On one hand, in the classical electrodynamics polarization of transparent media yields in lowering the speed of light by the factor of $n=\sqrt{\epsilon_r \mu_r}$ (refractive index).
On the other, in the quantum field theory the vacuum polarization does not decrease the speed of light. The thing it does is increase of strength of electromagnetic interaction. Why is it so?
My two guesses are the following:
- In principle the vacuum polarization does decrease speed of light ($\rightarrow 0$) but we implicitly put on-shell renormalization to keep $c$ constant.
- For some reason the process is completely different from polarization of transparent media.
Best Answer
In transparent media, there is matter present. The refractive index has its origin in the tree level process of $$e^- \gamma \to e^- \to e^- \gamma$$ (with $e^-$ either a part of the atom or free as in metals). There is no need for renormalization here, it's just a standard contribution to the probability amplitude of photon propagation. There are also other processes (interaction with nuclei and whatever loop corrections you can think of) but these are only second (and more) order corrections. In any case, what happens here is that photons gain effective mass via all those interactions, which is the ultimate reason they propagate slower.
In vacuum the relevant process is a loop-level one $$\gamma \to e^+e^- \to \gamma$$. You'll obtain an UV divergence here and to take care of it introduce a renormalization of the field which is what that process corresponds to in the QED Lagrangian (the relevant term being $F^2 = (\partial A)^2$. But this is just a kinetic term, not a mass term (which would be something like $m_A^2 A^2$). You could imagine adding such a term to the Lagrangian but that is of course inconsistent with the observation. We know that photons travel at the speed of light (in the stronger sense of special relativity that is built-in in every QFT).