 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ii/paper-2/15c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

F

\frac{d F}{d t} = \sum_{k} (F_{q_{k}} H_{p_{k}} - F_{p_{k}} H_{q_{k}}) = [F, H] .

Thus a constant of motion has $[F, H] = 0$ . For

F = \sum_{k} p_{k}

and the given Hamiltonian,

[F, H] = - \sum_{k} V_{q_{k}}

. Simultaneous translation

q_{k} \mapsto q_{k} + s

q_{i} - q_{j}

, and hence

V

, unchanged. Differentiation with respect to

s

gives

\sum_{k} V_{q_{k}} = 0

, so $[F, H] = 0$ also in this example. This is conservation of total momentum from translational symmetry.

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