The translation ${\displaystyle \mathbf {Q} (\mathbf {q} ,\mathbf {p} )=\mathbf {q} +\mathbf {a} ,\mathbf {P} (\mathbf {q} ,\mathbf {p} )=\mathbf {p} +\mathbf {b} }$ where ${\displaystyle \mathbf {a} ,\mathbf {b} }$ are two constant vectors is a canonical transformation. Indeed, the Jacobian matrix is the identity, which is symplectic: ${\displaystyle I^{\text{T}}JI=J}$.

Set ${\displaystyle \mathbf {x} =(q,p)}$ and ${\displaystyle \mathbf {X} =(Q,P)}$, the transformation ${\displaystyle \mathbf {X} (\mathbf {x} )=R\mathbf {x} }$ where ${\displaystyle R\in SO(2)}$ is a rotation matrix of order 2 is canonical. Keeping in mind that special orthogonal matrices obey ${\displaystyle R^{\text{T}}R=I}$ it's easy to see that the Jacobian is symplectic. However, this example only works in dimension 2: ${\displaystyle SO(2)}$ is the only special orthogonal group in which every matrix is symplectic. Note that the rotation here acts on ${\displaystyle (q,p)}$ and not on ${\displaystyle q}$ and ${\displaystyle p}$ independently, so these are not the same as a physical rotation of an orthogonal spatial coordinate system.

The transformation ${\displaystyle (Q(q,p),P(q,p))=(q+f(p),p)}$, where ${\displaystyle f(p)}$ is an arbitrary function of ${\displaystyle p}$, is canonical. Jacobian matrix is indeed given by

$${\displaystyle {\frac {\partial X}{\partial x}}={\begin{bmatrix}1&f'(p)\\0&1\end{bmatrix}}}$$

which is symplectic.

Symplectomorphism

Hamilton–Jacobi equation

Liouville's theorem (Hamiltonian)

Mathieu transformation

Linear canonical transformation

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