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?

  • $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Commented Aug 21, 2022 at 11:14
  • $\begingroup$ @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. $\endgroup$
    – Ryder Rude
    Commented Aug 21, 2022 at 11:17
  • $\begingroup$ @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. $\endgroup$
    – Ryder Rude
    Commented Aug 21, 2022 at 11:20
  • 3
    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Commented Aug 21, 2022 at 11:34

1 Answer 1


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.

  • $\begingroup$ 1. 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? $\endgroup$
    – Ryder Rude
    Commented Aug 21, 2022 at 11:39
  • $\begingroup$ @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 math.stackexchange.com/q/1804255 for example). $\endgroup$ Commented Aug 21, 2022 at 11:47
  • $\begingroup$ @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) $\endgroup$
    – ACuriousMind
    Commented Aug 21, 2022 at 11:50
  • $\begingroup$ 3. That is the Wilsonian aspect of renormalization, see e.g this answer of mine $\endgroup$
    – ACuriousMind
    Commented Aug 21, 2022 at 11:51
  • 1
    $\begingroup$ @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? $\endgroup$ Commented Aug 21, 2022 at 20:37

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