In this blog we have touched this question many times. If we let high-speed electrons collide head-on, then their scattering can be calculated in two ways:
1. from the Coulomb electric field, using classical physics, or
2. from the simple Feynman diagram, by integrating the "probability amplitude".
It is easy to calculate that the Coulomb model and the Feynman model approximately agree about the scattering probabilities. But why do they agree?
Also, what is the "virtual photon" γ which the electrons exchange?
In the Feynman diagram, the photon is a Fourier component of Green's function which describes the impulse response of the electromagnetic field. What does it mean that the other electron "absorbs" that component?
The classical limit
Nothing prevents the particles in Feynman diagrams from having macroscopic masses and charges.
If Feynman diagrams are correct, they must predict the behavior of macroscopic charges approximately right.
That is the case, at least in electron-electron scattering. It is easy to calculate that they correctly predict the scattering from the Coulomb field.
The macroscopic counterpart of the vertex correction is the interaction of a macroscopic charge with its own electric field. We have in this blog introduced the "rubber plate" model for the electric field: it is an elastic object which tries to keep up with the charge if the charge is accelerated.
With the rubber plate model, we may be able to calculate the classical vertex correction. Feynman diagrams should approximately replicate it. If that is the case, why do they calculate the same result? We do not know yet.
The classical limit of the magnetic moment μ is a more difficult case. We do not know what the classical limit is supposed to be. Does the charge move in the zitterbewegung loop at the speed of light? For a macroscopic particle, the loop length is extremely tiny,
h / (m c).
The rubber membrane model once again
In the rubber membrane model, we may imagine that the charge keeps hitting the membrane with a hammer, at short intervals, emitting Green's functions. The hammer also absorbs impulses which come back from previous hits, since the membrane wants to straighten up.
/
/ |
/ | γ virtual photon
\ |
\ |
\
\
e-
^ t
|
In this model, it is quite a natural assumption that the outgoing electron absorbs impulses which were emitted by the incoming electron. As if the incoming electron and the outgoing electron were different particles.
The Feynman diagram above looks sensible in this interpretation.
The "electric" F1 term in the vertex correction
In the utexas.edu link of the previous post, the vertex function has the form above. The term F₁ is the "electric" term, for which the loop correction
F₁(q²) → 0,
when q² → 0. In the rubber plate model of the electric field, we guess that F₁ corresponds to the effects of the elastic electric field "wobbling" when the electron meets another charge and changes course. If the electron passes by a nucleus, for example, at a large distance, then there is very little wobbling of the electric field. The wobbling does not affect much the momentum that the electron gains when it flies past the nucleus.
Question. Does the shape of the electric potential affect the electric vertex correction F₁? The electron can change course abruptly if it passes close to a very small charge. It it passes far from a large charge, then the change in the velocity vector is gradual. In the classical analogy, the wobbling in the first case should be significant, but in the second case insignificant.
In a Feynman diagram, a virtual photon q which contains just momentum, no energy, is a Fourier component wave of a type
exp(i p x / ħ).
It does not care about the shape of the Coulomb potential it was derived from.
Let the electric field originate from a macroscopic point mass M with an electric charge Q. Then only the coefficients of the Fourier components depend on Q. What happens in a Feynman diagram if we scale the coefficients by some constant C?
The "flux" of the virtual photon q depends linearly on C.
Does the scattering probability amplitude scale with the factor C²? Obviously, yes.
\ |
γ |\ |
| \ |
| ~~~~ q
| / |
|/ |
/ |
e- Q
^ t
|
Let the electron e- in the Feynman diagram be macroscopic, too. The diagram above contains the horizontal virtual photon q, and the vertex virtual photon γ.
But now we have a contradiction with the classical limit. In the classical limit, the elastic electric field of the e- behaves quite differently if Q is small and e- passes it quickly, than in the case where Q is large and e- uses a lot of time to pass it far away. In both cases, the momentum q is transferred, but for Q small, the electric field of e- wobbles much more.
Note that for a large Q we could use the Schrödinger equation to calculate the path of e- nonperturbatively. Any effect from the dynamics of the electric field of e- are negligible.
