= Solution
In four spacetime dimensions with $\hbar=c=1$, a <relativistic scattering cross-section> has <mass dimension> $-2$, since it is an area. The action is dimensionless, so the <Lagrangian density> has dimension four. The scalar <kinetic term> then gives $[\phi]=1$, and the cubic term gives $[\lambda]=1$.
Each term of the <four-point tree amplitude in phi cubed theory> has dimension $[\lambda^2]-[s-m^2]=2-2=0$. Hence $[|\mathcal M|^2]=0$. In the final integral,
$$
[dt]=2,\qquad [s]=[s-4m^2]=2,
$$
while the numerical factors $X,Y$ are dimensionless. Therefore
$$
\boxed{\left[\int dt\,\frac{|\mathcal M|^2}{s(s-4m^2)}\right]=2+0-4=-2=[\sigma].}
$$
Including the identical-particle factor changes only a dimensionless constant and preserves this check.
Back to article page
