Here, in page 11, you can see the so-called 'handbag' diagram that explains how a virtual photon emitted in a deep inelastic scattering (DIS) process interacts with a parton.
I'm going to use this diagram to compute the amplitude of the process $\gamma P \rightarrow X$, with $P$ the proton and $X$ a set of undetermined particles. So taking into account that the scattering matrix can be expanded in Taylor series as $S = 1 + A$, when wrting $A$ to first order you find that
$$ \sum_X \int \frac{d^3 \vec{p}_X}{(2\pi)^3 2E_X}\ |A(\gamma(q, \Lambda) P \rightarrow X)|^2 = 4\pi VT \epsilon^\mu_{q\Lambda} \epsilon^{\nu\ *}_{q\Lambda} W_{\nu \mu} $$
where the spacetime volume is $VT = (2\pi)^4 \delta(0)$, $W_{\nu \mu}$ is the hadronic tensor given as
$$ W_{\nu \mu} = \frac{1}{4\pi} \int d^4 z\ e^{iqz} \langle P| j_\nu^\dagger (z) j_\mu(0) |P\rangle, \quad j_\alpha = \sum_q e_q\bar{q}\gamma_\alpha q $$
and $q$ is the quark field with electric charge $e_q$.
If $S = 1 + A$ then the optical theorem reads
$$ \sum_X \int \frac{d^3 \vec{p}_X}{(2\pi)^3 2E_X}\ |A(\gamma(q, \Lambda) P \rightarrow X)|^2 = -2\Re [A(\gamma P \rightarrow \gamma P)] \tag1$$
RHS represents (-2 times) the real part of the handbag diagram depicted above (we should include the diagram with the photons exchanged) and can be written as
$$ -2\Re [A(\gamma P \rightarrow \gamma P)] = 2VT \Re \left( \sum_q e_q^2 \int \frac{d^4 k}{(2\pi^4)} \left[ \gamma_\nu \left\{ \frac{i(\not{k} + \not{q})}{(k + q)^2 + i0} + \frac{i(\not{k} - \not{q})}{(k - q)^2 + i0} \right\} \gamma_\mu \right]_{ij} [f(p, k)]_{ji} \epsilon^{\nu\ *}_{q\Lambda} \epsilon^{\mu}_{q\Lambda} \right) $$
$VT = (2\pi)^4 \delta(0)$, i.e., the spacetime volume, a quantity that cancels out in the equality (1). $\sum_{i, j}$ implicit and we have consider the approximation $m_q \ll k, q$ with $m_q$ the quark mass.
Now my problem is how to compute the real part ($\Re$) in the last formula since we know nothing about $f(p, k)$ and in principle the product of the polarization vectors is not necessarily real.
