 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1/section-b/b9/c/solution

For a normalized state, the expectation value is

⟨ Q ⟩ = ⟨ ψ ∣ Q ∣ ψ ⟩ = \int ψ^{*} (x, t) Q ψ (x, t) d x .

For a time-independent operator, differentiation gives

\frac{d}{d t} ⟨ Q ⟩ = ⟨ \dot{ψ} ∣ Q ∣ ψ ⟩ + ⟨ ψ ∣ Q ∣ \dot{ψ} ⟩ .

∣ \dot{ψ} ⟩ = - i H ∣ ψ ⟩ /ℏ

H

gives

⟨ \dot{ψ} ∣ = i ⟨ ψ ∣ H /ℏ

, so

\frac{d}{d t} ⟨ Q ⟩ = \frac{i}{ℏ} ⟨ H Q - Q H ⟩ = \frac{1}{i ℏ} ⟨[Q, H]⟩ .

In the integral derivation, the same self-adjointness step is integration by parts with vanishing boundary terms. An explicitly time-dependent observable would also contribute

⟨ \partial Q / \partial t ⟩

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