4
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • 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. $\endgroup$ – Ron Maimon Nov 8 '12 at 18:32
  • $\begingroup$ @RonMaimon:Yes, indeed. Corrected. $\endgroup$ – Arnold Neumaier Nov 8 '12 at 18:40
  • $\begingroup$ @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. $\endgroup$ – Arnold Neumaier Nov 8 '12 at 18:45
  • 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. $\endgroup$ – Diego Mazón Nov 8 '12 at 19:04
  • 1
    $\begingroup$ @ArnoldNeumaier: Ok, ok, I agree with you, and I deleted my answer. $\endgroup$ – Ron Maimon Nov 9 '12 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.