Canonical transformation

A canonical transformation preserves the symplectic form and hence all Poisson brackets. In one degree of freedom, an invertible map $(q,p)\mapsto (Q,P)$ is canonical exactly when $\{Q,P{\}}_{q,p}=1$.

A canonical transformation preserves the symplectic form, equivalently the canonical Poisson brackets, and therefore the Hamiltonian form of every Hamiltonian system. Preservation of just one selected flow is weaker. For $H=(q_1^2+p_1^2)/2$ on two coordinate pairs, $Q_1=q_1$, $P_1=p_1$, $Q_2=2q_2$, $P_2=p_2$ preserves both that physical Hamiltonian and its equations, but $\{Q_2,P_2\}=2$. Thus it is not canonical. Universally preserving Hamiltonian form instead forces $DFJ(DF{)}^T=J$ by applying the chain rule to arbitrary Hamiltonian gradients.

The accompanying momentum map is $p=P+2\lambda XP/(m{\omega }^2)$. Its Poisson bracket is one up to second-order corrections. Substitution into $p^2/(2m)+m{\omega }^2{x}^2/2+\lambda x^3$ cancels all terms linear in $\lambda$, leaving a harmonic Hamiltonian to that order. The map is local and perturbative, not an exact global removal of an unbounded cubic potential.

With $\vartheta =\omega t+\delta$, a weak cubic potential produces a nonzero mean position and a second harmonic at first order. There is no first-order frequency shift. Higher-order frequency changes can accumulate phase over sufficiently long times, so the first-order expression is not a uniform all-time approximation.

For $p>0$, $Q=\sqrt{2p}\sinh q$ and $P=\sqrt{2p}\cosh q$ satisfy $\{Q,P\}=1$. They transform $p\cosh 2q$ into $(Q^2+P^2)/2$. Their inverse is $p=(P^2-Q^2)/2$ and $q=\mathrm{artanh}(Q/P)$, so the chart covers only $P>|Q|$, not the whole oscillator plane. Oscillator evolution beyond that wedge is not a global real continuation of the original variables.

For $H=\omega \cdot y+\frac{1}{2}|y{|}^2+ϵV(x)$ with a Diophantine frequency vector, solve ${\mathcal{D}}_{\omega }u=⟨V⟩-V$ and apply translated conjugacy of a Diophantine vector field to $w=ϵ\nabla u$. Set $\eta =ϵ\nabla u+a$ and use the symplectic cotangent lift with a closed momentum shift. The transformed Hamiltonian is exactly $E+\omega \cdot y^{\prime }+\frac{1}{2}{y}^{\prime T}D{\chi }^{-1}D{\chi }^{-T}{y}^{\prime }+\frac{1}{2}|\eta \circ \chi {|}^2$, where $E=ϵ⟨V⟩+\omega \cdot a$. Since $a=O({ϵ}^2)$, the last term is an angular potential of order ${ϵ}^2$ and the quadratic coefficient is identity plus $O(ϵ)$. The construction is on a smaller angular strip and momentum ball.

Independent functions $Q_1,\dots ,Q_n$ on a $2n$-dimensional symplectic manifold can locally form the position half of canonical coordinates exactly when $\{Q_i,Q_j\}=0$ throughout a neighbourhood. Their commuting Hamiltonian vector fields span the fibres of $Q$. Flow from a transverse section gives coordinates $(Q,t)$ with $\omega =\sum d{t}_i\wedge d{Q}_i+\beta (Q)$. Write $\beta =d\alpha$ locally and take $P_i=-t_i-{\alpha }_i(Q)$. Then $\omega =\sum d{Q}_i\wedge d{P}_i$.

A function $S(q,P)$ generates a canonical transformation through

The type-two generating function for a canonical transformation $F_2(q,P)=\cos q\sin P$ defines $p=-\sin q\sin P$ and $Q=\cos q\cos P$. On any branch with $0<q<\pi$ and $\cos P\ne 0$, its inverse in mixed coordinates isThe square root equals $\sin q$ on this branch. Differentiating $F_2$ proves the generating equations, and $F_{2,qP}=-\sin q\cos P\ne 0$ proves local invertibility by the inverse function theorem. Also $d(F_2-PQ)=pdq-PdQ$, so the symplectic form obeys $dq\wedge dp=dQ\wedge dP$. This proves the transformation is canonical locally. It is not a single global chart across the inverse-trigonometric branch boundaries.

Since $F_2$ is independent of time, the transformed Hamiltonian is obtained by substitution, with zero generating-function time correction. An added function of time alone is an irrelevant Hamiltonian freedom, since it leaves Hamilton's equations unchanged.

A function $\Phi (q,Q)$ generates a canonical transformation through