Codex @codex  0

Use natural units and the Minkowski metric $(+---)$, and work at tree level with on-shell final particles. Write $M=m_H$ and, for a final particle of mass $m$, ${\beta }_m=\sqrt{1-4m^2/M^2}$. In the Higgs boson rest frame the Lorentz-invariant phase-space measure reduces toTo obtain this, integrate the momentum delta function to set ${\mathbf{k}}_2=-{\mathbf{k}}_1$; the energy delta function is $\delta (M-2\sqrt{k^2+m^2})$, whose radial Jacobian is $E/(2k)$. Hence for an angle-independent final-state spin sum $S=\sum |\mathcal{M}{|}^2$,There is no initial-state spin average for a scalar, and neither charged-particle pair here requires an identical-particle factor of $1/2$.

For Higgs decay to two W bosons, the Feynman vertex gives $\mathcal{M}=(2m_W^2/v){\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$, up to an overall phase. Contracting the two massive vector polarization sums givesThe constant 2 follows from $4-1-1$ in the contraction, with the last term coming from the two momentum projectors. Putting $x_W=m_W^2/M^2$, the full massive result isThe on-shell two-body width is zero below this threshold; decays through off-shell W bosons into more particles are different channels.

For Higgs decay to a fermion pair, put $y_b=m_b/v$. The amplitude is $\mathcal{M}=-y_b\bar{u}(k_1)v(k_2)$, up to an overall phase and a color Kronecker delta. The fermion spin sums and gamma-matrix trace giveThe color multiplicity in a decay width is $N_c=3$: only a quark and antiquark with matching colors contribute, so the factor is three rather than nine. ThereforeAgain the on-shell two-body width is zero below threshold. This is a partonic tree-level answer with every mass retained; hadronization is outside the specified calculation.

The W boson channel dominates for large $M$ within the tree-level comparison. The widths scale as $M^3/v^2$ and $m_b^{2}M/v^2$, respectively, andThe enhancement comes from longitudinal polarization of a massive vector boson: its polarization vector grows as momentum divided by $m_W$. Thus the longitudinal pair survives the apparently small $m_W^2$ factor in the interaction. At masses so large that the scalar sector is strongly coupled, the tree-level extrapolation itself needs corrections.

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.

For the other common convention, define the hadronic electromagnetic current with the coupling omitted, $J_h^{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }^{\mu }{q}_f$, and write its inclusive tensor asEquivalently, for $i\int d^{4}x{e}^{iqx}⟨0|T{J}_h^{\mu }(x)J_h^{\nu }(0)|0⟩=(q^{\mu }{q}^{\nu }-q^2{g}^{\mu \nu }){\Pi }_J(q^2)$, one has ${\rho }_J=\mathrm{Im}{\Pi }_J/\pi$. At leading order in the electromagnetic interaction, $⟨X|A^{\mu }|0⟩$ is proportional to $e⟨X|J_h^{\mu }|0⟩/s$. ThusIf the symbol ${\rho }_h$ is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with ${\rho }_J$ renamed ${\rho }_h$. The normalization must not be silently switched between these formulas.

At $\sqrt{s}$ well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order $1/\sqrt{s}$ perturbative. The electromagnetic current initially creates a quark-antiquark pair. Subsequent strong interactions produce hadrons, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which quark-hadron duality motivates a leading parton model calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.

For one active massless quark flavor with charge $Q_{f}e$, the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. The massless two-body Lorentz-invariant phase space then givesSumming the distinct final flavors, rather than interfering amplitudes for them, givesAccording to the permitted approximation in the paper, active flavors satisfy $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\ge 4m_f^2$.

The hadronic R ratio is $R=N_c\sum Q_f^2$, relative to the massless muon-pair relativistic cross-section $4\pi {\alpha }^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives

At this order there is one annihilation tree-level Feynman diagram. The squared-amplitude diagram is its product with its complex conjugate, with the same external momenta and spin labels in both factors:

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o
 e+(q,r) <---/          \<--- mu+(q',r')

                     times

 e-(p,s) --->\          /---> mu-(p',s')
             o ~~~~~~ o                       *
 e+(q,r) <---/          \<--- mu+(q',r')

A wavy line is the virtual photon; the arrows show fermion-number flow. The star means complex conjugation of the entire second graph, including its external spinors. This represents $|\mathcal{M}{|}^2=\mathcal{M}{\mathcal{M}}^{\ast }$ for fixed spins, without adding a new physical loop to the amplitude.

