 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-1/14c/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Write

E = - ℏ^{2} κ^{2} / (2 m)

. An even bound state is proportional to

cos (k x)

inside the well and to

e^{- κ ∣ x ∣}

outside, where

k^{2} = \frac{m}{a ℏ ^{2}} - κ^{2} .

Continuity of the logarithmic derivative at

a

gives

κ = k tan (ka) .

For sufficiently small

a

ka

lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.

a ↓ 0

κ \sim k^{2} a \to m / ℏ^{2}

. Hence

E_{0} = - \frac{m}{2 ℏ ^{2}} .

The wells converge distributionally to

V_{0} (x) = - δ (x)

, whose normalized even bound state is

ψ_{0} (x) = \frac{m}{ℏ ^{2}} e^{- m ∣ x ∣/ ℏ^{2}} .

Its derivative jump is precisely the one imposed by the delta potential.

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