- #1
- 3
- 0
(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-
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-
Last edited: