QED is an approximate description of reality. Even if it did give finite predictions in the continuum limit, those predictions would've been incorrect anyway! Newtonian gravity does give finite predictions for high gravity scenarios, but the numbers are incorrect.

QED can be defined in a mathematically rigorously (non-perturbative) way in the discrete spacetime approximation! Then why do we need to make the continuum QED mathematically rigorous, even when its predictions in the tiny length scales will be incorrect anyway?

I also read that a non-perturbative formulation of Yang Mills is a Millenium problem. But Yang Mills is already non-perturbative in discrete spacetime, i.e. in the length scales it works in the first place. In the tinier length scales, it will be incorrect anyway (Gravity effects will kick in).

So it shouldn't be a problem that QFTs at tiny length scales aren't rigorous / contain infinite quantities. Non-perturbative definitions of the theories exist in the discrete spacetime approximation, i.e. in the domain the theories hold in the first place.

  • 6
    $\begingroup$ We don't need anything. We want many things, fortunately. $\endgroup$ Apr 23, 2022 at 11:55
  • $\begingroup$ @AccidentalFourierTransform It's an illogical thing to want. You should build a new mathematical model for continuum spacetime, instead of wanting a fundamentally discrete model to be continuous. You are asking it to be something it isn't! $\endgroup$
    – Rain Deer
    Apr 23, 2022 at 11:59
  • 2
    $\begingroup$ why QFT might be in a better mathematical state than one gets the impression from the typical introduction to the topic: "How I Learned to Stop Worrying and Love QFT" arxiv.org/abs/1201.2714 $\endgroup$
    – Quillo
    Apr 23, 2022 at 13:08
  • $\begingroup$ Sorry to seem pedantic and do basic principles matter to you, or not? Nothing can be defined in a "mathematically rigorously (non-perturbative) way" in English; never. If that doesn't matter, why does the rest? Do you see words as less important than numbers, or what? "mathematically rigorously (non-perturbative) way" is simply wrong; prolly as wrong when translated into logic or maths as it is in English. "mathematically rigorous (non-perturbative) way", for instance, means something different. $\endgroup$ Apr 24, 2022 at 19:05

1 Answer 1

  1. The premise of the question is flawed - many practitioners of QFT definitely do not really care that people interested in rigor think QFT is lacking. They care that their procedures, whether "rigorous" or not, produce correct models of reality, as QED and the Standard Model demonstrably do. So, in turn, people who are interested in rigor are less interested in producing alternatives (like you seem to think discrete QFT is) and more interested in putting what people in practice actually do on a rigorous foundation.

    In this viewpoint, the Millenium problem is really that: It is a fact that physicists non-rigorously use the language of continuous QFTs to describe the world. The challenge is not "find something else that produces the same predictions within acceptable margins", it's "explain what they're doing". Note that we, historically, didn't start with lattice QFTs and then desparately tried to make them continuous - people started by quantizing a continuous classical field theory and got results, however non-rigorously.

  2. The claim that "QED can be defined in a mathematically rigorously (non-perturbative) way in the discrete spacetime approximation!" is certainly true in some senses, but the connection between lattice theories and continuum theories is much more subtle than it might first appear. For instance, $\phi^4$ theory is trivial in the continuum limit, fermion doubling complicates the treatment of fermions and there is the more general unease that a lattice inherently breaks continuous Lorentz symmetry, which is a feature of reality people would like to see reflected in their physical theory.

  3. "QED is an approximate description of reality." Perhaps, but this is not the only thing a physical theory is. Yes, physical theories are approximate models of reality. But they're often also supposed to be consistent and perfect models of an idealized world. As far as we can tell, space in reality is not discrete, at least not in the "there's a fixed lattice" way. So a "beautiful" physical theory shouldn't be claiming that it is, even if it produces correct approximative predictions.

Now, none of these points are absolutely convincing arguments. You can still say "I don't care, I like lattice theories and don't think we really need anything else". That's fine - it's probably the exact motivation of at least some people who have devoted their life to lattice theory.

  • $\begingroup$ But you can't really say that physicists are using continuum QFTs. They pretend that they're using continuum QFTs, just because the "ideal" reality that continuous QFTs describe is "beautiful" (exact Lorentz symmetry, etc). With a small enough lattice spacing, the implementation of lattice QFTs, in practice, is indistinguishable from continuous QFTs. So the Yang-Mills problem is already solved: The justification for what physicists are doing is that they're really using rigorous lattice QTF's (without any infinite bare paramters). They're just pretending to use continuous QFTs. $\endgroup$
    – Rain Deer
    Apr 23, 2022 at 13:23
  • $\begingroup$ The real world corresponds to neither the "ideal" world that continuous QFTs describe and nor to the one that lattice QFTs describe. The former is beautiful but the latter has solid mathematical ground. But what do we even do with the beauty of the former when it doesn't even correspond to the real world? The GUT will be beautiful, but it will be so without infinities. The lattice QFT is the true approximation of GUT, without any infinities $\endgroup$
    – Rain Deer
    Apr 23, 2022 at 13:26
  • 2
    $\begingroup$ @RainDeer I'm not going to argue with you. I'm just presenting the arguments that other people find convincing enough to want to do continuum QFT. $\endgroup$
    – ACuriousMind
    Apr 23, 2022 at 13:30
  • $\begingroup$ Then I can understand that... Every other approximate theory we've found has been able to deal with a continuous spacetime. Maybe "completing" QFT like this will get us closer to GUT. But we shouldn't rule out the possibility that QFT is perhaps an irreparably discrete mathematical model. That doesn't mean that reality is discrete. After all, QFT is only an approximation of reality for a reason. The reason might just be that it only makes sense as a discrete model. $\endgroup$
    – Rain Deer
    Apr 23, 2022 at 13:50
  • 3
    $\begingroup$ regarding point 2, see the famous no-go theorem by Nielsen and Ninomiya: en.wikipedia.org/wiki/Nielsen%E2%80%93Ninomiya_theorem $\endgroup$
    – Quillo
    Apr 23, 2022 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.