# Physical observability of running couplings

As far as I can tell, the running of the couplings in QFT is something like $\alpha(\mu)\propto\frac{1}{\ln{\frac{\mu}{\Lambda}}}$ (for QED) where $\mu$ is the unphysical renormalisation scale of a theory.

As $\mu$ is arbitrary, then this running of the coupling must be unobservable. How then does one interpret results such as?

The running coupling is easily observable. Consider, as a toy model, $\phi^4$ theory where we can fix the running coupling by demanding that a $2\to 2$ process has amplitude $$\mathrm{i}\mathcal{M} = \mathrm{i}\lambda(\mu)$$ for a process with Mandelstam variables $s^2 = \mu^2$ and $t^2 = u^2 = 0$ (or $s^2 = t^2 = u^2 = \mu^2$ or whatever, for discussion of this arbitrariness, see this question).

So, while the exact definition of $\lambda(\mu)$ depends on the renormalization scheme you chose, you can direct observe it by looking at the cross sections of the process at $\mu$ in an experiment. $\mu$ is "unphysical" only in the sense that it is an adjustable parameter of the theory that must in principle give the same results regardless of the chosen $\mu$. In practice, due to the large logs that appear when you try to compute processes whose energy scale is far from $\mu$, $\mu$ is very physical - it is the energy scale of the process, for whatever precise notion of "energy scale" you chose to adopt.

• Let me see if I understand correctly then. For your example, we could do a scattering experiment with a particular $s^2$. We could define $\mu^2=s^2$ or $\mu^2 = 3.5 s^2$ as we please. Then we measure $\lambda(\mu)$. Then we perform another experiment at a different $s^2$ and we would measure the $\lambda(\mu)$ that our RG flow equations tell us we should get (provided our definition of $\mu$ is consistent? – Kris Feb 13 '17 at 13:45