# What does mathematical consistency in QFT mean?

My question is more naive than Is QFT mathematically self-consistent?

Just when people talk about the mathematical consistency of QFT, what does consistency mean? Do people want to fit QFT into ZFC? For example (could be in a more general context) , if I refer to https://souravchatterjee.su.domains//qft-lectures-combined.pdf

29.3 Wightman axioms

Axiom 1. There exists a Hilbert space $$H$$ whose elements represent the possible states of the quantum system under consideration

Unfortunately, no one has been able to construct an interacting quantum field theory in $$\mathbb{R}^{1,3}$$ that satisfies the Wightman axioms

Do people want to prove the existence of the exact solution in interacting theories? (the exact solution as a possible state in Hilbert space)

If yes, the perturbative expansion is thought to be asymptotic, it may hardly be used to show the existence of the exact solution of interaction system (sum of infinity order of expansion possibly goes to diverge). If I use the lattice approach when the grids approach zero, QED may reach the Landau pole then may become problematic for even closer (not sure what would it be). But, QCD can be defined at the zero grid distance limit, due to the asymptotic freedom. Is QCD then mathematically consistent? Or am I not even wrong?

• Context? This can mean anything from "doesn't have a Landau pole" to "includes counter-terms for all divergences that can be generated". Nov 26 '21 at 13:18
• Not very specific context. In page 129 of the lecture note in the question, "Unfortunately, no one has been able to construct an interacting quantum field theory in R 1,3 that satisfies the Wightman axioms." Nov 26 '21 at 13:55
• Are you asking what "mathematical consistency" means or what "satisfies the Wightman axioms" means? You've given context for the one that needs less context. Nov 26 '21 at 14:09
• Both. I also heard from other people some while ago said QFT is not mathematical consistent, but I did not ask into details. Sorry I should have said clearer. Nov 26 '21 at 14:33

Lattice approaches are not plagued by issues with things like the path integral, but the continuum limit - both whether it exists and whether it converges against the theory we want it to (see, e.g. the triviality of $$\phi^4$$ theory for why you might be able to take a limit but it's not "the theory you want") - is a thorny issue for rigor as well.