 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/15a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Ψ (t) = e^{- i Et /ℏ} ψ .

Q

⟨ Q ⟩_{Ψ (t)} = ⟨ e^{- i Et /ℏ} ψ, Q e^{- i Et /ℏ} ψ ⟩ = ⟨ ψ, Q ψ ⟩ .

It is therefore independent of time.

E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m a ^{2}} .

The possible energy measurement results are

E_{1} = \frac{π ^{2} ℏ ^{2}}{2 m a ^{2}}, E_{2} = \frac{2 π ^{2} ℏ ^{2}}{m a ^{2}},

∣ c_{1} ∣^{2}

and

∣ c_{2} ∣^{2}

. Normalization gives

∣ c_{1} ∣^{2} + ∣ c_{2} ∣^{2} = 1

t

Ψ (x, t) = c_{1} e^{- i E_{1} t /ℏ} ψ_{1} (x) + c_{2} e^{- i E_{2} t /ℏ} ψ_{2} (x) .

The required matrix elements are

⟨ ψ_{1}, \overset{x}{^} ψ_{1} ⟩ = ⟨ ψ_{2}, \overset{x}{^} ψ_{2} ⟩ = \frac{a}{2}, ⟨ ψ_{1}, \overset{x}{^} ψ_{2} ⟩ = - \frac{16 a}{9 π ^{2}} .

Consequently

⟨ \overset{x}{^} ⟩_{Ψ (t)} = \frac{a}{2} - \frac{32 a}{9 π ^{2}} Re (c_{1}^{*} c_{2} e^{- iω t}),

ω = \frac{E _{2} - E _{1}}{ℏ} = \frac{3 π ^{2} ℏ}{2 m a ^{2}} .

Finally,

2∣ c_{1} c_{2} ∣ \leq ∣ c_{1} ∣^{2} + ∣ c_{2} ∣^{2} = 1

, so

⟨ \overset{x}{^} ⟩_{Ψ (t)} - \frac{a}{2} \leq \frac{16 a}{9 π ^{2}} .

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