Bhabha 散射

                     

贡献者: 待更新

   Consider the process $e^+(p)e^+(k)\rightarrow \gamma(p')\gamma(k')$

\begin{equation} \begin{aligned} i\mathcal{M} =(-ie)^2 \varepsilon^{*\lambda}_\mu(p') \varepsilon^{*\lambda'}_\nu(k') \left( \bar v^r(k)\gamma^\nu \frac{i({\not p}-{\not p}'+m)}{(p-p')^2-m^2} \gamma^\mu u^s(p) + \bar v^r(k)\gamma^\mu \frac{i({\not p}-{\not k}'+m)}{(p-k')^2-m^2} \gamma^\nu u^s(p) \right) \end{aligned} \end{equation}
Now we calculate the scattering amplitude. When we sum over $\lambda,\lambda'$, we use
\begin{equation} \begin{aligned} \sum_{\lambda=1,2}\varepsilon_\mu^{*\lambda} \varepsilon_{\mu'}^{\lambda}=-g_{\mu\mu'} \end{aligned} \end{equation}
\begin{equation} \begin{aligned} \overline{\sum}|\mathcal{M}|^2&=\frac{e^4}{4}\left[ \frac{1}{[(p-p')^2-m^2]^2}\text{Tr}[({\not k}-m)\gamma^\nu ({\not p}-{\not p}'+m)\gamma^\mu ({\not p}+m)\gamma_{\mu}({\not p}-{\not p}'+m)\gamma_{\nu}] \right. \\ &\quad + \frac{1}{[(p-k')^2-m^2]^2}\text{Tr}[({\not k}-m)\gamma^\nu ({\not p}-{\not k}'+m)\gamma^\mu ({\not p}+m)\gamma_{\mu}({\not p}-{\not k}'+m)\gamma_{\nu}]\\ &\quad + \frac{1}{[(p-p')^2-m^2][(p-k')^2-m^2]} \text{Tr}[({\not k}-m)\gamma^\nu ({\not p}-{\not p}'+m)\gamma^\mu ({\not p}+m)\gamma_\nu ({\not p}-{\not k}'+m)\gamma_\mu]\\ &\quad \left.+ \frac{1}{[(p-p')^2-m^2][(p-k')^2-m^2]} \text{Tr}[({\not k}-m)\gamma^\mu({\not p}-{\not k}'+m)\gamma^\nu({\not p}+m)\gamma_\mu ({\not p}-{\not p}'+m)\gamma_\nu]\right] \end{aligned} \end{equation}
Take the third line as an example:
\begin{equation} \begin{aligned} &\text{Tr}[({\not k}-m)\gamma^\nu ({\not p}-{\not p}'+m)\gamma^\mu ({\not p}+m)\gamma_\nu ({\not p}-{\not k}'+m)\gamma_\mu] \\ &=\text{Tr}[({\not k}-m)\gamma^\nu (2p^\mu - {\not p}'\gamma^\mu)({\not p}+m)(2p_\nu-\gamma_\nu {\not k}')\gamma_\mu]\\ &=-32(p\cdot k)(p'\cdot k') -16m^2p'\cdot k'+[32(k\cdot p)(k'\cdot p)-16m^2k\cdot k']\\ &\quad +[32(k\cdot p)(p'\cdot p)-16m^2k\cdot p'] +32m^2k'\cdot p+32m^2p\cdot p'+16m^2k\cdot p-16m^4\\ &=8m^2(p+k)^2-32m^4 \end{aligned} \end{equation}
Finally we have
\begin{equation} \begin{aligned} \overline {\sum}|\mathcal{M}|^2&= 8e^4 \left[ \frac{1}{4(p\cdot p')^2}((k\cdot p')(p\cdot p') +m^2(k\cdot p'+2p\cdot p'-k\cdot p)-2m^4) \right.\\ &\quad + \frac{1}{4(p\cdot k')^2} ((k\cdot k')(p\cdot k') +m^2(k\cdot k'+2p\cdot k'-k\cdot p)-2m^4)\\ &\left.\quad + \frac{1}{4(p\cdot p')(p\cdot k')} \left( m^2(p\cdot p'+p\cdot k')-2m^4 \right) \right]\\ &= 2e^4 \left[ \frac{k\cdot p'}{p\cdot p'}+\frac{m^2}{p\cdot p'}-\frac{m^4}{(p\cdot p')^2}\right.\\ &\quad +\frac{p\cdot p'}{p\cdot k'}+\frac{m^2}{p\cdot k'}-\frac{m^4}{(p\cdot k')^2}\\ &\left.\quad + m^2\left(\frac{1}{p\cdot k'}+\frac{1}{p\cdot p'}\right)-\frac{2m^4}{(p\cdot p')(p\cdot k')} \right]\\ &=2e^4\left[\frac{p\cdot k'}{p\cdot p'}+\frac{p\cdot p'}{p\cdot k'}+2m^2\left(\frac{1}{p\cdot p'}+\frac{1}{p\cdot k'}\right)-m^4\left(\frac{1}{p\cdot p'}+\frac{1}{p\cdot k'}\right)^2\right] \end{aligned} \end{equation}


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