I First excited state of Helium. Triplets and Singlet

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1. May 7, 2017

JamesHG

My question is the following one:

If i have the first excited state of helium, the possibilities for the two electrons are : 1s+,2s+ , 1s+2s-, 1s-,2s- and 1s-,2s+ , where + and - denotes spin up and down. If I use the slater's determinant to generate antisymmetric states, I get :

$$|u1> = \frac{1}{\sqrt{2}}[|1s,2s> - |2s,1s>]\otimes|1,1>$$
$$|u2> = \frac{1}{\sqrt{2}}[|1s+,2s-> - |2s-,1s+>]$$
$$|u3> = \frac{1}{\sqrt{2}}[|1s,2s> - |2s,1s>]\otimes|1,-1>$$
$$|u4> = \frac{1}{\sqrt{2}}[|1s-,2s+> - |2s+,1s->]$$

how i should i know which state (|u2> or |u4>) belows to singlet or triplet state if they can't be separed as tensorial product?

And why the energy from singlet and triplets are different?

Note : The |1,1> ... etc kets are referred to the coupled basis.

Thanks.

2. May 7, 2017

blue_leaf77

The eigenstates of He have a (approximately) definite spin, therefore its eigenstates must be a product between the spatial function and the spin function. Your first and third states already satisfy this requirement, but the rest do not. If you write those improper states correctly you should get one with spin function $|1,0\rangle$ which is a triplet and the other one with spin function $|0,0\rangle$ which is the only singlet. Hint: try forming linear combinations of $u_2$ and $u_4$.
The states you write there result from a Hamiltonian ignoring electron-electron repulsion. You can see how the incorporation of this Coulomb term affect the energies by treating it as a perturbation. You should see that the excited singlet states lie higher than the corresponding triplet state.

3. May 7, 2017

JamesHG

Thank you so much for answering my question!