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For a quantum system of two spin-1/2 particles, the spins are said to be aligned or parallel and anti-aligned or anti-parallel, if total $S=1$ (maximum value) and $S=0$ (minimum value) respectively, irrespective of the $M_S$ value. Therefore, the triplet state $|1,0\rangle$ is also considered a state with aligned spins even though $M_S=0$. Do I have the right impression?

Why did I ask this question? Let me set the context for why I am asking this question. One says that nuclear force favors the parallel alignment of nucleon spins. I wanted to understand what this really means. Does nuclear force favor all three $S=1$ states equally or only $S=1, M_S=\pm 1$ states for a system of two nucleons? Note that a term $\vec{S}_1\cdot\vec{S}_2$ in the nuclear potential will favour or disfavour all terms in a multiplet equally.

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  • $\begingroup$ are you asking if they really are parallel (no) or if we just say that? $\endgroup$
    – JEB
    Aug 31, 2023 at 0:39
  • $\begingroup$ Write all states involved, in terms of $\uparrow$s and $\downarrow$s. Apply spin raising and lowering operators on them. What do you conclude? $\endgroup$ Sep 1, 2023 at 0:34
  • $\begingroup$ @CosmasZachos I think I needed to set the context without which it seems a question of semantics. We know that nuclear force favors parallel alignment of nucleon spins. I wanted to understand what this really means. Does nuclear force favor all three $S=1$ states equally or only $S=1, M_S=\pm 1$ states? $\endgroup$ Sep 1, 2023 at 12:21
  • $\begingroup$ I have cleared the mess to make the question more pointed. $\endgroup$ Sep 1, 2023 at 12:29
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    $\begingroup$ Yes, rotationally invariant interactions cannot tell the difference among different M's. Lowering M cannot make a difference in symmetry! Parallel means symmetric, as it cannot be antisymmetric, so yes, then, S=1 . $\endgroup$ Sep 1, 2023 at 13:38

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You seem to be interested in the average of $<\vec{S}_1 \cdot \vec{S}_2>$ over all angles, not just the z-component. Indeed, the value of this quantity only depends on $S$, not $M_S$. There are various derivations of this fact, but those derivations don't give much of an insight as to the "alignment" or "anti-alignment" of the two spin vectors. Maybe the below discussion will help.

The easiest case to picture is $S=0$. In this case, knowledge of one spin vector $\vec{S}_1$ guarantees that the other spin vector is exactly opposite; $\vec{S}_1 = - \vec{S}_2$. Evidently the dot product will be negative. It's absolutely fair to call this ``anti-aligned''.

Starting with the anti-aligned $S=0$ state, the transformation which takes this entangled state to the $S=1$,$M_S=0$ state is to simply rotate one of the two spin vectors $180^{o}$ around the $z$-axis. If you do this, that will transform $S=0$ to $S=1$ (with $M_S$ still zero, since you haven't changed any of the $z$ components of the spin).

After this rotation, the question of whether the two spins end up pointing in generally in the same direction or generally the opposite direction depends on which way the spins were pointing in the first place. Evidently, if they were pointing in the $+z$ and $-z$ direction, that wouldn't change anything; they'd still be pointing opposite to each other after this rotation. But if the spins started in the $x$-$y$ plane, this rotation would indeed cause them to be aligned, pointing exactly in the same direction. Any other initial direction would end up at some relative angle; they wouldn't be parallel.

Since there is an entire plane of possible spins which end up aligned, but only one particular direction which ends up anti-aligned, on average the dot product is positive. (But because of the averaging, the magnitude of the dot product turns out to be 3 times less than the $S=0$ case.)

The "directional spin story" for $S=1$, $M_S=\pm1$ is quite different, because those states are not entangled, they're separable. But if you wanted to, you could construct entangled superpositions of these two states which are analogous to the above $S=1$ case. Those states could be acheived by starting with $S=0$ and rotating $180^o$ around the $x$-axis, or instead $180^o$ around the $y$-axis. Both of these would evidently yield the same averaged dot product as the $S=1$,$M_S=0$ case discussed above.

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  • $\begingroup$ I have cleared the mess to make the question more pointed. $\endgroup$ Sep 1, 2023 at 12:29
  • $\begingroup$ @Solidification : I've rewritten my answer accordingly! :-) $\endgroup$ Sep 1, 2023 at 14:38

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