# Canonical transformations in 2-d: rotations and polar coordinates

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercises 2.7.4 – 2.7.5.

Here are a couple of examples of canonical variable transformations.

Example 1 We rotate the 2-d rectangular coordinates through an angle ${\theta}$, giving the transformations

 $\displaystyle \overline{x}$ $\displaystyle =$ $\displaystyle x\cos\theta-y\sin\theta\ \ \ \ \ (1)$ $\displaystyle \overline{y}$ $\displaystyle =$ $\displaystyle x\sin\theta+y\cos\theta\ \ \ \ \ (2)$ $\displaystyle \overline{p}_{x}$ $\displaystyle =$ $\displaystyle p_{x}\cos\theta-p_{y}\sin\theta\ \ \ \ \ (3)$ $\displaystyle \overline{p}_{y}$ $\displaystyle =$ $\displaystyle p_{x}\sin\theta+p_{y}\cos\theta \ \ \ \ \ (4)$

To show this is a canonical transformation, we must evaluate the Poisson brackets. Here, ${q_{1}=x}$ and ${q_{2}=y}$. Remember that ${\theta}$ is a constant in these derivatives.

 $\displaystyle \left\{ \overline{x},\overline{y}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\overline{x}}{\partial q_{i}}\frac{\partial\overline{y}}{\partial p_{i}}-\frac{\partial\overline{x}}{\partial p_{i}}\frac{\partial\overline{y}}{\partial q_{i}}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

since neither coordinate depends on any momentum. Similarly ${\left\{ \overline{p}_{x},\overline{p}_{y}\right\} =0}$ since this Poisson bracket contains derivatives of ${\overline{p}_{i}}$ with respect to ${q_{i}}$ and these are all zero. The remaining Poisson bracket are of the form ${\left\{ \overline{q}_{i},\overline{p}_{j}\right\} }$. There are four of these, but we’ll work out only a couple. The other two have similar forms.

 $\displaystyle \left\{ \overline{x},\overline{p}_{x}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\overline{x}}{\partial q_{i}}\frac{\partial\overline{p}_{x}}{\partial p_{i}}-\frac{\partial\overline{x}}{\partial p_{i}}\frac{\partial\overline{p}_{x}}{\partial q_{i}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\overline{x}}{\partial x}\frac{\partial\overline{p}_{x}}{\partial p_{x}}+\frac{\partial\overline{x}}{\partial y}\frac{\partial\overline{p}_{x}}{\partial p_{y}}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\theta+\sin^{2}\theta\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (10)$ $\displaystyle \left\{ \overline{x},\overline{p}_{y}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\overline{x}}{\partial q_{i}}\frac{\partial\overline{p}_{y}}{\partial p_{i}}-\frac{\partial\overline{x}}{\partial p_{i}}\frac{\partial\overline{p}_{y}}{\partial q_{i}}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial\overline{x}}{\partial x}\frac{\partial\overline{p}_{y}}{\partial p_{x}}+\frac{\partial\overline{x}}{\partial y}\frac{\partial\overline{p}_{y}}{\partial p_{y}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sin\theta\cos\theta-\sin\theta\cos\theta\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (14)$

Similarly

 $\displaystyle \left\{ \overline{y},\overline{p}_{x}\right\}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (15)$ $\displaystyle \left\{ \overline{y},\overline{p}_{y}\right\}$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (16)$

Example 2 The transformation from 2-d rectangular to polar coordinates is given by

 $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{x^{2}+y^{2}}\ \ \ \ \ (17)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \arctan\frac{y}{x}\ \ \ \ \ (18)$ $\displaystyle p_{\rho}$ $\displaystyle =$ $\displaystyle \frac{xp_{x}+yp_{y}}{\sqrt{x^{2}+y^{2}}}\ \ \ \ \ (19)$ $\displaystyle p_{\phi}$ $\displaystyle =$ $\displaystyle xp_{y}-yp_{x} \ \ \ \ \ (20)$

For the Poisson brackets we have

 $\displaystyle \left\{ \rho,\phi\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\rho}{\partial q_{i}}\frac{\partial\phi}{\partial p_{i}}-\frac{\partial\rho}{\partial p_{i}}\frac{\partial\phi}{\partial q_{i}}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (22)$

because, again, the coordinates don’t depend on the momenta.

