Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 15D (Classical Dynamics)

For a canonical coordinate and momentum, the Poisson bracket isThe chain rule and Hamilton equations $\dot{q}=H_p$, $\dot{p}=-H_q$ giveFor the stated complex variables, direct differentiation givesSince $H=\omega a{a}^{\ast }$, the product rule for the Poisson bracket yieldsThe linear change of variables is invertible, so treating $a,a^{\ast }$ as independent coordinates when differentiating givesIn particular $a(t)=a(0)e^{-i\omega t}$ and $a^{\ast }(t)=a^{\ast }(0)e^{i\omega t}$. On a continuously chosen branch along a nonzero trajectory,are constant. The logarithm is undefined at the zero-energy equilibrium, and a fixed principal branch need not be continuous across its cut; the branch qualification is necessary.