Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  How to prove equivalence of RG flow of QFT coupling constant and diagrammatic resummation at fixed renormalization scale?

+ 6 like - 0 dislike
941 views

QFT books say that solving the RG equation $\frac {dg} {d\textbf{ln} \mu}=\beta(g)$, using the one-loop beta function, is to the "leading log" approximation equivalent to resumming infinitely many loop corrections arranged in Russian-doll fashion, at a fixed renormalization scale.

While I have no doubt about the validity of this statement, can someone point to me a direct diagrammatic proof? I feel that there may be lots of subtlety in a precise proof, and I'm not satisfied with plausibility arguments.

This post has been migrated from (A51.SE)
asked Jan 14, 2012 in Theoretical Physics by felix (110 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
I think the diagrams not only need to look like a series of loops, one encircling each others, but also need to be strongly ordered in loop momenta, with inner loops having much less momenta so appear as effective vertices to outer loops. In other words, we are looking at a reduced part of the phase space. The question is then how to prove it's this part of the phase space that gives leading behaviour. Another question is how to show that a generic N-loop diagram is "sub-leading" compared with diagrams that are reducible to N successive 1-loop diagrams.

This post has been migrated from (A51.SE)

1 Answer

+ 1 like - 0 dislike

It's not a "diagrammatic" proof, but you can see that this is in fact the "leading log" approximation from looking at what you get when you solve the Callan-Symanzik with the first loop Beta-function. Let's say I have some correlation function $\mathcal{G}(\lambda,\ell)$ which is a function of some marginal coupling $\lambda$ and $\ell \equiv \log \Lambda$ the log of the energy scale. Say the first loop $\beta$ function for $\lambda$ looks like

$\beta(\lambda) = b\lambda^2 + \mathcal{O}(\lambda^3)$

for some constant $b$. Just like scalar $\phi^4$ in $d=4$.

The CS equation looks like

$\left(\frac{\partial}{\partial\ell} - \beta(\lambda)\frac{\partial}{\partial\lambda}\right)\mathcal{G}(\lambda,\ell)$

Solving this equation with the lowest order $\beta$ function gives you $\mathcal{G}$ in the limit $\lambda \rightarrow 0$ but $\lambda\ell$ fixed. This is therefore the sum of the the terms leading in $\ell$ for every order of $\lambda$. You can see this by rewriting $\mathcal{G}$ in the following way:

$\mathcal{G}(\lambda,\ell) = \lambda\mathcal{G}^{(1)}(\lambda\ell) +\lambda^2\mathcal{G}^{(2)}(\lambda\ell) + \lambda^3\mathcal{G}^{(3)}(\lambda\ell) +\,\,...$

where the $\mathcal{G}^{(i)}$ are some unknown functions of a single variable. You can do this since there is a maximum level of divergence every order of perturbation theory. If you plug this in to the CS equation and keep track of the order of the terms you see you get a good differential equation for $\mathcal{G}^{(1)}$, but not for any of the higher order terms. If you went to the next order in $\lambda$ in the $\beta$ function that would give you a good equation for $\mathcal{G}^{(2)}$ which is the next to most divergent diagrams from every order of perturbation theory. So the RG procedure converts the limit $\lambda\rightarrow 0$, $\ell$ fixed that you get from standard perturbation theory, into the limit $\lambda\rightarrow 0$, $\lambda\cdot\ell$ fixed, which is frequently more useful.

This post has been migrated from (A51.SE)
answered Jan 18, 2012 by BebopButUnsteady (330 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...