# How is it that for a closed subshell configuration $L=S=0$?

For a closed subshell configuration of a many-electron atom, $$M_L=\sum_i m_{\ell_i}=0$$ and $$M_S=\sum_i m_{s_i}=0$$. But I do not understand why does it necessarily mean $$L=S=0$$. The values $$M_L=M_S=0$$ are compatible with nonzero values of L and S. Then how does $$M_L=M_S=0$$ enforce $$L=S=0$$? I looked at all my books (Bransden, Liboff etc), all of them did a poor job explaining this, in my opinion.

• The key is to notice that the $(2\ell+1)\times (2\ell+1)$ determinant of the angular momentum states has $L=0$. To see this note that Det is a 1-dim rep of the permutation group, and that only $L=0$ has dimension 1. – ZeroTheHero Oct 20 '20 at 3:45
• Unfortunately, I could not follow your explanation. – mithusengupta123 Oct 20 '20 at 4:00
• the $M_L=M_S=0$ does not force $L=0$ or $S=0$. It is a necessary but not sufficient for $L=0$, $S=0$ as you rightly guessed ... will drum up something more extensive later today. – ZeroTheHero Oct 20 '20 at 6:39
• @ZeroTheHero Thank you :-) – mithusengupta123 Oct 20 '20 at 15:03

As you say, for a closed subshell $$M_L=M_S=0$$

And this is true whatever direction you happen to have chosen for the $$z$$ axis.

If a vector has a $$z$$ component for any $$z$$ direction this can only be because it has length zero, as opposed to happening to be in a particular orientation where it lies entirely in the $$xy$$ plane, which is how the $$L>0, M=0$$ possibilities arise.

The reason $$S=0$$ is easy. Each electron can be thought as a bar magnetic with its field fully pointed along the $$z$$ axis. It can either be up or down. The occupancy of the shell goes as $$\propto 2\cdot(2\ell+1)$$ where $$\ell \in \mathbb{N}$$ so it's always going to be even. All the spins cancel out.

As for the $$L=0$$, maths is required for a detailed proof.

But let's consider an example, taking the $$p$$ shell.

If it's full, then there will be $$2$$ electrons with the angular momentum fully in the $$z$$ axis $$|\ell=1, m_\ell =1\rangle$$ and $$2$$ electrons with angular momentum fully in the $$-z$$ axis $$|\ell=1, m_\ell =-1\rangle$$. Adding these $$4$$ electrons vectorially, they cancel each other out and hence give zero total angular momentum.

So now let's look at the two (times 2 for spin) electrons with $$|\ell=1, m_\ell=0\rangle$$.
Their angular momentum is fully in the $$xy$$ plane, and the question here is: do they add together to make $$L_{\text{here}} \neq 0$$, or are they pointing the other way to cancel each other out?

Say we look at the $$L_x$$ basis, which will have a projection quantum number $$m_x$$. If they were pointing in the same direction hence reinforcing each other, you'd have either the two $$m_x$$ quantum numbers or the two $$m_y$$ to be same. But that would violate Pauli's exclusion principle. You need those to be equal and opposite as well, so they all cancel.