AQFT and the Standard Model

The German physicist Rudolf Haag presented a new approach to QFT that centralizes the role of an algebra of observables in his book "Local Quantum Physics". The mathematical objects known as operator algebras (C* and W*) seem to have a lot of importance from this perspective.

I have heard opinions for and against algebraic quantum field theory. A common argument for AQFT that I hear often is that it places QFT in a formal mathematical universe, but I don't really know what that even means. There are other approaches to quantum mechanics (using techniques from microlocal analysis, for example) that emphasize rigor, but do not necessarily deal with an algebra of observables.

I have heard more opinions rather than persuasive arguments against AQFT. The one argument that I am aware of is that AQFT does not deal very well with the Standard Model, which is one of the big empirical success of particle physics and plays a central role in mainstream (Lagrangian) Quantum Field Theory.

I am wondering how AQFT's treatment of the Standard Model (among other things) creates a problem for physicists.

• Hi Rohan, welcome to Physics Stack Exchange! Right now your question isn't really a question - you're looking for a list of arguments for and against, but that's more of a discussion topic, and not what this site is suited for. (See the FAQ) If you can formulate a more focused question that invites a specific answer, you can edit this post to reflect that and I'll reopen it. (I hope you do, since I'm pretty sure there's at least one good question in there somewhere.) – David Z Sep 7 '11 at 1:27
• Hi David, thanks for your comment. I guess my last sentence isn't very precise. I am really asking how AQFT's treatment of the standard model creates a problem for physicists. I hope (to your satisfaction), this is a bit more specific. – r.g. Sep 7 '11 at 1:37
• Sure, that's an improvement. I'll open this up again but I imagine you'll continue to receive feedback on it. By the way, I'd also suggest posting your clarification question about C* and W* algebras as a separate question. – David Z Sep 7 '11 at 1:52
• Thanks David. I'll delete that part to make my question a bit more focused. – r.g. Sep 7 '11 at 1:54
• @Rohan: I think you've misunderstood the argument that says AQFT does not play well with the Standard Model. AQFT is all about finding a "proper" mathematical model which satisfy some axioms relating to an interacting relativistic QFT --- it's a failure/stumbling block of the AQFT program that it cannot find such a thing in spacetime dimensions of 4 or greater. Therefore, AQFT has nothing to say about the Standard Model, because it doesn't know what the SM is! Jaffe has a very nice review of the whole endeavour: arthurjaffe.com/Assets/pdf/CQFT.pdf – genneth Sep 7 '11 at 9:09

The thesis

Epstein-Glaser Renormalization: Finite Renormalizations, the S-Matrix of Φ4 Theory and the Action Principle http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.7.1837 by Gudrun Pinter

discusses the standard model in the framework of AQFT. Of course, one gets only perturbative results, as there are not yet any constructions at all for interacting AQFT models in 4 spacetime dimensions. But apart from that, one can do everything rigorously that is done on a more cavalier basis in standard books on the standard model.

The Standard Model has not yet been cast in the language of AQFT, as far as I know. (In general, I don't think the Yang-Mills theory is put into the AQFT language yet). Therefore your question is rather vacuous at this stage. You need to ask it again later when the SM is properly formulated in (a certain extension of) AQFT.

In any case, as far as the practical calculation of physical quantities go, the final result should be independent of the formulation. For example, the canonical formalism and the path-integral formalism both give the same Feynman diagram and give the same numerical answer. The AQFT, once developed sufficiently to be able to treat the Standard Model, should give the same answer too.

So, the "new" formulation, when it becomes available, won't drastically change what physicists think. Of course a new perspective sometimes help to broaden your perspective, but that won't change the experimental prediction.

AQFT has advantages and disadvantages. First of all, it is a mathematically rigorous framework, that allows one to derive properties of, say, quantum fields from first principles. One can show for example that in 4d spacetime charges have to obey either bose or fermi (para)statistics, while on the other hand in low dimensional systems one can have things such as anyons. I would like to stress that this all follows from the basic assumptions and is not put in by hand.

From a physics viewpoint the largest criticism is probably that it is extremely difficult (in particular in 4d spacetime) to construct anything but free field theories. In any case, the standard mdoel is probably far out of reach at the moment. As far as I know, AQFT has not produced any measurable predictions so far.

Several points:

1) Algebraic QFT is not a new idea. Haag & collaborators have been thinking about these sorts of formalisms since the 60s, at least.

2) The basic idea of AQFT is very simple and certainly true in the Standard Model: the observables we measure are associated to regions in spacetime. And physics is at least approximately local, so the algebra of observables in a given region should be generated by observables localized in smaller subregions.

3) However, AQFT almost certainly does not describe the Standard Model. The Standard Model is an intrinsically effective field theory. All available evidence suggests that the Higgs field has a Landau pole, which would forbid a continuum limit. And AQFT is an attempt to axiomatize the idealization of continuum limits. It's not really the right language.

4) The last time I checked -- and I may be out of date -- the AQFT formalism struggled with gauge theories, where the algebra of observables is the gauge invariant part of a larger algebra generated by local fields like the vector potential $A_\mu$ which are not themselves gauge-invariant.