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 in the vertex correction.
If q² = 0, then the vertex function should be 1. We decide that the "counterterm" δ₁ must have the value:
With that value, the vertex function F₁^net(q²) has the right value 1 when q² = 0.
What is the logic in this? The idea is that the infinite value of the integral F₁^loops(q²) is "renormalized" to zero when q² = 0. We calculate a difference of the integral value when q² ≠ 0, compared to the integral value when q² = 0. The difference, defined in a reasonable way, is finite, even though the integral is infinite.
What is the relationship of this to our own idea in which destructive interference is used to make the integral to converge?
If q² = 0, then we claim that destructive interference cancels, for the Green's function, every Fourier component for which E ≠ 0. This is equivalent to the "renormalization" in the utexas.edu paper, where a "counterterm" δ₁ erases the entire Feynman integral.
What is the meaning of the difference
F₁^loops(q²) - F₁^loops(0)
for q² ≠ 0?
In the link, for q² << m²,
where λ is the "photon mass" used to regularize the infrared divergence. There is no rule of how we should choose λ. The formula is vague.
Is the electric form factor F1(q²) a microscopic quantum effect?
We are struggling to find the analogue of the electric form factor F₁(q²) in the classical limit. If the electron is a macroscopic charge, then the wobbling of its electric field will reduce the mass of the electron, since the far electric field of the electron does not have time to react.
If the mass of the electron is reduced, and it passes a negative charge X, then X will push the electron away a little bit more: the momentum exchange is reduced, and the cross section is less.
But if X is positive, then the reduced mass of the electron allows it to come a little bit closer: the momentum exchange is larger and the cross section is more.
In the literature, the form factor F₁(q²) only depends on the square q² – it does not differentiate between X being positive or negative.
The tree level diagram of e- X scattering only depends on q, not on the electron mass. Thus, the mass reduction would not even show in the Feynman integral cross section.
The quantum imitation principle, which we introduced on September 19, 2025, may solve the problem. When the electron meets X, the electron tries to "build" its electric field with a photon. But the resources of the electron only suffice to send one large photon (mass-energy ~ me) at a time. The electric form factor F₁(q²) would be a result of this shortage of resources.
In the classical limit, the electron is able to send many large photons simultaneously, and build its electric field at a high precision.
The Feynman integral may work correctly if the electron passes very close to X. Then the resources of the electron are severely limited. It may send a single large photon, attempting to build its electric field.
In the classical limit, the form factor F₁(q²) clearly is wrong. It does not describe the wobbling of the classical electric field.
We have to check if any empirical experiments have verified the factor F₁(q²). Does the anomalous magnetic moment depend significantly on F₁(q²)?
A practical calculation when e- is relativistic and meets a massive charge of size e-
Let us assume that the electron e- is relativistic and is deflected by X into a large angle. Let us try to estimate the magnitude of the electric vertex correction.
The southampton.ac.uk link above suggests that
F₁(q²) ~ 1 + α / (3 π).
That is, the cross section increases by ~ 1 / 1,300.
^ v ≈ c
/
/
e- • --------
● X
The mass-energy of the electric field of the electron at least a Compton wavelength
λe = 2.4 * 10⁻¹² m
away is α / (2 π) = 1/861 of the electron mass me.
Since the relativistic electron is scattered to a very large angle, its closest distance to X must be
~ re = 2.8 * 10⁻¹⁵ m.
If we reduce the mass of the electron by 1/861, then as it passes X, it will come closer to X. Let the time that the electron spends close to X be
t = 2 re / c.
The acceleration of the electron toward X is something like
a = c / r
= c / (2 re / c)
= c² / (2 re).
The acceleration takes the electron closer to X, very crudely:
Δr = 1/2 a t²
= 1/2 c² / (2 re) * (2 re)² / c²
= re.
If we reduce the mass of the electron by 1/861, the impulse that the electron receives is ~ 1/861 larger. We expect the scattering cross sections to grow something like 1/861.
However, we do not understand how the Feynman diagram could be able to calculate this. The Feynman diagram calculates the probability amplitudes for various scattering momenta p' ≠ p, if p is the initial momentum of the electron.
