The Wikipedia entry on Hilbert's sixth problem about QFT description is “Since the 1960s, following the work of Arthur Wightman and Rudolf Haag, modern quantum field theory can also be considered close to an axiomatic description.” So it is not completely axiomized.

Is the reason related to renormalization, or any other problem in current QFT?

  • $\begingroup$ Do we really know at this point that QFT is ''close to an axiomatic description''? The Wikipedia entry also says that ''Dirac formulated quantum mechanics in a way that is close to an axiomatic system'', but this seems debatable. $\endgroup$ Mar 2 at 22:11
  • $\begingroup$ @HollisWilliams, maybe only the part of interacting is not axiomized, so the wikipedia entry says so about the axiomization $\endgroup$ Mar 3 at 1:51
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    $\begingroup$ ''Maybe only the part of interacting is not axiomized'' Isn't that basically the only interesting part on the physics side? Only axiomatizing free field theory doesn't say much about whether it can be done for interacting fields. $\endgroup$ Mar 3 at 14:51

1 Answer 1


The reason is that there is no mathematically rigorous construction of any interacting quantum field theory in four space-time dimensions to this date. Because of that, one has not been able so far to verify whether or not such a theory satisfies either the Garding-Wightman or the Haag-Kastler axioms. The only 4-dimensional QFT models known to satisfy these axiom schemes are non-interacting and even the academic $\lambda\phi^4$ QFT model has recently been shown by Aizenman and Duminil-Copin to be non-perturbatively trivial (that is, non-perturbative renormalization makes the model non-interacting).

Perturbative QFT models cannot satisfy either axiom scheme because the perturbative series for QFT remains formal in most models even after perturbative renormalization. More precisely, we know for physically relevant QFT models (e.g. QED) that the renormalized perturbative series is divergent. There is partial evidence that, for some (not all) such models, this series may be asymptotic, but that is all. All that leaving aside the fact that only the short-distance (i.e. "ultraviolet") renormalization procedure for perturbative QFT is mathematically well understood. The long-distance (i.e. "infrared") divergences in the perturbative S-matrix affect the very definition of scattering in QFT when there are massless particles and there is no complete, rigorous solution to this problem yet.

To sum up, QFT could be said "not to be fully axiomatized" simply because we still do not have enough mathematical control over relevant QFT models in order to check any of the proposed axiom schemes for QFT. My MO answer goes a bit deeper on the meaning and relevance of these schemes in such a scenario.

  • $\begingroup$ What (and what models) do you have in mind when you say that the perturbative series is not asymptotic? (I'm referring to the "not all" parenthetical remark) $\endgroup$ Mar 3 at 2:20
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    $\begingroup$ For instance, the perturbative series for quantum chromodynamics in the low-energy regime (e.g. nuclear physics) doesn't seem to be asymptotic at all, due to quark confinement. Color charged states simply don't appear in scattering at that regime. $\endgroup$ Mar 3 at 3:09
  • $\begingroup$ Ah, I see. But then you're talking about an order of limits issue. If we take a fixed Schwinger function with fixed arguments, and look at it's perturbative expansion in a fixed scheme, then I do think it will be asymptotic even in QCD. If you ask about more general quantities such as low-energy spectrum, it is indeed a rather special situation to be able to compute them perturbatively. $\endgroup$ Mar 3 at 10:51
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    $\begingroup$ The point is what one is able to compute (or prove the existence thereof) for a given QFT model. Lattice QCD calculations assume both finite volume (i.e. infrared) and short-distance (i.e. ultraviolet) cutoffs, so there's no renormalization involved. We do know by the work by Balaban and Magnen, Rivasseau and Sénéor using weak-* compactness sorcery that the continuum limit of pure Yang-Mills Schwinger functions exist for finite volume, but we don't know anything about uniqueness, let alone its thermodynamical limit. For full, matter-coupled QCD we don't even have that. $\endgroup$ Mar 3 at 21:02
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    $\begingroup$ The perturbative QCD S-matrix (which, as you correctly pointed out, involves an infinite-time limit as well), on its turn, is physically reliable only in the high-energy limit, thanks to asymptotic freedom. $\endgroup$ Mar 3 at 21:06

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