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How do infrared divergences arise in QED? What is an example case of such a divergence and how do we usually deal with such divergences? Are they absorbed like ultraviolet divergences?

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Infrared divergences can arise in various ways. Usually we refer to two sources as most important:

  1. External particles with vanishingly small momenta (so called "soft particles"). These are particles, usually photons, that are not observable by detectors in the sense that their momenta are so small that instruments are not sensitive to them. It turns out that in the limit of vanishingly small momenta an IR divergence can arise.
  2. IR divergences in virtual loop integrals or, in 2 dimensions, built into the particle propagator itself. The IR divergences can arise due to the integral over internal loop momenta not determined by momentum conservation, that begins at one-loop order; part of this region contains soft internal particles. In 2 dimensional space it turns out that even the propagator of physical particles can be IR divergent.

Interestingly there is a great soft particle theorem due to Weinberg[1] and discussed in Peskin and Schroeder Chap. 6. It has been shown that the IR divergences in virtual loop integrals can be summed exactly and cancel the IR divergences from soft particles! In this way these two main forms of IR divergences are removed from the game!

[1] S. Weinberg, Phys. Rev. 140, B516 (1965).

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  • $\begingroup$ So what does it mean to have a "soft particle"? You cannot detect it even in principle? I'd assume this arises from parameters in the amplitude equation right? $\endgroup$ Jan 9 '20 at 11:06
  • $\begingroup$ No - if I improve my equipment I will be more and more sensitive to detecting ever softer particles. However, no matter how sensitive my equipment is, there will always be a continuum of photon momenta I can't observe $\endgroup$
    – lux
    Jan 9 '20 at 14:50
  • $\begingroup$ Okay that makes sense. Just one last question then, why would we say that there is always a continuum of photon momenta that we can't observe but we don't say that there is a continuum of momenta that we can observe that is always lower than the photon? Doesn't that argument go both ways ? $\endgroup$ Jan 9 '20 at 15:00
  • $\begingroup$ Not quite because the argument is as follows: you want to measure photons that are end products of your experiment / collision. Your equipment detects 3 photons of various momenta. The problem is that you don't know whether those were the only photons emitted (and hence whether your theoretical calculation should coincide with observation) because the process could have produced (as well as the photons you detected) an arbitrary, unknown, maybe even infinite number of photons that were undetected because their momenta were below your sensitivity.... $\endgroup$
    – lux
    Jan 9 '20 at 15:41
  • $\begingroup$ ... You must include all such possibilities in the calculation and it turns out that the sum over all such undetected emissions alongside your observed photons suffers an IR divergence as the unobserved momenta tend to zero $\endgroup$
    – lux
    Jan 9 '20 at 15:42

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