 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/15a/iv/solution

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

Normalizability requires

Re B > 0

. Since

∣ ψ ∣^{2} = ∣ A ∣^{2} e^{- (B + B^{*}) x^{2}},

the given Gaussian integral gives

⟨ \overset{x}{^}^{2} ⟩ = \frac{1}{2 ( B + B ^{*} )} = \frac{1}{4 Re B} .

Also

ψ_{x} = - 2 B x ψ

⟨ \overset{p}{^}^{2} ⟩ = ℏ^{2} \int ∣ ψ_{x} ∣^{2} d x = 4 ℏ^{2} ∣ B ∣^{2} ⟨ x^{2} ⟩ .

Therefore

⟨ \overset{p}{^}^{2} ⟩ = \frac{ℏ ^{2} ∣ B ∣ ^{2}}{Re B} .

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