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Epstein-Glaser causal perturbation theory

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Why does causal perturbation theory in the sense of Epstein Glaser fall under algebraic QFT rather than heuristic QFT in renormalization?

This post imported from StackExchange Physics at 2015-05-14 11:13 (UTC), posted by SE-user user41508
asked May 12, 2015 in Theoretical Physics by user41508 (15 points) [ no revision ]
retagged May 14, 2015
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Related: physics.stackexchange.com/q/6530/2451 and links therein.

This post imported from StackExchange Physics at 2015-05-14 11:13 (UTC), posted by SE-user Qmechanic

Such an approach uses heavily the results of the standard QED, but camouflaged.

 @VladimirKalitvianski: Causal perturbation theory is not just about QED but for all kinds of relativistic QFTs. See, e.g., http://arxiv.org/abs/hep-th/0001129

It is algebraic QFT since it is mathematically rigorous in the sense that it spells out the exact assumptions needed and the exact statements proved, and contains nowhere arguments that a mathematician would not approve of. Standard perturbative renormalization theory done in this way (e.g., in Salmhofer,s book) also counts to algebraic QFT. 

In contrast, heuristic QFT contains claims and arguments without bothering about figuring out in which framework the claims and arguments are rigorously valid.

 @VladimirKalitvianski: You are almost the only one handicapped by it. Causal perturbation theory is perfectly rigorous and can recover all results of renormalized perturbation theory that can be successfully compared with experiment. It does not claim to have solved the IR problem in QFT or the question of convergence of the perturbation series, so you shouldn't judge it be these standards (where all 4D QFTs are handicapped, your proposed alternative even more so).

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@sebastian for correct attribution your question has been merged into this duplicate imported from SE. If it was you who asked this question on SE, feel free to reclaim your imported account.

Let me consider your passage as a mathematically rigorous proof of hopelessness of my alternative approach.

1 Answer

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At the level of IR-finite, individual Feynman graphs, indeed EG is just as mathematically rigorous a renormalization scheme as the more usual BPH and BPHZ schemes. We are still doing formal perturbation theory (i.e. formal power series in the coupling constant) in either case. The reason why EG is associated with algebraic QFT is a matter of concept rather than rigor.

The main characteristic of the EG scheme is that it is formulated in position space, rather than in momentum space as BPH and BPHZ are. This allows a clear separation of the IR problem from the UV problem by means of adiabatic switching of the interaction by a smooth cutoff function with compact support in space-time. Doing this in BPH and BPHZ is very messy. This separation is important because the UV problem concerns the local observables of the theory, whereas the IR problem concerns the latter´s state space - the solution of the UV problem is IR-cutoff-independent, and the EG framework makes this very precise.

The separation of the UV problem from the IR problem by means of adiabatic switching allows one to define a net of local observables for each order of perturbation theory, by means of families of Bogolyubov´s S-matrices parametrized by all cutoff functions - the definition is such that the local algebras become cutoff-independent. It is in this sense that EG is nowadays more commonly associated to algebraic QFT - in fact, the branch of AQFT dealing with the perturbative construction of nets of local observables is usually called perturbative algebraic QFT since we are still dealing with formal power series in the coupling constant, and the EG scheme is simply the one that fits most naturally into the philosophy of (P)AQFT. One can go even further and write the EG scheme no longer in terms of individual graphs, but as a renormalization scheme for a formal deformation of the algebraic structure of the free net by the interaction.

At each order of perturbation theory, the net of observables can be shown to obey the Haag-Kastler axioms, apart from the vacuum axiom. The latter, of course, does not hold in an unmodified form since the cutoff interaction is not translation invariant, but this again is a property of the state space of the theory. Solving the IR problem amounts to building it for each order of perturbation theory by carefully removing the cutoff. Of course, here the devil is in the details, and the situation is no better than in standard QFT. In any case, the IR-cutoff theory is well-defined as formal power series for each choice of cutoff function.

answered May 13, 2015 by Pedro Lauridsen Ribeiro (580 points) [ revision history ]
edited May 13, 2015 by Pedro Lauridsen Ribeiro

Of course, here the devil is in the details, and the situation is no better than in standard QFT. In any case, the IR-cutoff theory is well-defined as formal power series for each choice of cutoff function.

Indeed, when I was a student, my professor told us that massless particles are radiated even at adiabatic switching since there are always modes for which the switching is "too fast" to not excite them. Introducing IR cutoff (finite mass, for example), formally helps, but, in my opinion, it does not solve the problem of better QFT formulation. Factually, the meaningful (non zero) result is inclusive where the interaction with soft modes is included rather than cut off. In the present formulation, unnecessary self-action (self-induction) is included and then removed with renormalization, and the necessary (strong and permanent) interaction with soft modes is treated perturbatively which later needs summation to all orders to obtain a meaningful result.

Putting aside the conceptual appeal of the EG scheme in the context of algebraic QFT (which actually was the main focus of my answer, but anyway), you are right. This formalism has the merit of clearly isolating the IR problem, but this does not make it that much easier to solve with our current knowledge. In QED, for instance, one still needs zero-recoil approximations (Bloch-Nordsieck, etc.) to deal with the IR problem, which involves partial resummation of the perturbative series, as you said. On the other hand, having a hard problem isolated from others which are not really relevant and bring unncecessary complications is always a good starting point. More importantly, the conceptual clarity gained by this separation makes the problem amenable to other lines of attack in the future.

@PedroLauridsenRibeiro I understand your line of reasoning, but I do not share your optimism about "conceptual clarity". As I said in another comment, the current QFT formulation is "handicapped". You may read my publications to see my points and really promising conceptual directions.

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