χe−iϕ
T̂ψ = −(ħ2/2I)d2(cos χeiϕ + sin χe−iϕ)/dϕ2
= (ħ2/2I)(cos χeiϕ + sin χe−iϕ)
Hence ψ is an eigenfunction with eigenvalue ħ2/2I .
(d) A function is a de�nite state of a particular variable if it is an eigenfunc-
tion of the corresponding operator. Hence, all four wavefunctions are
de�nite kinetic energy states, but only eiϕ and e−2iϕ are de�nite angular
momentum states.
P7F.4 �e rotational energy levels of a free molecule are given by [7F.10–287], E l =
ħ2 l(l + 1)/2I, where l = 0, 1, 2... and I is the moment of inertia, which for a
diatomic molecule is I = µR2, where µ is the reduced mass of the molecule and
R is the bond length.
For 1H127I, the reduced mass is
µ = (1 mu)(127 mu)/[(1 + 127) mu] = 0.992... mu
= (0.992... mu) × (1.6605 × 10−27 kg) = 1.64... × 10−27 kg
�is gives themoment of inertia as I = (1.64...×10−27 kg)×(160×10−12 m)2 =
4.21...×10−47 kgm2, and so the energy levels become E l = (1.0546×10−34 J s)2/[2×
4.21... × 10−47 kgm2] × l(l + 1) = l(l + 1) × (1.31... × 10−22 J).
Using this the energies of the �rst four levels are computed as
E0 = 0, E1 = 2.64 × 10−22 J, E2 = 7.91 × 10−22 J, E3 = 1.58 × 10−21 J
P7F.6 �e function Y1,+1 = − 12
√
3/2π sin θeiϕ .�e integral to evaluate is
∫
π
θ=0
∫
2π
ϕ=0
Y∗1,+1Y1,+1 sin θ dθ dϕ
Y1,+1 is complex , and so Y∗1,+1 = − 12
√
3/2π sin θe−iϕ . Hence, the integrand is
− 1
2
√
3/2π sin θe−iϕ × − 12
√
3/2π sin θeiϕ × sin θ
= (3/8π) sin3 θei(ϕ−ϕ) = (3/8π) sin3 θ