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"One electron is presumed to be in the ground state, the 1s state. An electron in an upper state can have spin antiparallel to the ground state electron ($S=0$, singlet state, parahelium) or parallel to the ground state electron ($S=1$, triplet state, orthohelium)." From HyperPhysics

When they say "parallel to the ground state electron" then is it assuming that they are both spin up, or both spin down? If so, isn't it then ignoring the $S=1$ state with spin up and spin down:

$$|1\rangle |0\rangle = \frac{|+\rangle|-\rangle + |-\rangle|+\rangle}{\sqrt 2}$$

Therefore, if one electron is presumed to be in the ground state, 1s, state, if the spins can be opposite, a second electron can also occupy the ground state in $S=1$ orthohelium. Is this correct?

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With such a two-electron spin wave function both electrons cannot be in the 1s state due to the electron parity (the wave function should be antisymmetric relative to exchange of two particles). When both electrons are in the orbital 1s, the only possible spin function reads

$$\frac{1}{\sqrt{2}} \left( |\uparrow \downarrow >-| \downarrow \uparrow> \right)$$

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  • $\begingroup$ The overall wavefunction must be anti-symmetric but then why can't it be that the spatial is the anti-symmetric bit, not the spin part? $\endgroup$ – SomePhysicsStudent Jun 9 '15 at 16:59
  • $\begingroup$ For example? psi=1s(1)1s(2)-1s(2)1s(1)? Than it is zero $\endgroup$ – freude Jun 9 '15 at 20:15
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Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric.

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