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Why is $\left|\uparrow\uparrow\right\rangle +\left|\downarrow\downarrow\right\rangle$ not discussed, despite having a total spin s = 0?

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    $\begingroup$ You've just calculated the z-component. What about <S^x>? $\endgroup$ Jul 10, 2016 at 1:12

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Because it doesn't have total spin $s=0$ - it has total spin $s=1$, with the spin component parallel to the $z$-axis being zero. If you looked at that state in a different basis (e.g. the $x-$ or $y-$ basis) it would very clearly not have spin 0.

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  • $\begingroup$ In that case, why is ∣↑↑⟩+∣↓↓⟩ not considered as part of the s = 1 triple state? $\endgroup$ Jul 10, 2016 at 1:27
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    $\begingroup$ @allidoiswin It is! The triplet space is not just a set of three vectors, it's an entire vector space - the eigenspace of the ${\bf S} \cdot {\bf S}$ operator with eigenvalue $s(s+1)$ for $s = 1$. So that state is a triplet state. $\endgroup$
    – tparker
    Jul 10, 2016 at 1:31
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Sometimes a picture can help. Its not hugely rigorous (especially for a physics site) but using the vector model, pictures can be constructed to illustrate the states that two spins $\alpha$ and $\beta$ can form.spins

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