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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