Suppose we consider the first excited state of the helium atom. We know that the first excited state of helium can exist as a triplet or singlet. The possible functions related to the spin of the two electrons in the triplet state are

$$\alpha(1)\alpha(2)$$ $$\beta(1)\beta(2)$$ $$\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) + \beta(1)\alpha(2) ] $$

while the one for the singlet state is

$$\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) - \beta(1)\alpha(2) ]$$

The triplet state predicts that the spins of the two electrons are parallel, but according to this equation

$$\dfrac{1}{\sqrt{2}} [\alpha(1)\beta(2) + \beta(1)\alpha(2) ] $$

there is a 50% probability that electron 1 is in the alpha state and a 50% probability that it is in the beta state: the same goes for electron 2. So, if this function predicts that the two spins are antiparallel, why is it part of one of the triplet states?

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    $\begingroup$ I think your confusion stems from the fact that "up" and "down" refer to the sign of the z component of the spin of individual electrons, whereas "triplet" and "singlet" describe the magnitude of the total spin of the combined system, not just the z components. Thus, one of the three triplet states has zero spin in the z-direction (one electron up and one down), but still has non-zero total spin, ie. the combined spin vector is within the x-y plane. The singlet state has zero overall spin, ie. the two spin vectors completely cancel each other, rather than just canceling in the z direction. $\endgroup$
    – Andrew
    Aug 18, 2021 at 20:19
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    $\begingroup$ @Andrew That fully answers the question I believe... I hope you invest the extra effort to add it as one? That would be appreciated by many, if you find the time. $\endgroup$
    – Stian
    Aug 19, 2021 at 14:01
  • $\begingroup$ @Andrew can you see this image?upload.wikimedia.org/wikipedia/commons/4/49/… If one of the triplet excited states has one alpha and one beta electron, shouldn't this be the configuration? 1s $\uparrow$ 2s $\downarrow$ $\endgroup$ Aug 19, 2021 at 19:58
  • $\begingroup$ This answer has a diagram that you should find informative. chemistry.stackexchange.com/questions/50277/… $\endgroup$
    – porphyrin
    Aug 20, 2021 at 6:56
  • $\begingroup$ @stian yttervik I didn’t write it as an answer because it seemed like a good answer should have images like in the answer porphyrin linked, and I didn’t have time to make them. I think that linked answer covers it though $\endgroup$
    – Andrew
    Aug 20, 2021 at 14:58


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