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.

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?

share|improve this question

1 Answer 1

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,

$$[\phi]=\frac{(d-2)}{2}$$

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.

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

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.