Interesting Q someone asked: quantum electrodynamics has an obvious classical theory it corresponds to, but quantum chromodynamics and electroweak theory don’t—why’s that? Possible answer: there’s electromagnetism in daily life, but no strong or electroweak force
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Replying to @litgenstein
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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