Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 9C / Solution

The Poisson bracket is $\{f,g\}={\sum }_a(f_{q_a}{g}_{p_a}-f_{p_a}{g}_{q_a})$. For a function with no explicit time dependence, the chain rule and Hamilton equations giveWith $L_b={ϵ}_{bcd}{x}_c{p}_d$, direct differentiation givesApply the product rule to $L_a={ϵ}_{acd}{x}_c{p}_d$ and substitute these two brackets. Contracting the epsilon symbols yields $\boxed{\{L_a,L_b\}={ϵ}_{abc}{L}_c}$. 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.