Helicity of the anti-fermions

  • Thread starter HZhang
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(Chiral Representation ##\gamma_5## is diagonal)
According to An 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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Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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