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Why terms non-analytical dependent on momenta in the effective action (in momentum space) are non-local? How to see this directly?

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Local terms always have fields/ operators at the same spacetime point, i.e.

$S = m^2\int d^4x \phi(x) \phi(x) = m^2\int d^4x \int d^4 y \delta^{(4)}(x-y) \phi(x) \phi(y)$

is local to-where-as

$S = m^2\int d^4x \int d^4 y f(|x-y|) \phi(x) \phi(y)$

is not local. Now we typically perform calculations in momentum space so a common term that might arise would be something like

$S = \int d^4p \hspace{1mm} p^2 \phi(p) \phi(-p) = \int d^4x \partial_\mu \phi(x) \partial^\mu \phi(x)$

which is local. Naively we might pick up other terms like

$S = \int d^4p \hspace{1mm} \frac{1}{p^n} \phi(p) \phi(-p) \sim \int d^4 x d^4 y (x-y)^{n-4} \phi(x) \phi(y)$

which is not local and where $0 < Re(n)<d$ where d is space-time dimension. Another possible non-local term that could conceivably arise that we wouldn't want is

$\int d^4 p \hspace{1mm} \log p^2 \hspace{1mm} \phi (p) \phi(-p) $.

and in fact terms of this form do arise at intermediate stages of calculations but are subtracted off by similarly non-local counter terms (see Peskin pg 335 - 338). I couldn't find the result for this integral in 4-dimensions but in 2 dimensions it's certainly non-local:

$\int d^2 p \hspace{1mm} \log \frac{p^2}{\Lambda^2} \hspace{1mm} \phi (p) \phi(-p) \sim \Lambda^4 \int d^2 x d^2 y \frac{1}{x^2+y^2} \phi (x) \phi(y)$

In general really all we ever want to see in our effective action in position space is derivatives with a delta function and the only thing that will give us this from momentum space is:

$ \int d^4 p p^n \phi (p) \phi(-p) \sim \int d^4 x d^4 y \delta^{(n)}(x-y) \phi(x) \phi(y) \rightarrow \sim \int d^4 x d^4 y \delta (x-y) \partial_x^n \phi(x) \phi(y) = \int d^4 x \partial_x^n \phi(x) \phi(x) $

where I have integrated by parts and n is taken to be a positive integer. Throughout I have been sloppy with numerical pre-factors, signs, dimensions and notation and lack any rigor (and grammar) what-so-ever, but hopefully this gets the point across.

To summarize: an effective action that is non-analytical in the momentum is an action that can't be written as a bunch of operators at the same space-time point or rather the Fourier transform from momentum space to position space isn't just delta function (plus possible derivatives).

A caveat: if you are working with a non-relativistic theory/action none of this holds since then its admissible to have action-at-a-distance etc. There is probably a better source for this but I have at least seen it in pg10 of http://arxiv.org/pdf/0905.4752v2.pdf .

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Though the fourier transform of non-analytic terms aren't delta functions in position space, it can be derivative of delta function. It is not obvious to tell whether such terms are local or not. –  Craig Thone Jun 24 '12 at 4:19
    
The Fourier transform of 1/p^2 in 4d is 1/x^2 up to singularities and constant. You can see it by dimensional analysis or by Gauss's law. –  Ron Maimon Jun 24 '12 at 7:43
    
EDIT: I cleaned this up a lot. Sorry for the prior sloppiness, and for any sloppiness that remains. –  DJBunk Jun 24 '12 at 17:29
    
Thanks for your answer. I still have some problems. I do not know how to calculate this integral$$\int d^4p \frac{1}{p^n}\phi(p) \phi(-p) \sim \int d^4x \int d^4y (x-y)^{n-1} \phi(x)\phi(y)~~~~$$ directly ? On the other hand, just as Ron said, dimensional analysis does not match. –  Craig Thone Jun 25 '12 at 2:38
    
Craig and Ron: Thank you very much for pointing out my errors, I edited some more to correct them. Craig: The power-law integral I got from wikipedia: en.wikipedia.org/wiki/Fourier_transform and the Log integral I got from mathematica –  DJBunk Jun 25 '12 at 14:17

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