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At tree level, the Feynman diagram has an electron-positron current and a muon-antimuon current joined by one virtual photon. In the massless limit, the spin average and final-state spin sum give $\overline{|\mathcal{M}{|}^2}=e^4(1+{\cos }^2\theta )$, and the integrated relativistic scattering cross-section is $4\pi {\alpha }^2/(3s)$.

The product of lepton-current matrix elements with a specified spin sum or spin average. Its normalization must be stated when quoting a differential scattering cross-section.

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:

Substitute the preceding spin average into the supplied differential scattering cross-section. Since $e^4=16{\pi }^2{\alpha }^2$,The angular integral is $2\pi {\int }_{-1}^1(1+x^2)dx=16\pi /3$. Consequently the relativistic scattering cross-section for the massless electron-positron annihilation into a muon pair isThe outgoing particles are distinct, so there is no identical-particle factor. Comparing with the requested monomial form givesThe integrated answer depends only on the incoming invariant $s$; the angular variables $t,u$ have been integrated out.

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.