# Does the total orbital angular momentum $L^{2}$ commute with the atomic Hamiltonian?

I would like to know if the total orbital angular momentum $$\mathbf{L}^{2}$$ commutes with the Hamiltonian of a many-electron atom when we consider the interactions between the electrons.

$$H = -\frac{1}{2} \sum_{i=1}^{N} \nabla_{i}^{2} - \sum_{i=1}^{N} \frac{Z}{r_{i}} + \sum_{i > j}^{N}\frac{1}{|\mathbf{r}_{i}- \mathbf{r}_{j}|} \\ \mathbf{L} = \sum_{i=1}^{N}\mathbf{l}_{i} \\ [\mathbf{L}^{2}, H] = 0\ \ (?)$$

This question arose when I was reading about the terms of multiplet of an atom. If I understood correctly, when we describe the state of an atom using the quantum numbers $$L$$, $$M_{L}$$, $$S$$ and $$M_{S}$$, we are considering the interaction of the electrons, but I thought it only would be possible if we adopt a central field approximation.