 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/14c/i/solution

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

⟨ H ⟩ = \int_{0}^{a} (\frac{ℏ ^{2}}{2 m} ∣ ψ^{'} (x) ∣^{2} + U (x) ∣ ψ (x) ∣^{2}) d x \geq 0.

Thus every energy eigenvalue is nonnegative.

For

0 < E < U_{0}

ψ (x) = {A sin (k x), B sinh (l (a - x)), 0 \leq x \leq a /2, a /2 \leq x \leq a,

where

k = 2 m E /ℏ

and

l = 2 m (U_{0} - E) /ℏ

. Continuity of

ψ

and

ψ^{'}

at the finite potential step gives

A sin (ka /2) = B sinh (l a /2),

A k cos (ka /2) = - Bl cosh (l a /2) .

Dividing and rearranging yields the bound-state quantization condition

\frac{1}{k} tan \frac{ka}{2} = - \frac{1}{l} tanh \frac{l a}{2} .

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