We may have found a case in which Feynman diagrams miscalculate the vertex correction.
The "magnetic" F2 term in the vertex correction
The magnetic term F₂(q²) multiplies a spin matrix σ. The photon q couples to the spin, or the magnetic moment μ, of the electron.
The term F₂(q²) does not go to zero when q² approaches zero. This indicates that the magnetic moment μ is roughly 1/861 times larger than the one predicted by the Dirac equation.
We have speculated that the "bare mass" of the electron is 1/861 times smaller than the full mass me. Then the Dirac equation gives a magnetic moment which is 1/861 larger.
Classically, the "bare mass" of a point charge should be minus infinite, because its electric field has an infinite energy.
We have to analyze the Feynman vertex integral, in order to understand what logic it uses to calculate the bare mass of the electron.
Let us only consider photons with a positive energy 0 < E < me c². The probability amplitude for the electron sending a photon whose energy is
~ 1/2 me c²
is quite small, about 1/1,000. Otherwise the magnetic momentum of the electron would be doubled.
What is the "meaning" of this 1/1,000? Let us try to "build" the electric field potential of the electron up to the distance 1/2 of the Compton wavelength λe. We need Fourier components whose wavelenght is λe.
We do have the energy me c² available in the process. In quantum mechanics, we can then describe things whose size is λe. We are able to describe the electric field of the electron down to the distance λe.
Hypothesis. Feynman diagrams are able to describe and understand the energy of the electric field of the electron down to the distance
h / (M c)
if the mass energy M is available in the process.
Since the available mass-energy for a static electron is me, Feynman diagrams can describe the electric field's energy down to the distance
λe = h / (me c).
Feynman diagrams reduce the mass of the electron by a factor 1/861, because that mass-energy is in the electric field of the electron, not in the electron itself.
This explains why our semiclassical zitterbewegung argument produces the same anomalous magnetic moment as Feynman diagrams. Both methods calculate the reduced mass of the electron in the same way, subtracting the mass-energy of the electric field farther than λe from the electron.
Why and how do Feynman diagrams understand the energy of the electric field?
Our hypothesis above claims that Feynman diagrams have a surprisingly high ability to understand and calculate things about the electric field.
Even better do Feynman diagrams describe electron-electron scattering: they can model strong scattering where the distance of the electrons is only the classical electron radius,
re = 2.8 * 10⁻¹⁵ m.
But we only claim that they understand the electric field energy at the scale
λe = 2.4 * 10⁻¹² m.
Incidentally,
re / λe = α / (2 π) ≈ 1/861.
The quantum mechanical principle: you cannot know things which are much smaller than the Compton wavelength?
This principle is broken by electron-electron scattering: we do get information down to the classical radius of the electron.
Why does quantum mechanics break its principle in this case?
And why does quantum mechanics obey the principle when it analyzes the field energy of the electron?
A possible solution: the electric 1 / r potential just "happens" to be such that one can describe scattering which takes place at a very small distance
A time-independent plane wave whose wavelength is the Compton wavelength of the electron, can describe an arbitrary electric potential down to the scale of the Compton wavelength.
But the formula for the electric potential, ~ 1 / r, happens to be such that this crude description produces the right scattering probability down to the classical electron radius.
It is a lucky coincidence.
To describe the energy of the electric field of the electron, we need quanta with an energy E. Since E is limited to me c² in the case of a slow electron, we only can describe the energy of the electric field down to the distance
λe
from the electron.
Quantum mechanics thinks that the electron has given up 1/861 of its mass-energy to its electric field. The rest of the mass is governed by the Dirac equation. This explains the anomalous magnetic moment.
Another interpretation: quantum mechanics thinks that in 1/861 of cases, the electron has emitted a real photon whose energy is ~ 1/2 me c². That doubles the magnetic moment of the electron in those rare cases.
Renormalization of the electron mass: the bare mass problem
We found a solution to the classic problem: what is the bare mass of the electron?
The bare mass is
860/861 me,
for a slow electron. The electric field of the electron only carries 1/861 of me.
We can say that electric field very close to the pointlike electron has no energy, if probed with another slow electron, or a slow positron.
