Daily Archives: Mon, 5 December 2016

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.