nikosbak said:
Oh I see , I think i got it , but still I don't 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
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.