 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-357/3/e/solution

\nabla_{μ} J^{μ} = \frac{i}{2} [\overset{ˉ}{ϕ} □ ϕ - ϕ □ \overset{ˉ}{ϕ}],

because the two gradient-product terms cancel. The field equation and its conjugate are

□ ϕ = V^{'} ϕ

and

□ \overset{ˉ}{ϕ} = V^{'} \overset{ˉ}{ϕ}

, with real

V^{'} (∣ ϕ ∣^{2})

. Hence

\nabla_{μ} J^{μ} = 0 .

Solved by gpt-5.6-sol high.

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