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I want to understand the relation between the Wilsonian RG and the usual QFT RG approach. Several questions have been asked, such as this and many others, yet I don't find a conceptual answer to what follows.

In the computation of loop integrals in QFT, in order to tame divergences a regulator $\Lambda$ is introduced. There are other ways to regularize the integrals of course, but this is the one I'm interested in to make a connection to the Wilsonian picture. The parameters in the Lagrangian - the bare couplings -depend on such regulator: their dependence on the cutoff is such that the renormalized parameters are finite at every order of perturbation theory. One may write a RG equation for both the running of the bare couplings with the cutoff or for the renormalized couplings with the renormalization scale, namely the Callan-Symanzik equation.

As an example, consider the $\phi^4$ theory $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi_0\partial^\mu\phi_0-\frac{m_0}{2}\phi_0^2-\frac{\lambda_0}{4!}\phi_0^4.\tag{1}$$ At the one-loop level, using the on-shell renormalization scheme we identify the measured mass with the renormalized mass $m_R$, defined as the pole of the dressed propagator. So $$m_R=m_0(\Lambda)+B(\Lambda)\tag{2},$$ where $B$ comes from the loop of the tadpole diagram and $m_0$ is adjusted so that $m_R$ does not depend on the cutoff.

While in this picture the role of both the bare and the renormalized coupling is clear, I'm not as convinced in the Wilsonian framework. First of all in this case we naturally have a built-in cutoff $\Lambda$. In this case step by step high momenta are integrated out and we are left with a Wilsonian effective action $$\begin{align} \exp&\left\{-\frac{1}{\hbar}W_c[\phi^<] \right\} \cr ~:=~& \int_{\Lambda/b\leq |k|\leq \Lambda} \! {\cal D}\phi^>~\exp\left\{ \frac{1}{\hbar} \left(-S[\phi^<+\phi^>]\right)\right\}.\end{align}\tag{3}$$ The (adimensional) couplings are changed and we have a new theory with a lower cutoff (before rescaling, at least).

I used to think that to get a "renormalized" coupling $m_R$ or $\lambda_R$ at a scale $\mu$ (in the sense of QFT), one would just reiterate the procedure until the new theory has cutoff $\mu$. For example, I thought that integrating out modes repeatedly, the initial coupling $m_0$ would eventually become $m_R$. Instead, it appears that in the Wilsonian picture the renormalized couplings retain a different role, since the renormalized mass is instead defined in term of cutoff and the adimensional correlation length $\xi$ $$m_R=:\Lambda/\xi.\tag{4}$$ Still, I don't see where in the Wilsonian picture the renormalized couplings would come, since the Wilsonian renormalization basically encodes the running of the coupling with the cutoff, which is already there and not just some regulator to perform subtractions. Can you explain where in the Wilsonian picture of RG $m_R$ appears, why is it necessary and how it differs from the bare mass? Of course the same question holds for any coupling.

References

  1. The renormalization group and the ϵ expansion, Wilson, Kogut 1974. Section 12.2.
  2. QFT & Condensed Matter: An Intro, R. Shankar. Section 14.7
Mr. Feynman
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1 Answers1

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  1. A main idea of the Wilsonian renormalization group (RG) flow is, as OP writes, that the renormalized coupling constants $g$ become the UV bare coupling constants $g_0$ when the renormalization scale $\Lambda=\Lambda_L$ approaches the UV cut-off $\Lambda_H$.

  2. One aspect [which might cause some confusion] is that there are different versions of the Wilsonian RG flow, specifically whether coarse-graining/blocking is included or not, cf. e.g. my Phys.SE answer here.

  3. Specifically, OP's eq. (4) seems to correspond to eq. (10.41) in Ref. 1 and eqs. (14.13) + (14.79) in Ref. 2, where the renormalized mass parameter $m=m_{\rm ph}$ is a physical mass in an on-shell renormalization scheme. We interpret the reciprocal correlation length $\xi^{-1}=\frac{m}{\Lambda_L}$ as a dimensionless counterpart.

Qmechanic
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  • Hello, Qmechanic. You mention two different versions of the RG, which I hadn't heard about before. Upon reading the other answer that you've linked, it appears to me that the difference is the rescaling, not the coarse graining itself. To be clear, I added two references to my answer, one of which is Wilson's well-known paper about the $\epsilon$ expansion, where such distinction is apparently made. It's not clear to me, even after reading your other answer what is playing the role of $m_R$ and what is playing the role of $m_0$. – Mr. Feynman Mar 13 '24 at 09:50
  • Could you please expand your answer in that regard, possibly comparing to the ideas in my refs? – Mr. Feynman Mar 13 '24 at 09:51
  • Hello @Qmechanic. I've been thinking really hard about this and I found other sources (e.g. Huang QFT) proposing an idea more akin to mine, so I might ask other questions on the site. Point 3 is something I understand, the problem is why such mass would appear as a distinct quantity in Wilsonian theory. Is it correct to see in this context $m_\text{phys}$ as the IR limit of the RG flow of the $\phi^2$ coupling? In such case, the equation in point 3 would just be fixing this value, right? – Mr. Feynman Mar 19 '24 at 08:08
  • In other words, I'm sorry in case the question was not clear about it, the main point is how (in Wilsonian renormalization) such "new" mass is related to the one changing according to the RG flow. – Mr. Feynman Mar 19 '24 at 08:11