Particle in the Yukawa potential - Showing that the $z$-component of the angular momentum is conserved

I'm sorry for this homework question but I'm sitting a really long time now on this rather "easy" looking problem and I can't find a way to solve it.

I'm given the Hamiltonian of the particle: $$H=\frac{1}{2m}(p_r^2+\frac{p_{\theta}^2}{r^2}+\frac{p_{\phi}^2}{r^2\sin^2\theta})-\alpha Q^2\frac{e^{-\beta mr}}{r}.$$

I have to explicitly show that the z-component of the angular moment $$L_z=mr^2\sin^2(\theta)\dot{\phi}=p_{\phi}$$ is conserved by showing that $$\{L_z,H\}=0$$

I thought that this should be reltaive straightforward cause a lot of terms of the poisson bracket will be zero (e.g. $$\frac{\partial L_z}{\partial \phi},\frac{\partial L_z}{\partial p_r},\frac{\partial L_z}{\partial p_{\theta}},\frac{\partial H}{\partial \phi},$$). Calculating the nonzero terms I get: $$\{L_z,H\}=2r\sin^2(\theta)\dot{\phi}p_r+2r\cos(\theta)\sin(\theta)\dot{\phi}p_{\theta}$$

Maybe it's obvious that this should be zero and I can't see it or I did a calculating error but if I did not I have to assume something in the problem description is wrong.

I would be awesome if anyone could help me out a little bit and maybe push me in the right direction. Thanks!

I think that the problem with your derivation is that you are not working in the phase space basis of Hamiltonian mechanics. For instance, when we work in these Mechanics, our basis vectors are position and momentum (being a 2N dimensional space, with N the number of dimensions of our system).

Therefore, if we want to do calculations with poisson brackets, it doesn't make sense to express the angular momentum in the direction of z-axis as $$L_z=mr^2sin{\theta}^2\dot{\phi}$$, because the $$\dot{\phi}$$ is not a basis vector, so we can't perform the partial derivative on it.

The correct way of working with it is to treat $$L_z$$ as $$p_\phi$$, and 'forget' that inside that quantity there exists an $$r,\theta$$ dependance. Therefore, we would find that $$\begin{equation} \frac{\partial{p_\phi}}{\partial{r}}=0\\ \frac{\partial{p_\phi}}{\partial{\theta}}=0, \end{equation}$$ just in the same way we say $$\frac{\partial{x}}{\partial{z}}=0$$; because in this system, the momentum space is part of the base.

• This makes a lot of sense, thank you for this explanation! Jan 10 '21 at 18:32

The angular momentum in the direction of $$z$$-axis given by $$L_z=mr^2\sin^2\theta\ \dot {\phi}$$ which we can express in term of canonical variable using $$\dot{\phi}=\frac{\partial H}{\partial p_\phi}=mr^2\sin^2\theta \ \dot{\phi}$$ $$L_z=p_\phi$$ from here it follows that $$\{L_z,H\}=\{p_\phi,H\}=0$$ all the partial derivative are zero!

• Ah I see, thank you! That's a way to do it. So my approach would maybe work too but we don't know what $p_r$ and $p_{\theta}$ are? (maybe they are just in the form that my calculated expression above will be zero?) Jan 10 '21 at 18:14
• Actually, I don't know How do you get there? So I did this way :) Jan 10 '21 at 18:15
• I just calculated the poisson bracket of $\{L_z,H\}$ but with $L_z=mr^2sin^2(\theta)\dot{\phi}$. Doing the partial derivatives brings you to my expression. Jan 10 '21 at 18:18