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 for 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?
Let X be static. The spatial momentum p of the electron changes its direction as the electron absorbs q, but the energy and p² do not change.
The ultraviolet divergence in the classical limit would come from the abrupt momentum change q?
Suppose that we have a classical charge which makes an instant turn. We ignore the fact that it will slow down as it radiates electromagnetic energy.
The power of the radiation is, according to Larmor,
P ~ a².
If we let the charge to accelerate to a speed v in half a time, the power is fourfold and the the time a half: the radiated energy is double.
In this way, we get an "ultraviolet divergence" in the classical limit. In the classical picture, this could show up as an ultraviolet divergence of the opposite sign in the absorption of waves sent by a Green's function.
~~~~ "cut-off" for a real photon
/ in tree level diagram
e- ----------------
~~~
/ \ no cut-off in elastic
e- ---------------- diagram
The requirement of energy conservation implements an ultraviolet "cut-off" to the tree level Feynman diagram. But there is no cut-off in the elastic Feynman diagram. This explains why there is no matching ultraviolet divergence in the tree level diagram.
The classical model suggests that the electron rarely radiates a large real photon
Let us have a mildly relativistic electron and let X have a charge -e. We assume that the electron is scattered to a large angle. We calculated in the previous blog post from the Larmor formula that the energy radiated by the electron is
W = 10⁻⁵⁷ / R³,
where R is the minimum distance, e.g., 10⁻¹⁴ m. The radiated energy is then
W = 10⁻¹⁵ J
= 6 keV.
The electron can, in principle, radiate away most of its kinetic energy, which for a mildly relativistic electron is ~ 500 keV.
This suggests that at most in ~ 1/100 of cases does the electron radiate a large photon. In most cases, the encounter is "mostly" elastic.
If the tree level Feynman diagram correctly calculates the probability of radiating a large photon, then that (small) probability should be deducted from the corresponding "mostly" elastic encounter probability.
This result is hard to map to the traditional interpretation of Feynman diagrams. There are no totally elastic encounters at all: an infinite number of small real photons are always radiated.
Most encounters are "mostly" elastic: there is no large photon radiated.
A small portion of encounters are very much inelastic: a large photon is radiated.
Suggested solution: we do not need the elastic Feynman diagram at all – use only tree level diagrams
e- --------------------------
| q
X --------------------------
The simplest Feynman diagram approximates the scattering of the electron quite well. Bremsstrahlung typically contains very little energy and momentum.
~~~~~~~~
/
e- --------------------------
| q
X --------------------------
The simple tree level bremsstrahlung diagram can be used to calculate real photons which are emitted. In rare cases, a very large real photon is emitted, and that significantly alters the 4-momentum of the outgoing electron. Here we assume that the diagram fairly well imitates the classical spectrum. That should be checked.
If no large photon is emitted, then the process is almost classical. An infinite number of very small real photons are emitted. A classical approximation probably is much more accurate than a Feynman diagram. That also holds for the scattering amplitudes of the electron: they should obey the Schrödinger equation, which in turn must obey the classical limit.
There is no need to do anything about the infrared divergence. The divergence simply is a sign that an infinite number of small photons are emitted.
What about the ultraviolet divergence? There is no ultraviolet divergence in tree level diagrams. No need to do anything.
Analysis of the ultraviolet renormalization in textbooks
Why would the counterterm
δ₁ = -F₁^loops(0)
work at all in the elastic Feynman diagram?
Let us once again refer to the rubber membrane and the sharp hammer model. If q² = 0, then then the entire Green's function from a hammer hit will be absorbed. But if q² ≠ 0, that means that the hammer position is accelerated. Not every wave produced by the previous hammer hit will be absorbed. The difference
F₁^loops(q²) - F₁^loops(0)
will then contain the escaped waves, the sign flipped.
Now we see a problem here. If a real photon contains more energy than the kinetic energy of the incoming electron, then the photon cannot escape.
But the elastic Feynman diagram still thinks that such a photon produced by the Green's function can avoid being absorbed? That would be illogical. That photon would contribute negatively to the difference above.
The textbook renormalization may have an error here. If a virtual photon is such that it cannot escape, then we have to assume a 100% absorption for it. This differs from the textbook treatment.
We suggested above that we can avoid renormalization altogether by using tree level Feynman diagrams. Thus, it may be unnecessary to worry about the renormalization of the elastic Feynman diagram.
More complicated diagrams with loops
Our analysis above suggests that tree level diagrams are sufficient for the analysis and we can skip the simplest Feynman diagram with a loop: the elastic scattering diagram. What about more complex diagrams?
k virtual photon
~~~~~~
/ \
e- ---------------------
| \
| ~~~
| q
X ---------------------
Above we have a complicated diagram. Classically, some of the field of the electron does not have time to take part in the movement of the electron e- as is passes by the charge X. If the virtual photon contains a lot of mass-energy E, then the effective mass of the electron is greatly reduced. The reduced mass classically should increase the radiation of the electron a lot as it passes by X.
This effect should be significant and measurable in experiments. In a few cases, the electron sheds a lot of mass to the virtual photon.
Now we have to deal with the ultraviolet divergence. What prevents the electron from losing temporarily all of its mass, or even more?
If q is small, then we can appeal to the classical limit: virtual photons cannot affect the path of the electron much. We simply can ignore the diagram with a loop.
How to prevent huge virtual photons in a loop?
Suppose then that q is large and k carries more energy than the electron possesses in the first place. Can we then extract that energy and break conservation of energy?
A Feynman diagram enforces conservation of energy for outgoing particles. Is this enough? It seems very odd if we probe the electron with another particle, and that particle scatters from a photon which has much more energy than the electron.
Suppose that we have a particle Y which can jump over a high potential wall and keep the energy if it absorbs a very large photon. If the virtual photon k "really can exist", then it can push Y over the potential wall, and conservation of energy is broken. This suggests that large virtual photons cannot exist. Virtual photons have to obey conservation of energy.
Let us then look at the wave model of a free electron whose 4-momentum is p. If the electron wave ψ(p) meets an electromagnetic wave k, the k wave will disturb the electron wave, according to the interaction in the lagrangian. A part of the k wave is "absorbed" by the electron, producing a
ψ(p + k)
wave.
The emission of a virtual photon by the electron wave is the reverse of this process.
___________ ψ(E, 0)
___________
___________
^ t
|
------> x
Why would the electron wave ψ(p) emit a photon wave k? In the diagram we have the electron wave for a static electron with the energy E = me c².
If the electron emits a virtual photon k = (E, 0), then the photon, obviously, will inherit its phase from the electron.
If the photon has a larger energy than E, is there anything which could determine the phase of the "extra energy" in the photon?
If not, then we found a cut-off which is like the conservation of energy cutoff for the outgoing particles in a Feynman diagram.
Conclusions
If an electron scatters from a charge X, we showed that we do not need the elastic scattering Feynman diagram at all. There is no need to handle the ultraviolet divergence in the simplest elastic diagram.
However, for more complex Feynman diagrams, the ultraviolet divergence still appears. In the next blog post we will study if destructive interference imposes a cut-off which removes the divergence.
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