 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-4/4d/a/solution

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

⟨ O ⟩_{ψ} = ⟨ ψ ∣ O ∣ ψ ⟩ .

Differentiating and retaining the possible explicit time dependence of the observable gives

\frac{d}{d t} ⟨ O ⟩_{ψ} = ⟨ \dot{ψ} ∣ O ∣ ψ ⟩ + ⟨ ψ \frac{\partial O}{\partial t} ψ ⟩ + ⟨ ψ ∣ O ∣ \dot{ψ} ⟩ .

i ℏ∣ \dot{ψ} ⟩ = H ∣ ψ ⟩, - i ℏ ⟨ \dot{ψ} ∣ = ⟨ ψ ∣ H,

where the Hamiltonian operator is Hermitian. Substitution yields

\frac{d}{d t} ⟨ O ⟩_{ψ} = \frac{i}{ℏ} ⟨ ψ ∣ H O ∣ ψ ⟩ - \frac{i}{ℏ} ⟨ ψ ∣ O H ∣ ψ ⟩ + ⟨ \frac{\partial O}{\partial t} ⟩_{ψ} = \frac{i}{ℏ} ⟨[H, O] ⟩_{ψ} + ⟨ \frac{\partial O}{\partial t} ⟩_{ψ} .

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