References: Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Section 2.7; Exercises 02.07.06 – 02.07.07, 02.07.08(4).

Here are a few more examples of canonical variable transformations.

Example 1First, we revisit the two-body problem, in which we simplified the problem by transforming from the coordinates and of the masses and to two new position vectors:

Here is the total mass, is the relative position, and is the position of the centre of mass. The conjugate momenta in the original system are

The conjugate momenta transform according to

where is the reduced mass.

To check that this is a canonical transformation, we need to calculate the Poisson brackets. To make things easier, note that the new coordinates depend only on the old coordinates (and not on the momenta), and conversely, the new momenta depend only on the old momenta (and not on the coordinates). Since the Poisson brackets and all involve taking derivatives of coordinates with respect to momenta (in the first case) or momenta with respect to coordinates (in the second case), all these brackets are zero. We need, therefore, to check only the mixed brackets between coordinates and momenta.

Because we’re dealing with 3-d vector equations, there are 3 components to each vector and to be thorough, we need to calculate all possible brackets between all pairs of components. However, if we do the component of each, it should be obvious that the and components behave in the same way.

First, consider

In the RHS, the term stands for all 6 components of the original position vectors, that is and the term in the denominators refers to all 6 components of the original momentum vectors. The in the numerators refers to the component of in 6. Hopefully this won’t cause too much confusion.

The second term on the RHS is zero because it involves derivatives of coordinates with respect to momenta (and vice versa). In the first term, depends only the components of and , and depends only on the components of and , so we have

The same result is obtained for the and components. If we look at mixing two different components, we have, for example

This is zero because each term in the sum contains a derivative of an component with respect to a component (or vice versa), all of which are zero.

For the centre of mass components, we have

where the last bracket is zero for the same reason as : we’re mixing and in the derivatives. Again, it should be obvious that the brackets for the other combinations of , and components work out the same way.

Example 2A bizarre transformation of variables in one dimension is given by

To show this is canonical, we need calculate only (since the Poisson bracket of a function with itself is always zero, we have ). We need one rather obscure derivative of a trig function.

We get

Thus the transformation is canonical.

Example 3Finally, we return to the point transformation, which is given in general by

In this case, the coordinate transformation to is completely arbitrary, but the momentum transformation must follow the formula given. The derivatives in the formula for are taken at constant . As in the earlier examples, since the coordinate formulas depend only on the old coordinates, and the momentum formulas depend only on the old momenta, the Poisson brackets satisfy

For the mixed brackets, we have

The second term in the first line is zero (mixed derivatives again). We used 28 to calculate the derivative and get the second line and then notice that the sum is an expansion of the chain rule for the derivative in line 3. Since and are independent variables, the result is that given in the last line. Thus a point transformation is a canonical transformation.