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The renormalized charge/coupling in QFT is usually phrased as renormalization scale $\mu$ dependent $e(\mu)$ in the renormalization group setting. But can we take the more elucidating angle of "momentum $p$ dependent" $e(p^2)$? 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 $$ x(t) = ln(t/t_0) + x_0. $$ (we can imagine that $x$, $t$, and $t_0$ correspond to renormalized coupling/vertex/charge $e$, momentum $p$, renormalization scale $\mu$, 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 $$ \beta '(x) = dx(t_0)/dln(t_0) = -1, $$ 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 $\mu$.

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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.

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  • $\begingroup$ "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. $\endgroup$ – MadMax Oct 22 '18 at 16:41
  • $\begingroup$ @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. $\endgroup$ – knzhou Oct 22 '18 at 17:17

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