Why are $L=S=0$ in full shells? Spectroscopic notation

Question

There are some threads on this question but I'm still not understanding this considering the following scenario: If we take for example Helium in its ground state so $1s^2$. Here obviously $L=0$, but coupling the two electron spins $s=\frac{1}{2}$ yields the following total spin possibilities $S=0,1$. What happens with $S=1$? In general it seems that in spectroscopic notation the $z$-component is used for total spin number. I'm a little bit confused about this, can someone help me please?

Answer 1

$S=1$ is not an allowed quantum state for the $1s^2$ configuration, because of the Pauli Exclusion Principle, which says no two identical fermions can occupy the same quantum state.

Generalising further: this is why all electrons come in pairs, spin up and spin down, in each orbital. So in "closed" shells, all spins are cancelled out and $S=0$ always.

Answer 2

Why do full shells have $L=S=0$?

Electrons are fermions, so they obey the Pauli exclusion principle: a shell is not full until electrons are in every possibly allowed state. If they have spin quantum number $s=1/2$, the available states are $m_s = \pm 1/2$, so the shell will be "full" when both these states are occupied. But then a full shell also entails a zero net (spin) angular momentum because $1/2 + (-1/2) = 0$. Granted this is technically only true for the $z$ projection ($m$) but if you think of this vectorially then it makes sense that summing all available states in a full shell gives you a zero net orbital/spin/total angular momentum.

What about the $S=1$ combination of two spin $s=1/2$?

Yes, two spin $1/2$ can combine into a singlet $S=0$ or a into a triplet $S=1$ state.
The total wavefunction for the electronic structure of the atom, however, also includes a spatial term. If both electrons are in the $1s$ state (ground state), then the spatial term is the same both, and the overall antisymmetry of the (fermionic) wavefunction causes it to vanish unless they are in the $S=0$ state. It's not as easy as to say that the two $m_s$ must vanish, since even for the $S=1$ of the triplet substates has a zero $z$ projection - if you do the maths, you can see that there is a key minus sign in the $S=0$ state ($=|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle $), which guarantees its antisymmetry.

You can indeed have electrons combining into the $S=1$ triplet, they just need to belong to different $\ell$ shells ($\ell$ controls the parity of the spatial part of the wavefunction). Like you can have a $1s$ electron and another $2p$ one. Indeed, Helium separated into Orthohelium and Parahelium corresponding to triplet and singlet states (respectively), though this division only affects excited states and the ground state is the same.

So, in conclusion, the Pauli exclusion principle is causing the overall (fermionic) wavefunction to be antisymmetric. Electrons in the same $\ell$ subshell have the same spatial wavefunction, which is symmetric, so they need to be in a singlet (antisymmetric) spin state, which is the $S=0$. This means that the individual electrons have to "add" together so as to cancel the net spin, which goes back to the vectorial picture I mentioned in the first paragraph.