In literature, the calculation of the vertex function requires the removal of both an ultraviolet divergence and an infrared divergence.
Matthew Schwartz (2012) in the link writes about how to remove the infrared divergence.
Our semiclassical "rubber plate" model of the static electric field of the electron explains the origin of both divergences, and why they can be removed.
e- ------->
momentum exchange p
● Z+
Let us have an electron passing close to a nucleus Z+. Let us assume that the collision is elastic, that is, no photons are radiated out in the process.
The particles exchange some amount of spatial momentum p.
Assume that the particles exchange the momentum p instantaneously. The electron receives a Dirac delta impulse. The classical spectrum of vibrations of the rubber plate, or the static electric field, around the electron would contain a significant amount of waves from arbitrarily low frequencies to arbitrarily high frequencies.
This reflects the fact that the (generalized) Fourier transform of the Dirac delta distribution δ(0) is a constant function:
δ-hat(ξ) = 1.
The response of the rubber plate to a pointlike infinitely short impulse (= hitting the rubber plate with a sharp hammer) would contain a very wide spectrum of frequencies.
virtual photon q
~~~~~~~~
/ \
e- --------------------------------------------
| virtual
| photon
| p
Z+ --------------------------------------------
Above we have the Feynman diagram of the vertex function. The electron makes a sharp turn when it receives the momentum p from the nucleus Z+. The electron sends a virtual photon q to itself. In a classical model, the photon q is a "summary" of the complex interaction which the electron has with its static electric field.
Removing the ultraviolet divergence
In a classical model we can remove extremely high frequencies from our calculations - that is, remove the ultraviolet divergence.
The electron does not really receive a Dirac delta impulse from the nucleus. Rather, the electron receives a smooth impulse spread over a considerable time which it spends close to the nucleus.
This may be the reason why we can regularize away the infinity in the Feynman integral formula.
The Feynman model assumes that the electron makes a Dirac delta impulse to a massless Klein-Gordon field and produces a Green's function response in the field. The electron later may absorb a component of the Green's function, that is, the photon q above. We sum all such absorption histories to form a picture of the whole process.
The electric form factor depends on the Planck constant h
Do Feynman formulas assume anything about the size of h? The rules and calculations in literature use natural units where h and c are set equal to 1. It is hard to follow where h would appear in the formulas. We know that h does not appear in the differential cross section of elastic Coulomb scattering, but h does appear in the cross section for bremsstrahlung.
In the link above, in formulas (88) and (93), the electric form factor depends on the value of the fine structure constant α:
F₁(q²) = 1 - α / (2 π)
* (1/2 log² (-q² / m²) + divergent part + constant).
Since α is inversely proportional to h, the magnitude of the Planck constant strongly affects the electric vertex function.
When we used our rubber plate model on March 13, 2021 to calculate the Lamb shift, we put a "cutoff" at the zitterbewegung radius. That radius is
λ / (2 π)
where
λ = h / (m c) = 2.4 * 10⁻¹² m
is the Compton wavelength of the electron. It is not clear if the cutoff should be dropped down to the classical radius of the electron.
There has to be a "vertex function" in the classical limit, and it cannot depend on the Planck constant
The bremsstrahlung formula from a Feynman diagram depends strongly on the value of h. We explained that by the fact that the radiation is born in a very small volume of spacetime, and it depends on the wavelength, that is, h, if waves can escape from such a small volume to the environment.
However, a radio transmitter produces electromagnetic waves through a similar mechanism as bremsstrahlung, but the output is classical: it does not depend on h.
In the vertex function, waves do not need to escape to the environment. Should the vertex function depend on h if the momentum exchange is small?
There almost certainly exists a classical "vertex function". If we take a macroscopic amount of charge in our hand and swing our hand, there is some interaction between the charge and its far electric field. The interaction does not depend on the value of h. Why would this be different for a single electron in our hand? Why would its interaction depend on h?
We are not certain if the dependency of the electric form factor on the Planck constant is correct in literature.
Removing the infrared divergence
Consider classical bremsstrahlung. Let the electron pass the nucleus quite far away. The classical radiation in the process is very small, and its energy is roughly evenly divided on all frequencies from zero to about f = 1 / Δt, where Δt is the time the electron is "close" to the nucleus.
The classical wave contains an infinite number of soft photons. We see it this way:
Classical number of photons. The classical wave energy in the frequency range f / 2 ... f suffices for a small fraction of a photon, say, 0.001 photons. Similarly, the energy in the range f / 4 ... f / 2 suffices for another 0.001 photons. We can continue this reasoning and get an infinite number of soft photons. However, only a few of the photons are observable. The total energy of the rest of the photons is infinitesimal.
In the Feynman formula for bremsstrahlung there is an infrared divergence. The formula thinks that just 1 photon is emitted at a time. Each emitted photon is a separate case. The classical model removes this divergence: an infinite number of photons are emitted at a time. The cases which the Feynman formula thinks are separate, actually are overlapping.
Literature uses an erroneous trick to remove the infrared divergences in the vertex function and bremsstrahlung
Both in the Schwartz link and the other link it is claimed that infrared divergences "cancel" each other in the vertex function and bremsstrahlung.
The idea is that the vertex correction for a differential cross section σ of Coulomb scattering is something like
σ * (1 - 1 / ε + a)
and the correction from very low-energy photons (which we cannot observe) is
σ * (1 + 1 / ε + b).
There ε is a positive real number which goes to zero.
In the sum of corrections, the divergent term 1 / ε does not appear.
If we make ε small, then the first correction claims that the cross section is negative. That does not make sense. The trick does not solve the mathematical problem.
Could it be that the "canceling" really is destructive interference? No. In quantum mechanics, interference only happens if we cannot distinguish between two histories. But even an infinitesimal photon is, in principle, observable. There is no interference between elastic Coulomb scattering histories and bremsstrahlung histories.
Bremsstrahlung contains an infinite number of soft photons for a single electron?
Suppose that we hold a macroscopic amount of charge in our hand and swing our hand. The classical electromagnetic wave then contains an infinite number of photons, like we argued above.
But Feynman diagrams, after regularization, for an individual electron claim that sending a single photon is quite rare, and sending two photons very rare.
How can we reconcile these two views?
We believe that the classical picture is correct. Even an individual electron always emits an infinite number of soft photons. This means that elastic Coulomb scattering never happens. Large photons are rare because a large momentum transfer requires the electron to come very close to the nucleus, and also because the large value of the Planck constant reduces the number of large real photons which can escape from there.
Conclusions
We are not sure if the electric form factor F₁ really should depend on the Planck constant for low-momentum transfers. We need to study this more.
The Lamb shift depends on h. In our semiclassical rubber plate model that is because the zitterbewegung radius depends on h.
Classical models explain why divergences can be removed in the electric form factor. But are these the real reason why regularization works in the QED case?
Literature uses a trick to "cancel" the infrared divergence in the electric form factor. The trick does not make sense mathematically.
When looking at the Lamb shift in literature we noticed that vacuum polarization reduces the shift by some 2.5%. Thus, vacuum polarization has to be a real phenomenon. We will study this in detail.
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