# Deep inelastic scattering (DIS) and handbag diagram

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.

This is a higher order diagram to figure 1 in your link, using the parton model of the proton, which is modeled with a great number of quarks antiquarks and gluons, on which the three valence quarks are a part. See this article .:

The fact that the proton is a bag of partons brings the level of calculations of deep inelastic scattering one level lower, electron - parton. The handbag diagram is part of the higher order diagrams entering the calculation.

The virtual photon of fig 1 interacts with one of the partons, and there is a (small) probability that the proton remains intact, the second photon being real, or a gluon jet, for example . To see the complexity of higher order diagrams in DIS have a look .

• This is not an answer to my question. Maybe I didn't explain correctly what I wanted to understand. Let me rephrase it: how do you reconcile Fig 1 from OP's paper with Fig. 8? In principli, $X$ in Fig. 1 could be the proton, when it's not broken, but nevertheless you get a photon in the final state according to Fig. 8 that do not appear in Fig. 1... How handbag diagram appears and why it is useful are my questions. – Vicky Oct 12 '20 at 9:51