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 $S=1-1=0$?
Best Answer
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.