Consider electron-positron interaction :
$$e^-e^+\rightarrow\mu^+\mu^-$$
when peskin book come to compute Non relativistic limit of this process said that, because we are in Non relativistic limit we have conservation of total spin(In picture below,And page 147 of the book). My question is that, what is special thing about Non relativistic case that we have conservation of spin?I think it's also true when we are in relativistic limit. But if you see picture in the end of the page 145, it should appear that we don't have spin conservation in relativistic limit. So when we have conservation of spin at all?
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$\begingroup$ I think the answer relates to masses and helicity though I can't recall exact relation and how it connects here. $\endgroup$– self.grassmanianCommented Jun 10, 2020 at 16:56
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Peskin just explained it at the bottom of page 147.
From the perspective of scattering theory, since we're considering the non-relativistic limit or low energy approximation, the first term $l=0$ (s-wave) is often enough to solve the problem. (you can find more in the section 6.4 of Modern Quantum Mechanics-Sakurai) That means $\vec{L}=0$ like Peskin said. The total angular momentum $\vec{J}=\vec{L}+\vec{S}$ is conserved so that the spin $\vec{S}$ is conserved.
But when we care about the relativistic limit like what we did in section 5.2, the contribution of higher partial wave would become important, for the spin-orbit coupling we don't have spin conservation in general.
That's what I understood right now. I hope it would help you
