Singlet and Triplet states: Why is the $S=0$ state defined as it is? I'm working on some exercise regarding the spin coupling of two electrons. There we have the wavefunctions corresponding to the S values as
$$\begin{align}
S = 1: &\begin{array}{c}\uparrow\uparrow \\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm]
S = 0: &\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow)
\end{align}$$
I understand that the three wavefunctions for $S=1$ are symmetric and the one for $S=0$ is antisymmetric. My question is, why is the combination with the minus sign is the one for $S=0$?
My thinking is that in a $\uparrow \downarrow$ or $\downarrow\uparrow$ combination the $S_z$ components would already add up to $0$ so that the minus sign would not change anything in the fact that $S=0$. 
Or is it that $\downarrow\uparrow$ or $\uparrow\downarrow$ each represent a state with $S=1$ and $S_z = 0$ so that subtracting the one from the other gives $S=1-1=0$?
 A: It's important to remember that when we say $S=1, 0$ we're really thinking of the eigenstates and eigenvalues of $\mathbf{S}^2 = (\mathbf{S}_1 + \mathbf{S}_2)^2$. Try applying this operator, and you'll see that the symmetric states have eigenvalue $S=1$, and the antisymmetric will have eigenvalue $S=0$. If you're simply applying $\mathbf{S}_z$, which is what you are doing in your question statement, then definitely both the mixed symmetric and antisymmetric states have eigenvalue $j=0$, as you've pointed out.
A: Just a slightly different view using the interchange operator $\hat P|m_1m_2 \rangle $ = $|m_1m_2 \rangle $

My question is, why is the combination with the minus sign is S=0?

Another way to look at this is:
$$\begin{align}
S = 1: &\begin{array}{c}\uparrow\uparrow 
\\ \dfrac{1}{\sqrt 2}(\uparrow\downarrow+\downarrow \uparrow) \\ \downarrow\downarrow\end{array} \\[5mm]
S = 0: &\ \frac{1}{\sqrt{2}} (\uparrow\downarrow-\downarrow\uparrow)
\end{align}$$
The triplet states are not affected by the interchange of $ m_1\iff m_2$
whereas the $S=0$ changes sign under interchange.
Using  $\hat P|m_1m_2 \rangle $ = $|m_1m_2 \rangle $ as the interchange  operator.
This results in
$$\hat P \textrm{(triplet)} = \textrm{triplet} $$
 But $$\hat P\textrm{(singlet)} = -~\textrm{ singlet} $$
A: You are correct that one of the $S = 1$ states has zero angular momentum along the $z$ axis. However, the $S = 0$ state has zero angular momentum along any direction, and this follows directly from the presence of the minus sign.
To see why, consider the operator for spin along a general direction, $S_{\hat{n}}$. Since the spins are identical, it doesn't matter which spin is what, so
$$\langle \uparrow \downarrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \alpha(\hat{n}).$$
Moreover, the operator $P$ that interchanges the spins doesn't affect $S_{\hat{n}}$, because
$$P S_{\hat{n}} = P \left(S_{\hat{n}}^1 + S_{\hat{n}}^2 \right) = S_{\hat{n}}^2 + S_{\hat{n}}^1 = S_{\hat{n}}$$
which further implies that
$$\langle \uparrow \downarrow | S_{\hat{n}} | \downarrow \uparrow \rangle = \langle \downarrow \uparrow | S_{\hat{n}} | \uparrow \downarrow \rangle = \alpha(\hat{n}).$$
This is all the information we need to evaluate the spin along the $\hat{n}$ direction for the two states you give. For the $S = 1$ state, we find
$$\frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} + \frac{\alpha}{2} = 2\alpha$$
while for the $S = 0$ state we find
$$\frac{\alpha}{2} + \frac{\alpha}{2} - \frac{\alpha}{2} - \frac{\alpha}{2} = 0.$$
The cross terms pick up a minus sign, due to the minus sign in the superposition. This shows the singlet has zero spin along any direction.
