mildly relativistic
e- • --------------------------------------
|
| q virtual photon
|
/ \ k + q virtual electron-
\ / positron pair e- e+
|
| q virtual photon
|
● ---------------------------------------
e+ static
Let us analyze purely classical waves which interact. The electron and the positron are presented as waves in the Dirac field. These waves meet each other.
There is an interaction between the electromagnetic field and the Dirac field.
The waves disturb the electromagnetic field, causing waves there. The waves in the electromagnetic field, in turn, disturb the Dirac field.
The vacuum polarization loop is the disturbance in the Dirac field.
If these all are classical waves, then there cannot be any divergence. A divergence would break conservation of energy.
Deep question. Do the quantum waves in the physical world, also in a Feynman diagram, behave like classical waves, or can they behave like in the Feynman integrals, where divergences occur?
If the answer to the question is that the waves must behave like classical waves, then we have a solution to the divergence problem of QED: there is no divergence if the calculations are done in the classical way.
Photon versus a laser beam. A laser beam is a classical wave. A photon behaves just like a laser beam until we measure the photon, and the wave "collapses". That is, a photon must be described as a classical wave until the measurement. Does the same hold for any intermediate state of a Feynman diagram, until the outgoing particles are measured?
Divergences are a result of "breaking into more degrees of freedom"
If one interprets the diagram above as a Feynman diagram, then the phase of the outgoing waves is not affected by k at all. There is a constructive interference for all k. This leads to the notorious divergence problem.
The divergence of the vacuum polarization loop happens when the system "breaks into more degrees of freedom". The value of the the 4-momentum k can be chosen freely – it is a new degree of freedom.
Classical waves also can break into more degrees of freedom. But for them, this does not create any divergences. Why?
The obvious answer is that destructive interference wipes out any classical waves which have high momenta k (short wavelength). If a wave is wiped out, then it cannot pass a disturbance forward.
A classical analogue of the vacuum polarization loop
Let us have two elastic metal plates. They correspond to the electron and the positron in the loop.
Let us model the electromagnetic wave with a wave propagating in a rubber membrane. The rubber membrane is somehow loosely attached to the plates, maybe via very elastic rubber blocks. This constitutes the interaction.
metal plate (e+)
------------------
interaction
/\/\/\/\/\/\/\/\/\/\/\ rubber membrane
interaction
------------------ (e-)
metal plate
--> wave propagation direction
As the rubber wave meets the plates, it interacts with them and creates waves into the plates. Later, these created waves can be absorbed back into the rubber membrane.
Does this mean that the rubber wave "breaks into more degrees of freedom"?
We can model the interaction by assuming that each small area element of the rubber membrane hits with a "sharp hammer" both metal plates. This is the Green's function approach to analyze the process.
The impulse from the sharp hammer generates waves of varying momenta k. There is no limit on how large |k| can be.
Let us analyze different momenta. Does it make sense that the e+ plate receives k and the e- plate -k? If the rubber wave pushes the plates apart (like an electric field pushes e+ and e- apart), then this assumption is reasonable. An area element of the rubber membrane hits both plates with a sharp hammer: one upward and the other downward.
Later, the process can happen time-reversed: the rubber membrane absorbs some waves in the plates.
We seem to have a process analogous to what happens in the Feynman diagram vacuum polarization loop. We can freely choose k, and it will contribute to the waves absorbed back into the rubber membrane.
Why does this not lead to a divergence in the classical system?
One aspect is conservation of energy. Hitting with an infinitely sharp hammer would consume an infinite amount of energy? Then all hits must be done with a "blunt hammer". Destructive interference wipes away all high |k|.
Why should Feynman diagrams allow a sharp hammer? Why not require a blunt hammer?
*** WORK IN PROGRESS ***
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