 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/35b/b/i/solution

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

The condition

2 m = ℏ^{2}

makes

H = - \frac{d ^{2}}{d x ^{2}} - V_{0} e^{- x^{2}} .

For

ψ_{a} (x) = e^{- a x^{2} /2}, a > 0,

⟨ ψ_{a} ∣ ψ_{a} ⟩ = \int_{R} e^{- a x^{2}} d x = \frac{π}{a} .

Since

ψ_{a}^{'} = - a x ψ_{a}

\frac{\int ψ _{a} ( - ψ _{a}^{''} ) d x}{\int ψ _{a}^{2} d x} = \frac{\int ∣ ψ _{a}^{'} ∣ ^{2} d x}{\int ψ _{a}^{2} d x} = a^{2} \frac{\int x ^{2} e ^{- a x^{2}} d x}{\int e ^{- a x^{2}} d x} = \frac{a}{2} .

The potential quotient is

- V_{0} \frac{\int e ^{- (a + 1) x^{2}} d x}{\int e ^{- a x^{2}} d x} = - V_{0} \frac{a}{1 + a} .

E_{0} \leq E (a) = \frac{a}{2} - V_{0} \frac{a}{1 + a} .

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