In 2021 we tried to solve the electron mass renormalization problem by claiming that its electric field has no energy at all, since the field can be Fourier decomposed into plane waves for which p ≠ 0 and E = 0. We ended up in problems trying to explain the anomalous magnetic moment.
Our solution also resolves the classical renormalization problem for the electron mass if we replace the classical description with the quantum description.
The classical renormalization problem suggested that a classical point charge is an inconsistent concept. By replacing it with the quantum description, we get rid of the inconsistency.
Conservation of the center of the mass in the electron self-energy diagram
_____ γ virtual photon
/ \
e- ------------------------------
--> t
In 2020 we claimed that the virtual photon γ cannot be carrying energy E > 0 in the Feynman self-energy diagram above. We argued in the following way:
Let an initially static electron emit the photon with E > 0. The electron starts to move to the opposite direction from the photon γ. Let the electron then somehow (magically) absorb γ. How does the electron now know where it must "jump" in order to keep the center of the mass of the system static?
A possible solution to the paradox: the energy of the electric field is always outside the electron. There really is no need for the electron to jump anywhere.
In the rubber membrane & the sharp hammer model, the hammer made a pit into the rubber membrane. There is some elastic energy in the stretched membrane, and that energy never comes back to the hammer.
Thus, the self-energy diagram above only describes one step in a continuous process. The continuous process does not involve any energy transfer.
This means that 860/861 me really is the "true" reduced mass of the electron. It never absorbs the remaining 1/861 me.
The remaining renormalization problem: how to get rid of regularization?
In 2021 we claimed that we can appeal to destructive interference to make Feynman integrals to converge absolutely. Any wave with a very short wavelength is canceled by destructive interference if the wavelength is much shorter than the "geometric scale" of the process.
We are not sure if our claim is correct. We have to study it further. If the electron emits a virtual photon with a very short wavelength, can we really claim that destructive interference cancels it? The system now contains two particles: the electron and the virtual photon. Does the destructive interference apply to this pair?
Why does the Green's function calculate right the field energy farther than the Compton wavelength? A time-independent Green's function
This is another crucial question which we have not solved yet. If the Feynman diagram correctly handles a classical limit, then it has to know what the field energy of the Coulomb field is.
Why does Green's function produce correct results?
Let us consider a time-independent Green's function for the Coulomb potential of a point charge. A point charge can be seen as a point "impulse" which distorts the electric field. The Coulomb field is the "impulse response".
Maxwell's equations without the charge are the "homogeneous" equation:
H(r) = 0.
The point charge is the perturbation which we write on the right side, a Dirac delta function:
H(r) = δ(r₀).
What is the energy required to build the Coulomb electric field?
We can build a crude 3-dimensional potential well of a radius R by combining plane waves whose wavelength is
λ > R,
and the formula
Real( exp(i k • r / ħ) ),
where Real takes the real part of the complex number, and k is the spatial momentum of the virtual photon.
What is the Fourier decomposition (transform) of the 3-dimensional function 1 / r?
It would be the familiar Feynman propagator for the photon, if k would be a 4-momentum.
How much energy is "carried" by each component wave, to build the electric field?
The energy of the electric field is
∞
~ ∫ 1 / r⁴ * r² dr
R
~ 1 / R,
down to a radius R.
We obtain a reasonable model if each component exp(i k • r) carries the same energy.
To construct the potential down to a radius R, we need momenta |k| which are at most ~ 1 / R. The energy down to a radius R is then
r < 1 / R
~ ∫ 1 / r² * 4 π r² dr
0
~ 1 / R.
Analysis of the time-dependent Green's function for the electric field
The Feynman integral in the anomalous magnetic moment, in the fnal.gov link of the previous blog post, is
The photon propagator (Green's function) looks like the formula in the 3-dimensional case, but this time k is a 4-momentum
k = (E, px, py, pz).
For a 4-momentum k, the square k² is defined
k² = -E² + p²
in the metric signature (- + + +).
We should figure out what the "impulse response" in this case looks like.
