In dimensional regularization, we introduce an arbitrary energy scale $\mu$. Naively, it plays the role of another parameter of the theory that needs to be fixed experimentally, but actually it is not since the RG-flow equations allow a change in $\mu$ to be compensated by a change in renormalized couplings.

The RG-flow is then to be analyzed, and based on its behavior in the UV we can either say that in the UV limit our theory reaches a fixed point and therefore is equivalent to some conformal field theory (like in QCD), or some couplings blow up in which case perturbative expansions make no sense anymore (like in QED).

My question is: why can't we just fix some $\mu$ and go from there?

In cut-off regularization scheme, for example, it is obvious that $\Lambda$ defines a "bounding energy scale" beyond which the regularized theory can not work (since the corresponding degrees of freedom simply do not exist). It is therefore natural to seek a higher $\Lambda$ whenever our current $\Lambda$ is not good enough, i.e. whenever we need to explain some higher-energy phenomena.

In contrast, in dimensional regularization we have already taken the $d \rightarrow 4$ limit, so the theory with $\mu$ and renormalized couplings evaluated at $\mu$ is already exact (well, at least perturbatively) and I can not imagine why one would need a higher $\mu$ after all.

Whilst I completely understand that in both schemes the resulting RG-flow equations are the same and define a beautiful and concise mathematical structure, I would nevertheless like to understand this conceptual point.

UPDATE: I also understand that $\Lambda$ plays a role of regularizer in the cut-off scheme, just like $d$ does in dimensional regularization (and not $\mu$). But it does not answer my question: why can't we simply take some $\mu$ and forget about the RG flow?

UPDATE 2: in case the answer to my question turns out to be "sure, why not, take some $\mu$ and go from there" $-$ then I have another one. How come that our theory can have different regimes in the UV and in the IR if I could just take some $\mu$? Maybe it has something to do with the validity of perturbative expansions?


1 Answer 1


We have to change the renormalization scale to a typical scale of the process we are considering because otherwise perturbation theory becomes rather useless, since process far away from the renormalization scale get large higher order corrections due to the "large log" $\ln(-p^2/\mu^2)$, where $p$ is the process scale and $\mu$ the renormalization scale.

The standard perturbative expansion is not only an expansion in $\lambda$, but also in $\ln(-p^2/\mu^2)$. As an illustrative example, consider massless 4D $\phi^4$ theory, where we might fix the renormalization scale at any $\mu$ by imposing $$ \mathrm{i}\mathcal{M} = -\mathrm{i}\lambda$$ for the 2-to-2 scattering $p_1 p_2\to p_3 p_4$ at $s=t=u=-\mu^2$ where $s,t,u$ are the usual Mandelstam variables. It's clear that we defined $\lambda(\mu)$ by this to be the relevant coupling at the scale $\mu$ - and to see that it becomes a bad choice for the coupling at scales far away from $\mu$, we have to consider the full amplitude $$ \mathrm{i}\mathcal{M} = -\mathrm{i}\lambda(\mu) - \mathrm{i}\frac{\lambda(\mu)^2}{32\pi^2}\left(\ln(-s/\mu^2) + \ln(-t/\mu^2) + \ln(-u/\mu^2)\right) + \mathcal{O}(\lambda(\mu)^3)$$ where the 1-loop contribution of order $\lambda(\mu)^2$ grows larger as the ratio of the Mandelstam variables (which are good measures for the energy scale of the process) to the renormalization scale goes away from unity.

It's not that it is not possible to compute the amplitudes of processes far away from the renormalization scale, but the perturbation series becomes inefficient, and the coupling $\lambda(\mu)$ is no longer really the "physical coupling" which should be pretty straightforwardly related to the vertices. "Running the coupling" is essentially restoring the coupling to a value where it is again a good parameter to consider the Feynman diagram perturbation series as being expanded in. This also explains why one would speak of the breakdown of perturbation theory as soon as the coupling becomes of the order of unity - you can't get around this by staying at a lower scale and perturbing that, since then the logarithms appear large and blow up the series, anyway.

  • $\begingroup$ I think I understood this. Did you mean $O(\lambda(\mu)^3)$ instead of $(\lambda(\mu)^3)$ in the second formula? $\endgroup$ Aug 20, 2015 at 13:18
  • $\begingroup$ So basically, everything is OK with fixing $\mu$ if we have a nonperturbative solution? And problems arise only when one considers perturbation theory? $\endgroup$ Aug 20, 2015 at 13:19
  • $\begingroup$ @Hindsight: Yes, I forgot to type the O. I'm not sure if you would need to fix $\mu$ in the first place if you have non-perturbative solutions, but yes, I would see no issue with that (although one should perhaps, as a sanity check, test whether the non-perturbative solution changes with the RG flow or not). $\endgroup$
    – ACuriousMind
    Aug 20, 2015 at 13:22
  • $\begingroup$ and what if I want to consider some process with one Mandelstam variable much greater than the other? For example, $s >> t$? Then I would have two different logs: $\ln (-s/\mu^2)$ and $\ln (-t/\mu^2)$ and one of them will surely "blow up" if I take $\mu^2$ to be close to either $-s$ or $-t$. If I take $\mu^2 \sim -s$ then the $\ln (-t/\mu^2)$ log will blow up even quicker (because the log function quickly descends when its argument is much less then one and tends to zero). How could I calculate such an amplitude then? $\endgroup$ Aug 21, 2015 at 6:59
  • $\begingroup$ @Hindsight: No one forces you to do the renormalization condition at $s=t=u=\mu^2$, you can also set the condition for e.g. $s=\mu^2 \wedge t=u=0$, and, in general, at $s=\mu\wedge t=a\mu \wedge u=b\mu$ for some parameters $a,b$, so choose the $s,t,u$ ratios that are appropriate to the process you want to calculate. $\endgroup$
    – ACuriousMind
    Aug 21, 2015 at 13:15

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