= Solution
The real <scalar field> has <canonical momentum> $\pi=\dot\phi$. Its <Legendre transform> gives
$$
\boxed{H=\int d^3x\left[\frac12\pi^2+\frac12(\nabla\phi)^2+\frac12m^2\phi^2+\frac{\lambda}{4!}\phi^4\right].}
$$
The first three terms constitute $H_0$, the free massive <Hamiltonian operator>. The remaining term is $H_I=\lambda\int\phi^4d^3x/4!=-\int\mathcal L_I d^3x$. This sign is a consequence of having no time derivatives in the interaction: the <canonical momentum> is unchanged, while the interaction enters the <Hamiltonian density> with the opposite sign to the <Lagrangian density>.
In the <interaction picture>, operators evolve with $H_0$ and states evolve with $H_I(t)$ constructed from the free interaction-picture field. Their evolution operator satisfies
$$
i\partial_tU_I(t,t_0)=H_I(t)U_I(t,t_0),\qquad U_I(t_0,t_0)=I.
$$
Integrating once gives $U_I=I-i\int_{t_0}^tH_I(t_1)U_I(t_1,t_0)dt_1$. Repeated substitution orders the integrations as $t\geq t_1\geq\cdots\geq t_n\geq t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a <time-ordered product>, which proves the <Dyson series>
$$
U_I(t,t_0)=\sum_{n=0}^\infty\frac{(-i)^n}{n!}\int_{t_0}^t dt_1\cdots dt_n\,
\mathrm T\{H_I(t_1)\cdots H_I(t_n)\}.
$$
Taking the asymptotic limits with the usual adiabatic or regulated scattering prescription yields
$$
\boxed{S=\mathrm T\exp\left[-i\int_{-\infty}^{\infty}H_I(t)dt\right]
=\mathrm T\exp\left[i\int d^4x\,\mathcal L_I(\phi_I(x))\right].}
$$
The field in this expression is the free field evolving in the <interaction picture>, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.
Use relativistically normalized two-particle states $|p_1,p_2\rangle=a^\dagger(p_1)a^\dagger(p_2)|0\rangle$, with $\langle p|q\rangle=(2\pi)^3 2E_p\delta^3(\mathbf p-\mathbf q)$. The connected first-order scattering contribution is
$$
\langle p_3,p_4|(S-I)|p_1,p_2\rangle_{\rm conn}
=-\frac{i\lambda}{4!}\int d^4x\,
\langle p_3,p_4|:\!\phi_I(x)^4\!:|p_1,p_2\rangle.
$$
Here the connected <S-matrix> uses a normalized vacuum and physical external one-particle states. At this order, <normal ordering> the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.
Two fields annihilate the incoming particles and two create the outgoing ones. There are $\binom42$ choices of the annihilating fields, two assignments of the incoming <momenta> and two assignments of the outgoing <momenta>. Hence the total factor is $\binom42\,2!\,2!=4!$, the <external-leg factorial cancellation at a quartic scalar vertex>. Each attachment contributes its plane-wave factor, so
$$
\langle p_3,p_4|:\!\phi_I(x)^4\!:|p_1,p_2\rangle_{\rm conn}
=4!e^{i(p_3+p_4-p_1-p_2)\cdot x}.
$$
The spacetime integral is the <four-momentum conservation> delta function, giving
$$
\boxed{\langle p_3,p_4|(S-I)|p_1,p_2\rangle_{\rm conn}
=i(2\pi)^4\delta^4(p_3+p_4-p_1-p_2)T,\qquad T=-\lambda+O(\lambda^2).}
$$
In particular, the tree <scattering amplitude> is independent of angle.
For completeness, derive the <relativistic two-body phase space> in the centre-of-mass frame. Write $\sqrt s=2E_*$, $p_*^2=E_*^2-m^2$. The spatial delta function sets the two final <momenta> opposite, and the <energy> delta function is $\delta(\sqrt s-2\sqrt{p^2+m^2})$. Its radial Jacobian is $E_*/(2p_*)$. Thus
$$
d\Phi_2=\frac1{16\pi^2}\frac{p^2dp}{E_p^2}\delta(\sqrt s-2E_p)d\Omega
=\frac{p_*}{16\pi^2\sqrt s}\,d\Omega.
$$
The <invariant flux factor> is $\mathcal F=4p_*\sqrt s$ for equal initial masses. Using $d\sigma=|T|^2d\Phi_2/\mathcal F$, the labeled angular density is $\lambda^2/(64\pi^2s)$. The outgoing quanta belong to one real <scalar field>, so they are identical. A full-sphere integration counts each event twice and requires the <identical final-state symmetry factor> $1/2!$. Consequently
$$
\boxed{\sigma=\frac12\int_{S^2}\frac{\lambda^2}{64\pi^2s}d\Omega
=\frac{\lambda^2}{32\pi s}+O(\lambda^3),\qquad s>4m^2.}
$$
Equivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong <elastic scattering from a quartic scalar contact interaction> cross-section.
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