# The μ in dimensional regularization

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• erikasan
In summary, the value of the ##\mu## in dimensional regularization appears to determine a "reference value" of an observable.
erikasan
TL;DR Summary
Does μ have to be a relevant energy scale to the scattering process considered or is it arbitrary?
I have a question about the ##\mu## in dimensional regularization and how it is related to renormalization conditions. I follow the same notation and conventions as in Schwartz. Take QED as an example:
$$\mathcal{L} =-\frac{1}{4}\left( F_{0}^{\mu \nu }\right)^{2} +\overline{\psi }_{0}\left( i\gamma ^{\mu } \partial _{\mu } -m_{0}\right) \psi _{0} -e_{0}\overline{\psi }_{0} \gamma _{\mu } A_{0}^{\mu } \psi _{0}$$ where the subscript ##0## indicates that it is the bare field/mass/charge. The ##\mu## arises when writing the bare charge on the form $$e_{0} =Z_{e} \mu ^{\frac{4-d}{2}} e_{R}$$ where ##e_R## is the renormalized charge, ##Z_e## is the dimensionless charge renormalization and ##\mu## is chosen to have mass dimension ##1## such that ##e_R## is dimensionless in any spacetime dimension.

Now I'm wondering about the importance of the value of ##\mu##. My professor seems to be insisting that, in spite of the theory being independent of ##\mu## and observables being independent of ##\mu##, the value of ##\mu## must be some relevant energy scale to some scattering process considered. To me this is odd because I can consider several different processes in QED and vary the center of mass energy in each of them independently however I like, and I won't necessarily be able to set ##\mu## equal to some energy scale that is relevant to all of them at once. And in the end, if observables are independent of ##\mu## then it sounds to me like the value is utterly arbitrary. But ##\mu## being some relevant energy scale is still what's being insisted and it makes me doubt whether I've really understood what renormalization conditions and renormalized couplings really are. So I want to get a second opinion to double-check this.

To me the ##\mu## is similar to the gauge-fixing parameter ##\xi## that arises from the Faddeev-Popov procedure; it determines the form of the gauge boson propagator but I can pick whatever value of ##\xi## I like. Similarly, assuming that some subtraction scheme has been chosen for ##Z_e##, the value of ##\mu## determines the value of ##e_R## but I can pick whatever value of ##\mu## I like. If I vary ##\mu## then ##e_R## must vary in such a way that the bare charge ##e_0## remains unchanged, which to me is just the statement of the renormalization group equation $$\mu \frac{d}{d\mu } e_{0} =0.$$ But while I in principle could have chosen ##\mu## to be anything (I think), in practice ##\mu## is often chosen such that ##e_R## becomes sort of a "reference value" of an observable at some particular energy scale. As an example, say I have calculated the (Fourier transformed) Coulomb potential $$\tilde{V}\left( p^{2}\right) =\frac{e_{R}^{2}}{p^{2}}\left[ 1-\Sigma \left( p^{2}\right) -\delta \right]$$ where ##\Sigma## are corrections from loop diagrams and ##\delta## is a counterterm. To one-loop the ##\Sigma## is something like $$\Sigma \left( p^{2}\right) =e^{2}_R\left[\frac{2}{\epsilon } +\ln\frac{\mu ^{2}}{p^{2}}\right]$$ where ##\epsilon## stems from taking ##d=4-\epsilon## and where I've omitted uninteresting factors and an integral over Feynman parameters. In minimal subtraction I set ##\delta =-2e^{2} /\epsilon##, but that still seemingly leaves an unphysical degree of freedom ##\mu## in my observable ##\tilde{V}\left( p^{2}\right)##. Now I claim that I can choose ##\mu## such that ##e_R## satisfies $$\tilde{V}\left( p_{0}^{2}\right) =\frac{e_{R}^{2}}{p_{0}^{2}}$$ at some arbitrary energy scale ##p^2_0##. This is what I mean by ##e_R## being a "reference value" of my observable, because I have defined it to be ##e^2_R=p_{0}^{2} \tilde{V}\left( p_{0}^{2}\right)##. This is my renormalization condition, and corresponds to setting $$-\Sigma \left( p_{0}^{2}\right) -\delta =0.$$ Now I can write my observable on a form that is explicitly independent of ##\mu##: \begin{align*} \tilde{V}\left( p^{2}\right) & =\frac{e_{R}^{2}}{p^{2}}\left[ 1-\Sigma \left( p^{2}\right) -\delta \right]\\ & =\frac{e_{R}^{2}}{p^{2}}\left[ 1-\Sigma \left( p^{2}\right) -\delta -0\right]\\ & =\frac{e_{R}^{2}}{p^{2}}\left[ 1-\Sigma \left( p^{2}\right) -\delta +\Sigma \left( p_{0}^{2}\right) +\delta \right]\\ & =\frac{e_{R}^{2}}{p^{2}}\left[ 1-\Sigma \left( p^{2}\right) +\Sigma \left( p_{0}^{2}\right)\right]\\ & =\frac{e_{R}^{2}}{p^{2}}\left[ 1-e_{R}^{2}\ln\frac{p^{2}}{p_{0}^{2}}\right] \end{align*} which apart from the not very illuminating numerical factors and integral over Feynman parameters agrees with Schwartz Eq. (23.4).

