I don't know if this question had been asked before.

By Haag's theorem, interaction picture for quantum field theory cannot exist under some condition. While it is proven that under renormalization, those conditions are not met so interaction picture can still be used.

Now causal perturbation theory is a rigorous way of treating infinities in QFT without renormalization. So in Causal perturbation theory, can interaction picture be applied in this case? I mean when one cannot use renormalization, there is a chance that Haag's theorem is valid in this case.

Edited: My question arise when I read the statement "renormalisation bypasses Haag’s theorem " in Klaczynski's work (https://arxiv.org/abs/1602.00662) on page 71.

The main argument is that the renormalized Lagrangian is not unitary equivalent to original one . I am not familiar with Causal perturbation theory, so the thing I want to know is in the treatment of causal perturbation theory, will unitarily break in some way? Thus making Haag's theorm not hold?

  • 3
    $\begingroup$ I thought causal perturbation theory is exactly the formalization of renormalization and perturbation theory in QFT using the formalism of Epstein and Glaser? $\endgroup$
    – NDewolf
    Commented Feb 17, 2020 at 15:43
  • $\begingroup$ This is an interesting question, although anything that relies on the non-unitarity of regulated theories is likely to be unphysical. It's hard to get around the capsule version of Haag's Theorem: The interaction representation exists only the absence of interactions. $\endgroup$
    – Buzz
    Commented Feb 18, 2020 at 0:48
  • $\begingroup$ The question mixes apples and oranges. Causal perturbation theory is just one among many approaches for doing renormalization in the sense of formal power series. On the other hand the (in my opinion overhyped) Haag's Theorem is nonperturbative. $\endgroup$ Commented Feb 18, 2020 at 14:35
  • $\begingroup$ So there are other approaches treating renormalization in rigorous sense. I would like to know some(all) of of them because I want to know how Haag's theorem interacts with renormalization. $\endgroup$
    – Ken.Wong
    Commented Feb 18, 2020 at 19:55
  • 1
    $\begingroup$ @Ken.Wong: As far as perturbative renormalization, there is BPHZ renormalization, renormalization with flow equations a la Wilson-Polchinski, Epstein-Glaser renormalization and that's about it (for the main ideas, there are of course variants, hybrids, etc.). Haag's Thm just tells you how not to learn renormalization (by following the old interaction picture canonical formalism). In other words, Haag's Thm shows that some naive approach to renormalization is well...naive. $\endgroup$ Commented Feb 21, 2020 at 18:19

1 Answer 1


Haag's theorem is a result about ground states (more generally, pure states that are invariant under the Poincaré group) and their induced GNS representation of canonical (anti)commutation relations.

It says that two invariant states are either equal or disjoint, where disjoint means that there is no unitary map relating the two corresponding irreducible representations of the algebra of canonical spacetime fields.

The physical consequence is that free and interacting theories, having usually distinct ground states (vacua), are represented in inequivalent ways (do not have "the same Hilbert space", stated in a rougher way).

In order to do scattering theory, in particular to write the S-matrix, one has therefore to do something different from what the naïve ideas suggest. In principle, this could be a serious obstacle to a mathematically rigorous approach to perturbative QFT, but actually it has already been overcome long ago by Haag himself. The so-called Haag-Ruelle scattering theory (for a reference, one may consult the books by Reed-Simon, vol 3 or 4 I don't remember, and references therein contained) takes into account this problem successfully: an S-matrix is written, and also LSZ formulas are proved.

Of course in Haag-Ruelle scattering theory it is assumed that both the free and the interacting theories are well-defined in a rigorous way (i.e. satisfying the so-called Wightman axioms, or the Osterwalder-Schrader euclidean version).

I am not familiar with the so-called causal perturbation theory. But to me it seems just a way to try and understand the correlation functions of the interacting theory, knowing which the whole theory can be reconstructed (as shown also by Wightman). This is the aim of constructive QFT that was developed mostly in the 60s-70s by Glimm, Jaffe, Simon and others. Using this approach, some interacting theories could be well-defined, such as $P(\phi)_2, \phi^4_3$ and Yukawa in 3 spacetime dimensions. However, the relevant theories in 4 spacetime dimensions have not yet been rigorously dealt with.

To sum up, once the interacting theory is known rigorously, either by tools of causal perturbation theory or by other means, then applying Haag-Ruelle scattering theory would overcome the mathematical difficulties of having inequivalent representations and allows to construct the S-matrix and derive the LSZ formulas, thus putting on solid grounds the perturbation theory used by physicists. Of course, one may see things turned around, and say that the success of predictions based on perturbation theory assures that there is a rigorous, unambiguous, underlying theory. I am sure that this was and still is the point of view driving people to study the rigorous aspects of QFT.

  • $\begingroup$ Does that mean once the interacting theory is constructed(rigorously), the S-matrix obtained corresponding to the original Lagrangian, not the Lagrangian in interaction picture? $\endgroup$
    – Ken.Wong
    Commented Feb 18, 2020 at 20:12
  • $\begingroup$ @Ken.Wong I am not sure I understand your comment. From a rigorous standpoint, it is always tricky to utilize path integral methods (and thus the lagrangian explicitly). Haag-Ruelle scattering uses Wightman axioms, and operator considerations, taking into account the inequivalence of free and interacting representations of the algebra of observables. $\endgroup$
    – yuggib
    Commented Feb 19, 2020 at 7:07

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