# Can non-central hamiltonians commute with $\vec{L}^2$?

Central potentials $V(r)$ trivially commute with the operator $\vec{L}^2$ in quantum mechanics because the latter is a function of the angular coordinates $(\theta,\phi)$ only. Non-central potentials, may or may not commute with $\vec{L}^2$. It is not obvious. For example, in atomic physics, the Darwin term $H_{\rm Darwin}=C\delta(\vec{r})$ is a non-central Hamiltonian because it depends on $\vec{r}$ rather than $r$. But it still commutes with $\vec{L}^2$. Can anyone help me understand how $[\vec{L}^2,H_{\rm Darwin}]=C[\vec{L}^2,\delta(\vec{r})]=0$?

Isn't the Darwin potential central? We could rewrite it as:

$$H_\text{Darwin}=C\delta (\vec{r})=\frac{C}{4\pi r^2}\delta (r)$$

Disclaimer: I tried to add a comment instead of directly answering, but I apparently need a higher reputation to do so.

• Yep. Commenting is a privilege for gaining reputation through providing sufficient answers and asking good questions. Basically being a good part of this community. Before then you shouldn't post comments as answers. With that being said, this does answer the part of the question asking about this specific commutator, so I could see this being counted as a good answer. Aug 22, 2018 at 10:42
• I undertstand that it encourages new users to take part in the site and avoid potential vandalism. I think that earning privilages is a good method of awarding users, but I particularly find this privilege quite frustating since I can't help other users. Aug 22, 2018 at 10:47
• You can help other users through useful answers :) Comments are not the only way to be helpful. I would argue answers are much more helpful than comments. Aug 22, 2018 at 10:50
• @TheAverageHijano Can you explain how $\delta(\vec{r})=\frac{1}{4\pi r^2}\delta(r)$? It is not obvious to me. Aug 22, 2018 at 15:13
• @mithusengupta123 give a look at physics.fme.vutbr.cz/~komrska/Eng/DodFA.pdf and fen.bilkent.edu.tr/~ercelebi/mp03.pdf. I'll also think about it during the weekend, let me know if you discover anything. Aug 23, 2018 at 22:20