Can someone explain what is non-local structure of field theory? I know you cannot have $\phi(x) \phi(y)$ term in Lagrangian which indicates the non-locality. However, why I cannot have the non-local terms as long as I have causality maintained? In QFT, one should not write an operator like $\phi(x)^2$ which will yield singularities like $\delta (x-x)$ if one does OPE? How should I understand the locality in field theory and OPE sense consistently?
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3 Answers
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The situation is more subtle than suggested by the other two answers as the following example shows.
In $d\ge 2$ dimensions, consider the Euclidean Gaussian field with propagator given in momentum space by $$ \frac{1}{p^{d-2\Delta}} $$ where $\Delta$ is in the interval $\left(\frac{d-2}{2},\frac{d}{2}\right)$. This satisfies the unitarity bound and in fact all the Osterwalder-Schrader axioms. Therefore, by analytic continuation to Minkowski space, this results in a QFT that satisfies all the Gårding-Wightman axioms including locality: $$ [\phi(x),\phi(y)]=0 $$ if $x-y$ is space-like.
On the other hand, the Lagrangian for this model is nonlocal.
answered Mar 3, 2019 at 19:50
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$\begingroup$ Just want to add that this is also called generalised free theory, or mean filled theory in some contexts. It however it's not completely accurate (in the most common sense) to say that it is local. The axiom you refer to is about causality. This theory can be described as the boundary dual of free massive scalar in rigid AdS space, and causality of this theory is due to causality of the bulk theory. However, this theory is not local in the usual sense of this word, e.g. it does not have a stress-energy tensor. $\endgroup$ Apr 4, 2019 at 7:04
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$\begingroup$ Is there really a problem with $\Delta>d/2$? $\endgroup$ Apr 4, 2019 at 7:05
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$\begingroup$ @PeterKravchuk: "usual" in usual sense of locality depends on which community one belongs to. For CFT folks, I know that having a local stress tensor is usually included in the definition of locality. For axiomatic QFT folks, locality is usually just the space-like separation commutation above. It is also the main ingredient is the definition of "local relative to" leading to the notion of Borchers class. Also, I am used to worry about contact terms, so I took $\Delta<d/2$ to make my life easier. But otherwise a larger $\Delta$ is fine. $\endgroup$ Apr 4, 2019 at 13:37
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It is impossible to maintain causality with an operator that is non-local. The reason is very simple:
If you have non-local operators, the equation of motion will include fields at a different spacetime event. There is no way of imposing that information can only be transmitted by the speed of light, because the communication from that other spacetime event to your position is manifestly instantaneous.
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$\begingroup$ So should I say, the localness implies the condition of causality. Or are they equivalent(in the sense of local if and only if causal) in this context? $\endgroup$ May 24, 2014 at 1:24
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1$\begingroup$ If you integrate out a field (eg. the photon) you get a non-local interaction (eg. the Coulomb force) but causality is maintained. $\endgroup$ Mar 3, 2019 at 20:15
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When you introduce $\phi(x) \phi(y)$ for $x \ne y$, you postulate an action at a distance, whichever the interval between said events is: time-like, null, or what. In other words, you admit some essence that isn’t a field, but propagates through the spacetime directly, in a point-to-point fashion. I am not sure you can’t maintain causality is such theory, but it will be not a QFT, but a hybrid theory. It would join two competing paradigms: one of a field, and another of an action at a distance. It might violate the Occam’s razor principle before other problems with it would appear.
