 Solution

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

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

U (x)

H = \frac{p ^{2}}{2 m} + U (x) .

[x, p] = i ℏ

gives

[H, x] = \frac{1}{2 m} [p^{2}, x] = - \frac{i ℏ}{m} p .

O = x

therefore gives

m \frac{d}{d t} ⟨ x ⟩_{ψ} = ⟨ p ⟩_{ψ} .

Since

[p^{2}, p] = 0

and direct action on a wavefunction gives

[U (x), p] = i ℏ U^{'} (x)

\frac{d}{d t} ⟨ p ⟩_{ψ} = - ⟨ U^{'} (x) ⟩_{ψ} .

Finally

H

has no explicit time dependence and

[H, H] = 0

, so

\frac{d}{d t} ⟨ H ⟩_{ψ} = 0 .

The first two equations are the expectation-value counterparts of momentum equals mass times velocity and Newton's second law; the last is conservation of energy. They become the classical equations directly when the force varies negligibly across the wave packet, so that

⟨ U^{'} (x)⟩ ≃ U^{'} (⟨ x ⟩)

. These conclusions are summarized by the classical equations from Ehrenfest theorem.

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