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  Anomalous dimension for bare actions with a standard kinetic term

+ 3 like - 0 dislike
550 views

In this paper on p42, it is explained that when starting with a bare action that contains a standard kinetic term, this kinetic term attains a correction in the course of the RG flow which can be denoted by $1/Z_{\Lambda}$, such that the effective kinetic term at a scale $\Lambda$ can be written as

$$ \frac{1}{2Z_{\Lambda}}\int\frac{d^dp}{(2\pi)^d}\phi(-p,\Lambda)p^2\phi(p,\Lambda) $$

the effective four point interaction would be

$$ \frac{1}{4!Z_{\Lambda}^2} \,\, \int\limits_{p_1,\ldots,p_4}\phi(p_1,\Lambda)\ldots\phi(p_4,\Lambda)\hat{\delta}(p_1+ ... p_4) $$

and analogously for higher order interactions. To get rid of the change done to the action described by the $Z_{\Lambda}$ dependent corrections, on redifine the field $\phi$ such as

$$ \phi \rightarrow \phi\left( 1- \frac{\eta}{2}\frac{\delta\Lambda}{\Lambda}\right) $$

with the anomalous dimension

$$ \eta = \Lambda\frac{d\ln Z_{\Lambda}}{d\Lambda} $$

I dont see why the anomalous dimension has to be defined like this in this case in order for the rescaling of the field leading to a cancellation of the changes to the action due to a renormalization step and like to see an explanation. How generally valid the above expression for the anomalous dimension anyway, is it "only" valid whan starting with bare actions that have a (standard) kinetic term? How can the anomalous dimension be calculated generally? I always thought that roughly speaking, when rescaling after the course graining step, the anomalous dimension "parameterizes" just the deviations from the canonical (some call them engineering) scaling dimensions that have to be applied.

asked May 28, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
retagged Mar 9, 2014

1 Answer

+ 1 like - 0 dislike

I will give some hints: It is anomalous scaling dimension. scaling dimension is defined as $$x \rightarrow \lambda x,\\ \phi(x) \rightarrow \lambda^\Delta \phi(\lambda x) $$ From the foemula (3.45) (maybe it is better to be $\phi(x) \rightarrow \Lambda^\frac{d-2}{2}\phi(\Lambda^{-1 }x)$), we know that the classical dimension of $\phi$ is $\Delta$. Due to $Z_{\Lambda}^{-1/2}$ in front of $\phi$, there will be a correction to classical dimension which gives the anomolous scaling dimension $\delta$ satisfying $$ \Lambda^{\delta}= Z_{\Lambda}^{-1/2}=e^{-\frac{1}{2} \ln Z_\Lambda }$$ So the anomalous scaling dimension is $\delta =-\frac{1}{2}\Lambda \frac{ d\ln Z}{d \Lambda}$.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Craig Thone
answered Nov 25, 2013 by Craig Thone (40 points) [ no revision ]
Nice answer, and +1, unfortunately Dilaton can't accept it yet because of his suspension : (

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Dimensio1n0
Thanks. Hope it be helpful to him.

This post imported from StackExchange Physics at 2014-03-09 16:17 (UCT), posted by SE-user Craig Thone

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