My question is the following one:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# I First excited state of Helium. Triplets and Singlet

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