 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/32a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

From

z_{j} = q_{j} + i p_{j}

and

\overset{z}{ˉ}_{j} = q_{j} - i p_{j}

, the chain rule gives

\frac{\partial}{\partial q _{j}} = \frac{\partial}{\partial z _{j}} + \frac{\partial}{\partial z ˉ _{j}}, \frac{\partial}{\partial p _{j}} = i \frac{\partial}{\partial z _{j}} - i \frac{\partial}{\partial z ˉ _{j}} .

Substituting these identities into the canonical Poisson bracket and cancelling the

f_{z_{j}} g_{z_{j}}

and

f_{\overset{z}{ˉ}_{j}} g_{\overset{z}{ˉ}_{j}}

terms gives

{f, g} = - 2 i \frac{\partial f}{\partial z _{j}} \frac{\partial g}{\partial z ˉ _{j}} + 2 i \frac{\partial g}{\partial z _{j}} \frac{\partial f}{\partial z ˉ _{j}} .

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