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Mar 3 at 21:06 comment added Pedro Lauridsen Ribeiro 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.
Mar 3 at 21:02 comment added Pedro Lauridsen Ribeiro 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.
Mar 3 at 10:51 comment added Peter Kravchuk 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.
Mar 3 at 9:00 vote accept XL _At_Here_There
Mar 3 at 3:09 comment added Pedro Lauridsen Ribeiro 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.
Mar 3 at 2:20 comment added Peter Kravchuk 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)
Mar 2 at 12:02 review Suggested edits
Mar 2 at 13:41
Mar 2 at 3:50 vote accept XL _At_Here_There
Mar 3 at 1:50
Mar 2 at 3:39 history answered Pedro Lauridsen Ribeiro CC BY-SA 4.0