Hi everyone. I've got some doubts about the interpretation of the running coupling. I'll expose to you some of the results and comments i read in the lectures notes of my QFT professor. As an example, we are studying ##\phi^4## and the 4-point amplitude accounting for the next to leading order correction, the 1-loop diagram. In order to deal with the divergence we apply renormalization and we choose the dimensional regularization and the on-shell prescription. This choice has the advantage of canceling the dipendence on ##\mu##, the renormalization scale. So we find that the effective coupling is:(adsbygoogle = window.adsbygoogle || []).push({});

$$\lambda_{eff} = \frac{\lambda}{1-\frac{3\lambda}{16\pi^2}\ln{\frac{k}{m}}}$$

This result has a quite straightforward interpretation. All the parameters have a direct physical meaning and this encodes the running of the constant when ##k## increases.

After that he decides to do again the calculation for the running constant using the group renormalization equation. To do so, he changes the renormalization scheme to MS. In this scheme ##\mu## does not go away and we find the running:

$$\lambda_{eff}(\mu) = \frac{\lambda(\mu_0)}{1-\frac{3\lambda(\mu_0)}{16\pi^2}\ln{\frac{\mu}{\mu_0}}}$$

Now, ##\mu## has not any physical meaning and neither has ##\lambda##, in this scheme the parameters of the lagrangian are not the physical ones. Nevertheless, the professor states: "If we choose ##\mu = k## and ##\mu_0 = m## then we get exactly the same result".

I don't understand why we want to compare the 2 coupling, one is physical and the other is not. In addition, ##\mu## is arbitrary, the observables don't depend on it, so why do we have to choose ##\mu = k##? What am i missing here?

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# A Running Coupling in MS scheme

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