I have a question about the usage of terminology for "parallel" and "anti-parallel" spins for a two electron system as described here:
That is we have a system consisting of two unpaired electrons. That is both have $s_1=s_2=\dfrac 12$ spins with $z$-quantization quantum numbers $m_{s_i}= \pm \dfrac12$ for $i =1,2$.
The text says:
we obtain a triplet state in a situation of parallel spin of the electrons ($S = 1$) and a singlet state where with electrons of anti parallel spin ($S = 0$). Decisive with respect to ...
I'm not able to understand it.
Why does a triplet state coincide with situation of parallel spins and singlet with anti-parallel spins?
The triplet state $s=1$ is $3$ times degenerated and thus imposes $3$ symmetric eigen-states with respect to $z$-axis quantum number $m_S=-1,0,+1$. The eigen-functions are:
$$|↑↑\rangle \text{ for } m_s=1$$
$$\frac{1}{\sqrt{2}} (|↓↑\rangle +|↑↓\rangle )\text{ for }m_s=0$$
$$|↓↓\rangle\text{ for }m_s=-1$$
Why does $s=1$ correspond to parallel spins?
For example $\dfrac{1}{\sqrt{2}} (|↓↑\rangle +|↑↓\rangle )$ has only anti-parallel components.
What I don't understand is how we identify $s=1$ with parallel spins or $s=0$ with anti-parallel spins.