There definitely is a classical Lagrangian for those theories. It’s just that quantum effects are very important in the presence of confinement and SSB, not so much when everything’s perturbative.
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Replying to @seanmcarroll
Good stuff! Do you think the classical lagrangian is unique? Or that it depends on how you take the classical limit?
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Replying to @litgenstein @seanmcarroll
In non perturbative QFTs there actually exist observables for which we don't have Lagrangians .. Do you mean unique modulo gauge symmetries & conserved global symmetries? Even the Dirac equation gives the same physics depending on if you use +-m
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Kinda curious what kind of observable we could find in a non-perturbative QFT that doesn't have a Lagrangian. Maybe something from QCD?
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consider the operator e^(-1/φ^2)
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Honestly, I'd have no idea how to treat that operator to get any reasonable number. Therefore, imma take your word for it! Thank you!
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we treat all functions of field operators as their taylor series expansion (hint)
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Well sure, but what's it mean to apply a field with a negative exponent to a field with a positive exponent. Can you treat the operator expansion as inverse matrices? ( I'm still in the habit of expanding fields into c&a operators)
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yea the reason we don't know how to treat those is because we never learn .. we never learn them becuase they diverge about 0 or small couplings and hence don't admit a perturbative expansion ..
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But you can still yield real eigenvalue s using these types of operators or am I pigeonholing the term observerable too hard.
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I honestly don't understand this field all too well .. i understand it when i'm talking to the non perturbative guy here at UCI, but i forget exactly how it's defined non perturbatively
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I hear that, in parastatistics observables have at least two definitions which still trips me up tremendously. Probably just trying to latch onto one definition too hard. Thanks for the info tho it's cool to see operators like that. Things I usually dont think about!
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