In our previous blog post we raised the question what exactly is the virtual photon which in Feynman diagrams mediates electric repulsion or attraction.
/ \
/ \
~~~~~~~~ virtual photon
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e- Z nucleus
Time flows upward in the diagram. The virtual photon apparently is in no way an oscillating wave of the electromagnetic field, like a real photon is.
The virtual photon symbolizes the electric attraction of the nucleus. But why should we call it a photon at all and how can it transform itself into a virtual pair in the vacuum polarization diagram?
Why does Feynman use the Klein-Gordon propagator for the virtual photon? The static electric field is present at all times. What propagates?
The answer may lie in the fact that Feynman studies what happens if the electric field of the nucleus is switched on in a small spacetime patch. We assume that the electric field is otherwise switched off. We want to find the perturbation which happens if we switch the field briefly on.
When the electric field is switched on, an electromagnetic wave will start to spread from the patch. That wave will exert an electric force on the electron. Furthermore, that wave will try to tunnel into real or virtual electron-positron pairs.
It is as if the nucleus Z would hold a capacitor from a handle, put it around the electron, and switch the electric field on for a short time interval.
+ -
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Z ----------------| e- |
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Z uses the capacitor to simulate the electric field it would have if it were not switched off.
If e- acquires a momentum p from the capacitor, the handle will take care that Z acquires -p.
The virtual photon in the Feynman diagram symbolizes the capacitor setup and the real photon(s) which the capacitor sends when it is switched on. We kind of use real photons to reconstruct for a brief time the electric field which Z would have at the electron e-. These real photons will have vacuum polarizarion diagrams like any real photon.
What remains is that we should show our capacitor setup reproduces the Feynman formula for a virtual photon.
UPDATE: Does the capacitor approach make much sense for the static field of the nucleus Z?
We know from experiments the electric attraction that the field exerts on an electron. If our calculation of vacuum polarization would show that the field is weaker, we should just adjust the Coulomb force formula to match the measured value? Or is it so that the Coulomb field is the underlying field, but the effect of the vacuum polarization has to be added anyway?
Vacuum polarization can have an effect on a dynamically changing configuration. Specifically, the electron flies past the nucleus and its momentum will change somewhat. We could make a Fourier decomposition of the changing electric field caused by the electron. That decomposition will show real photons being produced in the flyby. And those real photons cause vacuum polarization. But how does this relate to the Feynman formula?
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