# How to tell local and non-local in QFT?

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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–  Arnold Neumaier Nov 8 '12 at 17:46
This provide additional information contributing to this subject. Thanks very much! –  Simon Jan 19 '13 at 4:03

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.

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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. –  Ron Maimon Nov 8 '12 at 18:32
@RonMaimon:Yes, indeed. Corrected. –  Arnold Neumaier Nov 8 '12 at 18:40
@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. –  Arnold Neumaier Nov 8 '12 at 18:45
Almost everyone says "local action functional" meaning "the action functional corresponding to a local Lagrangian density". It is abuse of language, just semantic. –  drake Nov 8 '12 at 19:04
@ArnoldNeumaier: Ok, ok, I agree with you, and I deleted my answer. –  Ron Maimon Nov 9 '12 at 5:20