In this case, however, the new momenta do depend on the old coordinates, so we need to actually do some calculation.

 $\displaystyle \left\{ p_{\rho},p_{\phi}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial p_{\rho}}{\partial q_{i}}\frac{\partial p_{\phi}}{\partial p_{i}}-\frac{\partial p_{\rho}}{\partial p_{i}}\frac{\partial p_{\phi}}{\partial q_{i}}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(-\frac{x\left(xp_{x}+yp_{y}\right)}{\left(x^{2}+y^{2}\right)^{3/2}}+\frac{p_{x}}{\sqrt{x^{2}+y^{2}}}\right)\left(-y\right)-\frac{x}{\sqrt{x^{2}+y^{2}}}p_{y}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left(-\frac{y\left(xp_{x}+yp_{y}\right)}{\left(x^{2}+y^{2}\right)^{3/2}}+\frac{p_{y}}{\sqrt{x^{2}+y^{2}}}\right)x-\frac{y}{\sqrt{x^{2}+y^{2}}}\left(-p_{x}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{y^{2}\left(yp_{x}-xp_{y}\right)}{\left(x^{2}+y^{2}\right)^{3/2}}-\frac{x}{\sqrt{x^{2}+y^{2}}}p_{y}-\frac{x^{2}\left(yp_{x}-xp_{y}\right)}{\left(x^{2}+y^{2}\right)^{3/2}}+\frac{y}{\sqrt{x^{2}+y^{2}}}p_{x}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{y^{3}p_{x}+x^{3}p_{y}}{\left(x^{2}+y^{2}\right)^{3/2}}+\frac{y^{3}p_{x}+x^{3}p_{y}}{\left(x^{2}+y^{2}\right)^{3/2}}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (27)$

Finally, we need to work out the mixed brackets.

 $\displaystyle \left\{ \rho,p_{\rho}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\rho}{\partial q_{i}}\frac{\partial p_{\rho}}{\partial p_{i}}-\frac{\partial\rho}{\partial p_{i}}\frac{\partial p_{\rho}}{\partial q_{i}}\right)\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x^{2}}{x^{2}+y^{2}}-0+\frac{y^{2}}{x^{2}+y^{2}}-0\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (30)$ $\displaystyle \left\{ \rho,p_{\phi}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\rho}{\partial q_{i}}\frac{\partial p_{\phi}}{\partial p_{i}}-\frac{\partial\rho}{\partial p_{i}}\frac{\partial p_{\phi}}{\partial q_{i}}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{xy}{x^{2}+y^{2}}-0+\frac{xy}{x^{2}+y^{2}}-0\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (33)$ $\displaystyle \left\{ \phi,p_{\rho}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\phi}{\partial q_{i}}\frac{\partial p_{\rho}}{\partial p_{i}}-\frac{\partial\phi}{\partial p_{i}}\frac{\partial p_{\rho}}{\partial q_{i}}\right)\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{y}{x\left(1+\frac{y^{2}}{x^{2}}\right)\sqrt{x^{2}+y^{2}}}-0+\frac{y}{x\left(1+\frac{y^{2}}{x^{2}}\right)\sqrt{x^{2}+y^{2}}}-0\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (36)$ $\displaystyle \left\{ \phi,p_{\phi}\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\phi}{\partial q_{i}}\frac{\partial p_{\phi}}{\partial p_{i}}-\frac{\partial\phi}{\partial p_{i}}\frac{\partial p_{\phi}}{\partial q_{i}}\right)\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{y^{2}}{x^{2}\left(1+\frac{y^{2}}{x^{2}}\right)}-0+\frac{1}{1+\frac{y^{2}}{x^{2}}}-0\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{y^{2}}{x^{2}+y^{2}}+\frac{x^{2}}{x^{2}+y^{2}}\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (40)$

Thus all the Poisson brackets are correct, so the transformation is canonical.

# 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.