I'm taking QFT course in this term. I'm quite curious that in QFT by which part of the mathematical expression can we tell a quantity or a theory is local or non-local?
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3$\begingroup$ related: physics.stackexchange.com/q/13624/7924 $\endgroup$– Arnold NeumaierCommented Nov 8, 2012 at 17:46
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$\begingroup$ This provide additional information contributing to this subject. Thanks very much! $\endgroup$– SimonCommented Jan 19, 2013 at 4:03
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1 Answer
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A quantity is local if it is a finite linear combination $\sum_k g_k P_k(x)~~$ of products $P_k(x)$ (or other pointwise functions, such as $\sin \Phi(x)~$ for sine-Gordon theory) of field operators or their derivatives at the same point $x$.
A quantum field theory is local if its classical Lagrangian density is local. (By abuse of terminology, an action or a Lagrangian may also be called local if the corresponding Lagrangian density is local.)
Since in QFT fields are only operator-valued distributions, a local quantum field product is not well-defined without a renormalization prescription, which involves an appropriate limit of nonlocal approximations. In 1+1D, normal ordering is sufficient to renormalize the field products, while in 3D and 4D more complicated (mass and wave function) renormalizations are needed to make sense of these products.
answered Nov 8, 2012 at 17:51
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1$\begingroup$ When you say "it's Lagrangian is local", you mean "it's Lagrangian density is local". In your definition, the integral of a local thing is not local. As I said in my answer, it is standard abuse of language to call the integral of a local term local. $\endgroup$ Commented Nov 8, 2012 at 18:32
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$\begingroup$ @RonMaimon:Yes, indeed. Corrected. $\endgroup$ Commented Nov 8, 2012 at 18:40
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$\begingroup$ @RonMaimon: even with the traditional abuse, not any sum or integral of a local term deserves the label local. The only abuse allowed is to call the Lagrangian local when in fact only the Lagrangian density is local. $\endgroup$ Commented Nov 8, 2012 at 18:45
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1$\begingroup$ Almost everyone says "local action functional" meaning "the action functional corresponding to a local Lagrangian density". It is abuse of language, just semantic. $\endgroup$ Commented Nov 8, 2012 at 19:04
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1$\begingroup$ @ArnoldNeumaier: Ok, ok, I agree with you, and I deleted my answer. $\endgroup$ Commented Nov 9, 2012 at 5:20