[NOTES/CM-10008] Examples --- Canonical Transformations

Example 1
Let us find the transformation generated by type 1 function \(F_1(q,Q)=qQ\).
We then get
\begin{equation}
p =\pp[F_1]{q}= Q, \qquad P =-\pp[F_1]{Q}= q
\end{equation}
Thus we get the transformation \(q,p \to Q=p, P=-q\) and \(K= H\).

Example 2
If we choose type 1 generator as \(F_1(q,Q)= q \sin Q\), we get
\begin{eqnarray} p=\pp[F_1]{q}= \sin Q &\Longrightarrow& Q=\sin^{-1}p \label{EQ13}\\ P =-\pp[F_1]{Q}= - q\cos Q &\Longrightarrow& P = -q\sqrt{1-p^2}\label{EQ14} \end{eqnarray} and \(K=H\).

Example 3
Use Poisson bracket text to verify that the transformation given by \eqref{EQ13} and \eqref{EQ14} is indeed a canonical transformation.

Method 1 :Express new variables \(Q,P\) as functions of old variables \(q,p\) and compute the Poisson brackets. Remember that for one degree of freedom, it is sufficient to check the Poisson bracket of \(Q,P\) because other Poisson brackets of \(Q,Q\) and \(P,P\) are trivially zero. The given transformation equations are \begin{equation} Q= \sin^{-1}p, \qquad P = -q\sqrt{1-p^2}. \end{equation} Using these equations we get

\begin{eqnarray} \{Q,P\}_\text{PB} &=& \pp[Q]{q}.\pp[P]{p} -\pp[Q]{p}.\pp[P]{q}\nonumber\\ &=& 0 - \frac{1}{\sqrt{1-p^2}}(-\sqrt{1-p^2})\nonumber\\ &=&1.\nonumber \end{eqnarray}

Method 2 : Express old variables \(q,p\) as functions of new variables \(Q,P\) and compute the Poisson brackets. \begin{equation} q= -\frac{P}{\cos Q}, \qquad p= -\sin Q. \end{equation} Hence we get

\begin{eqnarray} \{q,p\}_\text{PB} &=& \pp[q]{Q}.\pp[p]{Q} -\pp[q]{P}.\pp[p]{Q}\nonumber\\ &=&\pp[q]{Q}\times 0 - \Big(\frac{-1}{\cos Q}\Big) \times \cos Q\nonumber\\ &=&1\nonumber. \end{eqnarray}