Conditions for a transformation to be canonical

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercise 2.7.3.

We’ve seen that the Euler-Lagrange equations are invariant under canonical transformations, but in the Hamiltonian formalism where the system moves in a {2n}-dimensional phase space with {n} coordinates {q} and {n} momenta {p}, more general transformations are possible:

\displaystyle   \overline{q}_{i} \displaystyle  = \displaystyle  \overline{q}_{i}\left(q,p\right)\ \ \ \ \ (1)
\displaystyle  \overline{p}_{i} \displaystyle  = \displaystyle  \overline{p}_{i}\left(q,p\right) \ \ \ \ \ (2)

In order for such a transformation to be canonical, we require that the new variables {\overline{q}} and {\overline{p}} satisfy Hamilton’s equations, that is

\displaystyle   \frac{\partial H}{\partial\overline{p}_{i}} \displaystyle  = \displaystyle  \dot{\overline{q}}_{i}\ \ \ \ \ (3)
\displaystyle  -\frac{\partial H}{\partial\overline{q}_{i}} \displaystyle  = \displaystyle  \dot{\overline{p}}_{i} \ \ \ \ \ (4)

In principle, then, we could check the Hamiltonian in the new coordinates to see if these equations are valid, but it would seem that whether or not a set of coordinates and momenta is canonical should be determinable from the variables themselves, and not depend on the specific Hamiltonian. Here we derive a set of conditions on the {\overline{q}} and {\overline{p}} that determine whether or not the transformation is canonical.

The time derivative of any function {\omega} can be written as a Poisson bracket:

\displaystyle  \dot{\omega}=\left\{ \omega,H\right\} \ \ \ \ \ (5)

For the transformed velocities, we have

\displaystyle   \dot{\overline{q}}_{j} \displaystyle  = \displaystyle  \left\{ \overline{q}_{j},H\right\} \ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\overline{q}_{j}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right) \ \ \ \ \ (7)

Here, {H} is written as a function {H\left(q,p\right)} of the original variables. If we write it as a function of the transformed variables, we can find the two derivatives of {H} in 7 by using the chain rule:

\displaystyle   \frac{\partial H\left(\overline{q},\overline{p}\right)}{\partial p_{i}} \displaystyle  = \displaystyle  \sum_{k}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}\right)\ \ \ \ \ (8)
\displaystyle  \frac{\partial H\left(\overline{q},\overline{p}\right)}{\partial q_{i}} \displaystyle  = \displaystyle  \sum_{k}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right) \ \ \ \ \ (9)

Inserting these into 7 we get

\displaystyle   \dot{\overline{q}}_{j} \displaystyle  = \displaystyle  \sum_{i}\sum_{k}\left[\frac{\partial\overline{q}_{j}}{\partial q_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}\right)-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\right]\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\sum_{i}\left(\frac{\partial\overline{q}_{j}}{\partial q_{i}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}\right)+\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\sum_{i}\left(\frac{\partial\overline{q}_{j}}{\partial q_{i}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}-\frac{\partial\overline{q}_{j}}{\partial p_{i}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\left\{ \overline{q}_{j},\overline{q}_{k}\right\} +\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\left\{ \overline{q}_{j},\overline{p}_{k}\right\} \ \ \ \ \ (12)

In order for this result to satisfy 3, we must have

\displaystyle   \left\{ \overline{q}_{j},\overline{q}_{k}\right\} \displaystyle  = \displaystyle  0\ \ \ \ \ (13)
\displaystyle  \left\{ \overline{q}_{j},\overline{p}_{k}\right\} \displaystyle  = \displaystyle  \delta_{jk} \ \ \ \ \ (14)

We can repeat the calculation for {\dot{\overline{p}}_{i}}:

\displaystyle   \dot{\overline{p}}_{j} \displaystyle  = \displaystyle  \left\{ \overline{p}_{j},H\right\} \ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\left(\frac{\partial\overline{p}_{j}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (16)
\displaystyle  \displaystyle  = \displaystyle  \sum_{i}\sum_{k}\left[\frac{\partial\overline{p}_{j}}{\partial q_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}\right)-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\left(\frac{\partial H}{\partial\overline{q}_{k}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}+\frac{\partial H}{\partial\overline{p}_{k}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\right]\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\sum_{i}\left(\frac{\partial\overline{p}_{j}}{\partial q_{i}}\frac{\partial\overline{q}_{k}}{\partial p_{i}}-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\frac{\partial\overline{q}_{k}}{\partial q_{i}}\right)+\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\sum_{i}\left(\frac{\partial\overline{p}_{j}}{\partial q_{i}}\frac{\partial\overline{p}_{k}}{\partial p_{i}}-\frac{\partial\overline{p}_{j}}{\partial p_{i}}\frac{\partial\overline{p}_{k}}{\partial q_{i}}\right)\ \ \ \ \ (18)
\displaystyle  \displaystyle  = \displaystyle  \sum_{k}\frac{\partial H}{\partial\overline{q}_{k}}\left\{ \overline{p}_{j},\overline{q}_{k}\right\} +\sum_{k}\frac{\partial H}{\partial\overline{p}_{k}}\left\{ \overline{p}_{j},\overline{p}_{k}\right\} \ \ \ \ \ (19)

Requiring this to satsify 4, we have

\displaystyle   \left\{ \overline{p}_{j},\overline{p}_{k}\right\} \displaystyle  = \displaystyle  0\ \ \ \ \ (20)
\displaystyle  \left\{ \overline{p}_{j},\overline{q}_{k}\right\} \displaystyle  = \displaystyle  -\delta_{jk} \ \ \ \ \ (21)

The last equation is equivalent to

\displaystyle  \left\{ \overline{q}_{j},\overline{p}_{k}\right\} =\delta_{jk} \ \ \ \ \ (22)

which agrees with 14. Thus in order for the transformation to be canonical, the conditions are

\displaystyle   \left\{ \overline{q}_{j},\overline{q}_{k}\right\} \displaystyle  = \displaystyle  \left\{ \overline{p}_{j},\overline{p}_{k}\right\} =0\ \ \ \ \ (23)
\displaystyle  \left\{ \overline{q}_{j},\overline{p}_{k}\right\} \displaystyle  = \displaystyle  \delta_{jk} \ \ \ \ \ (24)

Note that these Poisson brackets require calculating the derivatives of the new variables {\overline{q}} and {\overline{p}} with respect to the original ones {q} and {p}, but they don’t involve any particular Hamiltonian. Thus it’s possible to determine whether or not a transformation is canonical entirely from the transformation equations 1 and 2.