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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4/33c/b/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 33C / b / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

Let

g = a U_{0} ≫ 1

and

z = ka

. A denominator zero away from the removable zero at

z = 0

obeys

2 z + i g - i g e^{2 i z} = 0, e^{2 i z} = 1 - \frac{2 i z}{g} .

Near

z = π

, write

z = π + δ

and choose the logarithm branch with

δ \to 0

. Expanding gives

δ = - \frac{z}{g} - \frac{i z ^{2}}{g ^{2}} + O (g^{- 3}), z = π - \frac{π}{g} + \frac{π}{g ^{2}} - \frac{i π ^{2}}{g ^{2}} + O (g^{- 3}) .

Consequently

α = - \frac{π}{a U _{0}} + O ((a U_{0})^{- 2}), γ = (\frac{π}{a U _{0}})^{2} + O ((a U_{0})^{- 3}) .

The lower-half-plane pole is a scattering resonance: an odd standing wave is nearly confined between the two high repulsive barriers but leaks out slowly. If

E = E_{R} - i Γ/2 = ℏ^{2} k^{2} / (2 m)

, then

Γ ≃ 2 ℏ^{2} πγ / (m a^{2})

and its probability lifetime is

ℏ/Γ

. It is not a genuine bound state of the repulsive potential.

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