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One common way to analyze a quantum field theory is to regularize it by introducing an ultraviolet cutoff. After perhaps renormalizing the theory, one would hope that you can remove the cutoff by taking a limit. In the constructive quantum field theory literature I often see the related term "ultraviolet stability". On p.1 in the review paper by Gallavotti, he says

existence of the functional integrals defining the generating functions of the probability distribution of the interacting fields in finite volume: the ultraviolet stability problem.

yet in other sources I've seen ultraviolet stability and the ultraviolet limit stated as two separate problems, with UV stability sometimes referring to bounds that are uniform in the cutoff but not necessarily the removal of the cutoff.

Is there really a difference between the ultraviolet limit problem and ultraviolet stability, and if so, how do they differ? Second, if we do have bounds that are uniform in the cutoff how does this not necessarily imply the existence of a UV complete theory with no cutoffs or regularization?

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  • $\begingroup$ Can you not see that one smart way to do things is to completely side-step this confusion by finding a treatment that does NOT introduce an ultraviolet cut-off? Then there is no "remove the cut-off by taking a limit" and then all arguments about limits will be actually some physical limiting behaviour being discussed, and all discussions about ultraviolet stability problem will be actually about stability of some form. It does not even have to be dimensional regularisation; it could be Pauli–Villars, or it can be Causal Perturbation, where then there is no infinities needing regularisation. $\endgroup$ Commented yesterday
  • $\begingroup$ @naturallyInconsistent I had in mind lattice gauge theories where the cutoff is in the form of the lattice spacing. I thought UV stability was referring to a theory being mathematically well behaved in the UV, I'm not aware of the physical consequences. Would you mind elaborating? $\endgroup$ Commented yesterday
  • $\begingroup$ @naturallyInconsistent There are no rigorous approaches to constructive quantum field theory that do not involve a cutoff. I don't think it's even conceivable, as without it you cannot distinguish between $\phi^4_4$, which is just Gaussian, and 4D Yang-Mills, which is asymptotically free and so there is ostensibly a nongaussian model that has yet to be constructed. Perturbation theory is blind to these sorts of issues as both models are perturbatively renormalizeable. $\endgroup$ Commented yesterday
  • $\begingroup$ @Prox I do not think you are correct. For example, the Wiki page for Pauli–Villars state that it works well for Abelian QFT like QED but does not work well for Yang–Mills, and that 't Hooft and Veltman, famous for dimensional regularisation, also introduced unitary regulators that are generalisations of Pauli–Villars for Yang–Mills. I mean, I am not sure why you think constructive QFT cannot have these alternative regularisations? $\endgroup$ Commented yesterday
  • $\begingroup$ @naturallyInconsistent Sorry I missed your earlier reference to Pauli-Villars. PV is a kind of ultraviolet cutoff so i don't see why you're suggesting it as an alternative. It's certainly usable in rigorous field theory although it has its downsides in that it kills reflection positivity. Dimension regularization on the other hand is virtually useless outside of perturbation theory. That being said it really is just another kind of UV cutoff even in the perturbative setting. Finally causal perturbation theory is a purely perturbative formalism so it's not very useful either. $\endgroup$ Commented yesterday

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The two are related, but the more precise meaning for ultraviolet stability is a bound on either the Hamiltonian in the Hilbert space approach or the partition function in the functional integral approach. If $Z_{\kappa, \Lambda}$ is the partition function on a domain of volume $|\Lambda|$ with UV cutoff $\kappa$, then the ultraviolet stability bound is $$Z_{\kappa, \Lambda}\leq \exp(c|\Lambda|)$$ Where $c$ is a constant independent of $\kappa, \Lambda$. Generally an ultraviolet stability bound is the first step in constructing the continuum limit because it is essential for controlling correlation functions. However it absolutely is not equilvalent to the existence of a continuum limit. For example, it (without further work) doesn't say anything about correlation functions.

There is a complementary set of bounds for correlation functions that can also be considered to be ultraviolet stability bounds, and if you have those then your correlation functions are uniformly bounded in $\kappa,\Lambda$. But a bounded sequence is not necessarily a convergent one! For example $(-1)^n$ is a bounded sequence in $\mathbb{R}$ but obviously it does not converge to anything. So even more work is needed to get from boundedness to the existence of the field theory. Sometimes this is too hard and one is forced to rely on compactness arguments to an extract a convergent subsequence, but then you lose out on symmetries, uniqueness, and properties of this nature.

However from an informal point of view once you have ultraviolet stability you can probably be somewhat confident that the model exists. This is why it's fine to casually conflate the two notions despite the technical gulf that exists between them, and I think that's all that Gallavotti is doing.

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  • $\begingroup$ Thank you! So putting it another way, UV stability says that the measure exists for all values of $\kappa, \lambda$, but nothing about the limiting measure nor the moments of any of these measures (including the cutoff ones). From this I assume that a continuum theory is defined as both the the limiting Hamiltonian/functional integral along with finite correlation functions of all orders? This makes sense given how important correlation functions are in defining a QFT. $\endgroup$ Commented yesterday
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    $\begingroup$ UV stability says more than that the measure exists. That's generally pretty easy to show. It's a uniform bound independent of the UV cutoff, so it tells you that the measure is not blowing up as the cutoff is being removed. A priori we do not know if a limiting functional integral exists. If it exists and has good correlation functions that are well behaved, we win. $\endgroup$ Commented yesterday

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