= Solution
Label the initial particle and antiparticle $p_1,p_2$ and the corresponding final momenta $p_3,p_4$, with $s=(p_1+p_2)^2$ and $t=(p_1-p_3)^2$. Charge flow permits an $s$-channel heavy-scalar annihilation graph and a $t$-channel heavy-scalar exchange graph. It excludes a third $u$-channel graph for these fixed charge assignments. The effective theory has one contact graph.
\Image[/past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2016-heavy-scalar-matching-diagrams.png]
{title=Particle-antiparticle scattering by heavy scalar annihilation and exchange, compared with its low-energy quartic contact interaction; external charge labels are held fixed}
The two cubic <Feynman vertices> and the heavy <propagator> give
$$
i\mathcal M_{\rm full}=(-ig)^2\left[\frac i{s-M^2+i0}+\frac i{t-M^2+i0}\right].
$$
For $|s|,|t|\ll M^2$,
$$
\mathcal M_{\rm full}=\frac{2g^2}{M^2}+\frac{g^2(s+t)}{M^4}+\cdots,
\qquad
\mathcal M_{\rm eff}=-4\lambda=\frac{2g^2}{M^2}.
$$
This independently checks the sign and the factor $1/2$ in <8/c/solution>. The next difference is the derivative correction of order $M^{-4}$ in the amplitude.
Let $E=\sqrt s$ denote the total center-of-mass <energy>, with $m\ll E\ll M$. The final particle and antiparticle are distinct, so no identical-particle factor is needed. The leading isotropic <differential scattering cross-section> and total <cross section> are
$$
\frac{d\sigma}{d\Omega}=\frac{|\mathcal M|^2}{64\pi^2s},
\qquad
\boxed{\sigma\simeq\frac{|\mathcal M|^2}{16\pi s}
=\frac{g^4}{4\pi M^4E^2}=\frac{\lambda^2}{\pi E^2}.}
$$
Corrections are relatively of order $E^2/M^2$ and, in a high-energy expansion, $m^2/E^2$, in addition to loop corrections when they matter. If instead $E$ denotes each beam particle's energy, substitute $s=4E^2$ before comparing numerical prefactors.
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