Editing Canonical transformation (section)

== Examples ==
* The translation <math>\mathbf{Q}(\mathbf{q}, \mathbf{p})= \mathbf{q} + \mathbf{a}, \mathbf{P}(\mathbf{q}, \mathbf{p})= \mathbf{p} + \mathbf{b}</math> where <math>\mathbf{a}, \mathbf{b}</math> are two constant vectors is a canonical transformation. Indeed, the Jacobian matrix is the identity, which is symplectic: <math>I^\text{T}JI=J</math>.
* Set <math>\mathbf{x}=(q,p)</math> and <math>\mathbf{X}=(Q,P)</math>, the transformation <math>\mathbf{X}(\mathbf{x})=R \mathbf{x}</math> where <math>R \in SO(2)</math> is a rotation matrix of order 2 is canonical. Keeping in mind that special orthogonal matrices obey <math>R^\text{T}R=I</math> it's easy to see that the Jacobian is symplectic. However, this example only works in dimension 2: <math>SO(2)</math> is the only special orthogonal group in which every matrix is symplectic. Note that the rotation here acts on <math>(q,p)</math> and not on <math>q</math> and <math>p</math> independently, so these are not the same as a physical rotation of an orthogonal spatial coordinate system.
* The transformation <math>(Q(q,p), P(q,p))=(q+f(p), p)</math>, where <math>f(p)</math> is an arbitrary function of <math>p</math>, is canonical. Jacobian matrix is indeed given by <math display="block">\frac{\partial X}{\partial x} = \begin{bmatrix} 1 & f'(p) \\ 0 & 1 \end{bmatrix}</math> which is symplectic.