Show that the parton model diagram, Fig. 10.1, gives
$$
\begin{aligned}
& \frac{\hat{\sigma}_T}{\hat{\sigma}_0}\left(z, Q^2\right)=e_i^2 \delta(1-z) \\
& \hat{\sigma}_L\left(z, Q^2\right)=0 .
\end{aligned}
$$
We outline the various stages of the calculation. First, show that for $\gamma^*(q) \mathrm{q}(p) \rightarrow \mathrm{q}\left(p^{\prime}\right)$
$$
\overline{|\mathscr{\Re}|^2}=2 e_i^2 e^2 p \cdot q,
$$
where we have averaged over transverse polarization states of the incoming $\gamma^*$. From Section 4.3, we have
$$
F d \dot{\sigma}_T=\overline{|\mathscr{T}|^2}(2 \pi)^4 \delta^{(4)}\left(p^{\prime}-p-q\right) \frac{d^3 p^{\prime}}{2 p_0^{\prime}(2 \pi)^3} .
$$
where $F$ is the $\gamma^* \mathrm{q}$ flux factor. Calculate $F \hat{\sigma}_T$ by making use of (6.47). Use $F \hat{\sigma}_0=8 \pi^2 \alpha$, see $(10.5)$.
To determine the parton model prediction for $F_2 / x$ of (10.4), we input in (10.8) the following cross section ratio for $\gamma^* \mathrm{q} \rightarrow \mathrm{q}$ :
$$
\frac{1}{\hat{\sigma}_0}\left(\hat{\sigma}_T+\hat{\sigma}_L\right)=e_i^2 \delta(1-z) \text {, }
$$
see (10.10) and (10.11). After substitution, we obtain
$$
\frac{F_2\left(x, Q^2\right)}{x}=\sum_i e_i^2 \int_x^1 \frac{d y}{y} f_i(y) \delta\left(1-\frac{x}{y}\right)=\sum_i e_i^2 f_i(x) .
$$
An identical expression is found for $2 F_1$. The parton model results of (9.13) and (9.14) are indeed reproduced.