The relation of gamma matrix between field operator change and chirality in Peskin and Schroeder (page 165) I'm reading Compton scattering in Peskin's book (page 165) and there is a sentence I can't understand.

The third sentence in the above paragraph says that three $\gamma$-matrix between field operator guarantee that if the initial electron is right-handed, the final electron is right-handed. Additionally, in the next page, there is a sentence says two $\gamma$-matrix will convert a right handed-electron into a left-handed electron. Therefor I think in the amplitude between field operator the even number of $\gamma$-matrix will change the chirality while odd number $\gamma$-matrix will keep it.
I can't understand why it is the case that the even number of $\gamma$-matrix in the amplitude between field operator will change the chirality while odd number $\gamma$-matrix will keep it.
In fact, somebody had asked this question Why three gamma matrix between initial and final electron lead to the same helicity? but nobody gave an answer, so I ask this again.
 A: Refer to the original equation (5.97), write $\gamma^i$ except $\gamma^o$ for $\overline{u}$, explictly
$$iM=...\overline{u}(p')\gamma^u\frac{\gamma^\alpha p_\alpha-\gamma^\beta k'_\beta+m}{(p-k')^2-m^2}\gamma^\nu u(p)$$
observe:
For right hand spinor, $\begin{pmatrix}0 \\ *\\ \end{pmatrix},$ we have $\gamma^\mu \begin{pmatrix}0 \\ *\\ \end{pmatrix} = \begin{pmatrix}0 & \sigma^\mu \\ \overline{\sigma}^\mu & 0\\ \end{pmatrix} \begin{pmatrix}0 \\ *\\ \end{pmatrix} =\begin{pmatrix}\sigma ^\mu *\\ 0\\ \end{pmatrix}$
For left hand spinor, $\begin{pmatrix} *\\ 0\\ \end{pmatrix},$ we have $ \gamma^\nu \begin{pmatrix}*\\ 0\\ \end{pmatrix} = \begin{pmatrix}0 & \sigma^\nu \\ \overline{\sigma}^\nu & 0\\ \end{pmatrix} \begin{pmatrix}*\\ 0\\ \end{pmatrix} =\begin{pmatrix}0\\ \overline{\sigma}^\nu *\\ \end{pmatrix}$
Thus every $\gamma^\alpha$ "flip" $\begin{pmatrix} *\\ .\\ \end{pmatrix}$ once and pick either $\sigma^\alpha$ or $\overline{\sigma}^\alpha$, and
in the original equation, two the dominate term had three $\gamma$, as $m$ is negligible in relativistic limit.
We begin with $u(p)=\begin{pmatrix}0 \\ u_R\\ \end{pmatrix}$, terms like
$$\overline{u}(p')\gamma^u \gamma^\alpha p_\alpha \gamma^\nu \begin{pmatrix}0 \\ u_R\\ \end{pmatrix}$$
$$=\overline{u}(p')\gamma^u \gamma^\alpha p_\alpha \begin{pmatrix} \sigma^\nu u_R\\ 0\\\end{pmatrix}$$
$$=\overline{u}(p')\gamma^u \begin{pmatrix} 0\\ \overline{\sigma}^\alpha p_\alpha \sigma^\nu u_R\\ \end{pmatrix}$$
Here, we write $\overline{u}(p')$ in terms of $\gamma^o$ but it does not pick up any $\sigma$ but just flip once more.
$$=u^+(p') \gamma^o \begin{pmatrix} \sigma^u \overline{\sigma}^\alpha p_\alpha \sigma^\nu u_R\\ 0\\ \end{pmatrix}$$
$$=u^+(p') \begin{pmatrix}0\\ \sigma^u \overline{\sigma}^\alpha p_\alpha \sigma^\nu u_R\\ \end{pmatrix}$$
$u^*(p')$ should be something like $\begin{pmatrix}0 \\ u_R^*\\ \end{pmatrix}$ for non-zero result.
$$=u_R^+(p') \sigma^u \overline{\sigma}^\alpha p_\alpha \sigma^\nu u_R(p) $$
$$=u_R^+(p') \sigma^u (\overline{\sigma}. p) \sigma^\nu u_R(p) $$
Thus both $u(p)$ and $u(p')$ are right hand and similarly you collect term by term, (5.99) is derived.
A: I am going to show an answer of my question based on my personal understanding.  The changing of chirality is in fact the result of below relation.
$\overline{u}(p^{\prime})\gamma^{\mu}(\frac{1+\gamma^{5}}{2})u(p)= u^{\dagger}(p^{\prime})(\frac{1+\gamma^{5}}{2})\gamma^{0}\gamma^{\mu}(\frac{1+\gamma^{5}}{2})u(p)=\overline{u}_{R}(p^{\prime})\gamma^{\mu}u_{R}(p)$
$\overline{u}(p^{\prime})\gamma^{v}\gamma^{\mu}(\frac{1+\gamma^{5}}{2})u(p)= u^{\dagger}(p^{\prime})(\frac{1-\gamma^{5}}{2})\gamma^{0}\gamma^{v}\gamma^{\mu}(\frac{1+\gamma^{5}}{2})u(p)=\overline{u}_{L}(p^{\prime})\gamma^{\mu}u_{R}(p)$
If the chirality of initial particle is determined, the chirality of the final particle is determined automatically according to the number of $\gamma$ matrix between field operator.
This is my current understanding, and I hope it could help people who is confused about it.
