I'm studying different tools for regularization and renormalization. Until now I vaguely understand

1) the wilson approach to renormalization where one thinks of the theory as essencially effective and sees how the couplings evolve with the scale $\Lambda$ as one integrates high energy modes and 2) The cut-off approach (I haven't seen a better name yet) that can be found in Chapter 3. of Zee's QFT in a nutshell where he calculates n-point correlation functions using a cut-off $\Lambda$ and then rewrites the results in terms of physical couplings instead of the bare ones.

Both of this approaches end up with physical coupling constants that depend on the scale. However, when I looked at dimensional regularization and counterterm renormalization I found out that altough the infinities are removed there doesn't seem to be any running for the coupling constants. Is this correct?

The question would be: Does dimensional regularization (and counterterm renormalization) give rise to running coupling constants as the other regularization methods? And if this is so, does every renormalization scheme implies running coupling constants?

  • $\begingroup$ A side note: "Running" is actually in terms of the tangible scale of the scattering/physical process, which is $p^2$ rather than some obscure "renormalization scale $\mu$" (text book QFT view) or "cutoff scale $\Lambda$" (Wilsonian view), See here: physics.stackexchange.com/questions/436173/… $\endgroup$
    – MadMax
    Commented Jun 25, 2019 at 21:14

2 Answers 2


Does dimensional regularization (and counterterm renormalization) give rise to running coupling constants as the other regularization methods?

Yes. In Dimensional Regularization (DR) schemes, you always introduce a scale $\mu$, for (mass!) dimensional analysis consistency. Renormalized couplings depend on this scale $\mu$ as dictated by the RG equations: $$\mu \frac{\text d}{\text d \mu} g^i = \beta ^i(g).$$

In practice, the scale $\mu$ is introduced by requiring the action to be dimensionless. Take for instance massless $\phi ^4$ theory: $$\mathscr L = \frac{1}{2}(\partial \phi_0)^2-\frac{g_0}{4!}\phi_0 ^4,$$ where the subscript $0$ denote the bare couplings. In $d=4-\epsilon$ dimensions, you can easily see that $g_0$ has mass dimension $$[g_0]=\epsilon.$$ You can define a dimensionless renormalized coupling $g$ by: $$g_0 (\epsilon)= Z_g(\mu ,\epsilon) g(\mu,\epsilon) \mu ^{\epsilon}.$$ By requiring $g_0$ to be independent of $\mu$, you can derive the RG equation satisfied by $g(\mu,\epsilon)$ (in arbitrary space-time dimension, in particular in the limit $\epsilon \to 0$).

In the above example, you are somehow forced to introduce a new parameter $\mu$, but the same procedure can be applied if your original four-dimensional theory already contains some mass scale at the classical level. For instance, if the scalar $\phi _0$ had physical mass $m$, you may as well define: $$g_0 = Z g m^{\epsilon},$$ without having to introduce a new scale $\mu$. This is perfectly consistent with dimensional analysis, but also less useful from the practical point of view, because it does not allow you to tame "large-logs" by a clever choice of $\mu$.

And if this is so, does every renormalization scheme implies running coupling constants?

As I hope is clear from the above discussion, a running coupling is completely a matter of definition. You can do without it in dimensional regularization, by simply fixing it once for all.

However, in modern particle physics, the phenomena of interest range from the $\text {GeV}$ scale of hadronic physics to the $10^{19} \text {GeV}$ scale of quantum gravity (?). In this context, using a running coupling allows you to trust the results of leading order computations without worrying of large-logs.

So, I would dare to say that every useful renormalization scheme implies running couplings.


Firstly, the running of the coupling captured in the beta function is a fundamental property of a theory and must be independent of any scheme. If $\beta = 0$ for all couplings, this would imply a scale invariance.

Now, if we are performing renormalization just to quickly compute a diagram, one can apply regularisation along with Zimmermann's forest formula to do so, without ever introducing a scale, $\mu$ as it can be introduced at the end using dimensional analysis.

However, if one wants to formally perform renormalization, re-writing the Lagrangian, then it becomes natural to introduce a scale $\mu$ which abosrbs the dimensionality of the coupling, so that one can work with a dimensionless coupling in the perturbative treatment.

The factor relating the bare and renormalized coupling is what contributes to the beta function. Dimensional regularization is simply a method to keep the infinites in an unevaluated form, such that one can subtract them. There are an infinite number of ways to regularize, but the principle of renormalization is the same.

  • 2
    $\begingroup$ I think that your first sentence is highly misleading. You are probably aware that different schemes give different beta functions, both in the UV (maybe starting at some $\ell >1$ loop order) and in the IR. The simplest example which comes to my mind is dim'reg vs. some mass dependent scheme, such as "momentum subtraction". In the latter scheme, particles automatically stop contributing to the runnings below their mass thresholds, while in the former they need to be explicitly integrated out from the theory. $\endgroup$
    – pppqqq
    Commented Jul 20, 2018 at 16:43
  • $\begingroup$ @pppqqq I think that the intended notion in the first sentence is that it should be possible to translate calculations made using renormalization of any physical theory into real world observables and that the real world physical observables should end up being the same no matter what scheme is used even if the way that you get there with different schemes is not the same (subject to theoretical uncertainties in the respective calculations based upon how many terms the answer is calculate to in a particular scheme). (Not sure this is mathematically true for all nonphysical theories.) $\endgroup$
    – ohwilleke
    Commented Jun 24, 2019 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.