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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

http://ist-socrates.berkeley.edu/~phylabs/adv/ReprintsPDF/ATM%20Reprints/02%20-%20Atomic%20Spectra%20and%20Atomic%20Structure.pdf

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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.

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    $\begingroup$ Seems correct. See here. $\endgroup$
    – Martin Ueding
    Jul 12, 2016 at 19:12
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    $\begingroup$ 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. $\endgroup$
    – Emilio Pisanty
    Jul 12, 2016 at 23:34
  • $\begingroup$ Oh, cool! Thanks a lot for NIST link and for ideas! $\endgroup$
    – MsTais
    Jul 14, 2016 at 14:49

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