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.


1 Answer 1


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 .

Another link.

  • $\begingroup$ 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. $\endgroup$
    – Vicky
    Oct 12, 2020 at 9:51
  • $\begingroup$ I expanded and added links $\endgroup$
    – anna v
    Oct 12, 2020 at 13:09
  • $\begingroup$ I have precised my question $\endgroup$
    – Vicky
    Oct 18, 2020 at 20:49
  • $\begingroup$ why dont you try the question at physicsoverflow.org which is mainly theoretical $\endgroup$
    – anna v
    Oct 19, 2020 at 4:41
  • $\begingroup$ what's the difference between the 2 platforms? $\endgroup$
    – Vicky
    Oct 19, 2020 at 12:35

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