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During a QFT class, my professor briefly explored the fact that infrared (IR) divergences can be removed from QED by considering "soft" photons that have an energy inferior to the detector sensitivity, arguing that such diagrams then cancel the IR divergence and make it unobservable. So in that example, we considered the IR divergence of the electron propagator, and added diagrams with one or more external photon(s).

This argumentation left me pretty unsatisfied, because we are adding there diagrams of a different kind (i.e. with external photon legs), and that means that if we had a perfect detector (such as nature has), the divergence would be observable if QED would be a perfect theory of nature.

I know we can remove the IR divergence by putting the system in a box, but I was wondering if there exists another way to remove IR divergences in continuous QFT (so no box)?

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    $\begingroup$ Remember that IR divergences come from Feynman diagrams and these are just a convenient way to split the process in easily computable parts. When you combine the diagrams these IR divergences cancel. This is well explained in many textbooks, e.g. Peskin & Schroeder. $\endgroup$ – Oбжорoв Jul 15 '19 at 12:25
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    $\begingroup$ Perturbation theory is great for some things, but it tends to obscure other basic things. An oscillating current in a wire produces an outgoing EM wave. How many external legs would the corresponding diagram have? We can't jiggle (or scatter) an electron without producing EM radiation. Nature and QED both know that. If we ask perturbation theory "What is the scattering amplitude with no EM radiation at all," perturbation theory answers with nonsense (IR divg), as it should. Such a thing simply doesn't exist in QED, just like an electron without an electric field doesn't exist in QED. $\endgroup$ – Chiral Anomaly Jul 15 '19 at 13:02
  • $\begingroup$ @ChiralAnomaly Mm so you are saying that the soft photons are not just a convenient way to get rid of the IR divergence, but a reflection of the fact that, in reality, we can never have an electron propagate as a closed system, i.e. without emitting radiation that is not reabsorbed? I can get convinced by that, maybe you would like to turn your comment into an answer? $\endgroup$ – Jxx Jul 15 '19 at 14:38

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