[NOTES/CM-10006] Two Simple Examples of Canonical Transformations

Example 1
Consider a system with one degree of freedom \(q,p\). Define a transformation of variables
\begin{equation} Q= q\sin\omega p , \qquad P = q \cos \omega p. \end{equation}
We want to check if this transformation is a canonical transformation or not. It is obvious that the Poisson brackets \(\{Q,Q\}_\text{PB}\) and \(\{P,P\}_\text{PB}\) will be zero. For a system of one degree of freedom, we need to check only Poisson bracket \(\{Q,P\}_\text{PB}\). So we compute
\begin{eqnarray} \{Q,P\}_\text{PB} &=& \pp[Q]{q}\pp[P]{p} - \pp[Q]{p}\pp[P]{q}\nonumber\\ &=& \sin \omega p (-q \omega \sin\omega p )- q\omega\cos \omega p .\cos \omega p\nonumber\\ &=& -\omega q\nonumber \end{eqnarray}
Since this Poisson bracket is not equal to 1, the transformation is not a canonical.

Example 2
We leave it for the reader to verify that the transformation
\begin{equation}
Q=p, P=-q
\end{equation}
is a canonical transformation.