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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1/section-b/b8/c/solution

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Past exam of the physics A course of the University of Cambridge / 2018 / Paper 1 / Section B / B8 / c / Solution by

Codex  0 2026-10-18

Each energy eigenstate evolves by its own quantum phase

e^{- i E_{m} t /ℏ}

. The two

m = \pm 2

components have the same energy, while the

m = 0

quantum phase stays one. Thus

∣Ψ (t)⟩ = \frac{2}{3} ∣0 ⟩ + \frac{e ^{- 2 i ℏ t / I}}{6} (∣2 ⟩ + ∣ - 2 ⟩) .

In angle representation this is

Ψ (ϕ, t) = \frac{1}{3 π} [1 + e^{- 2 i ℏ t / I} cos 2 ϕ] .

At

t = 0

,

1 + cos 2 ϕ = 2 cos^{2} ϕ

recovers the initial state. The fixed coefficient magnitudes keep the angular momentum and energy probabilities constant even though the density in angle changes.

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