Running Couplings and Renormalization Explained

[ledamage]

Hi there!

I'm struggling a bit with running couplings. Srednicki introduces dimensional regularization and the \overline{\mathrm{MS}} scheme, then calculates a squared transition amplitude for some reaction in \varphi^3 theory. Eventually, he calculates the beta function for the coupling and solves the renormalization group equation. Clear so far.

In the transition amplitude, a factor \ln(s/\mu^2) occurs, where s is a Mandelstam variable and \mu is the factor arising from dimensional regularization. Now he says, to avoid large logarithms, we should put \mu^2 \sim s (which we can do since physics must be independent of \mu). Then, according to the beta function, the coupling runs with the involved momenta, the well-known behavior.

Now, \mu^2 \sim s certainly is a convenient choice, but what if I choose \mu to be constant (or something else), which I am free to do? Then, the coupling doesn't run at all. It seems to me that the running coupling is just a way of viewing things (here in particular by using the \overline{\mathrm{MS}} scheme and \mu^2 \sim s)?

Another quick question: I read everywhere that renormalization is related to the behavior of the theory at small distances/large momenta. How are small distances and large momenta related? And what has renormalization to do with it?

 

[Avodyne]

Now, \mu^2 \sim s certainly is a convenient choice, but what if I choose \mu to be constant (or something else), which I am free to do? Then, the coupling doesn't run at all.

Correct. If you use a fixed \mu, then the coupling doesn't run. The expansion parameter governing perturbation theory then appears to be \alpha(\mu)\ln(s/\mu^2), and if this is large, perturbation theory breaks down. The advantage of using the running coupling is that we see that the expansion parameter is really \alpha(s^{1/2}), which might be small even if \alpha(\mu)\ln(s/\mu^2) is not. Indeed, this is the case at high energies in QCD.

How are small distances and large momenta related?

Fourier transform.

And what has renormalization to do with it?

With small distances = large momenta, nothing. As for the connection between behavior at small distances and renormalization, see section 29 of Srednicki.