# Cyclic coordinates and Poisson brackets

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 2.7; Exercises 2.7.1 – 2.7.2.

Hamilton’s canonical equations are:

 $\displaystyle \frac{\partial H}{\partial p_{i}}$ $\displaystyle =$ $\displaystyle \dot{q}_{i}\ \ \ \ \ (1)$ $\displaystyle -\frac{\partial H}{\partial q_{i}}$ $\displaystyle =$ $\displaystyle \dot{p}_{i} \ \ \ \ \ (2)$

If a coordinate ${q_{i}}$ is missing in the Hamiltonian (that is, ${H}$ is indepedent of ${q_{i}}$), then

$\displaystyle \dot{p}_{i}=-\frac{\partial H}{\partial q_{i}}=0 \ \ \ \ \ (3)$

Thus the conjugate momentum ${p_{i}}$ is conserved. Such a missing coordinate ${q_{i}}$ is known as a cyclic coordinate. [I’m not sure of the origin of this term. Again Google doesn’t provide a definitive answer.]

There is a general method for calculating the rate of change of some function ${\omega\left(p,q\right)}$ that depends on the momenta and coordinates, but not explicitly on the time (${\omega}$ is allowed to depend implicitly on time since ${p}$ and/or ${q}$ can depend on time). The time derivative can then be written using the chain rule:

 $\displaystyle \frac{d\omega}{dt}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\dot{q}_{i}+\frac{\partial\omega}{\partial p_{i}}\dot{p}_{i}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \left\{ \omega,H\right\} \ \ \ \ \ (6)$

where in the second line we used Hamilton’s equations 1 and 2. The last line defines the Poisson bracket of the function ${\omega}$ with the Hamiltonian ${H}$. We can see that if ${\left\{ \omega,H\right\} =0}$, the function ${\omega}$ is conserved.

Since ${\left\{ H,H\right\} =0}$ automatically, the total energy (represented by the Hamiltonian) is conserved, provided there is no explicit time dependence. Such a time dependence can arise if the system is subject to some external force, for example.

From the definition 5 we can derive a few fundamental properties of Poisson brackets. We’ll consider a general Poisson bracket between two arbitrary functions ${\omega\left(p,q\right)}$ and ${\lambda\left(p,q\right)}$. Then

 $\displaystyle \left\{ \omega,\lambda\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sum_{i}\left(\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}-\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sum_{i}\left(\frac{\partial\lambda}{\partial q_{i}}\frac{\partial\omega}{\partial p_{i}}-\frac{\partial\lambda}{\partial p_{i}}\frac{\partial\omega}{\partial q_{i}}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left\{ \lambda,\omega\right\} \ \ \ \ \ (10)$

A Poisson bracket is distributive, in the sense that

 $\displaystyle \left\{ \omega,\lambda+\sigma\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\left(\lambda+\sigma\right)}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\left(\lambda+\sigma\right)}{\partial q_{i}}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\left[\frac{\partial\lambda}{\partial p_{i}}+\frac{\partial\sigma}{\partial p_{i}}\right]-\frac{\partial\omega}{\partial p_{i}}\left[\frac{\partial\lambda}{\partial q_{i}}+\frac{\partial\sigma}{\partial q_{i}}\right]\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)+\sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ \omega,\lambda\right\} +\left\{ \omega,\sigma\right\} \ \ \ \ \ (14)$

One more identity is useful, which we can derive using the product rule:

 $\displaystyle \left\{ \omega,\lambda\sigma\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\left(\lambda\sigma\right)}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\left(\lambda\sigma\right)}{\partial q_{i}}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i}\sigma\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\lambda}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\lambda}{\partial q_{i}}\right)+\sum_{i}\lambda\left(\frac{\partial\omega}{\partial q_{i}}\frac{\partial\sigma}{\partial p_{i}}-\frac{\partial\omega}{\partial p_{i}}\frac{\partial\sigma}{\partial q_{i}}\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\{ \omega,\lambda\right\} \sigma+\left\{ \omega,\sigma\right\} \lambda \ \ \ \ \ (17)$

The Poisson brackets involving the coordinates ${q_{i}}$ and momenta ${p_{i}}$ turn up frequently, so it’s worth deriving them in detail. We have

$\displaystyle \left\{ q_{i},q_{j}\right\} =\sum_{k}\left(\frac{\partial q_{i}}{\partial q_{k}}\frac{\partial q_{j}}{\partial p_{k}}-\frac{\partial q_{i}}{\partial p_{k}}\frac{\partial q_{j}}{\partial q_{k}}\right)=0 \ \ \ \ \ (18)$

This follows because, in the Hamiltonian formalism, the ${q_{i}}$s and ${p_{i}}$s are independent variables, so ${\frac{\partial q_{j}}{\partial p_{k}}=\frac{\partial p_{j}}{\partial q_{k}}=0}$ for all ${j}$ and ${k}$. For the same reason, we have

$\displaystyle \left\{ p_{i},p_{j}\right\} =\sum_{k}\left(\frac{\partial p_{i}}{\partial q_{k}}\frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial p_{i}}{\partial p_{k}}\frac{\partial p_{j}}{\partial q_{k}}\right)=0 \ \ \ \ \ (19)$

