 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-301/2/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

L_{1} = \frac{1}{2} \partial_{μ} ϕ_{1} \partial^{μ} ϕ_{1} - \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} = \frac{1}{2} \dot{ϕ}_{1}^{2} - \frac{1}{2} ∣\nabla ϕ_{1} ∣^{2} - \frac{1}{2} m_{1}^{2} ϕ_{1}^{2} .

The field's conjugate momentum is

π_{1} = \partial L_{1} / \partial \dot{ϕ}_{1} = \dot{ϕ}_{1}

. The positive sign of the time-derivative term uses the metric convention fixed in Question 1.

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