http://bohr.physics.berkeley.edu/classes/221/notes/emdirac.pdf
In "classical" physics we have a crude model for electron-positron annihilation: the particles come so close in the Coulomb potential that the sum
mass-energy + kinetic energy
doubles.
The acceleration makes them to radiate away all their energy (mass + kinetic + potential) in electromagnetic radiation. What is left is zero energy, zero momentum, and zero charge - the particles have annihilated.
https://www.hep.phy.cam.ac.uk/~thomson/partIIIparticles/handouts/Handout_4_2011.pdf
The crude classical model gives a total cross section of 0.6 nanobarns for a 10 GeW collision
e- e+ ---> μ- μ+,
while the correct, experimental, one is 1 nanobarn. The pair annihilates into one virtual photon, and the photon produces the muon pair.
The simplest Feynman diagram for the above reaction contains the virtual photon. The formula for the diagram contains a photon propagator. The photon propagator is "aware" of the Fourier transform of a static Coulomb potential.
The exact form of the Coulomb potential well is relevant here. If we modify the potential in a way where the deep part of the well becomes very narrow, then we expect the annihilation probability to become very small.
But the simplest Feynman diagram for the production of two photons,
e- e+ ---> γ γ
contains an electron propagator, not a photon propagator. The information about the Coulomb field is not too detailed in the Feynman integral formula: just the coupling constant α! The Dirac field propagator is not related to the Coulomb field (or is it?).
How can the Feynman diagram then know the cross section for the production of two photons?
Maybe it does not know? If the initial momenta for e- and e+ are small, then a positronium "atom" tends to form. Feynman diagrams do not handle bound states.
The photon and electron propagators are schematically
1 / (E^2 - p^2)
1 / (E^2 - p^2 - m^2)
where E is the total energy, p is the spatial momentum, and m is the electron mass.
If E and p are large (like in the 10 GeV experiment), then the effect of m is small - the propagators have almost the same value.
The simple annihilation diagram is equivalent to a Compton scattering diagram rotated in spacetime 90 degrees, such that time becomes a spatial dimension. Has anyone found out why a rotation by 90 degrees, or by 180 degrees, is allowed? The 1949 papers by Feynman just mention that we can rotate, but on what grounds?
The original 1930 annihilation paper by Paul Dirac calculates the probability that an electron under the influence of two photon beams falls into a negative energy state, that is, it becomes a positron.
The crude classical model, which we described above, does not reproduce the Dirac annihilation cross section, which for large energies is ~ 1 / E, while the classical model would predict ~ 1 / E^2. Why is this? Could it be that the emission of the first photon makes the electron to fall closer to the positron, since the electron loses kinetic energy? The cross section is larger because of energy dissipation. The electron loses its kinetic energy as it speeds up.
Remo Ruffini et al. (2009) have written 223 pages about the history:
Electron-positron pairs in physics and astrophysics: from heavy nuclei to black holes
https://arxiv.org/abs/0910.0974
Ruffini et al. mention that the inverse reaction: a collision of two photons which results in pair production, has yet to be observed in a laboratory.
The Schwinger effect has neither been observed. In this blog we have stressed that the Schwinger process is not possible because momentum would not be conserved if an extremely strong coherent flux of photons (= a single laser beam) would produce massive particles like an electron or a positron.
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