# How can one recover/obtain information from the renormalization group procedure?

I know the basic idea behind the renormalization group approach as it is used in mathematical physics to study both QFT and statistical mechanics. However, I have trouble understanding how can one recover the information one was trying to obtain using this technique. Let me elaborate.

Although there is no such thing as a 'general approach' to RG, I want to try to sketch the ideas from a generic model. As stated in Brydges & Kennedy's article, one starts with integrals of the form: $$\begin{eqnarray} Z(\varphi) = \int d\mu_{C}(\psi) e^{-V_{0}(\psi+\varphi)} = (\mu_{C}*e^{-V_{0}})(\varphi) \tag{1}\label{1} \end{eqnarray}$$ where $$\varphi = (\varphi_{x})_{x\in \Lambda}$$ is a Gaussian process with joint distribution $$\mu_{C}$$, mean zero and covariance $$C$$. Suppose we can write $$C$$ as a sum $$C=C_{1}+C_{2}$$. Then: $$\begin{eqnarray} Z(\varphi) = \int d\mu_{C_{1}+C_{2}}(\psi)e^{-V_{0}(\psi+\varphi)} = \int d\mu_{C_{2}}(\zeta)\int d\mu_{C_{1}}(\psi) e^{-V_{0}(\psi+\varphi+\zeta)} = \int d\mu_{C_{2}}(\zeta) (\mu_{C_{1}}*e^{-V_{0}})(\varphi+\zeta) = \int d\mu_{C_{2}}(\zeta)e^{-V_{1}(\varphi+\zeta)} \tag{2}\label{2} \end{eqnarray}$$ where: $$\begin{eqnarray} V_{1} = -\ln \mu_{C_{1}}*e^{-V_{0}} \tag{3}\label{3} \end{eqnarray}$$ Thus, we can define a map on the (informal) space of actions, called renormalization group map and denoted by $$RG$$, such that $$RG: V_{0} \to V_{1}$$. Analogously, if $$C=C_{1}+\cdots +C_{n}$$, $$n \ge 2$$, then sucessive applications of (\ref{2}) lead to: $$\begin{eqnarray} Z(\varphi) = \int d\mu_{C_{n}}(\zeta_{n})(\mu_{C_{n-1}}*e^{-V_{n-1}})(\varphi+\zeta_{n}) = \int d\mu_{C_{n}}(\zeta_{n})e^{-V_{n}(\varphi+\zeta_{n})} \tag{4}\label{4} \end{eqnarray}$$ where $$V_{n}= RG(V_{n-1})=\cdots = RG^{n-1}(V_{0})$$. We thus defined a 'trajectory' $$V_{0}\to V_{1}\to V_{2}\to \cdots \to V_{n}$$.

All these being said, I believe the main idea of the process is to (luckily) prove that the above trajectory ends up in a fixed point. In other words, luckily we have $$RG^{n}(V_{0}) = V^{*}$$ for every $$n$$ suficiently large.

The above scenario, although very generic, appears in some discussions on the topic. As an example, see Salmhofer's book. Now comes my questions.

(1) How can one recover the information about $$Z(\varphi)$$ once we was lucky and obtained $$V^{*}$$? See, $$Z(\varphi)$$ was our object of study in the first place, right? But I don't see how to get back and obtain it.

(2) One situation in which the covariance splits into a sum of covariances is when one is trying to approach the continuum limit from a scaled lattice. This can be done in either QFT and statistical mechanics, but I believe it is more common for QFT models. But when we think about statistical mechanics, can one obtain critical temperatures, critical exponents and other thermodynamics entities from the above process? Is it possible to ilustrate how it could be done considering this very generic model?