Note. The electric field, presumably, is always present around the electron, and is time-independent. Why does the Feynman diagram study a time-dependent impulse response? The Feynman integral would converge absolutely, if we would drop the time dimension out. No need to do the dimensional regularization 4 - 2 ε. Our note about conservation of the center of mass above suggests that a virtual photon in the self-energy diagram cannot contain energy E at all.
The photon propagator is quite similar to the massless Klein-Gordon propagator. The massless Klein-Gordon equation is the usual wave equation.
#
# ======= sharp hammer
v
____ ____ tense rubber membrane
\_/
The "impulse response" for the massless Klein-Gordon equation should be similar to the one when we hit a rubber membrane with a sharp hammer. But how is this related to the elastic energy of the rubber membrane when a pointlike static weight is sitting on it?
Bringing energy into the integral. Technically, the Feynman integral needs a parameter containing energy, since it has to reduce the mass-energy me of the electron. If we would use a time-independent Fourier decomposition of the electric field, how could we introduce energy there? This is a technical explanation, but does not help us to understand the physics.
Creating the Coulomb field around an electron, time-dependently
Suppose that the potential is initially flat around an electron. We want to create a pit into the potential, such that the pit would resemble the Coulomb potential around the electron.
Let the potential suddenly change from flat to the Coulomb 1 / r potential. Let us take the Fourier decomposition of this temporally changing potential.
The Fourier decomposition probably is somewhat similar to the Green's function of Maxwell's equations, or alternatively, the Green's function of the Klein-Gordon equation.
We assume that we cannot use quanta whose mass-energy is > me to create the potential.
This means that we can only imitate the Coulomb field down to a distance of about λe from the electron. The resolution of quanta < me c² is no better.
But the energy of the Coulomb field at distances > λe is only
α / (2 π) ≈ 1/861
of me c². The Coulomb field would have 861 times the required energy, if we would always use a quantum roughly of the size ~ me c². Let us create the potential pit in such a way that in < 1/861 of cases we use the energy of a large quantum whose size is ~ me c².
The time-independent impulse response of the electric field is the Coulomb field. One can guess that a time-dependent impulse response roughly creates a Coulomb field, too. This is a heuristic reason why the propagator in the Feynman integral is the Green's function for Maxwell's equations.
How is the time-dependent Green's function related to the energy of the electric field? We restrict the Green's function only to use quanta < me c².
<-- 2 λe -->
___ ___ potential
\________/
• e-
The Fourier decomposition of the time-dependent potential pit will have energies E up to me c², and momenta |p| up to me c.
The coefficients in the Fourier decomposition may be interpreted as probabilities of measuring a certain energy E if we measure the energy of the time-dependent potential pit. The probability of seeing the energy as roughly me c² must be < 1/861. The expectation value of the energy should be me c² / 861.
In the anomalous magnetic moment experiment, the virtual photon of the magnetic field B is "measuring" the reduced mass of the electron. It is a good guess that the Fourier decomposition of the pit, or the Green's function of Maxwell's equations give probabilities for various reduced masses.
We found the connection between the Green's function and the energy of the Coulomb field at a distance > λe.
Conclusions
We may have solved the century-old problem about what is the mass-energy of the electric field of the electron. It is
α / (2 π) * me ≈ 1/861 me.
We now have a qualitative understanding what the anomalous magnetic moment Feynman diagram calculates: it determines the mass-energy of the electric field of the electron, and calculates the magnetic moment based on the reduced mass of the electron.
A brief Internet search did not reveal prior art. This may be the first time someone gives an intuitive explanation for the diagram.
We introduced a new general idea: quantum mechanics tries to imitate classical mechanics. The resolution of the imitation is restricted by the Compton wavelength associated with the energy available in the process. The imitation may in some cases be more accurate, if there is a lucky coincidence. The imitation is further restricted by quantization. Measurements of energy must return whole quanta, not fractions.
We still have to study the Feynman integral in detail. Why is the logarithmic divergence canceled by symmetric positive and negative values? What is the role the energy E versus the momentum p? How do gamma matrices treat me, E, and p?
Our most important remaining task is to eliminate regularization and renormalization from Feynman diagrams. In 2021 we came up with some ideas: high-energy virtual particles may be canceled by destructive interference. But do those ideas really work?
