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}
致读者： 小时百科一直以来坚持所有内容免费无广告，这导致我们处于严重的亏损状态。 长此以往很可能会最终导致我们不得不选择大量广告以及内容付费等。 因此，我们请求广大读者
热心打赏 ，使网站得以健康发展。 如果看到这条信息的每位读者能慷慨打赏 20 元，我们一周就能脱离亏损， 并在接下来的一年里向所有读者继续免费提供优质内容。 但遗憾的是只有不到 1% 的读者愿意捐款， 他们的付出帮助了 99% 的读者免费获取知识， 我们在此表示感谢。