# Understanding renormalizability

I want to understand when a given theory is renormalizable and how to find renormalizable theories for different dimensions (the latter will become clearer later on).

To do so, we work through an example: the following real scalar field theory in $$d$$ spacetime dimensions

$$\begin{equation*} \mathcal{L} = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac 1 2 m^2 \phi^2 - \frac{\lambda_3}{3!} \phi^3 - \frac{\lambda_4}{4!} \phi^4 - \frac{\lambda_5}{5!} \phi^5 - \frac{\lambda_6}{6!} \phi^6 \end{equation*}$$

Let us work with $$3 \leq n \leq 6$$. We first determine the coupling constants

\begin{align*} [\mathcal{L}] = d, \quad [\partial_{\mu}]=1, \quad [m] = 1,\quad \Rightarrow \quad& [\phi^n] = \frac 1 2 n(d-2), \quad [\lambda_n] = d- \frac 1 2 n(d-2) \\ &[\phi] = \frac 1 2 (d-2), \quad [\phi^2] = d-2 \end{align*}

I learned that a theory with $$[\lambda_n] < 0$$ is nonrenormalizable (from M. Srednicki's beautiful book, chapter 18). Hence, from $$[\lambda_n] = d- \frac 1 2 n(d-2)$$, we get the following condition

$$\begin{equation} [\lambda_n]<0 \iff n> \frac{2d}{d-2} \tag{*} \end{equation}$$

From $$(*)$$ we see that the renormalizability of the theory depends on the powers of $$\phi$$ and the dimension $$d$$. In particular, we see that our theory is renormalizable if we work with $$d = 1$$ and $$d=3$$. For $$d > 3$$ our theory becomes nonrenormalizable

Let me ask some questions

• How to check whether our theory is renormalizable when $$d=2$$ or not? Our formula $$n> \frac{2d}{d-2}$$ breaks down for such a case.

• Let's say we want to find renormalizable theories for higher dimensions. Based on $$(*)$$, for $$d=4$$ we see we would need to drop $$\phi^5$$ and $$\phi^6$$ terms. For $$d=5,6$$ we see we would need to drop $$\phi^4$$, $$\phi^5$$ and $$\phi^6$$. For $$d=7,8,9,10,..,10^6,...$$ (I did not check further than $$d=10^6$$ but it seems that the relation holds for further dimensions) we see we would need to drop $$\phi^3$$, $$\phi^4$$, $$\phi^5$$ and $$\phi^6$$. Does this mean that the simplest scalar field theory $$\mathcal{L} = \frac 1 2 \partial_{\mu} \phi \partial^{\mu} \phi - \frac 1 2 m^2 \phi^2$$ is renormalizable for all dimensions?

PS: Please note that I am aimed at understanding under which conditions is a theory renormalizable and not solving this one in particular.

• $\frac{2d}{d-2}$ is monotonic decreasing so no need to check further than $d=10^6$ -- you've done more than enough. – AccidentalFourierTransform Apr 1 at 12:01

## 1 Answer

Your statement (*) is valid only for $$d > 2$$ (you have to take care when dividing an inequality by a non-positive number). Based on the original formula for $$[\lambda_n]$$, renormalizability holds for all $$n$$ if $$d \le 2$$.

You are correct that the "simplest" (free) scalar field theory is renormalizable for all $$d$$.

• Oh so that means that we can only state that the sample theory is renormalizable for $d=3$. How could we check if our theory is renormalizable for $d=1,2$? – JD_PM Apr 1 at 10:54
• @JD_PM Just directly compute $[\lambda_n] = d- \frac 1 2 n(d-2)$ and check that it's non-negative. – nanoman Apr 1 at 10:57