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The Hamiltonian is Hermitian. That should've been enough to make it unitary. But infinite amplitudes mean it's not even unitary. One could say that this is because we're dealing with a crazy Hilbert space having a continuum of variables. But the theory isn't unitary even after discretization! The amplitudes after regularization/discretization become finite but are still larger than 1. So again, why isn't it unitary?

Qmechanic
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Ryder Rude
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  • I have no idea what you're talking about here. Are you perhaps looking at probability densities being greater than 1 and mistaking them for probabilities? Do you have any reference for this claim that "QFT is not unitary"? In any case, please be more explicit what you actually want to know here. – ACuriousMind Aug 21 '22 at 11:14
  • @ACuriousMind I'm talking about the infinite probabilities that we get prior to renormalisation. How can it not be unitary despite the Hamiltonian being Hermitian? At least discretization of fields should've made it unitary. But the amplitudes are larger than 1 even after that. – Ryder Rude Aug 21 '22 at 11:17
  • @ACuriousMind I'm not claiming that "Qft is not unitary". I'm only asking why it is not unitary prior to renormalisation, when exponentials of Hermitian matrices are guaranteed to be unitary. – Ryder Rude Aug 21 '22 at 11:20
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    I see. I would suggest making that clearer in the question - the word "renormalization" doesn't even appear in the body of the question, and it's a bit stream-of-consciousness how we jump from a sentence about a "crazy Hilbert space" to discretization. I feel like there's a lot of fragments of thoughts in this question that aren't really spelled out - please remember that other people don't have the same context as you do when you ask a question, and that we can only understand the parts of your thought process that you actually write down. – ACuriousMind Aug 21 '22 at 11:34

1 Answers1

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The core problem that arises in QFT is that, if you want to be rigorous, expressions like $\phi(x)^4$ do not actually exist when the $\phi(x)$ is a quantum field, i.e. an operator-valued distribution, because there is no general unique theory of multiplying distributions. The UV divergences that we usually have to renormalize away can in some formulations (causal perturbation theory, also sometimes called Epstein-Glaser renormalization) be directly related to a choice of how to define the point-wise product of such distributions.

So an interacting Hamiltonian of e.g. $\phi^4$-theory isn't "Hermitian", it's nonsense, $+\lambda\phi^4$ isn't a description for a self-adjoint operator, it's a chiffre for $\phi^4$-theory and the usual ways (including renormalization!) to extract QFT predictions from such a formal Hamiltonian that isn't actually an operator on anything in a mathematically rigorous sense.

The way physicists often work with QFT is contradictory - on the one hand we pretend this is mostly like quantum mechanics, and on the other hand we must avoid fallacious statements like "the Hamiltonian is self-adjoint so time evolution cannot possibly diverge" - and that is probably a large part of why it is so hard to put QFT on mathematically rigorous footing.

ACuriousMind
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  • Why isn't the product defined? 2. Why isn't QFT unitary even after regularization? Regularization should've made it just another quantum theory because of finite number of variables. 3. I read that renormalisation is a necessity even without infinities, because interactions change effective mass and charge. So wouldn't we have needed renormalisation even if it was already unitary, i.e. the product was well-defined?
    • – Ryder Rude Aug 21 '22 at 11:39
  • @RyderRude regarding the first question: Quantum fields need to be thought of as Operator valued distributions. That means e.g. the field operator becomes an operator only after integrating over a region in space. Now the problem is that there is no canonical way to define the product of a distribution (see https://math.stackexchange.com/q/1804255/ for example). – Quantumwhisp Aug 21 '22 at 11:47
  • @RyderRude 1. That's just a mathematical fact about the way distributions work, cf. e.g. nlab. This is the starting point for causal perturbation theory 2. What exactly do you mean by "regularization"? If you mean "putting the theory on a finite lattice", then the relationship between lattice QFT and continuum QFT is rather subtle and you should ask a specific question about that. (In particular you should exhibit a specific Hamiltonian that you think is actually self-adjoint in a rigorous sense and yet produces non-unitary evolution) – ACuriousMind Aug 21 '22 at 11:50
  • That is the Wilsonian aspect of renormalization, see e.g this answer of mine
    • – ACuriousMind Aug 21 '22 at 11:51
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    @ACuriousMind: It sounds then to me it's not that (R)QFT is "hard to put on firm ground", because there do exist ways to define distributive products, it's just that each product defines a different interacting (R)QFT, and what we don't have is some way to choose which of those is the best one. But that sounds like an experimental problem, not a math problem - so what does it mean to say it is thus "hard to make mathematically rigorous"? Can we use experimental data to constrain the form of the Hamiltonian enough, or not? – The_Sympathizer Aug 21 '22 at 20:37
  • That is, what criteria would need to be met for a claimed "mathematically rigorous solution" to be accepted as "the" solution? I.e. how would we know it was the solution and not just some solution? Also, the linked answer says "the product of distributions is not generally a distribution". What if we try to just run with that - that is, that $\phi^4$ is not an OVD, but something more general than an OVD, just as a distribution is more general than a function? – The_Sympathizer Aug 21 '22 at 20:45
  • @The_Sympathizer 1. I'm not saying this is the one issue that means QFT is not mathematically rigorous. It is one of the issues that makes QFT as practiced by most physicists non-rigorous. There are other issues blocking a full rigorous formulation (but what exactly these issues are depends on your specific attempt at formalization). 2. I'm not sure where you took the idea of "'the' solution" from - I never use the word "solution". You are correct this just means that there are many different versions of the same QFT for different choices for the point-wise product - – ACuriousMind Aug 21 '22 at 20:53
  • related by renormalization flow, this is completely compatible with the standard story of renormalization. 3. If you want to say the product is something "more general than a distribution", then you would have to provide a mathematical framework for how to work with that more general thing. Just saying "pretend this thing exists" doesn't mean anything - you have to define the properties of such a generalized object. – ACuriousMind Aug 21 '22 at 20:53
  • @ACuriousMind : So then what exactly would it take to "resolve" the issue? What would count as a "resolution"? – The_Sympathizer Aug 21 '22 at 23:28
  • @The_Sympathizer Resolve what issue? – ACuriousMind Aug 21 '22 at 23:31
  • @AcuriousMind: The issue of (R)QFT not being on mathematically rigorous footing. – The_Sympathizer Aug 21 '22 at 23:36
  • @The_Sympathizer As I said above, "what exactly these issues are depends on your specific attempt at formalization". If you want to discuss specific aspects of rigorous QFT, you should probably ask a separate question about that. We already have e.g. https://physics.stackexchange.com/q/6530/50583, https://mathoverflow.net/q/393826, https://physics.stackexchange.com/q/673224/50583, https://physics.stackexchange.com/q/16142/50583 – ACuriousMind Aug 21 '22 at 23:47