Spectroscopy and Russell-Saunders coupling

Amok

So, this has been bugging me for years (seriously). So my spectroscopy course says the following:

1) In an atom, for a closed subshell, for each electron possesing an $$m_l$$ quantum number there is another electron possesing $$-m_l$$, which means that the sum $$M_L = \sum {m_l} = 0$$. I have no problem with this, it is pretty obvious. However, my course concludes from this that the magnitude of the total angular momentum (magnitude of L) has to be 0! How can they know that, I would only conclude that the projection of the vector L on the z-axis is 0 and nothing else!

2)The same rule applies to spin angular momenta.

I guess my more general doubt is how to determine the possible values for the magnitude of L (the sum of angular momenta of electrons) by knowing only the magnitudes of the angular momentum of each individual electron as well as the projection of angular momenta of electron on the z-axis.

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DrClaude

Mentor
If, for example, $L=1$, then there must exist states with $M_L= \pm 1$ in addition to $M_L=0$.

If only $M_L=0$ can exist, then the only total angular momentum possible is $L=0$.

"Spectroscopy and Russell-Saunders coupling"

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