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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.

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  • $\begingroup$ 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. $\endgroup$ – ZeroTheHero Oct 20 '20 at 3:45
  • $\begingroup$ Unfortunately, I could not follow your explanation. $\endgroup$ – mithusengupta123 Oct 20 '20 at 4:00
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    $\begingroup$ 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. $\endgroup$ – ZeroTheHero Oct 20 '20 at 6:39
  • $\begingroup$ @ZeroTheHero Thank you :-) $\endgroup$ – mithusengupta123 Oct 20 '20 at 15:03
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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.

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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.

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