What is the probability of QED interactions?

Question

In Feynman diagrams of QED vertices of interaction are often labelled by the amplitude sqrt(alpha) where alpha is the e.m. fine structure constant. When higher order diagrams are constructed, each time when a virtual photon is created and absorbed the probability of the interaction process gets 1/137 time lower. Is this rule also applicable to real photons, eg in the Compton scattering diagram which has two vertices, is the interaction probability then equal to alpha? And, if an electron and a positron collide emitting two photons, what is the probability of this process?

Answer 1

Since there are two vertices in the lowest-order diagrams for these processes, the amplitude is proportional to $\alpha$ and the probability (such as the differential cross section) is proportional to $\alpha^2$.

Answer 2

We use Fermi's golden rule to calculate the probability of a given interaction per unit time:

$$\Gamma_{i\to f}=\frac{2\pi}{\hbar}|\mathcal{M}_{i\to f}|^2\rho$$

In this formula, $\rho$ is a Lorentz-invariant phase space factor; the nature of this factor depends on the particular interaction configuration (i.e. is it a $1\to 2$ reaction, a $2\to 2$ reaction, a $2\to 3$ reaction, etc.). Meanwhile, all of the "real" physics happens in the matrix element $\mathcal{M}_{i\to f}$, also known as the amplitude. In QED, this is the part where you have terms that are proportional to (not in general equal to) powers of $\alpha$.

Once you have your matrix element for a particular process, to get the total probability of the interaction happening, you must sum (or integrate) over all relevant initial and final states. This is fairly involved, but most QFT textbooks include several examples of this.

For Compton scattering, your matrix element is ultimately proportional to $\alpha$, which means your actual transition probability will be proportional to $\alpha^2$.