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 $\{Q,P{\}}_{q,p}=1$.

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

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