• RG flow loses information. You can have multiple UV theories that give rise to the same IR fixed point. Nov 15, 2020 at 15:01
• Oh, right. This is an important point indeed. But, considering that one such decomposition is chosen, the above scenario is accurate, right? My point is that, usually, it is stated that, because of the above, the problem becomes the study of the RG map I defined in my post. But I was confused on how to recover the initial information once this map is studied. Nov 15, 2020 at 16:02
• It's been a while since I looked the BK article and it is now behind a paywall I can't access, but the intent of the article was not to explain the RG. Rather, they were inspired by the RG and led to the discovery of a new expansion method for infinite volume limits in statistical mechanics, and constructive QFT. As a result, the description in the MO question is not really the RG, because there is no rescaling at each step. This rescaling is important, in order to have the right notion of fixed point. Nov 15, 2020 at 20:59
• About (2) vs (1). In statistical mechanics (1), one has a definite starting point $V_0$, say a lattice model at a given temperature and, e.g., zero magnetic field. Then one runs the RG forward from there $V_0\rightarrow V_1\rightarrow V_2\cdots$ and then eventually converge to a fixed point $V_{\ast}$. In QFT or the use of the RG to construct a model in the continuum (2), the story is much more complicated conceptually, let alone technically. See: physics.stackexchange.com/questions/372306/… Nov 15, 2020 at 21:03
• @AbdelmalekAbdesselam your first comment maybe clarifies something it is not very clear to me yet. As you know, I study your answers frequently and in one of them you discuss a more concrete (yet, rather general) model and you proceed to reescale to a unit lattice. There, the initial integral ($Z$ in my post) becomes $Z$ itself but with $V_{0}$ replaced by $RG(V_{0})$. This is a very nice approach, and the ideas are clear to me. In my post, however, after each iteration, one ends up with different object, since at each step the covariance matrix changes $C_{n}\to C_{n+1}$. (Continues) Nov 15, 2020 at 22:22

1. The limiting function $$V^\ast$$ is such that any further convolutions of $$e^{-V^\ast}$$ with $$\mu$$ return $$e^{-V^\ast}$$, so $$Z^\ast(\phi)=e^{-V^\ast(\phi)}$$.
2. To obtain critical properties, you need the correlator $$K(x,x')=\langle\phi(x)\phi(x')\rangle$$. The decay length of the correlator diverges at the critical temperature $$T_c$$ as a power law $$(T-T_c)^{-\alpha}$$ and the power $$\alpha$$ is the critical exponent. The correlator is obtained by adding a source term $$\lambda\psi(x)\psi(x')$$ to the exponent in the definition of $$Z(\phi)$$ and then evaluating $$dZ/d\lambda$$ at $$\phi=0$$.
• Thanks for the answer Carlo! Just a clarification: what is $Z^{*}(\phi)$? You mean $Z(\varphi) = \int d\mu_{C}(\psi)Z^{*}(\psi+\varphi)$? Nov 15, 2020 at 14:51
• $Z^\ast$ is the functional $Z$ that you obtain at the end of the renormalization procedure, by taking the exponent of minus the functional $V^\ast$. Nov 15, 2020 at 15:05
• Oh, I think I understand your point now. Because of (\ref{4}), $Z(\varphi) = Z_{n}(\varphi)$ where $Z_{n}(\varphi) = \int d\mu_{C_{n}}(\zeta_{n})e^{-RG^{(n-1)}(\varphi+\zeta_{n})} = (\mu_{C_{n}}*e^{-RG^{(n-1)})}(\varphi) = e^{-RG^{(n)}(\varphi)}$, so for sufficiently large $n$ the fixed point is attained and $Z(\varphi) = e^{-V^{*}(\varphi)}$, right? Nov 15, 2020 at 15:56
• yes; of course the final $Z$ which you obtain is not the $Z$ you started out with, that is why I used a different symbol $Z^\ast$; but that is the whole point of the renormalization procedure: the initial $Z$ contains small-distance details that become irrelevant near a critical point, when the correlation length diverges, and the final $Z^\ast$ no longer contains these details. Incidentally, this why the renormalization group is more accurately referred to as a semi-group. Nov 15, 2020 at 16:14
• yes indeed; $Z$ could be defined on a lattice and $Z^\ast$ could then be the continuum limit Nov 15, 2020 at 17:22