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For an unpolarized initial fermion, average over its two spin states and sum over the unobserved final spin:The initial scalar has only one spin state. One can evaluate this sum directly from normalized Dirac spinors, or use consistent fermion spin sums to express it as a trace. It is the squared sum of both tree scattering amplitudes, not the sum of their separate squares.

The relativistic scattering cross-section is obtained by integrating the Lorentz-invariant phase-space measure and dividing by the invariant flux factor:There is no identical-final-particle factor, because the outgoing scalar and fermion are distinct. All energies are positive, with $E_q^2=|\mathbf{q}{|}^2+M^2$ and $E_l^2=|\mathbf{l}{|}^2+m^2$.

Equivalently, in the centre-of-momentum frame set $s=-(p+k{)}^2$. Integrating the energy delta function in the relativistic two-body phase space givesFor this elastic process $k_f=k_i$, so integrate $\overline{|T{|}^2}/(64{\pi }^{2}s)$ over the full solid angle. This supplies the requested prescription without evaluating the angular integral.

Use the momentum-space Feynman rules with the scalar Feynman propagator $i/(p^2-m^2+i0)$. The four-leg vertex of a factorial-normalized scalar interaction has weight $i\lambda$ for the positive interaction sign printed here. There are $4!$ Wick contractions assigning four external legs to its four fields, cancelling the factorial in its coefficient. A negative interaction sign would give $-i\lambda$; its squared tree amplitude is the same.

At each vertex include $(2\pi {)}^4{\delta }^4(\sum p)$ with all incident momenta taken incoming. Assign an internal momentum to each line and integrate each independent loop with $\int d^4\ell /(2\pi {)}^4$. Divide a graph by its Feynman-diagram symmetry factor, sum the graphs at the chosen order, and omit disconnected vacuum graphs from normalized amplitudes. For an S-matrix element, amputate external propagators and put the external momenta on shell as in the LSZ reduction formula; the external one-particle residues are one at tree level.

Define the invariant amplitude by the relativistically normalized matrix elementwith $⟨p|p^{\prime }⟩=2E_{\mathbf{p}}(2\pi {)}^3{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The lowest-order connected four-point graph is one contact vertex:There is no exchange graph at this order because there is no three-field interaction.

For the elastic scattering from a quartic scalar contact interaction, write $\sqrt{s}$ for the total centre-of-mass energy and $E_p=\sqrt{s}/2$ for the energy of each incoming particle. The incoming and outgoing spatial momentum magnitudes both equal $k=\sqrt{E_p^2-m^2}$, with $E_p>m$. The invariant incident flux isThe Lorentz-invariant phase-space measure for two outgoing particles isIn the centre-of-mass frame, the spatial delta function sets ${\mathbf{p}}_4=-{\mathbf{p}}_3$, while the energy delta function has radial derivative $2k/E_p$. Therefore the relativistic two-body phase space satisfiesThe two outgoing real-scalar particles are identical. Integrating over the full solid angle counts each unordered pair twice, so include the identical final-state symmetry factor $1/2!$. This givesThis is isotropic. When $E$ denotes each particle's energy, $s=4E^2$ and the full-sphere event density is ${\lambda }^2/(512{\pi }^2{E}^2)$. If $E$ denotes the total energy of the pair, $s=E^2$ and it is ${\lambda }^2/(128{\pi }^2{E}^2)$. Stating the answer in $s$ removes that energy-label ambiguity.

An equally valid angular convention selects one outgoing particle in a hemisphere, so each event is represented once. In that convention omit $1/2!$ and use $d\sigma /d\Omega ={\lambda }^2/(64{\pi }^{2}s)$ on the hemisphere, or ${\lambda }^2/(256{\pi }^2{E}_p^2)$. Both conventions give ${\sigma }_{\mathrm{event}}={\lambda }^2/(32\pi s)$ at this order. The identical-state factor concerns counting final states and is separate from the vertex factorial.