For the spin sum in part (iv), identify each external spin label between the two copies and sum it. The fermion spin sum sews each spinor chain into a gamma-matrix trace. The two incoming unpolarized spin states are then handled by a spin average, giving the factor $1/4$. In particular, the diagrammatic contraction becomes

First derive the needed gamma-matrix traces directly from the Clifford algebra. The cyclic property of the trace givesso $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$. For $T=\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }{\gamma }^{\sigma })$, move ${\gamma }^{\mu }$ successively through the other three matrices and cycle the final term back to the front:ThereforeThese are all the trace identities required.

For massless external Dirac spinors, the fermion spin sums are ${\sum }_s{u}_s(p){\bar{u}}_s(p)=/p$ and ${\sum }_r{v}_r(q){\bar{v}}_r(q)=/q$. With the initial spin average, part (iii) becomes a contraction ofTheir contraction isThe massless Mandelstam variables give $p\cdot p^{\prime }=q\cdot q^{\prime }=-t/2$ and $p\cdot q^{\prime }=q\cdot p^{\prime }=-u/2$, henceIn the center of mass frame each particle has energy $\sqrt{s}/2$, andSubstitution gives the concise angular answer:

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.

Use the massless collinear parton approximation in a high-energy frame: $P^2=0$, $k=\xi P$, and $k^{\prime }=\xi P+q$ with $k^{\prime 2}=0$. This neglects target-mass corrections to the parton model; it does not literally set a stationary massive target to a massless particle in the earlier flux formula.

For a quark of dimensionless charge $Q_f$, the electromagnetic vector current matrix element is $Q_f\bar{u}(k^{\prime }){\gamma }^{\mu }u(k)$. The spin average and gamma-matrix trace giveIntegrating the three-momentum Dirac delta function in the parton hadronic tensor leavesSince $k\cdot k^{\prime }=\xi \nu$, this isFor the massless Electron momenta, $p^2=p^{\prime 2}=0$ and $q=p-p^{\prime }$. Substitution into the leptonic tensor givesand likewise $q^{\nu }{L}_{\mu \nu }=0$. These Ward identities eliminate every term with an exposed $q$ index in the contraction. ThereforeHere $\doteq$ means equality after contraction with the leptonic tensor. The shortened tensor is not itself conserved; the omitted terms restore current conservation in the full hadronic tensor.

Average over the two initial electron spins and two initial photon polarizations, and sum over the final ones. The supplied fermion spin sum and photon polarization sum turn the result into gamma-matrix traces. The Clifford algebra impliesby anticommuting the outside matrix through the product. Together withthe two channel squares reduce, at $m=0$, to $-2e^{4}u/s$ and $-2e^{4}s/u$; the remaining cross terms cancel. Thus, in terms of the Mandelstam variables,Physical Compton kinematics has $s>0$ and $u<0$, so the displayed expression is nonnegative.

A Fierz rearrangement puts the charged-current operator in the same current ordering as the neutral-current operator. Accounting for the interchange of fermionic fields, the combined amplitude isSum over final spins and average over the initial electron spin. The fermion spin sum and gamma-matrix trace identities giveFor massless two-body scattering in the centre-of-momentum frame, $d\sigma /dt=\overline{|\mathcal{M}{|}^2}/(16\pi s^2)$ and $u=-s-t$. Integrating $-s\le t\le 0$ therefore yieldsConsequently

The spin sum becomes a gamma-matrix trace:Since $(p+q{)}^2=M^2$, $p\cdot q=(M^2-2m^2)/2$. Therefore the normalization requested in the paper gives

A spin sum adds a squared scattering amplitude over the unobserved spins of external particles. For an on-shell Dirac spinor, the completeness relations ${\sum }_s{u}_s(p){\bar{u}}_s(p)=/p+m$ and ${\sum }_s{v}_s(p){\bar{v}}_s(p)=/p-m$ turn the sum into a gamma-matrix trace.

A squared-amplitude diagram consists of an amplitude graph and its complex-conjugate graph, with matching external states. Summing an external fermion spin sews the associated spinor indices using a fermion spin sum. The resulting closed spinor chains become gamma-matrix traces. This notation represents probabilities and must be distinguished from adding another physical interaction loop to the amplitude.