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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 soon 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 '17 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 '17 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 '17 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 '17 at 18:05
  • $\begingroup$ The ground state is the state of lowest energy. An exact computation is not easy, but not strictly necessary either. The electrostatic Hamiltonian can be treated as a perturbation of the sum of two Hamiltonians for a Helium ion with a single electron, where the perturbation corresponds to the interaction between the electrons. The Helium ion states can be computed exactly, and as you know the lowest energy state is called 1s. The state space of two electrons is the tensor square of that of one, and a basis of eigenfunctions is a perturbation of the tensor product basis. $\endgroup$ – doetoe Feb 15 '17 at 18:21

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