Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In Jaume Gomis's lecture 5 on CFT at Perimeter Institute, he says (at 27:40 minute mark) that the beta function, classically, of the $m^2$ parameter in massive $\lambda \phi^4$ theory is

$$\beta(m^2) = -2m^2\,.$$

The intuitive reasoning he gives is that since the dimension of $m^2$ is 2, then the beta function should be -2 times m-squared.

I always thought that the beta functions for parameters of the Lagrangian only start at one-loop and are classically zero. What is the rigorous definition of a beta function (including the classical part)?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

The theory is (even classically) not scale invariant. Just by dimensional analysis, you can note that the scalar field has scaling dimension 1, and the mass (as the name suggests) must also have a scaling dimension of 1. So $m^2$ has a ascaling dimension of 2, which suggests the RG equation which you've written in the question. That essentially says that the $m^2$ parameter decreases exponentially, but with scaling dimension 2 as you go to the UV.

Only when you have couplings which are classically dimensionless, do the RG equations have no tree level contributions. eg: coefficients of dimension 4 operators: like $\lambda$ in $\lambda \phi^4$ or $g$ in $g \phi \bar{\psi} \psi $

FYI: Tree level is equivalent to classical field theory and loop contributions are like quantum corrections. You can see that by restoring powers of $\hbar$ in the path integral.

share|improve this answer
    
Thanks! but I already understand it qualitatively. I want to know the formal definition: why the minus sign? how to connect it to the 't Hooft parameter $\mu$. And it's relation to the 'anomalous dimension' $\gamma(m^2)$... Thanks! –  QuantumDot May 17 '13 at 14:42
    
By the `t Hooft parameter $\mu$, do you mean the anomalous scale you get while doing dim reg? That is a quantum phenomenon (since you get it from dim reg on loop integrals). In the classical theory, $m^2$ does not have anomalous scaling. Quantum (loop) corrections to the scaling dimension are what we call anomalous. The minus sign is kinda by definition, because something with a positive mass dimension increases as you flow to the IR and decreases as you flow to the UV. Since increasing $\Lambda$ corresponds to going towards the UV, we need a negative sign. –  Siva May 17 '13 at 16:17
    
Yes $\mu$ is dim-reg parameter. So, if I understand you correctly? Is it true, in all cases the classical beta function for a parameter, $x$, of mass-dimension $[x]$ is ALWAYS $\beta(x) = -[x]\times x$? Thank you again! So how do I write out the full thing with quantum corrections (involving the anomalous part $\gamma$)? Is it $\beta(x) = -[x]\times x + \gamma(x)$? Is that the exact correct formula? –  QuantumDot May 18 '13 at 2:46
    
Yes, the loop-corrected RGE will become (small correction to what you wrote) $\beta(x) = - ([x]+\gamma(x)) x$. $\gamma(x)$ is called the anomalous scaling dimension, since it's a correction to $[x]$. Here, the RHS shows only the leading order in $x$; in general there could also be higher order terms. –  Siva May 18 '13 at 3:16
    
Excellent! Thanks very much –  QuantumDot May 18 '13 at 3:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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