 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/33a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Insert the stated expansion into the Time-dependent Schrodinger equation

i ℏ \frac{d}{d t} ∣ ψ (t)⟩ = (H_{0} + v (t)) ∣ ψ (t)⟩ .

When the derivative acts on each free phase, its energy factor is cancelled by the action of

H_{0}

on the corresponding eigenstate. The remaining terms obey

i ℏ [\overset{a}{˙} (t) e^{- i E_{0} t /ℏ} ∣0 ⟩ + \int_{0}^{\infty} e^{- i E_{k} t /ℏ} (\dot{b}_{k} (t) ∣ k, + ⟩ + \overset{c}{˙}_{k} (t) ∣ k, - ⟩) d k] = - F e^{- iω t} x ∣ ψ (t)⟩ .

i ℏ

gives

\overset{a}{˙} (t) e^{- i E_{0} t /ℏ} ∣0 ⟩ + \int_{0}^{\infty} e^{- i E_{k} t /ℏ} (\dot{b}_{k} ∣ k, + ⟩ + \overset{c}{˙}_{k} ∣ k, - ⟩) d k = \frac{i F}{ℏ} e^{- iω t} x ∣ ψ (t)⟩ .

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