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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.

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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. –  felix Jan 15 '12 at 8:50

2 Answers 2

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

$\left(\frac{\partial}{\partial\ell} - \beta(\lambda)\frac{\partial}{\partial\lambda}\right)\mathcal{G}(\lambda,\ell) = 0$

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.

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Theres no equal sign in the second "equation"... –  kηives Dec 28 '12 at 22:06

If you solve the equation to second order you obtain:

$g=\frac{g_0}{1-a g_0 log\lambda}=g_0 +\sum_n g_0^{n+1}a^nlog^n \lambda$

the second part is the contribution of powers of the first loop diagram, for instance try to write g at 1-loop first and then sum all the powers of the first order diagram.

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