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With relativistically normalized external states, $\mathcal{M}$ is the reduced S-matrix element after removing the four-momentum conservation distribution. A relativistic scattering cross-section combines $|\mathcal{M}{|}^2$ with the Lorentz-invariant phase-space measure and the invariant flux factor. Identical-leg factorials in Feynman vertices and the identical final-state symmetry factor in the event integral have different roles.

The Källén function controls two-body relativistic kinematics. In the rest frame of total invariant mass $\sqrt{s}$, the magnitude of each final momentum is $\sqrt{\kappa (s,m_1^2,m_2^2)}/(2\sqrt{s})$. It also determines the invariant flux factor in a two-particle collision.

The real scalar field has canonical momentum $\pi =\dot{\varphi }$. Its Legendre transform givesThe first three terms constitute $H_0$, the free massive Hamiltonian operator. The remaining term is $H_I=\lambda \int {\varphi }^4{d}^{3}x/4!=-\int {\mathcal{L}}_I{d}^{3}x$. 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 satisfiesIntegrating once gives $U_I=I-i{\int }_{t_0}^t{H}_I(t_1)U_I(t_1,t_0)d{t}_1$. Repeated substitution orders the integrations as $t\ge t_1\ge \cdots \ge t_n\ge t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a time-ordered product, which proves the Dyson seriesTaking the asymptotic limits with the usual adiabatic or regulated scattering prescription yieldsThe 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 $|p_1,p_2⟩=a^{†}(p_1)a^{†}(p_2)|0⟩$, with $⟨p|q⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. The connected first-order scattering contribution isHere 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 $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=4!$, the external-leg factorial cancellation at a quartic scalar vertex. Each attachment contributes its plane-wave factor, soThe spacetime integral is the four-momentum conservation delta function, givingIn 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 $\sqrt{s}=2E_{\ast }$, $p_{\ast }^2=E_{\ast }^2-m^2$. 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 $E_{\ast }/(2p_{\ast })$. ThusThe invariant flux factor is $\mathcal{F}=4p_{\ast }\sqrt{s}$ for equal initial masses. Using $d\sigma =|T{|}^{2}d{\Phi }_2/\mathcal{F}$, the labeled angular density is ${\lambda }^2/(64{\pi }^{2}s)$. 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 $1/2!$. ConsequentlyEquivalently 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.

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 metric $(+---)$, ${ϵ}^{0123}=+1$ and relativistically normalized external spinors. The Fermi interaction givesup to an irrelevant phase. The fermion spin sums replace $u\bar{u}$ and $v\bar{v}$ by the slashed massless momenta. Write $C=P^2+Q^2$. The gamma matrix trace identities yieldThe symmetric-antisymmetric mixed contractions vanish. The symmetric contraction is $2[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$, while the two epsilon tensors contract to $-2[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Including $G_F^2/2$ and the $1/4$ initial-spin average therefore givesThe singlet color charge contractions ${\delta }_{ij}{\delta }_{kl}$ have net factor one after summing final colours and averaging initial color charge states. Thus this is also the colour-averaged partonic result; no extra factor three is required.

In the centre-of-momentum frame, let $c_{\theta }=\cos \theta$. The Mandelstam variables satisfy $t=-s(1-c_{\theta })/2$ and $u=-s(1+c_{\theta })/2$, giving $(p_2\cdot k_1)(p_1\cdot k_2)=s^2(1+c_{\theta }{)}^2/16$ and $(p_2\cdot k_2)(p_1\cdot k_1)=s^2(1-c_{\theta }{)}^2/16$. HenceThe massless invariant flux factor is $2s$ and $d{\Phi }_2=d\Omega /(32{\pi }^2)$, so the weak charged-current quark scattering cross-section isFor a purely left-handed weak charged current, $P=1$, $Q=-1$, this reduces to $d\sigma /d\Omega =G_F^{2}s(1+\cos \theta {)}^2/(32{\pi }^2)$, fixing the sign of the forward term. Purely right-handed weak charged currents at both vertices have the same unpolarized angular law. The factorization convention for $F,H_1,H_2$ could be rescaled by a common numerical factor; the displayed product fixes the normalization.

Use the Lorentz-invariant phase-space measure from the original PDF, including the $1/(2E_{p^{\prime }})$ and $1/(2E_{q^{\prime }})$ factors missing from the local TeX. In the center of mass frame put $E=\sqrt{s}/2$ and $k=\sqrt{s-4m^2}/2$, assuming $s>4m^2$. The spatial Dirac delta distribution sets ${\mathbf{q}}^{\prime }=-{\mathbf{p}}^{\prime }$, leavingSince $d(2E_{k^{\prime }})/d{k}^{\prime }=2k^{\prime }/E_{k^{\prime }}$, the radial integral givesThe invariant flux factor is $2\sqrt{\kappa (s,m^2,m^2)}=2\sqrt{s(s-4m^2)}$, using the Källén function. Dividing the phase space by this flux yields $d\sigma /d\Omega =|\mathcal{M}{|}^2/(64{\pi }^{2}s)$, with the final-state labels retained as in the printed formula.

Let $x=\cos \theta$. Of the Mandelstam variables, $t$ isThe azimuthal integral contributes $2\pi$, and the angular endpoints are thereforeChanging variables gives the requested expression from the formula supplied in the paper:There is a normalization qualification: the displayed starting formula integrates over labeled final momenta and contains no $1/2!$. For the physical cross-section of two indistinguishable outgoing quanta of this real scalar field, the identical-particle factor in a final-state phase-space integral divides the full integral by two. With that convention $X=1/(32\pi )$; equivalently, integrate only one representative of each exchanged pair. The boxed value $1/(16\pi )$ follows the given formula exactly.

The deep inelastic scattering process contains a virtual photon exchanged between the electron and the hadron:

Use an electromagnetic vector current $J^{\mu }$ containing the dimensionless quark charges, with the coupling $e$ factored out. The scattering amplitude, up to an irrelevant phase, isTo match the printed prefactor, define the leptonic tensor with a spin sum over both electron spins and keep the initial spin average $1/2$ outside it. The gamma-matrix trace givesFor a stationary target and massless electron, the invariant flux factor is $4ME$. The inclusive final-state Lorentz-invariant phase-space measure and target spin average are contained in $4\pi W_H^{\mu \nu }$. Thus the differential scattering cross-section isHere $q^4$ means $(q^2{)}^2$. If the initial spin average is instead built into the leptonic tensor, its normalization is $2(\cdots )$ and the displayed cross-section prefactor must be doubled. The two conventions give the same observable.

With relativistically normalized external states, a two-particle initial state has differential cross-section $d\sigma =|\mathcal{M}{|}^{2}d{\Phi }_n/\mathcal{F}$, where $d{\Phi }_n$ is the Lorentz-invariant phase-space measure and $\mathcal{F}$ is the invariant flux factor. For identical unobserved final particles, divide the full labeled phase-space integral by their permutation multiplicity.