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The inclusive product of hadronic-current matrix elements, summed over final states with a Lorentz-invariant phase-space measure. Electromagnetic current conservation makes both indices transverse to the exchanged four-momentum.

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.

Assume a complete relativistically normalized basis of scalar momentum eigenstates transforms asThis states both reversal of spatial momentum and preservation of the normalization of basis quantum states. Antilinearity alone would not suffice: multiplying a conjugation operator by two gives an antilinear operator that multiplies squared norms by four.

Expand $|\varphi ⟩=\int d{\Pi }_p\varphi (\mathbf{p})|\mathbf{p}⟩$ and similarly for $|\psi ⟩$, with $d{\Pi }_p=d^{3}p/((2\pi {)}^{3}2E_p)$. This Lorentz-invariant phase-space measure is unchanged under $\mathbf{p}\mapsto -\mathbf{p}$. The antilinearity of the quantum time-reversal operator gives conjugated expansion coefficients. Orthogonality of the momentum eigenstates and cancellation of their unit-modulus phases yieldSince momentum reversal is a bijection of the complete basis, $\hat{T}$ is also onto. ThusThis proves antiunitarity, with the normalization hypothesis explicitly included. For scalar multiparticle quantum states the same argument uses the complete occupation-state basis and reverses all momenta; it is not restricted to a single-particle wave packet.

The authoritative PDF has $m_Z^2/v$, not the dimensionally incorrect $m_Z^2/v^2$ in the TeX conversion. The two identical Z bosons give the vertex factor two, so the scattering amplitude for the decay isup to an irrelevant overall phase. Apply the massive vector polarization sum, identifying the hint's $M_Z$ with $m_Z$:Since $k_1\cdot k_2=(m_H^2-2m_Z^2)/2$, putting $x=m_Z^2/m_H^2$ givesNo initial-spin average is needed for a Higgs boson. In its rest frame the two-body momentum is $|\mathbf{k}|=(m_H/2)\sqrt{1-4x}$. Integrating the energy and momentum delta functions in the Lorentz-invariant phase-space measure gives $d{\Phi }_2=\sqrt{1-4x}d\Omega /(32{\pi }^2)$, hence $\int d{\Phi }_2=\sqrt{1-4x}/(8\pi )$.

The printed generic width formula treats the daughters as labelled. Here the identical final-state symmetry factor is $1/2!$, without which the same physical configuration is counted twice. Thus the Higgs decay to two Z bosons hasEquivalently, $1/v^2=\sqrt{2}{G}_F$ gives the prefactor $G_F{m}_H^3/(16\sqrt{2}\pi )$. The threshold limit is zero and the large-mass limit is $m_H^3/(32\pi v^2)$. The additional factor for identical daughters is a required specialization of the PDF's generic phase-space formula, not an extra factor in the vertex.

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.