Timeline for How can one recover/obtain information from the renormalization group procedure?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2020 at 1:03 | history | bounty ended | CommunityBot | ||
| Nov 15, 2020 at 17:22 | comment | added | Carlo Beenakker | yes indeed; $Z$ could be defined on a lattice and $Z^\ast$ could then be the continuum limit | |
| Nov 15, 2020 at 16:30 | vote | accept | JustWannaKnow | ||
| Nov 15, 2020 at 16:29 | comment | added | JustWannaKnow | 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. | |
| Nov 15, 2020 at 16:22 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 450 characters in body
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| Nov 15, 2020 at 16:14 | comment | added | Carlo Beenakker | 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 15:56 | comment | added | JustWannaKnow | 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 15:05 | comment | added | Carlo Beenakker | $Z^\ast$ 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 14:51 | comment | added | JustWannaKnow | 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 14:15 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |