Problem statement: The Hamiltonian of a system is given by the formula:
\begin{equation*}
H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + V(r,\theta).
\end{equation*}
Under what condition is $f=p_\theta^2$ an integral of motion?
Attempted solution:
In order for $f$ to be an integral of motion, according to Poisson theorem, is that:
\begin{equation*}
[f, H] =0,
\end{equation*}
where $[,]$ is the Poisson bracket.
Therefore:
\begin{align*}
\left[\frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2} + V(r,\theta), p_\theta^2\right] &= 0 \Leftrightarrow \\
\left[\frac{p_r^2}{2m}, p_\theta^2\right] + \left[\frac{p_\theta^2}{2mr^2}, p_\theta^2\right] + \left[V(r,\theta), p_\theta^2\right] &= 0.
\end{align*}
But how do I proceed from here on?
EDIT1: Regarding the first term:
\begin{align*}
\left[\frac{p_r^2}{2m},p_\theta^2\right]&=\frac{1}{2m}\left[p_r^2,p_\theta^2\right]=\frac{1}{2m}[p_r p_r, p_\theta^2]=\frac{1}{2m}\left(p_r[p_r,p_\theta^2]+p_r[p_r,p_\theta^2]\right)\\
&=\frac{1}{m}[p_r,p_\theta^2]=
\frac{1}{m}\left(p_\theta[p_r, p_\theta]+p_\theta[p_r,p_\theta]\right) = \frac{2p_\theta[p_r,p_\theta]}{m}
\end{align*}
So it boils down to what is $[p_r, p_\theta]$ equal to ? I'm not sure whether it can be answered directly without resorting to the definition of Poisson bracket.
The coordinates in our problem are $q_i = \{r, \theta\}, p_i = \{p_r, p_\theta\}$. Therefore:
\begin{align*}
[p_r, p_\theta] &= \sum_{i=1}^2 \left(\frac{\partial p_r}{\partial q_i}\frac{\partial p_\theta}{\partial p_i} – \frac{\partial p_r}{\partial p_i}\frac{\partial p_\theta}{\partial q_i}\right)\\ &=
\left(\frac{\partial p_r}{\partial r}\frac{\partial p_\theta}{\partial p_r} – \frac{\partial p_r}{\partial p_r}\frac{\partial p_\theta}{\partial r}\right) +
\left(\frac{\partial p_r}{\partial \theta}\frac{\partial p_\theta}{\partial p_\theta} – \frac{\partial p_r}{\partial p_\theta}\frac{\partial p_\theta}{\partial \theta}\right)\\
\end{align*}
And now I'm stuck again.
The way I understand it $p_r, p_\theta$ are two components of momentum along $r, \theta$ respectively. Does that imply that $p_r$ does not depend on $\theta$ and that it only depends on $r$ ? If the answer is yes, then the result for the first term is zero.
EDIT 2:
Based on the answers in Hamiltonian formalism all the canonical variables are taken independent to each other, therefore $[p_r, p_\theta]=0$. Similarly it may be shown that the 2nd term is also zero.
Best Answer
I think it is pretty obviously in your last line. So what are the results of the commutator $[p_r,p_\theta]$ and $[p_\theta,p_\theta]$? Note that $p_\theta$ is not commute with $\theta$. Then you should get the answer.