Question about RG from QFT perspective, in particular Weinbergs book

Question

This must be fairly basic I fail to understand. According to Weinberg, QFT Vol2, Ch.18 (The preamble)

When we replace bare couplings and fields with renormalized couplings and fields defined in terms of matrix elements evaluated at a characteristic energy scale $\mu$, the integrals over virtual momenta will be effectively cut-off at energy and momentum scales of order $\mu$.

Where does this statement come from? How are the loop momenta greater than this scale (squared) integrated out? Or in what sense are they integrated out?

If I understand correctly, when we calculate counterterms to relate bare parameters to renormalized ones (at a certain scale they are equal) then the counterterms are essentially those that if taken under the regulated loop integral (which suppresses in one way or other the high momenta modes) gives me the result which is "physical" (in the sense that one can relate it to your favourite scheme to calculate actual observables)

But take e.g. the hard cut-off as regulator. The modes close to the cut-off do not get suppressed at all, no matter how far away I am from my renorm. scale, since the loop integral is unchanged in these intervals (up to a very large constant).

So what does Weinberg mean?

The linked question all have answers pointing toward the opposite fact: The cut-off is the scale that (trivially) suppresses the high energy modes. Not $\mu$. But clearly, this seems weird because following Weinbergs line of thought, he then proceeds to let all parameters flow by RG flow, very similar to how one would to in classical statistical field theory. Without the previous fact though there's no actual interpretation of what changing $\mu$ does (it does not get rid of unnessecary DOF).

Answer 1

Lets take for example QED. We know we need to include renormalization constants on the gauge field $A_\mu$, mass of the fermions $m$, the fermion field $\psi$, and the coupling constant $e\rightarrow e_0 = Z_1/(Z_2\sqrt{Z_3})$. We also know from the QED Ward identity that $Z_1 = Z_2$ so we can write $e^2 = e_0^2 Z_3$. Then under dimensional regularization (dim-reg) $\alpha(\mu)^\epsilon = Z_3\alpha_0$. We can find $Z_3$ is related to the 1PI as $Z_3\simeq 1 - \frac{\alpha}{3\pi}\frac{1}{\epsilon}+\cdots$. Renormalization group flow then tells us that there is a corresponding $\beta$-function since \begin{equation} \frac{\partial\alpha}{\partial\ln \alpha} = -\epsilon\left(1 - \frac{\alpha}{3\pi}\frac{1}{\epsilon}\right)\alpha_0(\mu^2)^{-\epsilon} = \frac{\alpha^2}{3\pi} \end{equation} which gives us a differential equation for $\alpha$.

So, to answer your first question, the derivation above is a primitive (and not compelte) reason to where the statement comes from: the new scale at which the coupling constant is defined to change the loop integrals to be non-divergent and handle the large logarithms (so if the divergence is $\propto \ln(E/\mu)$ then we should take $\mu$ to be on the order of $E$). This makes the higher energy DOF to be effectively integrated out since as we are lowering the energy scale $\mu$ from the UV to something that is measurable, we are performing a course graining (from the viewpoint of the path integral).

Let me also say that the view of course graining the theory via changing the energy scale of the theory from $\alpha\rightarrow \alpha(\mu)$ eliminates the language of "surpressing higher energy modes,'' since it does not do that, it physically eliminates (renormalization group flow and thus the beta function telling us this) those DOF at the UV scale. As far as I was taught, we physically can not access those UV DOF after integrating them out via RG-flow (hence course graining). Hopefully that clears up some of the physics behind Weinberg's chapter 18 (if I recall correctly, towards the end of chapter 18.2 he talks about the sliding scale that I sort of elaborated to above).