2
$\begingroup$

We start with a Lagrangian $L$, which is valid up to some scale $\Lambda$ and includes couplings $g,m$.

In the Wilsonian perspective, we note that the contributions from fluctuations at scales between some arbitrary scale $\lambda$ and the cutoff $\Lambda$ are $\propto \ln(\lambda /\Lambda)$. The key idea is then that we can use the cutoff $\lambda$ instead of $\Lambda$ if we include counter terms $\delta L(\lambda,\Lambda)$. In particular, we end up with a renormalized effective Lagrangian $$ L_R = L + \delta L(\lambda,\Lambda) ,$$ which is valid up to the scale $\lambda$. One way to understand this effective Lagrangian is by defining new parameters $g_R, m_R$ that are scale dependent. Specifically, we have $g_R (\lambda) := g +\frac{g^2}{12 \pi^2} \ln(\lambda /\Lambda)$, where $g$ is the "bare" coupling that appears in the original Lagrangian which is valid for cutoff $\Lambda$.

It is then usually argued (see, for example, page 84 here) that this implies that if we measure $g_R$ at some scale $\mu_0$, we can predict its value at any other scale $\mu$ by using the analogous evolution equation $$g_R (\mu) = g(\mu_0) +\frac{g(\mu_0)^2}{12 \pi^2} \ln(\mu /\mu_0)$$

This seems to make sense at first glance, but raises many questions upon closer inspection. So how can this actually be derived? My main issue is that the scales $\mu$ and $\mu_0$ are physical scales relevant to the energy in a given process, whereas $\lambda$ and $\Lambda$ are cutoffs. Moreover, $g$ has no scale dependence. It's the charge that appears in the original Lagrangian $L$ which is valid up to the cutoff scale $\Lambda$. But there is no reason why this $g$ should depend on the energy scale in a given process.

$\endgroup$
  • $\begingroup$ The coupling $g$ is energy scale independent at the tree/classical level. Taking into account the quantum/loop/renormalization corrections, $g$ is inevitably incoming-momentum/energy dependent for a particle-particle scattering process. $\endgroup$ – MadMax Jul 8 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.