# QFT and its non-rigorous assumptions

I have been trying to figure out all the non-rigorous assumptions of QFT (as performed in an operator theory) that allow it to function as it currently is. So far, the three big candidates I found are these :

1. The product of distributions is allowed in some sense (this runs afoul of Schwartz's impossibility theorem, although it is possible to define it rigorously but it is rarely done in QFT)
2. The interacting field operator is related to the free field operator by some unitary transformation (Wrong in most cases, as shown by Haag's theorem and other such things)
3. The vacuum of the interacting theory is related to the vacuum of the free theory. In particular, we have that $\langle 0 \vert \Omega \rangle \neq 0$. According to Haag (p. 71), this is also wrong. The usual formula for relating the vacuas fail to converge.

Are those the only common wrong assumptions? I'm also thinking that derivatives of the field operators may be ill defined as derivatives on operators are usually well defined for bounded operators, but I'm not quite sure.

• This feels like a big-list question where each person can provide one item. – AHusain May 4 '17 at 19:16
• I sense negative energy. How would the detractors of QFT explain away the 1965, 1999, 2004, 2013, and, arguably, partially, 1963, 1979, 1982 Nobel prizes? – Cosmas Zachos May 4 '17 at 19:24
• Obligatory reminder that comments are for improving and clarifying the question, not to give partial answers or expressing how much you like/hate this question. If you want to discuss this question more freely, please take it to chat. – ACuriousMind May 4 '17 at 19:28
• I am asking what the assumptions of QFT are, beyond the well known ones. All things related to Hilbert spaces, Poincaré invariance, gauge theory, commutation relations, etc, are well known, but there are a few hidden assumptions that many textbooks do not point out explicitely, since they are not 100% rigorous. How many such assumptions form the basis of most QFT (for instance, the QFT of the standard model)? – Slereah May 4 '17 at 22:35
• I don't think "lies" would be an appropriate term in this context. I have essentially 0 knowledge of QFT; but I highly doubt that the theory is based off of lies. Assumptions, sure, invalid assumptions, perhaps, but lies; I don't think that is the proper term for it. As far as I'm aware the intent of the theory is not to provide incorrect information. – JMac May 5 '17 at 13:41

## 1 Answer

I'm not sure there is a definite answer, since nobody seems to agree as to what QFT really is (at least so says Nate Seiberg in his 2015 Breakthrough Prize talk), but the problems you mention are well known and accounted for.

Quantum Mechanics gets loads of issues with infinite degrees of freedom. So in reality what we actually do is to regularize the theory, both infrared (e.g. putting in finite volume) and ultraviolet (e.g. putting in a lattice). When properly regularized QFT has finite degrees of freedom, and therefore all Haag's objections go away. You have a perfectly defined interaction picture and a mapping between free and interacting ground states, so everything goes smoothly.

This has the drawback of breaking Lorentz Symmetry, but the hope is that once we renormalize, by taking appropriate limits, we are left with sensible answers and recover Poincaré invariance.

Since the couplings in QFT are very singular and ill-defined we should have done the regularization in the first place just to know what the hamiltonian is, so this is a key part of QFT.

The problem with Haag's approach, and other axiomatic ones, is that they work directly with already renormalized fields, and therefore run on multiple problems. This is why more recent textbooks, starting with Weinberg (I think), emphasize that regularization and the renormalization group are key concepts one cannot do QFT without, even in the absence of interactions.