# Super-renormalizable theory and $\beta$-function

There is the statement that $\beta$-function vanishes for super-renormalizable theories. In $D=2$, scalar field has mass dimension zero. So any polynomial interaction is super-renormalizable. Then shouldn't all of them have vanishing $\beta$-functions? But there are many theories (e.g, sine-Gordon) in $2D$ which have nontrivial $\beta$-function. I must be missing something very basic here.

• Where is the statement that $\beta$-functions vanish in superrenormalizable theories? I've not heard such a general claim? – ACuriousMind Sep 10 '16 at 12:33
• First line of Page 770 of this book by Zinn-Justin (4th edition) : amazon.in/Quantum-Critical-Phenomena-International-Monographs/… Statement reads : "The theory is super-renormalizable and thus the $\beta$-function vanishes." – Physics Moron Sep 10 '16 at 12:39
• I don't have access to that book, but I'm pretty sure you've overlooked some crucial additional property in the context. – ACuriousMind Sep 10 '16 at 12:41
• Yes, I understand that I am missing something as I have made it clear in my question itself. Can you tell me some related (possibly $weaker$) statement about $\beta$-function of super-renormalizable theories.. That would help. Thanks! – Physics Moron Sep 10 '16 at 12:47

In a qft, it may be possible to redefine other parameters than coupling to absorb the infinities coming from higher order corrections. In this way, coupling constant does not get renormalized and hence the beta function vanishes. It is a possibility in super-renormalizable theory as fewer diagrams are divergent and the condition may be satisfied.

As an example for the Sine- Gordon model, the action is

$$\mathcal{S}(\theta)=\int d^2x [\frac{1}{2}(\partial_\mu\theta(x))^2-\frac{m^2}{k^2}cosk\theta(x)]$$ Redefining, $\theta=k\theta$ gives $$\mathcal{S}(\theta)=\frac{1}{t}\int d^2x [\frac{1}{2}(\partial_\mu\theta(x))^2-m^2cos\theta(x)]$$ with $t=k^2$. Perturbative expansion in the power of k only modifies the $cos\theta$ term as a self interaction and the divergences arising can be absorbed by a redefinition of m. In this way, coupling constant does not get renormalized and hence beta function vanishes.

This property is not true in general as the vanishing of beta function to all orders implies a finite theory ($\mathcal{N}=4$ SYM) which is a result need to be obtained from a non-perturbative analysis unless it is trivially true as in the former case. Most qft exist perturbatively and the existence of fixed points is not known non-perturbatively. A super-renormalizable theory does not have a vanishing beta function generally as can be seen from $\phi^3$ theory beta function which in d dimension reads,

$\beta(g)=(d/2-3)g-\frac{3g^3}{256\pi^3}+O(g^5)$ ( Collins "Renormalization," eqn. 7.3.7)

$\phi^3$ theory is super-renormalizable for $d<6$ but $\beta$ function is not zero. It however shows asymptotic freedom which is a property of super-renormalizable theories ( I am not aware of the proof though).

• But $\beta$-fn for sine-Gordon theory is $non$-$trivial$ (see e.g, arxiv.org/abs/hep-th/0003258). – Physics Moron Sep 12 '16 at 13:39
• They are using a two parameter expansion with a theory which has a UV cut-off. So it is clear there that coupling constants are getting renormalized and the divergence is absorbed by these coupling. In the example given above divergences has been absorbed in a redefinition of mass term and coupling constant k is not affected and this results in a vanishing beta function. There is no general rule for determining beta function vanishing in a qft but a super-renormalizable theory has a better chance because of the reason in the answer given above. – ved Sep 13 '16 at 6:13