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I have a basic question about the logic of renormalization in quantum field theory (QFT). We met the ultraviolet (UV) divergence in loop corrections. The standard argument is, our current field theory is incorrect at high-energy scale, we regularize and renomalize the theory to make it work.

My question is, why do we expect a correct theory will be capable for loop corrections without regularization? Is that possible that nature or saying a correct theory at high energy scale is non-perturbative from free particles?

P.S. non-perturbative from free particles mean (i) if one uses perturbative expansion from free particle, the "correct" theory still gives infinity; and (ii) if one uses perturbative expansion from a special interacting solvable system or by certain non-perturbative technique, the results from that theory always agrees with experiments.

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  • $\begingroup$ Can you explain what you mean by "non-perturbative from free particles"? $\endgroup$ – JeffDror Feb 14 '14 at 1:52
  • $\begingroup$ I mean, (i) if one uses perturbative expansion from free particle, a "correct" theory still gives infinity; and (ii) if one uses perturbative expansion from a special interacting solvable system or by certain non-perturbative technique, the result from that theory agrees with experiment $\endgroup$ – user26143 Feb 14 '14 at 1:55
  • $\begingroup$ We propose an answer in M. Ribarič and L. Šušteršič, An open problem of theoretical physics Modification of the QFTs so as to make their diagrams convergent. arXiv:1503.06325 (2015). $\endgroup$ – user103001 Jan 5 '16 at 15:49
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The UV divergences have (most the time) nothing to do with perturbation theory (or, stated otherwise, free particles). They are also present in non-perturbative approach (see for example Non-Perturbative Renormalization Group, or Exact Renormalization Group).

Divergences, or better, cut-off dependence of observables means that the quantity you are looking at depends on the high energy physics, and is, in some sense, a free parameter of the theory. By that, I mean that stating that the system is described at low energy by a $\phi^4$ theory is not enough to completely define the theory, you also need to give the value of the mass (for example).

One main thing to keep in mind when studying QFT, especially in the context of high-energy physics, is that there is no reason that having a cut-off dependence is problematic. In the case of statistical mechanics and condensed-matter, the cut-off dependence is representative of the microscopic physics. In QED, say, there are also no good reason to think that the cut-off is unphysical, since we know that in fact at high energy, QED is a subpart of the electro-weak force, etc.

One of the QFT where people really want that there are no infinities is the case of General Relativity (if you think that a QFT should be enough, and you don't need fancy strings or who knows what). That's the famous case of the non-renormalizability of GR. But that's a perturbative statement, which does not mean that GR in non-renormalizable, implying that you have to invent something else. Indeed, the asymptotic safety scenario assumes that in fact GR is non-pertubatively renormalizable in the UV, meaning that there is a UV fixed point that allows to send the cut-off to infinity without divergences (you still have to fix one or two quantities to be on the critical surface, but that's much better than the infinity of coupling constant to be fixed in the perturbative approach). Interestingly, there are good evidences that such UV fixed point indeed exists.

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