1
$\begingroup$

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?

enter image description here

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Kris Feb 13 '17 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.