= Solution
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 \b[$i\lambda$] 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 element
$$
\langle p_3p_4|S-1|p_1p_2\rangle=i(2\pi)^4\delta^4(p_1+p_2-p_3-p_4)\,\mathcal M,
$$
with $\langle p|p'\rangle=2E_{\mathbf p}(2\pi)^3\delta^3(\mathbf p-\mathbf p')$. The lowest-order connected four-point graph is one contact vertex:
$$
\boxed{\mathcal M=\lambda+O(\lambda^2),\qquad |\mathcal M|^2=\lambda^2+O(\lambda^3).}
$$
There is no exchange graph at this order because there is no three-field interaction.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-43-scalar-contact.png]
{title=Tree-level contact diagram for two incoming and two outgoing real scalar particles}
{height=360}
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 is
$$
F=4\sqrt{(p_1\cdot p_2)^2-m^4}=8E_pk=4k\sqrt s.
$$
The <Lorentz-invariant phase-space measure> for two outgoing particles is
$$
d\Phi_2=(2\pi)^4\delta^4(p_1+p_2-p_3-p_4)\prod_{j=3}^4\frac{d^3p_j}{(2\pi)^3\,2E_j}.
$$
In 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> satisfies
$$
\frac{d\Phi_2}{d\Omega}=\frac{k}{16\pi^2\sqrt s}.
$$
The 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 gives
$$
\boxed{\frac{d\sigma_{\rm event}}{d\Omega}=\frac1{2!}\frac{|\mathcal M|^2}{64\pi^2s}=\frac{\lambda^2}{128\pi^2s}+O(\lambda^3).}
$$
This is isotropic. When $E$ denotes each particle's energy, $s=4E^2$ and the full-sphere event density is \b[$\lambda^2/(512\pi^2E^2)$]. If $E$ denotes the total energy of the pair, $s=E^2$ and it is \b[$\lambda^2/(128\pi^2E^2)$]. 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 $d\sigma/d\Omega=\lambda^2/(64\pi^2s)$ on the hemisphere, or $\lambda^2/(256\pi^2E_p^2)$. Both conventions give $\sigma_{\rm event}=\lambda^2/(32\pi s)$ at this order. The identical-state factor concerns counting final states and is separate from the vertex factorial.
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