# "Running with momentum $p$" v.s. "running with renormalization scale $\mu$"

The renormalized charge/coupling in QFT is usually phrased as renormalization scale $$\mu$$ dependent $$\alpha(\mu)$$ in the renormalization group setting. But can we take the more elucidating angle of "momentum $$p$$ or $$p^2$$ dependent" $$\alpha(p^2)$$? The renormalization scale $$\mu$$, as it is taught in most QFT text books (often introduced un-intuitively as the scale parameter in dimensional regularization), is baffling to new learners rather than clarifying.

Let's shed some light on the renormalization scale $$\mu$$ with a simple example of $$x(t) = ln(t/t_0) + x_0.$$ (in the physics context, translated to $$\alpha(p) = ln(p/\mu) + \alpha_0$$ with $$\alpha$$ being the coupling constant, $$p$$ being momentum , $$\mu$$ being renormalization scale, respectively)

The variable $$x$$ is the solution to a first-order differential equation ($$\beta$$-function) of $$\beta (x) = dx(t)/dln(t) = 1,$$ with the initial condition $$x(t)|_{t = t_0} = x_0.$$

The "running with renormalization scale $$\mu$$" approach is tantamount to regarding $$x(t, t_0, x_0)$$ as the solution to an alternative differential equation (differentiating against the initial condition point $$t_0$$, which is $$\mu$$ in physics context) $$\beta '(x) = dx(t_0)/dln(t_0) = -1,$$ with the initial condition $$x(t_0)|_{t_0 = t} = x_0.$$ Is this wicked and naughty way of looking at the original differential equation really helpful (or just add to the confusion)?

Let's take a look at another example of self-energy $$\Sigma(\not{p})$$ in the fermion propagator $$G = \frac{i}{\not{p}-m_0 - \Sigma(\not{p})+i\epsilon}$$ where self-energy $$\Sigma(\not{p})$$ can be generally expressed as $$\Sigma(\not{p}) = a(p^2) + b(p^2)\not{p}.$$ To simplify our discussion, let's assume that (which means there is no wave function renormalization) $$b(p^2) = 0.$$ If we further expand self energy as $$\Sigma(p^2) = a(p^2) = m_0' + c_1p^2 + c_2p^4 + ...$$ we will find out that $$m_0'$$ is divergent, while $$c_1$$ and $$c_2$$ are finite. The whole (mathematically shady) mass renormalization business is hinging on the assumption that $$m_r = m_0 + m_0'$$ is finite (or equivalently, $$m_0 = m_r - m_0'$$, regarding $$m_0'$$ as mass counter term), so that the fermion propagator $$G = \frac{i}{\not{p}-m_0 - \Sigma(p^2)+i\epsilon}$$ $$= \frac{i}{\not{p}- (m_r + c_1p^2 + c_2p^4 + ...) + i\epsilon}$$ is finite and well defined.

Note that while $$m_0$$ and $$m_0'$$ are divergent, finite $$m_r$$ (it's not the physical pole mass $$m_p$$, unless $$c_1= c_2 = 0$$) can be determined by experiment.

On the other hand, the finite coefficients $$c_1$$ and $$c_2$$ can be calculated ($$d\Sigma(p^2)/dp^2$$ and $$d^2\Sigma(p^2)/(dp^2)^2$$ are finite, is that cool! It has to do renormalizability/local counter terms of renormalizable QFT), so that we know how self-energy $$\Sigma(p^2)$$ (or more precisely, the finite and well defined $$m_0 + \Sigma(p^2) = m_r + c_1p^2 + c_2p^4 + ...$$) runs with momentum/energy $$p^2$$.

The whole discussion above about running of $$\Sigma(p^2)$$ does NOT depend on the renormalization scale $$\mu$$ at all!

Update:

"Can you use renormalization schemes without $$\mu$$"? Surely one can, without resorting to any kind of RG (be it Wilsonian/Polchinskian/Wetterichian RG or perturbative QFT RG). Just resume the geometric series (that is how Landau pole was found by Landau!) of Feynman diagrams a la, 1/N (t'Hooft), rainbow/ladder approximation, etc. There are tons of alternative ways of achieving this so called RG enhancement without invoking RG accompanied by the illusive $$\mu$$.

No, you can't simply identify the renormalization scale $$\mu$$ with the momentum $$p$$.

To recap, many renormalization schemes depend on a parameter $$\mu$$ with the dimensions of energy/momentum. The quantity $$\mu$$ need not have any physical interpretation. However, it turns out that if the typical momentum scale of a process is $$O(\mu)$$, then higher-order contributions (loop diagrams) will be smaller.

Hence the seemingly useless and confusing parameter $$\mu$$ is actually one of the greatest advantages of continuum RG over Wilsonian RG. By choosing $$\mu$$, we can make the calculation of a physical observable much more efficient. For instance, the coupling $$e^2(\mu)$$ describes the generic strength of all interactions involving particles with momentum $$O(\mu)$$. (For more detail, see this question.) That's why continuum RG is also called "resummation". It moves around the terms within a series to put most of the contribution in the leading terms.

You can't just say $$\mu$$ is "the momentum" because even the simplest processes have multiple momentum scales. For example, consider your typical $$2 \to 2$$ QED scattering, where particles with momenta $$p_{1i}, p_{2i}$$ scatter to momenta $$p_{1f}, p_{2f}$$. Which of these four momenta is supposed to be $$\mu$$? Actually, none of them! It's usually taken to be the momentum of the exchanged photon, i.e. $$p_{1f} - p_{1i}$$ for $$t$$-channel scattering.

Picking $$\mu$$ is a seriously nontrivial issue. Hundreds of papers have been written on the topic of "scale setting in QCD", which is the question of how to pick $$\mu$$ for QCD processes. This is extremely important for getting accurate results and completely opaque. I was told once that for any given $$\mu$$ you should treat the results you get for $$\mu' \in [\mu/2, 2 \mu]$$ as "theoretical uncertainty".

Can you use renormalization schemes without $$\mu$$? Absolutely, just use Wilsonian RG (for an overview, see here). It is indeed conceptually clearer, but it's never used for precision calculations in particle physics for exactly the reasons above.

• "Can you use renormalization schemes without $\mu$"? Surely one can, without resorting to any kind of RG. Just resume the geometric series (that is how Landau pole was found by Landau!) of Feynman diagrams a la, 1/N, rainbow/ladder approximation, etc. There are tons of ways of doing this. Oct 22 '18 at 16:41
• @MadMax Of course you can avoid having the letter $\mu$ in your formulas if you want, I'm just saying why it is useful to have it. Oct 22 '18 at 17:17
• "No, you can't simply identify the renormalization scale $\mu$ with the momentum $p$." No, I am NOT identifying the renormalization scale $\mu$ with the momentum $p$. Instead, I am identifying the renormalization scale $\mu$ with the initial condition at $p \rightarrow p_0 = \mu$. Jan 12 at 16:52