Is this thinking correct or is there a flaw in the argument somewhere? Is ##\mu## truly arbitrary and is a renormalization condition essentially something that fixes all unphysical degrees of freedom introduced when rewriting the bare parameters in the Lagrangian? To me this sounds exactly like what's happening in the on-shell renormalization scheme for the fermion mass in QED: I set the infinite parts of the counterterms to exactly cancel any UV divergences, but I also introduce an arbitrary unphysical finite part to the counterterms which I can choose in such a way that the renormalized mass is exactly equal to the pole mass.

vanhees71, jim mcnamara and malawi_glenn
I agree with you

erikasan
In calculations, you will get ##\log(\mu / E)##-factors, where ##E## is some energy related to that process. If you wanna do perturbuation theory - you want these terms to be small. Thus you often choose ## \mu \approx E## for practical purposes.
Perhaps this is what your professor meant?

Last edited:
vanhees71 and erikasan
Maybe, but at the same time it seems we're perfectly happy doing pertubation theory with terms that are literally infinite, but that's okay because the infinities cancel. Similarly, can't I just choose ##\log\left( \mu /E\right)## as large as I want knowing that all ##\mu## dependence cancels out anyway?

vanhees71
erikasan said:
Maybe, but at the same time it seems we're perfectly happy doing pertubation theory with terms that are literally infinite, but that's okay because the infinities cancel. Similarly, can't I just choose ##\log\left( \mu /E\right)## as large as I want knowing that all ##\mu## dependence cancels out anyway?
No, you can't (well, of course you CAN, but you should not).

While it is true that any ##\mu##-dependence will cancel out at any given fixed order of perturbation theory, it still has an impact on the convergence of your series, because the higher order terms that you include are different.

To see this look for example at the LL running of the strong coupling

If you expand this out in a series in ##\alpha_s## you get

Now consider some perturbative series of the form

Because we set ##\mu = E ## we do not see any logarithms of the form ##\log(E/\mu)## anywhere.
If you truncate your series after the nth order, the terms you neglected are of order ##\mathcal{O}(\alpha_s\alpha_s^n)##. Given that we are in the perturbative regime of QCD and ##\alpha_s## is not too large, this should be a good enough expansion.

Now you could say "but I can set ##\mu## to whatever I want, because it will cancel out to any fixed order in perturbation theory". If you do that, instead of (3) you will get

Given the relation between ##\alpha_s(E)## and ##\alpha_s(\mu)## in (2), truncated after the first order, (3) and (4) are identical up to the second order in perturbation theory as written here.

But look at the higher order terms. If we truncate the series after the nth order, now they are of the form ##\mathcal{O}(\alpha_s(\alpha_s L)^n)##, with ##L=\log(E/\mu)##. If you choose ##\mu## too far from ##E##, these logs will become larger and your series expansion will get worse.

But if you choose ##\mu\approx E##, all these higher order logarithms will be effectively included in the evolution of your coupling in (1). So even though (3) and (4) are equivalent to the order of perturbation theory shown here, they differ in the higher order terms (which are either included in the coupling or not, depending on your choice of ##\mu##) and this can make a huge difference.

vanhees71, erikasan and malawi_glenn

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