 Solution

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

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

[H_{1}, ϕ_{1} (t, x)] = \frac{1}{2} \int d^{3} y [π_{1} (t, y)^{2}, ϕ_{1} (t, x)] = - i π_{1} (t, x) .

\dot{ϕ}_{1} = π_{1}

i m_{1}^{2} ϕ_{1} (x)

. The gradient term gives

\frac{1}{2} \int d^{3} y [(\nabla ϕ_{1} (y))^{2}, π_{1} (x)] = i \int d^{3} y \nabla ϕ_{1} (y) \cdot \nabla_{y} δ^{(3)} (y - x) = - i \nabla^{2} ϕ_{1} (x),

[H_{1}, π_{1}] = i (m_{1}^{2} - \nabla^{2}) ϕ_{1}, \overset{π}{˙}_{1} = (\nabla^{2} - m_{1}^{2}) ϕ_{1} .

(\partial_{t}^{2} - \nabla^{2} + m_{1}^{2}) ϕ_{1} = (□ + m_{1}^{2}) ϕ_{1} = 0.

Suitable decay or spatial smearing removes the boundary term in the integration by parts.

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