Could it be that the two electron propagators in the integral become larger if we reduce the mass of the electron:
No. The value of the propagator measures "how much" the electron is off-shell. Reducing the mass of the electron does not bring the electron closer to on-shell.
far field
• ---> v
|
| rubber band
|
e- • ---> v
● X
Classically, the electric field of the electron becomes distorted when the electron bounces from X. If we treat the electron and its inner field as a single particle, that single particle (reduced electron) is "on-shell" after the bounce.
After the bounce, the electron must supply the missing momentum to the far field of the electron, and must get rid of the excess kinetic energy that the electron has. In this sense, the electron is "off-shell" after the bounce, if we treat the electron and its entire electric field as a single particle. The excess kinetic energy escapes as electromagnetic radiation.
The bounce puts the electron and its far field into an "excited state". The process can be understood like this:
- The bounce from X converts kinetic energy of the electron into an excitation of its electric field. The excitation decays by emitting electromagnetic radiation.
The process will always radiate real photons which have large wavelengths. Elastic scattering is a process which does not happen at all. If Feynman diagrams claim that it is possible, then they miscalculate the process.
Feynman diagrams calculate the energy radiated in an inelastic collision, and subtract that energy from the elastic path
Now we figured out what the inelastic and elastic Feynman diagrams calculate. (Inelastic = a real photon is radiated. Elastic = no photon is radiated.) This interpretation is inspired by the rubber membrane model.
1. If the electron is not under an acceleration, it will always reabsorb all the real photons which it emitted when it hit the electromagnetic field with a Green's function.
2. The diagram which concerns a real photon emission, calculates the probability amplitude of that photon being emitted. It does not calculate the electron flux. We cannot expect the sum of these amplitudes be 1 – rather it is infinite, since an infinite number of photons are always emitted. The diagram, loosely, calculates radiated energy.
3. The elastic Feynman diagram simply reflects the fact that if energy is radiated out, then the electron cannot reabsorb that energy.
The above interpretation explains why the infrared divergent parts of the tree level radiation diagram and the elastic diagram cancel each other exactly: they both calculate how much energy is radiated out!
Since the electron will always radiate real photons when it passes by a charge X, the elastic Feynman diagram is useless? It does not describe any process in nature. It does not calculate anything useful.
Radiated photons reduce the energy and change the momentum of the electron. The photons, all of them, must be taken into account when we calculate the momenta of the electrons coming out from the experiment.
What does this mean concerning the ultraviolet divergence? We discard the purely inelastic Feynman diagram, but there are Feynman diagrams which contain both the emission and reabsorption of a virtual photon and an emission of a real photon. The ultraviolet divergence can dive up there.
~~~~~~
/ \
e- ---------------------
| \
| ~~~
|
X ---------------------
Does the elastic collision Feynman diagram calculate anything useful?
Let an electron pass by a charge X. We pretend that there is no radiation out. Then the elastic Feynman diagram is relevant. If q² varies, what does the integral calculate?
k
~~~~
/ \
e- ---------------------
The diagram for the free electron is above.
k
~~~~
/ \
e- -----------------------
| 2
| q
X -----------------------
The diagram for the colliding electron differs from the free electron, because we have the vertex and the electron propagator marked with 2 in the diagram, as well as the photon propagator for q. The photon q propagator is factored out from F₁(q²).
The electron propagator measures how much the electron is off-shell: being more off-shell means that the absolute value of the propagator is smaller.
It is obvious that q² affects the Feynman integral, but does it have any intuitive meaning?
In the rubber membrane and the sharp hammer model, if we assume that there is no energy loss, then the hammer must hit at the same place as it did earlier. The process is the same as for the free electron. If we make the (unrealistic) assumption that there is no radiation loss, then the Feynman integral should have the same value as for the free electron.
Classically, we can think like this: we strip the far electric field of the electron off, to remove radiation losses. We keep the rigid inner field. Since the inner field is rigid, we can assume that the electron has no electric field at all, and the mass-energy of the field is in the mass of the pointlike electron. The elastic Feynman diagram is useless: we can just look at the tree-level diagram.
But could there be some microscopic effect which comes from the quantization?
*** WORK IN PROGRESS ***