A photon that spends part of its life as a pair

Everything in Part 3 so far has been preparation. Lesson 3.1 counted powers and told you which diagrams diverge. Lesson 3.2 built the analysis that dimensional regularization runs on. Lesson 3.3 derived the recipe, in the abstract, on an integral with no process attached to it.

This is the calculation.

A photon propagating through empty space can, for a while, stop being a photon. It splits into an electron and a positron, the pair travels, and it closes back into a photon. The amplitude for that — the one-loop correction to the photon propagator — is called the vacuum polarization, and it is written $i\Pi^{\mu\nu}(q)$ :

i\Pi_2^{\mu\nu}(q) \;=\; -(-ie)^2\!\int\!\frac{\mathrm{d}^d\ell}{(2\pi)^d}\;\mathrm{Tr}\!\left[\gamma^\mu\,\frac{i(\not{\ell}+m)}{\ell^2-m^2+i\varepsilon}\,\gamma^\nu\,\frac{i(\not{\ell}+\not{q}+m)}{(\ell+q)^2-m^2+i\varepsilon}\right]

Three things in that expression are worth naming before any of it is evaluated. There is a trace, because the loop is a closed cycle of fermion propagators and the spinor indices have nowhere to go but back onto themselves. There is an overall minus sign, which every closed fermion loop carries and no bosonic loop does. And there are two free indices, $\mu$ and $\nu$ , which means the answer is not a number but a rank-2 tensor — sixteen functions of $q$ , apparently, where every calculation in Part 2 produced one.

It will turn out to be one function after all, and you will know that before you integrate anything.

draw — the amplitude assembles here, term by term

The one-loop photon self-energy

i\Pi^{\mu\nu}(q)

. The fermion circulates around the loop, so the spinor indices close on themselves — hence the trace, and hence the

-1

every closed fermion loop carries.