 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.

The two cubic Feynman vertices and the heavy propagator giveFor $|s|,|t|\ll M^2$,This independently checks the sign and the factor $1/2$ in 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 areCorrections 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.