# Why is QFT not even unitary prior to renormalisation?

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?

• 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. Commented Aug 21, 2022 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. Commented Aug 21, 2022 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. Commented Aug 21, 2022 at 11:20
• 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. Commented Aug 21, 2022 at 11:34

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.