 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-3/9c/solution

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

{f, g} = \sum_{a} (f_{q_{a}} g_{p_{a}} - f_{p_{a}} g_{q_{a}})

. For a function with no explicit time dependence, the chain rule and Hamilton equations give

\frac{df}{d t} = \sum_{a} (f_{q_{a}} \overset{q}{˙}_{a} + f_{p_{a}} \overset{p}{˙}_{a}) = \sum_{a} (f_{q_{a}} H_{p_{a}} - f_{p_{a}} H_{q_{a}}) = {f, H} .

With

L_{b} = ϵ_{b c d} x_{c} p_{d}

{p_{a}, L_{b}} = - ϵ_{ba d} p_{d} = ϵ_{ab d} p_{d}, {x_{a}, L_{b}} = ϵ_{ab c} x_{c} .

Apply the product rule to

L_{a} = ϵ_{a c d} x_{c} p_{d}

and substitute these two brackets. Contracting the epsilon symbols yields

{L_{a}, L_{b}} = ϵ_{ab c} L_{c}

. Equivalently, the two product terms are precisely the infinitesimal rotation of the vector

x \times p

about the

b

axis, with the same bracket convention as the position and momentum vectors.

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