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As we know, we can remove infrared divergences by summing over all final states with arbitrary number of soft photons. But in QCD this does not work, since gluons are not "neutral" because they carry color charge. What's more, they are massless. So can the proof in QED for infrared divergences also apply to QCD? If not, how to deal with this problem?

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  • $\begingroup$ This might be of interest to you. Schroder (arxiv.org/abs/hep-ph/0410130) suggests dimensional reduction to solve the QCD infrared problem. $\endgroup$
    – Clever
    Feb 1, 2015 at 10:45

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To add some detail to drashunk's answer, the Kinoshita-Lee-Nauenberg theorem states that in any unitary theory infrared divergences will cancel when all possible final and initial states in a finite energy window are summed over. So in QCD, in general we need to include $n\to m$ scattering where we sum over $n$ and $m$.

The KLN theorem is pretty general; its proof uses nothing except unitary quantum mechanics. The initial state sum turns out to be much stronger than needed in QCD. In general for IR finiteness at the cross-section level the only initial states you need in QCD are the ones obtained by so-called "double cut diagrams." To construct these diagrams one takes a Feynman diagram and its mirror image, joining the final states lines along the middle. Then one orders the vertices in some way and vertically cuts the diagram, summing over all vertex orderings and cuts. Cutting in a different location than the middle has the neat interpretation of specifying a different final / initial state. Crucially, if your process involves only gluons for instance, you don't need to sum over intial states with quarks. This QCD-specific version of the KLN theorem is proved in George Sterman's book on QFT using old-fashioned perturbation theory. I don't think it's very well-known.

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