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Just a simple question for lattice QCD experts, is continuum limit of lattice field theory a relativistic quantum field theory?

Because i heard that lattice QCD is done in imaginary time, producing a ground-state (lowest-energy state) in infinite temporal extend limit. And that two-point correlation functions computed in imaginary time can be analytically continued to real-time. But for massive particles (such as quarks) to be relativistic, their kinetic energies should be equal to or greater than $mc^2$.

Qmechanic
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Peter
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1 Answers1

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This isn't really specific to lattice theory - doing stuff in "imaginary time" means you're doing Euclidean QFT. This isn't so much "relativistic" or "non-relativistic" as it is just a formal computation - no one thinks this Euclidean theory directly describes any physical setting.

However, analytic continuation of the results to "real time" then gives you quantities for a relativistic Minkowskian theory. It is not at all obvious that analytically continuing the results of the a priori unphysical Euclidean theory should have anything to do with what the physical Minkowskian theory does, but this is the miracle of Wick rotation - it turns out that this actually works (although in many cases you need to think a bit carefully about what Euclidean theory corresponds to what Minkowskian theory, see e.g. this question and its answers), and the formal statement of this is called the Osterwalder-Schrader theorem.

ACuriousMind
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