Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S7 / 3

The canonical Poisson bracket isThe coordinate derivatives are Kronecker deltas, so direct substitution gives

For a phase-space function with no explicit time dependence, the chain rule and Hamilton's equations giveAn explicitly time-dependent function would instead satisfy $df/dt={\partial }_{t}f+\{f,H\}$.

The Jacobi identity is $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0$. If $f,g$ have no explicit time dependence and are conserved, $\{f,H\}=\{g,H\}=0$. Taking $h=H$ in that identity yields $\{H,\{f,g\}\}=0$, orThus the Poisson bracket of the two conserved functions is another conserved function; it need not be independent of the original ones.

For $L_i={ϵ}_{ijk}{q}_j{p}_k$, the canonical Poisson brackets give $\{L_i,L_j\}={ϵ}_{ijk}{L}_k$. Consequently projections onto fixed axes $n,m$ obeyIf the two original projections are conserved and the axes are linearly independent, the preceding Jacobi identity argument makes the projection along $n\times m$ conserved as well. The vectors $n,m,n\times m$ form a basis of three-dimensional space, so every fixed-axis projection is a linear combination of these conserved quantities. Angular momentum is therefore conserved along every axis.

For the standard meaning of a canonical transformation, the new coordinates must retain Hamiltonian form for every Hamiltonian, with the same physical function expressed in the new coordinates. Let $z=(q,p)$, $Z=F(z)=(Q,P)$, $S=DF$ and $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. With $K(Z)=H(F^{-1}(Z))$, the chain rule givesAgreement with $\dot{Z}=J{\nabla }_{Z}K$ for arbitrary $K$ forces $SJ{S}^T=J$. Its matrix entries are precisely the Poisson brackets of the new coordinates, soThe qualification for every Hamiltonian matters. Preserving one Hamiltonian flow does not imply canonicality. For example, with two coordinate pairs and $H=(q_1^2+p_1^2)/2$, the invertible change $Q_1=q_1$, $P_1=p_1$, $Q_2=2q_2$, $P_2=p_2$ preserves the equations and the same physical Hamiltonian, but $\{Q_2,P_2\}=2$. Thus a definition requiring only one chosen Hamiltonian flow is too weak to prove the claimed bracket identities. The universal definition supplies the missing condition.