Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 15C / Solution

A canonical transformation is a differentiable invertible change of phase-space coordinates preserving the symplectic form, equivalently the canonical Poisson brackets. At each fixed time, let $D$ be its Jacobian in the ordering $(q,p)\mapsto (Q,P)$ and put $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. The condition iswhich is equivalent to $D^{T}JD=J$. Time dependence changes the Hamiltonian by a generating-function time derivative, but not this fixed-time symplectic condition.

The Hamilton-Jacobi equation is $S_t+\frac{1}{2}[(S_q{)}^2+{\omega }^2{q}^2]=0$. Separating $S=W(q,E)-Et$ gives $(W_q{)}^2=2E-{\omega }^2{q}^2$. On the positive-momentum branch, therefore,Choose the integration constant so that the given angle relation is obtained. For $E>0$, $\beta =S_E$ yields $q=\sqrt{2E}\sin [\omega (t+\beta )]/\omega$, while $p=S_q=\sqrt{2E-{\omega }^2{q}^2}$ on this branch. Continuing the angle across the turning points gives the full orbitThe square-root expression is locally the positive branch; the cosine formula correctly changes sign on the rest of the orbit.

The generating function has differential $dS=pdq+\beta d\alpha -Hdt$. At fixed time, taking an exterior derivative gives $dq\wedge dp=d\beta \wedge d\alpha$. Thus the canonical new coordinate is $Q=\beta$ and its conjugate momentum is $P=\alpha$. Directly, $\alpha =(p^2+{\omega }^2{q}^2)/2$ and $\beta =atan2(\omega q,p)/\omega -t$ locally implyThe angle is a local chart or a periodic variable, and is undefined at the zero-energy point. The source's phrase “variables $(\alpha ,\beta )$” lists the new momentum before its coordinate: if interpreted instead as the conventional coordinate-then-momentum ordering, that ordered pair would have $\{\alpha ,\beta \}=-1$ and be anti-canonical.