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I am not familiar with the notation used on wikipedia (http://en.wikipedia.org/wiki/Triplet_state). Is there a more physcial way to explain the cause of the triplet state (maybe without referencing the wave function)?

In my book it says the folllowing:

'For the overall spin of the electrons exist the possibilities S = 0 and S = 1, which is >why the term scheme of the Helium atom splits in a singulet and a triplet system.'

I also found that J (total angular momentum quantum number) = L+-S can be 0,1 and 2 but I can't see why.

On a related note: does the intercombination prohibition mean that one helium atom can only be singulet or triplet forever?

Edit1: Thanks for your quick answer. But aren't you implying in your solution that L=0 (due to L-S <= J <= L+S)?

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  • $\begingroup$ The $1s$ triplet to singlet transition is forbidden but it will occur - just rather slowly. Also the transition is only forbidden for emission of a photon. The triplet state can decay in collisions with other atoms, or in the liquid or solid by transferring energy to the lattice. $\endgroup$ – John Rennie Jun 24 '14 at 11:45
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In the 1s 1s state (both electrons in the ground state), one electron must be spin up and the other spin down by the Pauli exclusion princple. In other words the electron spins are anti-parallel and the state is a singlet.

However, if only one electron is in the 1s state, and the other electron is in a higher energy level, the two electrons can have parallel or antiparallel spins. If the spins are parallel, it is a triplet state.

See Helium Energy Levels for more information.

The lifetime of the 1s 2s triplet state is 7859 seconds. It relaxes to the ground state by a magnetic dipole (M1) transition.

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  • $\begingroup$ I think this is not fully correct: the singlet state is the antisymmetric spin state, $|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle$ (up to a constant), whereas the complement in spin space is three dimensional and consists of symmetric states, including those in which the spins are parallel, but also $|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle$ $\endgroup$ – doetoe Jan 11 '15 at 15:30

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