贡献者： 待更新
Consider the process $e^-(p)\gamma(k)\rightarrow e^-(p')\gamma(k')$
\begin{equation}\begin{aligned}
i\mathcal{M} &= (-ie)^2 \varepsilon_\mu^*(k')\varepsilon_\nu(k) \left[\bar u(p')\gamma^\mu \frac{i(\not{p}+\not{k}+m)}{(p+k)^2-m^2} \gamma^\nu u(p)+
\bar u(p')\gamma^\nu \frac{i(\not{p}-\not{k}'+m)}{(p-k')^2-m^2} \gamma^\mu u(p)
\right]~.
\end{aligned}\end{equation}
this process is related to $e^+e^-\rightarrow \gamma\gamma$（
Bhabha 散射）by crossing symmetry, so we can simply do variable replacements:
\[
k\rightarrow -p',\quad p'\rightarrow -k~.
\]
Since we change a $e^+$ of 'in' state to $e^-$ of 'out' state, we should make the above substitutions, as well as add a factor $-1$.
\begin{equation}\begin{aligned}
\overline{\sum}|\mathcal{M}|^2
=2e^4\left[\frac{p\cdot k'}{p\cdot k}+\frac{p\cdot k}{p\cdot k'}+2m^2\left(\frac{1}{p\cdot k}-\frac{1}{p\cdot k'}\right)+m^4\left(\frac{1}{p\cdot k}-\frac{1}{p\cdot k'}\right)^2\right]~.
\end{aligned}\end{equation}
in the "lab" frame in which the electron is initially at rest:
\[
k=(\omega,\omega \hat z), \ p=(m, \boldsymbol{\mathbf{0}} ),\ p'=(E', \boldsymbol{\mathbf{p}} '),\ k'=(\omega',\omega'\sin\theta,0,\omega'\cos\theta)~.
\]
we will express the cross section in terms of $\omega,\theta$.
\begin{equation}\begin{aligned}
m^2&=(p')^2=(p+k-k')^2=p^2+2p\cdot(k-k')-2k\cdot k'=m^2+2m(\omega-\omega')-2\omega\omega'(1-\cos\theta)\\
&\Rightarrow
\frac{1}{\omega'}-\frac{1}{\omega}=\frac{1}{m}(1-\cos\theta)\\
&\Rightarrow \omega'=\frac{\omega}{1+\frac{\omega}{m}(1-\cos\theta)}~.
\end{aligned}\end{equation}
the phase space integral:
\begin{equation}\begin{aligned}
\int \,\mathrm{d}{\Pi} _2 &= \int \frac{ \,\mathrm{d}^{3}{ \boldsymbol{\mathbf{k}} '} }{(2\pi)^3} \frac{1}{2\omega'} \frac{ \,\mathrm{d}^{3}{ \boldsymbol{\mathbf{p}} '} }{(2\pi)^3}\frac{1}{(2 E')}(2\pi)^4 \delta^4(k'+p'-k-p)\\
&=\int \frac{(\omega')^2 \,\mathrm{d}{\omega} ' \,\mathrm{d}{\Omega} }{(2\pi)^3} \frac{1}{4\omega'E'}\times 2\pi \delta(\omega'+\sqrt{m^2+\omega^2+(\omega')^2-2\omega\omega'\cos\theta}-\omega-m)\\
&=\int \frac{ \,\mathrm{d}{\cos} \theta}{2\pi}\frac{\omega'}{4E'}\frac{1}{\left|1+\frac{\omega'-\omega\cos\theta}{E'}\right|}\\
&=\frac{1}{8\pi}\int \,\mathrm{d}{\cos} \theta\frac{(\omega')^2}{\omega m}~.
\end{aligned}\end{equation}
Using
\[
\,\mathrm{d}{\sigma} =\frac{1}{2E_A 2E_B|v_A-v_B|} \,\mathrm{d}{\Pi} _2 \times \overline{\sum}|\mathcal M|^2~.
\]
we finally obtain the Klein-Nishina formula:
\begin{equation}\begin{aligned}
\frac{ \,\mathrm{d}{\sigma} }{ \,\mathrm{d}{\cos} \theta} &= \frac{1}{2\omega 2m}\cdot \frac{1}{8\pi} \frac{(\omega')^2}{\omega m }
\left(\overline{\sum}|\mathcal M|^2\right)\\
&=\frac{\pi \alpha^2}{m^2}\left(\frac{\omega'}{\omega}\right)^2\left[\frac{\omega'}{\omega}+\frac{\omega}{\omega'}-\sin^2\theta\right],\quad \omega'=\frac{\omega}{1+\frac{\omega}{m}(1-\cos\theta)}~.
\end{aligned}\end{equation}