(Chiral Representation ##\gamma_5## is diagonal)(adsbygoogle = window.adsbygoogle || []).push({});

According toAn introduction to QFT -Peskin & Schroeder 3.3 : ##h=\hat{p}\cdot S##

and ##h=+1/2## is right-handed while ##h=-1/2## is left-handed.

It is quite easy for fermions. But I'm confused when it comes to anti-fermions.

In Pestkin's book 5.2 (##e^+e^-\rightarrow\mu^+\mu^-##) high energy limit, for a right-handed electron: ##\xi=(10)##, so ##u=\sqrt{2E}(0 0 1 0)##; while for a left-handed positron: ##\xi=(01)##, so ##v=\sqrt{2E}(0 0 0 -1)##. And both particles have##(\hat{p}\cdot\sigma)\xi=+\xi##. (The momentum of the electron is z+, positron is z-; both spin up in z+ direction).

I don' understand that since the helicity of the positron is left, why ##(\hat{p}\cdot\sigma)\xi=+\xi##.

And one more question: how can the bases be chosen for an anti-fermion when:

1. spin: z-; momentum: z-

2. spin: z-; momentum: z+

3. spin: z+; momentum: z+

4. spin: z+; momentum: z-

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# Helicity of the anti-fermions

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