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At tree level, the Feynman diagram has an electron-positron current and a muon-antimuon current joined by one virtual photon. In the massless limit, the spin average and final-state spin sum give $\overline{|\mathcal{M}{|}^2}=e^4(1+{\cos }^2\theta )$, and the integrated relativistic scattering cross-section is $4\pi {\alpha }^2/(3s)$.

When the full Lorentz-invariant phase-space measure treats $n$ identical final particles as labeled, each physical final configuration is counted $n!$ times. The unlabelled relativistic scattering cross-section therefore includes $1/n!$. For two identical scalars, this divides the full-angle result by two; equivalently, integrate over a region containing only one representative of each exchanged pair.

Substitute the preceding spin average into the supplied differential scattering cross-section. Since $e^4=16{\pi }^2{\alpha }^2$,The angular integral is $2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair isThe outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form givesThe integrated answer depends only on the incoming invariant $s$; the angular variables $t,u$ have been integrated out.

In four spacetime dimensions with $\hslash =c=1$, 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 $[\varphi ]=1$, 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,while the numerical factors $X,Y$ are dimensionless. ThereforeIncluding the identical-particle factor changes only a dimensionless constant and preserves this check.