# Intuition for parameter $\mu$ in dimensional regularization

In dimensional regularization, a dimensionless coupling $$g$$ is replaced by $$\mu^{4-d}g$$ where $$\mu$$ has dimension of mass, so that $$g$$ can remain dimensionless. $$\mu$$ is unphysical, though its choice affects the values of counterterms. By setting the derivatives of observables with respect to $$\mu$$ equal to zero, we get renormalization group equations.

I'm fine with this, but there's an additional claim slipped in: the renormalized couplings you get from using $$\mu$$ "describe physics at the energy scale $$\mu$$".

But when I first learned DR, $$\mu$$ was presented as a parameter that was just thrown in for convenience; it was totally meaningless, like the ghost particle mass in Pauli-Villars. Where does this interpretation of $$\mu$$ come from?

• Can you give a reference that actually claims this $\mu$ from dimensional regularisation is the same as the $\mu$ we write for the energy scale? Commented Dec 10, 2015 at 1:04
• See these lecture notes: ocw.mit.edu/courses/physics/… Commented Dec 10, 2015 at 1:12
• There's no claim they're the exact same thing. But something like that statement is implicit; for example, they say "$\alpha$ becomes stronger going into the infrared, as we decrease $\mu$", where $\mu$ is the one from DR, implying that $\mu$ is some kind of energy scale. Commented Dec 10, 2015 at 1:14

Dimension regularization (dim-reg) is not very intuitive. You could say MS is not a very physical renormalization scheme. There are however several ways in which $$\mu$$ is connected to an actual physical energy scale in applications:

• $$\mu$$ is arbitrary in general, however, in calculations you usually get logarithms of the form $$\log (\frac{\mu}{M})$$ where $$M$$ is some energy scale in your problem. Could be a momentum transfer for example. If you want your perturbative corrections to be small, you better choose $$\mu \sim M$$ otherwise the logs would be large and your perturbative correction would not be small. This is mainly the reason why $$\mu$$ is usually tought of as an energy scale in the problem, even though in principle it is arbitrary. If you follow this prescription for choosing $$\mu$$, you will find that it really is true that at high momentum transfer the QCD 2->2 scattering becomes weaker.

• In problems with several interesting scales this leads to a problem, as you get several logarithms, say $$\log(\frac{\mu}{m})$$ and $$\log(\frac{\mu}{M})$$, with say $$m \ll M$$. In this case you cannot choose a $$\mu$$ such that all logarithms are small. To solve problems of this kind with dim-reg, Effective Field Theory techniques are needed. That is you first construct an EFT valid for momenta smaller than $$M$$ and matching small momentum S matrix elements between the theories. For definiteness, let's say $$M$$ is some heavy particle mass. In this case you would match the S matrix elements for the light particle between the full theory and the EFT witout heavy fields at some scale, say $$\mu \sim M$$, implementing the decoupling of the heavy particle BY HAND. The MS scheme does not satisfy the decoupling theorem, but you can put it in by hand. Similarly as with the former case, you match the theories at $$\mu$$ of order $$M$$ to avoid large logs in the matching.

In both cases, you put in the "interpretation" of $$\mu$$ by hand, to make your life easier and make perturbative correction actually small. In this sense $$\mu$$ in applications is usually connected to some physical scale, even though in principle it could be arbitrary.

I try to give an answer which has a different attitude towards your question. In my very opinion, $$\mu$$ has a very important role, in simplification of amplitude computations using the Renormalization Group. For simplicity, we restrict ourselves to the $$\lambda \phi^4$$ theory.

Imagine that one is supposed to compute an amplitude using Feynmann diagrams, obviously one should include the loops up to the demanded precision.

Using Renormalization group one can prior to any computation choose a proper scale commensurate with the scale of the Mandelstam variables(Which determine the scale of scattering by three quantity for different channels namely s,t,u that are written in terms of the incoming and outgoing 4-momentums)

That chosen scale is literally $$\mu$$! And using the running formula for the couplings(like $$\alpha(\mu)={{\alpha(m)}/1-clog(\mu/m)}$$) we've derived using RG equations, one can find $$\alpha(\mu)$$ where now the coupling in computed at the new scale which is of the same order of the Mandelstam variables.

You might ask "what's the point of it?"

The point: Using the value of the coupling constant at the new scale, we have automatically increased the share of the first terms in the perturbative expansion and minimized the share of the higher-order terms that include lots of loops and make the computations very hard!

In other words, we're accumulating much of the loop corrections using a lower scale coupling, in the tree-level contributions of a perturbative expansion using a higher scale coupling!

Don't forget that the $$\mu$$-independence of amplitudes enables us to do this exchange of "diagram shares".

Unfortunately in MS scheme and also On-Shell renormalization "accompanied" by Dimensional Regularization one can't include all the loop corrections at low scales in a single tree level diagram at a higher scale, but in fact, we can only lower the importance the loop corrections by running $$\alpha(\mu)$$ up to scales $$\mu$$ ~ $$t$$ (or "$$s$$" or "$$u$$").

In the on-Shell scheme, one can imagine that $$\alpha=\alpha(s,t,u)$$ and by determining the value of $$\alpha$$ at a certain scale, like $$(s_0,t_0,u_0)$$ one can find the value of the coupling constant at another scale, like $$\alpha(s',t',u')$$ using RG equations(this time in a 3D space, of 3 Mandelstam variables rather than one variable called $$\mu$$ ) and then there's no need to do any loop computations at all. In a sense, we've made up a full classical effective field theory in which there's no need for loop computation any longer!

The complication rises when $$s,t,u$$ are all of, different orders! Then, in this case, one usually introduces a new variable like $$Q$$ that is the geometric average of Mandelstam variables: $$Q=(stu)^{1/6}$$ which is the desired variable $$\mu$$ in MS scheme.