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Helium has two electrons. The only configuration they can have (ground state) is spin $\frac{1}{2}$ for one and $-\frac{1}{2}$ for the other one, by the Pauli exclusion principle, which is definitely a singlet state. So I'm assuming it's something to do with excited states. If that's the case, there's an exercise in my textbook which asks for an explanation of the energy ordering of the spin triplet state of helium, but wouldn't that depend on which excited state it was in?

I have read the answers to the related question 'What causes the triplet state in helium?' and still don't understand what the 'energy ordering' would be.

user13948
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If we ignore spin-orbital coupling, helium atom's wavefunction can be represented by a product of spin part and spatial part. Triplet state is a state where spin part of the wavefunction is symmetric with respect to exchange of two electrons.

This means that the spatial part must be antisymmetric, since electrons are fermions. But this implies that there must be a node in the wavefunction — a hypersurface where the wavefunction vanishes. This can only happen in an excited state, since ground-state solution of non-symmetrized Schrödinger equation has no nodes and is always non-degenerate, so has strictly less energy. This gives you the answer about ordering of two lowest energy states with zero angular momentum.

Ruslan
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