[NOTES/CM-10003] Four Types of Canonical Transformations

The action principle requires that a transformation \( (q_k,p_k) \to (Q_k,P_k)\) will be a canonical transformation if there exists new Hamiltonian \(K\) and a function \(F\), called generator of canonical transformation, such that \begin{equation}\label{EQ001} \sum_k p_k\, {dq}_k - H dt = \sum_k P_k\,dQ_k -K dt + dF. \end{equation} The \(4N\) variable \(q_k,p_k\) and \(Q_k, P_k\) satisfy \(2N\) relations and \(2N\) variables can be taken to be independent. Selecting the independent variables in four particular ways, we will have four types of canonical transformations. These transformations will be described below. The discussion given here will be adequate guide to work with more general cases if required.

Our discussion here will be restricted to time independent transformations only.

Type 1 Transformations

The type one canonical transformations are those for which the old and new coordinates \((q_k,Q_k), k=1,2,\ldots,N\) are independent. These are conveniently described by generator \(F_1(q,Q)\) obeying

\begin{equation}\label{EQ01Z} \sum_k p_k  {dq}_k - H dt = \sum_k P_k dQ_k -K dt + dF_1(q,Q, t). \end{equation}

Substituting \begin{equation} dF_1(q,Q,t) = \sum _k \pp[F_1]{q_k} dq_k + \sum \pp[F_1]{Q_k} + \pp[F_1]{t}, \end{equation} in r.h.s. of \eqref{EQ001}, we get \begin{eqnarray} p_k = \pp[F_1]{q_k}, &\qquad& P_k = - \pp[F_1]{Q_k},\label{EQ01B}\\ \text{and } \qquad K &=& H + \pp[F_1]{t}.\label{EQ02}. \end{eqnarray}

Type 2 Transformations

A type 2 transformation corresponds to the situation where \(q_k, P_k\) are independent variables. Starting with equations \begin{equation}\label{EQ03} \sum_k p_k dq_k - H dt = \sum_k P_k dQ_k - K dt + dF_1 \end{equation} we use \begin{equation} d\big(\sum P_k Q_k\big) = \sum_k P_k dQ_k + \sum_k Q_k dP_k, \end{equation} and rewrite \(\sum_k P_k dQ_k\) as \begin{equation} \sum_k P_k dQ_k = d\big(\sum P_k Q_k\big) - \sum_k Q_k dP_k, \end{equation}

Eq.\eqRef{EQ03} then becomes \begin{equation} \sum_k p_k dq_k - H dt = -\sum_k Q_k dP_k - K dt + dF_1 + d\big(\sum P_k Q_k\big) \end{equation} Using the notation \(F_2(q,P) = F_1 + \sum_k(P_kQ_k)\) and writing \begin{equation} dF_2(q,P) = \sum_k \Big( \pp[F_2]{q_k} dq_k + \pp[F_2]{P_k} dP_k\Big) \end{equation} we get

\begin{eqnarray} \sum_k p_k dq_k - H dt &=& -\sum_k Q_k dP_k - K dt + dF_2\nonumber\\ \sum_k p_k dq_k - H dt &=& -\sum_k Q_k dP_k - K dt + \sum_k \Big( \pp[F_2]{q_k} dq_k + \pp[F_2]{P_k} dP_k\Big)\nonumber \end{eqnarray}

Since \(q_k,P_k\) are assumed to be independent, we get
\begin{eqnarray} p_k= \pp[F_2]{q_k}, && Q_k = \pp[F_2]{P_k}\\ K &=& H + \pp[F_2]{t}. \end{eqnarray}

An Example
The identity transformation \( Q_k=q_k,P_k=p_k\)is obviously a canonical transformation.One cannot use type 1 generator as \(q\) and \(Q\) are not independent. However this can be regarded as a type 2 transformation with generator given by \(F_2(q,P)= \sum_k q_kP_k\). Then we would get

\begin{eqnarray} p_j&=& \pp{q_j}\big(\sum_k q_k P_k\big)= P_j,\nonumber\\ Q_j &=& \pp{P_j}\big(\sum_k q_k P_k\big) = q_j.\nonumber \end{eqnarray}

This is just the identity transformation.

Type 3 transformations

In case \(\{p_k,Q_k\}\) form a set of independent variables we proceed as follows. Use \begin{equation} \sum p_kdq_k + q_k dp_k = d\big( \sum_k p_kq_k\big) \end{equation} to write \begin{equation}\label{EQ011} \sum_k p_k dq_k - H dt = \sum_k P_k dQ_k - K dt + dF_1 \end{equation} as

\begin{eqnarray} -\sum_k q_k dp_k + d\big( \sum_k p_kq_k\big)- H dt &=& \sum_k P_k dQ_k - K dt + F_1 \nonumber\\ - \sum_k q_k\, dp_k- H dt &=& \sum_k P_k dQ_k - K dt + F_1-\sum_k q_k dp_k \nonumber\\ \end{eqnarray}

Therefore we introduce \( F_3(p,Q) = F_1(q,Q) - \sum_kp_k q_k\) and write

\begin{eqnarray} - \sum_k\big( q_k dp_k\big)- H dt &-& \sum_k P_k dQ_k = - K dt + dF_3 \nonumber \end{eqnarray}

The variables \(q_k,P_k\) are then given by \begin{equation} q_k = -\pp[F_3]{p_k}, \qquad P_k = -\pp[F_3]{Q_k} \end{equation} and, as usual, the Hamiltonian \(K\) is given by \begin{equation} K = H + \pp[F_3]{t}. \end{equation}

Type 4 transformations

In this case \(\{p_k,P_k\}\) are assumed to form a set of independent variables and we proceed as follows. Use equations

\begin{eqnarray} \sum_k p_kdq_k = -\sum_k q_k dp_k &+& d\big( \sum_k p_kq_k\big)\nonumber\\ \sum_k P_kdQ_k = -\sum_k Q_k dP_k &+& d\big( \sum_k P_kQ_k\big)\nonumber \end{eqnarray}
to write \begin{equation}\label{EQ011A} \sum_k p_k dq_k - H dt = \sum_k P_k dQ_k - K dt + dF_1 \end{equation} as
\begin{eqnarray} -\sum_k q_k dp_k + d\big( \sum_k p_kq_k\big)- H dt &=& -\sum_k Q_k dP_k + \sum_k P_kQ_k - K dt + F_1 \nonumber\\ -\sum_k \big( q_k\, dp_k\big)- H dt &=& -\sum_k Q_k dP_k - K dt + \sum_k P_k dQ_k + F_1 - \sum_k q_k dp_k \nonumber \end{eqnarray}

Therefore, we introduce \( F_4(p,Q) =\sum_k P_kQ_k + F_1(q,Q) - \sum_k p_kq_k\) and write

\begin{eqnarray} - \sum_k\big( q_k dp_k\big)- H dt &-& \sum_k P_k dQ_k = - K dt + dF_4 \nonumber \end{eqnarray}
The variables \(q_k,Q_k\) are then given by \begin{equation} q_k = -\pp[F_4]{p_k}, \qquad Q_k = \pp[F_4]{P_k} \end{equation} and, as usual, the Hamiltonian \(K\) is given by \begin{equation} K = H + \pp[F_4]{t}. \end{equation}