 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/36e/c/solution

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

⟨ ϕ_{n} ∣ T ∣ ϕ_{m} ⟩ = δ_{nm} \frac{ℏ ^{2} π ^{2} n ^{2}}{2 m a ^{2}} .

For the linear potential,

⟨ ϕ_{n} ∣ V ∣ ϕ_{n} ⟩ = \frac{V _{0}}{2} = \frac{9 ℏ ^{2}}{2 m a ^{2}},

and

⟨ ϕ_{1} \frac{x}{a} ϕ_{2} ⟩ = 2 \int_{0}^{1} y sin (π y) sin (2 π y) d y = - \frac{16}{9 π ^{2}} .

Thus, in units

ℏ^{2} / (m a^{2})

H = ((π^{2} + 9) /2 - 16/ π^{2} - 16/ π^{2} (4 π^{2} + 9) /2) .

E_{0} \leq \frac{ℏ ^{2}}{m a ^{2}} [\frac{5 π ^{2} + 18}{4} - \frac{9 π ^{4}}{16} + \frac{256}{π ^{4}}] \approx 9.25936 \frac{ℏ ^{2}}{m a ^{2}} .

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