# 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 $[\vec{L}^2,H_{\rm Darwin}]=C[\vec{L}^2,\delta(\vec{r})]=0$?

$$H_\text{Darwin}=C\delta (\vec{r})=\frac{C}{4\pi r^2}\delta (r)$$
• @TheAverageHijano Can you explain how $\delta(\vec{r})=\frac{1}{4\pi r^2}\delta(r)$? It is not obvious to me. – mithusengupta123 Aug 22 '18 at 15:13