Label
$$
e^-(p_1,s_1)+e^+(p_2,s_2)
\longrightarrow e^-(p_3,s_3)+e^+(p_4,s_4).
$$
This is <Bhabha scattering>. There are two diagrams:

* the $s$-channel annihilation topology $e^-e^+\to\gamma^*(p_1+p_2)\to e^-e^+$;
* the $t$-channel exchange topology in which the electron and positron lines exchange $\gamma^*(p_1-p_3)$.

With $s=(p_1+p_2)^2$ and $t=(p_1-p_3)^2$, a consistent external-fermion ordering gives
$$
\begin{aligned}
i\mathcal M_s
&=[\overline v(p_2)(-ie\gamma^\mu)u(p_1)]
\frac{-i\eta_{\mu\nu}}s
[\overline u(p_3)(-ie\gamma^\nu)v(p_4)],\\
i\mathcal M_t
&=-[\overline u(p_3)(-ie\gamma^\mu)u(p_1)]
\frac{-i\eta_{\mu\nu}}t
[\overline v(p_2)(-ie\gamma^\nu)v(p_4)],\\
i\mathcal M&=i\mathcal M_s+i\mathcal M_t.
\end{aligned}
$$
The displayed relative minus sign is the <fermionic sign> from putting the external fermion operators into the common chosen order.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 301 / 4 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-301/4/i/solution