 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/15a/i/solution

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

For

ψ = A e^{- B x^{2}}

ψ_{t} = (\frac{A ^{'}}{A} - B^{'} x^{2}) ψ, ψ_{xx} = (- 2 B + 4 B^{2} x^{2}) ψ .

Substitution into the Schrodinger equation and comparison of the constant and

x^{2}

A^{'} = - i ℏ A B, B^{'} = - \frac{i}{2ℏ} - 2 i ℏ B^{2} .

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