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If field operators are really distributions then surely objects like the commutator, any correlation functions or even the free theory lagrangians are all ill-defined since products of distributions are not well defined?

Qmechanic
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ColourConfined
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    Sure, quantum field theory as practiced by many physicists is mathematically ill-defined. Do you have a more specific question about that? As it stands, it is hard to tell what exactly you expect as an answer here. – ACuriousMind Dec 27 '21 at 21:54
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    Related: https://physics.stackexchange.com/q/337423/50583 (on the general nature of quantum fields mathematically), https://physics.stackexchange.com/q/449676/50583 (on a specific issue with operator-valued distributions vs. the usual approach) – ACuriousMind Dec 27 '21 at 21:55
  • Apart from the links suggested by @ACuriousMind I also suggest you see https://physics.stackexchange.com/q/684960/ and https://physics.stackexchange.com/q/330536/ – Gold Dec 28 '21 at 00:30

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