can be a theory with an infinite number of divergent integrals of the form

$$ \int \frac{d^{p}k}{k^{m}} $$

for m=1 , 2 , 3 , 4 ,...... so the theory would be IR non renormalizable and you would need and infinite set of operations to cure all the IR divergences



Renormalization in the sense you use it does not cure IR divergences, only UV. This allows to parametrize the UV physics with (in)finite set of constants.

IR divergences are more physical, and tell you about a disease of the model or of the interpretation of the result. For instance, infrared divergences in critical phenomena are a signature that perturbation theory (in the coupling constant) fails, and that resummations are needed (giving rise, for instance, to the anomalous dimension).

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    $\begingroup$ To add to your comment: UV divergences tell you there's something wrong with way you've set up the path integral $\int e^{iS} d\phi$. IR divergences tell you that you've chosen an observable whose expectation value isn't well-defined in the phase you're studying. $\endgroup$ – user1504 Oct 28 '13 at 21:25
  • $\begingroup$ @user1504: I don't agree with your second sentence. IR divergences may just arise because the way you do the calculation is not the correct one, and the same observable is non-perturbatively well defined. For instance, see arxiv:1011.3324 for a discussion of the IR divergences (and the way to get rid of them) in the ordered phase of the O(N) model and non-relativistic bosons. $\endgroup$ – Adam Oct 28 '13 at 21:38
  • $\begingroup$ That's a fair point. My comment does cover 95% of the cases one encounters in learning the subject though. $\endgroup$ – user1504 Oct 29 '13 at 12:47
  • $\begingroup$ @user1504: I'm not sure... The cases I know that could correspond to what you said would be the radiation of deep IR photons and maybe IR longitudinal gluons. That's what I called the problem of interpretation (even though I'm not sure for gluons, I learned that a while ago). Do you have other examples in mind ? $\endgroup$ – Adam Oct 29 '13 at 13:21
  • $\begingroup$ Forgot to respond. The (probably canonical) example is 2d massless scalars, where the 2-point function of the basic field is divergent. It doesn't make sense to ask about the expection value of $\phi(x)$ in this theory; there is no such operator. $\endgroup$ – user1504 Dec 3 '13 at 20:53

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