 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-1/18d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

The attractive potential has

Δ > 0

- \frac{ℏ ^{2}}{2 M} χ^{''} (x) - \frac{ℏ ^{2}}{2 M Δ} δ (x) χ (x) = E χ (x) .

Integrate from

- ε

ε

. Continuity makes the delta term equal to

- ℏ^{2} χ (0) / (2 M Δ)

, while the integral of

E χ

tends to zero as the interval shrinks. The second-derivative term integrates to the difference of first derivatives. Hence

χ^{'} (0^{+}) - χ^{'} (0^{-}) = - \frac{χ ( 0 )}{Δ} .

This expresses the point interaction through a derivative jump while the wavefunction remains continuous.

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