Dependence of observables on renormalization scale

Question

Observables are supposed to be independent of the scale $\mu$ introduced during regularization. Srednicki says so and yet the derivatives $\frac{d\alpha}{d\ln \mu}$ and $\frac{dm}{d\ln \mu}$ are taken to be nonzero (Eq.s 28.20 and 28.27 in Srednicki's book), where $\alpha$ is the coupling constant and $m$, the physical mass, both of which are observable. What am I missing?

Answer 1

In perturbation theory, instead of an expansion purely in powers of the coupling constant $g$, you often find that the coupling constant is multiplied by "large logs" of the form $\log p/M$, where $M$ is some mass scale. If $p \ll M$ or $p \gg M$, then this log can be "large" (a very large or small number, not order 1). Since the "effective expansion parameter" is not really $g$ but $g \log p/M$, this can lead to a breakdown in perturbation theory if $\log p/M$ is large.

The renormalization group trick is that in some renormalization schemes, the parameter $M$ that appears in the log is a sliding renormalization scale $\mu$. Since this is an arbitrary parameter and nothing physical depends on it, we can choose $\mu$ so that the logs are under control. In other words, we choose $\mu$ so that perturbation theory converges as quickly as possible, for the given value of momentum transfer $p$.

The "cost" of using our freedom to choose $\mu$, is that the parameters in the Lagrangian become $\mu$ dependent, in such a way that physically observable quantities do not depend on $\mu$. The mass and coupling constant you refer to in your question are presumably these parameters in the Lagrangian, which are not directly measurable. The physical mass is defined by the pole in the propagator, which does not depend on $\mu$. Similarly the physical coupling is often defined in terms of a scattering amplitude; the electric charge can be defined in terms of the three point function $e\gamma \bar{e}$ in the limit that the photon momentum goes to zero.

Answer 2

You are right, observables are supposed to be independent of the arbitrary renormalization scale $\mu$. But, why is everybody talking about $\frac{d\alpha}{d\ln \mu}$?

In reality, $\alpha(p)$ is momentum $p$ dependent. For the example of a scattering process, $\alpha(p)$ is related to the scattering amplitude, and $p$ is the momentum transfer of the incoming/outgoing particles. In other words, $\alpha(p)$ is "running" with momentum $p$, in stead of "running" with the arbitrary renormalization scale $\mu$. So where is this wicked and naughty $\frac{d\alpha}{d\ln \mu}$ coming from?

Let's look at a simplified example of $$ \alpha(p) = \ln(p/p_0) + \alpha_0. $$

$\alpha(p)$ is the solution of the differential equation ($\beta$-function) $$ \beta (\alpha) = \frac{d\alpha(p)}{d\ln p} = 1, $$ with the initial condition $$ \alpha(p)|_{p = p_0} = \alpha_0. $$

The "running with renormalization scale $\mu$" approach is tantamount to regarding $\alpha(p, p_0, \alpha_0)$ as the solution to an alternative differential equation (differentiating against the initial condition point $p_0$, which is the renormalization scale $\mu = p_0$) $$ \beta '(\alpha) = \frac{d\alpha(p, p_0, \alpha_0)}{d\ln p_0} = -1, $$ with the initial condition $$ \alpha(p_0)|_{p_0 = p} = \alpha_0. $$

For historical reasons, the above un-intuitive way of "running with renormalization scale $\mu$" was first introduced as the byproduct of dimensional regularization. Conceptually, it's wrong to invoke "running with renormalization scale $\mu$". The QFT text books can get away from this butchering of physics concept, because mathematically $p$ and $\mu$ usually appear in the form of $\ln(p/\mu)$ therefore differentiating by $\ln p$ or $\ln \mu$ is basically the same with only a sign change.

In my humble opinion, "running with momentum $p$ approach" should be taught in QFT text books, whereas "running with renormalization scale $\mu$" approach is baffling to new learners rather than clarifying.

For more details, please see another post here.

---

Added note:

Why do we say that the renormalization scale $\mu$ is arbitrary and non-physical, while the momentum $p$ of a particle is physical?

For the above simplified example: $$ \alpha(p) = \ln(p/p_0) + \alpha_0. $$ the initial condition is set at $p_0$ (corresponding to setting the renormalization scale to $\mu = p_0$) $$ \alpha(p)|_{p = p_0} = \alpha_0. $$

The selection of the initial point at $p_0$ is totally arbitrary. We can change the initial condition to be at $p_1$ (corresponding to setting the renormalization scale to $\mu = p_1$) and have $$ \alpha(p) = \ln(p/p_1) + \alpha_1. $$ where $$ \alpha(p)|_{p = p_1} = \alpha_1. $$ and $$\alpha_1 = \alpha_0 + \ln(p_1/p_0).$$

As we can verify in the above example, even though the arbitrary renormalization scale is changed from $\mu = p_0$ to $\mu = p_1$, the measurable quantity of $\alpha(p)$ remains the same.