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Is there a difference in the electronic configuration for singlet and triplet states? For example, He atom has 1s2 configuration in its ground state (singlet state)

But what about when the He atom is excited. It can come in both states (Signlet and Triplet)

Can I tell from the excited electronic configuration whether the atom is in a singlet or a triplet state?

Thanks!

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I am not sure that I get your question right, but let me try to answer according to my understanding.

The spin part of the two electron wave function of the singlet state $|0,0\rangle=|l=0,m=0\rangle$ is

$|0,0\rangle=(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)/\sqrt{2}$

The three triplet states look like this:

$|1,0\rangle=(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle)/\sqrt{2}$

$|1,1\rangle=|\uparrow\uparrow\rangle$

$|1,-1\rangle=|\downarrow\downarrow\rangle$

If you are unfamiliar with the notation, $\uparrow$ denotes spin $+1/2$ and $\downarrow$ denotes spin $-1/2$; the numbers on the left denote: the total angular momentum $l$ of both electrons and its $z$ component $m$.

If the electrons are in the same energy state, they have to have different spins, that is, the state can be either $|0,0\rangle$ or $|1,0\rangle$, so in principle singlet and triplet are possible. If one of the electrons is excited, any of the four states is possible, since the spins don't have to be different any more. Therefore, if you know that both electrons have spin up or both have spin down, you can be sure, that it is a triplet state. If one spin is up and the other down, you cannot tell from the spin configuration whether it is a singlet or a triplet, since again both $|0,0\rangle$ or $|1,0\rangle$ are possible.

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    $\begingroup$ I'd suggest using $\uparrow$ and $\downarrow$ instead of $+$ and $-$ to denote spins, otherwise they look too mixed up when adding and subtracting kets. $\endgroup$
    – Ruslan
    Commented Apr 12, 2014 at 12:45
  • $\begingroup$ I was too lazy to look up the TeX commands for them and didn't know them by heart, to be honest. $\endgroup$
    – Photon
    Commented Apr 12, 2014 at 12:46
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    $\begingroup$ \uparrow and \downarrow $\endgroup$
    – Ruslan
    Commented Apr 12, 2014 at 12:46

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