Why does an annihilating pair usually emit just 2 photons? Why not 200?

Paul Dirac developed the hole theory of electrons. There is a sea of electrons in the energy state -511 keV. We can lift such an electron into an energy state +511 keV. It becomes a real electron, and the vacant state is a positron.

In semiconductors, things like this do happen. Holes are vacant states of electrons. Holes behave like positrons.

How many electrons in the -511 keV state there are in a cubic meter of vacuum? Very many. In pair production, such an electron is always available.

Annihilation happens when a +511 keV electron changes state to a -511 keV electron.

Dirac calculated the cross section for annihilation by assuming that the electron moves to the -511 keV state through an intermediate state. It usually emits two photons in the annihilation process.

Why just two photons? Classically, the electron would spiral into the positron, and radiate at very many frequencies until it is destroyed.

A Feynman diagram of a 3 photon emission contains more vertices than a the 2 photon diagram. The calculated probability amplitude is much lower for 3 photons.

Why does Nature favor a small number of photons?

In the classical model, probably most of the energy is radiated during the last few rounds in the spiral. Let the radius of the circle be the electron classical radius 3 * 10^-15 m. The electron moves at almost the speed of light. A round takes 6 * 10^-23 s. The frequency is 1.7 * 10^22 Hz, which corresponds to a photon of energy 10^-11 J  or 70 MeV.

The classical model seems to favor photons of very large energy.

But the electron might emit many "soft" photons of much lower energy at the initial stages of its downward spiral. Feynman diagrams do not handle such almost bound states. Have we seen soft photons in experiments?

The number of photons and the coupling constant

Suppose that we have a classical electron orbiting a proton at a fixed radius r.

Let us reduce the Coulomb constant to one half. This reduces also α.

The speed of the electron is reduced by a factor 1 / sqr(2), as well as the energy of a single photon it might emit.

The kinetic energy (or the negative potential energy) is reduced by a factor 1 / 2. If the electron were to radiate away its kinetic energy, fewer photons are needed now that α is smaller. A small α tends to favor a small number of photons.

Our simple model is consistent with Feynman diagrams: if we make the coupling constant α smaller, then diagrams with many photons will have a lower probability relative to diagrams with few photons.

The smallness of the coupling constant explains why annihilation usually produces just two photons.