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Let's have the Lagrangian $$ L = L_{0} + \lambda V , \qquad (1) $$ where $\lambda$ is constant which is small in the next senses: if $\lambda$ is dimensionless, it means that $\lambda < 1$; if it has dimension $[\lambda ] = l^{-n}$, it means that it is small in compare with characteristic energy of the free states: $\frac{\lambda}{ E^{n}} < 1$.

How to connect the dimension of $\lambda$ with renormalizability of theory which is given by $(1)$? Is a dimensionless of $\lambda$ the necessary condition for renormalizability of the theory?

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I will illustrate the general procedure with a real scalar field, governed by the Lagrangian,

$$\mathcal{L}=\frac{1}{2}\partial_\mu \phi \partial^\mu \phi -\frac{1}{2}m^2 \phi^2$$

If we work in natural units, wherein $\hbar = c=1$, the action is dimensionless (as it normally as units of $\hbar$, equivalent to units of angular momentum), and if we take, $$[\mathrm{d}^d x]=-d$$

the Lagrangian density must have dimension $[\mathcal{L}]=d$. We always have $[\partial_\mu]=1$, and from the kinetic term of the Lagrangian, we may deduce the dimension of the field,


Consider an interaction term of the form,

$$V\sim \lambda \phi^n$$

where $\lambda$ is the coupling constant. (Normally, the interaction term also features $1/n!$ to simplify the Feynman rules of the theory.) We find the dimensions of the potential are,

$$[V]=[\lambda] + n[\phi] = [\lambda] + \frac{n}{2}(d-2)=d$$

from which we may deduce that, providing the dimensions of $V$ are consistent, the coupling has,

$$[\lambda]=d-\frac{n}{2}(d-2)$$ If we consider $d=4$ dimensions, we find $[\lambda]=4-n$. I outline a few scenarios:

$$\phi^4 \implies [\lambda]=0 \implies \text{renormalizable}$$ $$\phi^3 \implies [\lambda] > 0 \implies \text{super-renormalizable}$$ $$\phi^6 \implies [\lambda] < 0 \implies \text{non-renormalizable}$$

The method may be applied to virtually any other Lagrangian, and provides a fast method to determine renormalizability in most circumstances. For additional information, see Peskin and Schroeder's text.

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It is "empirical" examples, while I rather had asked about strictly connection. – Andrew McAddams May 24 '14 at 17:27
@AndrewMcAddams: I've stated the relation between the dimension of the coupling and renormalizability... – JamalS May 24 '14 at 19:31

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