 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/35b/a/solution

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

Put

G_{k} (x - x^{'}) = e^{ik ∣ x - x^{'} ∣} .

Away from

x = x^{'}

, this satisfies

(d^{2} / d x^{2} + k^{2}) G_{k} = 0

2 ik

, so, distributionally,

(\frac{d ^{2}}{d x ^{2}} + k^{2}) G_{k} (x - x^{'}) = 2 ik δ (x - x^{'}) .

Applying

d^{2} / d x^{2} + k^{2}

to the proposed integral equation therefore gives

(\frac{d ^{2}}{d x ^{2}} + k^{2}) ψ (x) = \frac{2 m}{ℏ ^{2}} V (x) ψ (x) .

Since

k^{2} = 2 m E / ℏ^{2}

, this is precisely the stated Schrodinger equation. The term

e^{ik x}

is the incident wave and the choice

e^{ik ∣ x - x^{'} ∣}

gives outgoing waves on both sides of each source point.

 New to topics? Read the docs here!