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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.

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The terminology "parallel" and "antiparallel" is a bit loose, whereas the mathematical expressions for the states are exact. Rather than "parallel" and "antiparallel" it might be better to say "with total spin 1" and "with total spin zero", but that is a bit of a mouthful.

You are correct that the terminology does not apply straightforwardly to the state $\frac{1}{\sqrt{2}} (|↓↑\rangle +|↑↓\rangle)$. However, it is significant that that state does have total spin 1, therefore the two spins are in some sense adding up in a parallel as opposed to antiparallel fashion. If you express this state in terms of eigenstates of the x component of spin then you get $$ \frac{1}{\sqrt{2}} (|↓↑\rangle +|↑↓\rangle ) = \frac{1}{2\sqrt{2}} (R-L)(R+L) + \frac{1}{2\sqrt{2}}(R+L)(R-L) \\ = \frac{1}{2\sqrt{2}}\left( RR + RL - LR - LL + RR -RL + LR - LL \right) \\ = \frac{1}{\sqrt{2}} (RR + LL) $$ where I used the notation $$ R = \frac{1}{\sqrt{2}}( |↑\rangle +|↓\rangle ), \;\;\;\; L = \frac{1}{\sqrt{2}}( |↑\rangle -|↓\rangle ) $$ which are "right" and "left" in the sense of eigenstates of x component of spin. So in this basis the state does look a bit more like the spins are parallel to one another.

You could also explore what the other states look like in this basis. It will be instructive.

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You cannot identify parallel spins with the triplet and antiparallel spins with the singlet.

The addition of two spins is a special case of the the addition of angular momenta.

In this special case you end up with four states.

Three of the states -- ($|\uparrow\uparrow\rangle$, $|\downarrow\downarrow\rangle$, and $(1/\sqrt{2})(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle$ -- have the same quantum number $s=1$.

The singlet is $(1/\sqrt{2})(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$ and has $s=0$.

This set of lecture notes explains more.

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  • $\begingroup$ "You cannot identify parallel spins with the triplet and antiparallel spins with the singlet." This is exactly the intuition I have and which gave rise for this question after facing the notations in the linked text above. It seems that this confusing identifications "parallel spins with the triplet and antiparallel spins with the singlet" are "motivated" by the comment (iv) on page from your linked notes: it says: $\endgroup$ – katalaveino Feb 3 at 14:50
  • $\begingroup$ The singlet corresponds to $S = 0$; which means that the two spins $S_1$ and $S_2$ are antiparallel. The triplet corresponds to an angle $\alpha = 70 deg$ between the two vectors $S_1$ and $S_2$. This is the closest the two spins come to being parallel." That is I think that it might be better to say "with total spin 1" and "with total spin zero" instead of "parallel" and "antiparallel" as Andrew stated. while "parallel" and "antiparallel" in this context seems be done by somebody who prefered to sleep during quantum mechanic lecture :) $\endgroup$ – katalaveino Feb 3 at 14:54

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