# Dimension of interaction in a QFT theory

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1. May 27, 2015

### nikosbak

The problem statement.

When an exercises say " the interaction in a QFT has dimensions Δ" , what does it mean?, it means the field or the Lagrangian has this mass dimension?

In this exercise I'm trying to find the classical beta function (β-function) for the assciated couling.

2. May 27, 2015

### fzero

It sounds like they mean the power-counting dimension of the operator corresponding to the interaction in question, where the dimension of the field is determined from the kinetic term. So, in 4d for a scalar $\phi$, the dimension of $\phi^4$ would be 4, while the dimension of $\phi^2 (\partial \phi)^2$ would be 6, etc.

3. Jun 2, 2015

### nikosbak

Oh I see , I think i got it , but still I dont see how to compute the beta function to get to the form
$$\beta(g)=(d-\Delta)g+\mathcal{O}(g^2)$$
I really appreciate the help :)

The full exercise is :

Show that if the interaction in a QFT is has dimension Δ then there is a classical β function for the associated coupling given by ,
$$\beta(g)=(d-\Delta)g+\mathcal{O}(g^2)$$
Discuss what happens to interactions where Δ > d or Δ < d.

Any hits or insight will do , thank you very much :D

4. Jun 2, 2015

### fzero

I'll outline this, because once the concepts are put together, there really isn't a lot of work left for you to do.I will consider the case of a single field, but you should generalize the argument for multiple fields + allow for derivatives in the coupling.

Consider an interaction of the form

$$S_\text{int} = \int d^dx g_0 \Phi^r$$

and do the rescaling $x\rightarrow \lambda x$. Under this transformation, the fields transform according to their scaling dimension, $\delta$, namely $\Phi\rightarrow \lambda^{-\delta}\Phi$. In fact, we will find that

$$S_\text{int} \rightarrow \int d^dx g_0 \lambda^\kappa \Phi^r,$$

for some exponent $\kappa$ that you should relate to $d$ and $\Delta$. The theory is obviously only invariant under the scale transformation if the exponent $\kappa$ is zero for all of the terms in the action, but we can still define an effective coupling $g(\lambda) = g_0 \lambda^\kappa$. The corresponding beta function, defined as

$$\beta_g(\lambda) = \lambda \frac{dg(\lambda)}{d\lambda},$$

can then be computed.

5. Jun 2, 2015

### nikosbak

Ahaa I solved it . It was a lot easier that I thought !

thank for your time :)