5

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?

IamWill
  • 3,151
  • 2
    RG flow loses information. You can have multiple UV theories that give rise to the same IR fixed point. – Aaron Bergman Nov 15 '20 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. – IamWill Nov 15 '20 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. – Abdelmalek Abdesselam Nov 15 '20 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: https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability/375571#375571 – Abdelmalek Abdesselam Nov 15 '20 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) – IamWill Nov 15 '20 at 22:22
  • I've always thought that the difference was explained because you were addressing the continuum limit, whereas the toy model in my post is trying to sketch a more general idea. But now that you mentioned the work of BK, this difference might be explained by the fact that BK were not trying to explain RG, but using an idea. As you mentioned, there is no reescaling at each step and I believe this is another way to say what I mentioned that after each interaction one end up with a different object. I'm getting it right? – IamWill Nov 15 '20 at 22:26
  • For completeness, the forementioned answer was this one https://mathoverflow.net/questions/62770/what-mathematical-treatment-is-there-on-the-renormalization-group-flow-in-a-spac – IamWill Nov 15 '20 at 22:27
  • @IamWill: Short answer: Yes. However, until you look at the answer in physics.stackexchange I linked to, I am afraid this discussion will not be very productive. One of the main point I make there is the distinction between autonomous vs nonautonomous dynamical systems and the need for an autonomous RG, in order to have a clean notion of "fixed points". You said one ends up with different objects, but it is worse than that: one ends up with objects living in different spaces, and the successive maps are different. – Abdelmalek Abdesselam Nov 16 '20 at 19:20
  • @AbdelmalekAbdesselam amazing. I saved the link to read it later because I've been a little busy these past days and I want to read it carefully, and it demands time. I will read it as soon as I can and if something is not clear I reach you back. Anyway, thanks for the comments again, I've been learning so much from you! – IamWill Nov 16 '20 at 19:31

1 Answers1

2

  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 $\phi=0$.

In reference to the title of the post: "How can one recover/obtain information from the renormalization group procedure?" Information that depends on features that appear at small distances cannot be recovered, it is lost in the renormalization flow (which is not reversible). The information that remains refers to features that persist at large distances, such as a diverging correlation length and the critical exponents associated with it.

Carlo Beenakker
  • 177,695
  • Thanks for the answer Carlo! Just a clarification: what is $Z^{}(\phi)$? You mean $Z(\varphi) = \int d\mu_{C}(\psi)Z^{}(\psi+\varphi)$? – IamWill Nov 15 '20 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 $V^\ast$. – Carlo Beenakker Nov 15 '20 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? – IamWill Nov 15 '20 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. – Carlo Beenakker Nov 15 '20 at 16:14
  • Oh, right. $Z^{}$ is the fixed point for those $Z_{n}$. So, your answer made me realize I had some misconceptions. The whole point is not evaluate $Z$ by this iterating process, but, instead, each iteration is just a process of becoming closer to a critical system, right? At the end of the day, $Z$ is what you know from a well-behavior theory and $Z^{}$ is what you intent to know when the criticality makes things difficult to analyze. – IamWill Nov 15 '20 at 16:29
  • 1
    yes indeed; $Z$ could be defined on a lattice and $Z^\ast$ could then be the continuum limit – Carlo Beenakker Nov 15 '20 at 17:22