 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-84/1/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Treat

ψ

and its complex conjugate as independent variables in the functional derivative. Integrating the gradient variation by parts, with vanishing boundary terms, gives

δ E = \int [- \frac{ℏ ^{2}}{2 m} \nabla^{2} ψ + V_{0} ∣ ψ ∣^{2} ψ + W_{0} ∣ ψ ∣^{4} ψ] δ ψ^{*} d x + complex conjugate .

The factors follow from

\partial_{ψ^{*}} ∣ ψ ∣^{4} = 2∣ ψ ∣^{2} ψ

and

\partial_{ψ^{*}} ∣ ψ ∣^{6} = 3∣ ψ ∣^{4} ψ

i ℏ \partial_{t} ψ = - \frac{ℏ ^{2}}{2 m} \nabla^{2} ψ + V_{0} ∣ ψ ∣^{2} ψ + W_{0} ∣ ψ ∣^{4} ψ .

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