# Symmetries of atomic orbitals, s-state forms a triplet!

I have a trouble understanding how s-subshell electrons can form a triplet state ever.

In general isn't it true, that there are only two cases for s-state:

1. $\ell=0$, $s=1/2$, $J=1/2$ - doublet (one electron, hydrogen)
2. $\ell=0$, $s=0$, $J=0$ - singlet (two electrons, helium)

So, s-state is always a singlet state when it is not a hydrogen. But how on Earth it can form a triplet? Pauli exclusion principle wouldn't allow two spin-up electrons on s-subshell, no?

Essentially, I reference p.90 of this

I actually think I have figured it out: if we consider the situation when the principle quantum number for, let's say, excited helium is different, then we can have spin-up electron in $1S_1$ state and spin-up electron in $2S_1$ state. This is going to be a triplet. Please, correct me if I am wrong.
• Yeah, that's one way. A good way to play around is the NIST atomic levels database; for helium it lists exactly those configurations. However, you can also make a triplet $S$ state by combining $p$ orbitals, like e.g. the $1s^22s^22p3p\ ^3S$ state in carbon, which is essentially the superposition $$\frac{ |2,1,1⟩ - |3,1,-1⟩}{\sqrt{2}}$$ in the $|n,l,m_l⟩$ orbital basis, tensored with any triplet spin state. As an exercise, try figuring out the $^3S$ states in oxygen and titanium - NIST provides enough info to piece them together. – Emilio Pisanty Jul 12 '16 at 23:34