= Solution
The <deep inelastic scattering> process contains a virtual <photon> exchanged between the <electron> and the <hadron>:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-305-q4a.png]
{title=Inclusive electron-hadron scattering through a virtual photon}
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, is
$$
\mathcal M_X=\frac{e^2}{q^2}\bar u(p')\gamma_\mu u(p)\langle X|J^\mu|H\rangle.
$$
To 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> gives
$$
\boxed{L_{\mu\nu}=\operatorname{Tr}(\not p'\gamma_\mu\not p\gamma_\nu)
=4\left(p_\mu p'_\nu+p_\nu p'_\mu-g_{\mu\nu}p\cdot p'\right).}
$$
For 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> is
$$
\frac{d\sigma}{d^3p'}=\frac1{4ME}\frac1{(2\pi)^3 2E'}\frac12\frac{e^4}{(q^2)^2}L_{\mu\nu}(4\pi W_H^{\mu\nu})
=\boxed{\frac{e^4}{8(2\pi)^2 MEE'(q^2)^2}L_{\mu\nu}W_H^{\mu\nu}.}
$$
Here $q^4$ 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.
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