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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 261
+1
0
+2
−1
−2
+3
−3
z
+1
−1
+2
−2
+3
−3
0
Figure 7.12
(iii) l̂z(cos ϕ) = (ħ/i)d/dϕ(cos ϕ) = (ħ/i) × (− sin ϕ) = −(ħ/i) sin ϕ.
Hence cos ϕ is not an eigenfunction of the operator (ħ/i)d/dϕ.
(iv) l̂z(cos χeiϕ+sin χe−iϕ) = (ħ/i)(d/dϕ)(cos χeiϕ+sin χe−iϕ) = (ħ/i)×
(i cos χeiϕ − i sin χe−iϕ) = ħ(cos χeiϕ − sin χe−iϕ). Hence cos χeiϕ −
sin χe−iϕ is not an eigenfunction of the operator (ħ/i)d/dϕ.
(b) For ψ = cos ϕ,
⟨lz⟩ =
∫
2π
0 ψ∗ l̂zψ dϕ
∫
2π
0 ψ∗ψ dϕ
=
−(ħ/i) ∫
2π
0 cos ϕ sin ϕ dϕ
∫
2π
0 cos2 ϕ dϕ
= −(ħ/2i) sin 4π
∫
2π
0 cos2 ϕ dϕ
= 0
where Integral T.7 is used to evaluate the numerator.
For ψ = cos χeiϕ + sin χe−iϕ ,
⟨lz⟩ =
∫
2π
0 ψ∗ l̂zψ dϕ
∫
2π
0 ψ∗ψ dϕ
= ∫
2π
0 (cos χe−iϕ + sin χeiϕ)ħ(cos χeiϕ − sin χe−iϕ)dϕ
∫
2π
0 (cos χe−iϕ + sin χeiϕ)(cos χeiϕ + sin χe−iϕ)dϕ
=
ħ ∫
2π
0 cos2 χ − sin2 χ − cos χ sin χe−2iϕ + cos χ sin χe2iϕ dϕ
∫
2π
0 cos2 χ + sin2 χ + cos χ sin χe−2iϕ + cos χ sin χe2iϕ dϕ
Note that as einϕ is periodic over 2π for integer n, ∫
2π
0 einϕdϕ = 0. Hence
⟨lz⟩ =
ħ2π(cos2 χ − sin2 χ)
2π(cos2 χ + sin2 χ)
= ħ cos 2χ
(c) (i) For ψ = eiϕ
T̂ψ = −(ħ2/2I)d2(eiϕ)/dϕ2 = −(ħ2/2I)(i)2eiϕ = (ħ2/2I)eiϕ
Hence ψ is an eigenfunction with eigenvalue ħ2/2I .