For metric signature $(+---)$, the momentum-space rules for the <Yukawa interaction> are:

* an internal scalar line contributes $i/(k^2-m_\phi^2+i\epsilon)$;
* an internal fermion contributes $i(\not k+m)/(k^2-m^2+i\epsilon)$;
* each $\phi\bar\psi\psi$ or $\phi\bar\chi\chi$ vertex contributes $-ig$;
* incoming and outgoing fermions contribute $u_r(p)$ and $\bar u_r(p)$, while incoming and outgoing antifermions contribute $\bar v_s(q)$ and $v_s(q)$;
* every closed fermion loop contributes an additional minus sign.

At leading order the process has one $s$-channel scalar propagator. With $s=(p+q)^2$,
$$
i\mathcal M
=\bigl[\bar v_s(q)(-ig)u_r(p)\bigr]
\frac{i}{s-m_\phi^2+i\epsilon}
\bigl[\bar u_{r'}(p')(-ig)v_{s'}(q')\bigr].
$$
Thus, up to an irrelevant overall sign,
$$
\mathcal M=-\frac{g^2}{s-m_\phi^2+i\epsilon}
[\bar v_s(q)u_r(p)][\bar u_{r'}(p')v_{s'}(q')].
$$
The <fermion spin sums> and $\operatorname{tr}(\not a\not b)=4a\mathbin\cdot b$ give
$$
\sum_{r,s}|\bar v_s(q)u_r(p)|^2=4(p\mathbin\cdot q-m_\psi^2),
$$
and the analogous final sum is $4(p'\mathbin\cdot q'-m_\chi^2)$. Therefore
$$
X=\frac{4g^4(p\mathbin\cdot q-m_\psi^2)
(p'\mathbin\cdot q'-m_\chi^2)}{(s-m_\phi^2)^2}.
$$
Using $p\mathbin\cdot q=(s-2m_\psi^2)/2$ and its final-state analogue yields
$$
\boxed{X=\frac{g^4(s-4m_\psi^2)(s-4m_\chi^2)}{(s-m_\phi^2)^2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 301 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-301/3/solution