In handling two 1/2-spin particles, why is there only one singlet state? [closed]

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

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

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.
• @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. Jul 10, 2016 at 1:31
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.