The mixed Poisson bracket is a different story, however:

 $\displaystyle \left\{ q_{i},p_{j}\right\}$ $\displaystyle =$ $\displaystyle \sum_{k}\left(\frac{\partial q_{i}}{\partial q_{k}}\frac{\partial p_{j}}{\partial p_{k}}-\frac{\partial q_{i}}{\partial p_{k}}\frac{\partial p_{j}}{\partial q_{k}}\right)\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{k}\delta_{ik}\delta_{jk}-0\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta_{ij} \ \ \ \ \ (22)$

Hamilton’s equations 1 and 2 can be written using Poisson brackets by setting ${\omega}$ equal to ${q_{i}}$ and ${p_{i}}$ respectively in 6:

 $\displaystyle \dot{q}_{i}$ $\displaystyle =$ $\displaystyle \left\{ q_{i},H\right\} \ \ \ \ \ (23)$ $\displaystyle \dot{p}_{i}$ $\displaystyle =$ $\displaystyle \left\{ p_{i},H\right\} \ \ \ \ \ (24)$

Example In two dimensions, we have a Hamiltonian:

$\displaystyle H=p_{x}^{2}+p_{y}^{2}+ax^{2}+by^{2} \ \ \ \ \ (25)$

If ${a=b}$, then in polar coordiantes, the only coordinate appearing in ${H}$ is the radial distance from the origin ${r=\sqrt{x^{2}+y^{2}}}$, which means that the polar angle ${\theta}$ is a cyclic coordinate. This means that the conjugate momentum ${p_{\theta}}$ must be conserved. That is,

$\displaystyle \dot{p}_{\theta}=\left\{ p_{\theta},H\right\} =0 \ \ \ \ \ (26)$

However, ${p_{\theta}}$ is the angular momentum ${\ell_{z}}$, so this just says that angular momentum is conserved.

To see this explicitly, it’s easier to convert to polar coordinates. From Hamilton’s equations

 $\displaystyle \dot{x}$ $\displaystyle =$ $\displaystyle \frac{\partial H}{\partial p_{x}}=2p_{x}\ \ \ \ \ (27)$ $\displaystyle \dot{y}$ $\displaystyle =$ $\displaystyle 2p_{y}\ \ \ \ \ (28)$ $\displaystyle p_{x}^{2}+p_{y}^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\dot{x}^{2}+\dot{y}^{2}\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v^{2}}{4}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right) \ \ \ \ \ (31)$

where in the fourth line, ${v}$ is the linear velocity and in the fifth line we converted this to polar coordinates. Thus the Hamiltonian becomes, in the case where ${a=b}$:

$\displaystyle H=\frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)+ar^{2} \ \ \ \ \ (32)$

To find the conjugate momenta in polar coordinates, we can write out the Lagrangian. We use ${p_{x}\dot{x}=\frac{\dot{x}^{2}}{2}}$ and ${p_{y}\dot{y}=\frac{\dot{y}^{2}}{2}}$ and get

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \sum_{i}p_{i}\dot{q}_{i}-H\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)-\frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-ar^{2}\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\dot{r}^{2}+r^{2}\dot{\theta}^{2}\right)-ar^{2} \ \ \ \ \ (35)$

The conjugate momenta are thus

 $\displaystyle p_{\theta}$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{\theta}}=\frac{1}{2}r^{2}\dot{\theta}\ \ \ \ \ (36)$ $\displaystyle p_{r}$ $\displaystyle =$ $\displaystyle \frac{\partial L}{\partial\dot{r}}=\frac{\dot{r}}{2} \ \ \ \ \ (37)$

From this we can see that ${p_{\theta}}$ is indeed angular momentum as it’s proportional to the product of ${r}$ and the tangential velocity ${v_{\theta}=r\dot{\theta}}$. (‘Real’ momentum and angular momentum must, of course, also contain a factor of a mass, but from the definition of the Hamiltonian above, we see that the mass has been incorporated into the momentum parameters.)

Plugging these back into 32 we get

$\displaystyle H=p_{r}^{2}+p_{\theta}^{2}+ar^{2} \ \ \ \ \ (38)$

We can now calculate the Poisson brackets easily:

 $\displaystyle \left\{ p_{\theta},H\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial p_{\theta}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial p_{\theta}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0-\frac{\partial p_{\theta}}{\partial p_{\theta}}\frac{\partial H}{\partial\theta}=0\ \ \ \ \ (40)$ $\displaystyle \left\{ p_{r},H\right\}$ $\displaystyle =$ $\displaystyle \sum_{i}\left(\frac{\partial p_{r}}{\partial q_{i}}\frac{\partial H}{\partial p_{i}}-\frac{\partial p_{r}}{\partial p_{i}}\frac{\partial H}{\partial q_{i}}\right)\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0-\frac{\partial p_{r}}{\partial p_{r}}\frac{\partial H}{\partial r}\ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2ar \ \ \ \ \ (43)$

Thus ${p_{\theta}}$ (the angular momentum) is conserved, while ${p_{r}<0}$, so that the object is always being pulled in towards the origin.