Regularization or renormalization are not needed if one uses a mathematically correct approximation method. Ultraviolet divergences are a result of a Feynman diagram only "hitting" the electromagnetic field with one Green's function – which is a poor approximation of the process.
In the diagram above, we see an electron passing close to a very heavy negative charge X.
Let us switch to the classical limit. The electron is then a macroscopic particle with a very large charge, 1.8 * 10¹¹ coulombs per kilogram.
Feynman diagrams have no restrictions about the mass of the particles. The particles are allowed to be macroscopic.
As the large electron passes X, it emits a classical electromagnetic wave which has a huge number (actually, infinite) number of real photons.
The electric vertex correction is about the wobbling of the electric field relative to electron as it passes X. In particular, the far field of the electron does not have time to take part in the process. The electron appears to have a reduced mass as it passes X.
Classically, it is obvious that the inner electric field of the electron tracks the movement of the electron very accurately. The inner field is "rigid", and does not affect the movement of the electron much.
The rubber membrane model
#
#========= sharp hammer
v keeps hitting
______ _____ tense rubber membrane
\__/
• e- weight makes a pit
In the rubber membrane model of the electron electric field, we can imagine that the weight of the electron is implemented with a sharp hammer hitting the membrane at very short intervals.
|
|
|
|
• e-
^ t
|
-----> x
Let us analyze the Green's functions of the hammer hits if the electron stays static in space.
We see that if E ≠ 0, then there is a complete destructive interference for any
exp(i (-E t + p • r) / ħ).
That is expected, since the electric field is static.
Let us then assume that the charge X passes by the electron e-. The electron is accelerated, and gains some final velocity v.
For large |E|, the destructive interference still is almost complete. For what values of E is the destructive interference incomplete?
Let a be the acceleration of the electron. Let Δt be the cycle time of a wave with E ≠ 0.
During the cycle time, the electron moves a distance
R = 1/2 a Δt².
The wavelength is
λ = c Δt.
We see that if Δt is very short, then the electron moves negligibly during a cycle, compared to the wavelength λ. Intuitively, the destructive interference is strong then.
Let t be the time when the electron is accelerated. Intuitively, destructive interference is spoiled the most if the cycle time is t. That is, the wavelength is
c t.
In this blog we have claimed that the electric field "does not have time to follow the electron", if it is at a distance c t from the electron. Destructive interference matches this.
The ultraviolet divergence is due to the fact that a Feynman diagram only hits the electromagnetic field once with a Green's function. In reality, the electron keeps hitting all the time.
Regularization and renormalization of the ultraviolet divergence in the electric vertex corrention
Let us look at how Vadim Kaplunovsky handles the ultraviolet divergence.
*** WORK IN PROGRESS ***
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