Past exam of the theoretical physics I course of the University of Cambridge / 2013 / Paper 1 / 3

In canonical coordinates, the Poisson bracket isIn particular, $\{q_j,g\}={\partial }_{p_j}g$, $\{p_j,g\}=-{\partial }_{q_j}g$, $\{f,q_j\}=-{\partial }_{p_j}f$ and $\{f,p_j\}={\partial }_{q_j}f$. Hence $\{q_i,p_j\}={\delta }_{ij}$ and the coordinate–coordinate and momentum–momentum brackets vanish.

The chain rule along a Hamiltonian trajectory givesThe Poisson bracket of time-dependent conserved quantities follows by defining $D_t={\partial }_t+\{\cdot ,H\}$. The ordinary product rule and Jacobi identity implyTo see the Hamiltonian part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_{t}f=D_{t}g=0$ implies $D_t\{f,g\}=0$, even when the conserved functions depend explicitly on time.

A locally invertible time-independent canonical transformation is canonical whenThe proposed coordinates are locally independent only for $q_1\ne 0$, since their configuration-space Jacobian has determinant $2q_1$. Work on one sign branch of $q_1$.

The mixed brackets first requireThe last bracket condition isExpressing $A,B$ in the $Q$ coordinates turns this into ${\partial }_{Q_2}A={\partial }_{Q_1}B$. Thus their shift one-form is closed and locally is the differential of an arbitrary smooth function $S(Q_1,Q_2)$. The canonical momenta for a squared coordinate therefore have the general local formConversely, any such $S$ satisfies all bracket conditions. This does not give one nonsingular transformation through $q_1=0$, or identify the two sign branches globally.

Choose $S=Q_2^3/3$. Then