A question about the implication of UV divergence in QFT

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.

• Can you explain what you mean by "non-perturbative from free particles"? – JeffDror Feb 14 '14 at 1:52
• 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 – user26143 Feb 14 '14 at 1:55
• 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). – user103001 Jan 5 '16 at 15:49

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).