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I see that in the 1$s$ orbital the electrons have to have different spins, still there is two possible states,

  • a singlet state with an anti-symmetric superposition of spins and symmetric spatial part
  • a triplet state with symmetric superposition of spins and anti-symmetric spatial part

So, the question reduces to WHY the first one? I have read somewhere else that a ground state has to have a symmetric spatial wave function always but without further reasoning why this should be necessarily true.

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Think of the spatial part alone. The unique combined state where both are in the 1s state is symmetric.

For the total state to be antisymmetric we must have that the spin part is antisymmetric (because the full state space is the tensor product of the spatial and the spin part). This is the lowest possible energy state.

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  • $\begingroup$ Yes, I have understood this part of the deal. WHY should the spatial part be symmetric though??? $\endgroup$
    – Marsl
    Feb 15, 2017 at 16:38
  • $\begingroup$ There is no a priori reason why the spatial part should be symmetric. However, if for each electron individually there is a unique ground state, there is just one possibility for both to be in that ground state. That state just happens to be symmetric. $\endgroup$
    – doetoe
    Feb 15, 2017 at 17:19
  • $\begingroup$ If the individual ground state would be degenerate, then it should be possible to have a spatially antisymmetric ground state. Note that the interaction term in the Hamiltonian could a priori also change things (in fact it doesn't change what is the ground state, but it does influence its energy eigenvalue) $\endgroup$
    – doetoe
    Feb 15, 2017 at 17:22
  • $\begingroup$ Well if there is no a priori reason for the ground state to have a symmetric spatial wave function than how can we know its the symmetric one? Is this something we know from spectroscopy where we see that $\Delta s = 0$ selection rule should only yield transitions from singulets? $\endgroup$
    – Marsl
    Feb 15, 2017 at 18:05
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    $\begingroup$ Ok, I was rather looking for some evident reasoning other than computing it by pertubation theory. I was just asking myself wether I was missing something obvious here. $\endgroup$
    – Marsl
    Feb 15, 2017 at 22:20

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