Several authors have presented derivations which do not depend on Feynman diagrams, or on the quantum field theory.
On March 7, 2021 in this blog, we derived the anomalous magnetic moment from zitterbewegung. We assumed that the mass-energy of the electric field of the electron, at a distance > ~ 1 Compton wavelength, does not take part in zitterbewegung. That is, the effective mass of the electron is lower than me, which forces the zitterbewegung loop to be larger.
A larger current loop has a larger magnetic moment. The electron moves at the speed of light in the loop. The length of the loop is the Compton wavelength
λ = h / (me c).
The magnetic moment of the electron is
μ ~ A I,
where A is the area of the loop and I is the current around the loop. Suppose that me is reduced by a factor k < 1. Then the length of the loop grows by a factor 1 / k, the area by 1 / k², and I is reduced by a factor k. The magnetic moment grows by the factor
1 / k > 1.
Let us try to find the analogies between our zitterbewegung argument and the Feynman diagram calculation.
Feynman diagram calculations of the electron anomalous magnetic moment
The link contains the calculation with Feynman diagrams. In this link there is another calculation:
The electron self-energy calculation diverges. A crucial question is how the authors of the two previous texts are able to get the vertex integral to converge? The second text uses dimensional regularization, calculating in
4 - 2 ε
dimensions, where ε > 0 is a small number.
The formula above calculates the contribution of the vertex diagram. We have removed the extraneous subscript μ of m in the original text.
We should analyze what does the formula calculate, in semiclassical terms.
The calculation of the magnetic moment does not depend on the initial momentum p of the electron. We can assume that p is zero. Also, the calculation does not depend on q. We can assume that |q| is very small.
The 4-momentum of the virtual photon is
l + p + q / 2 ≈ l.
The value of l can be chosen freely, and it is integrated over. Symmetries cancel most of the integral.
If q would be zero, then the diagram would be the self-energy of the electron. The self-energy is "absorbed" into the electron mass me, and we can say that we know what the diagram produces: the familiar electron.
If the 4-momentum l is chosen to be pure energy, then l plays much the same role as m in the formula?
Superficially, the integral only diverges slowly. If we integrate over the range of the 4-momentum l where n is large and
R = 2ⁿ < |l| < 2ⁿ⁺¹ = 2 R,
Then the volume of integration is
~ R⁴,
but the value of the integral is
~ 1 / R⁴,
since l has the power 4 in the integrand in the formula above.
The integral, superficially, only diverges logarithmically.
Let us assume that p = 0 and |q| is very small. If we remove the photon l + p + q / 2 from the diagram, then the formula for i 𝓜 has no integral in it, and is the tree-level scattering amplitude.
The integral in the formula above adds to that amplitude slightly. The integral, apparently, does not depend much on q, and we can assume that q = 0.
The integral does depend on me, the mass of the electron. We can say that the Compton wavelength of the electron,
λ = h / (me c) = 2.4 * 10⁻¹² m
defines the "scale" of the integral. This might explain why our zitterbewegung argument works. But what exactly is the connection between the abstract integral and the classical notion of the energy of the electric field?
A semiclassical electron: it hits with a sharp hammer
We may imagine that the electron is a point particle which hits with a "sharp hammer" the electromagnetic field at very short intervals. That creates the static electric field of the electron. The sharp hammer creates an impulse response in the electromagnetic field: it is the Green function.
In the Feynman diagram, a hit with the hammer creates all the alternative photons in the vertex – all possible 4-momenta.
----- ----- rubber membrane
\•/
weight
An analogous system is a small weight sitting on a tense rubber membrane. We can imagine that the weight hits the membrane at short intervals. Each hit creates Green's function. Their interference creates the static pit where weight sits on the membrane.
Calculating with the electron as a wave packet: a nonperturbative approach
Feynman diagrams are normally calculated in "momentum space". That is, a particle is represented as a plane wave of a form
W = exp(i (p • r - E t)).
The plane wave extends to infinity. We can visualize the plane wave as crests of a wave which fill a large 4-cube. The cube might be one cubic meter which is studied for one second.
We can visualize a coupling in that cubic meter: the wave W disturbs another field, say, the electromagnetic field. The disturbance is modeled with Green's functions which hit the other field, in every point of the cubic meter.
Destructive interference cancels every Fourier component of those Green's functions, except those which "resonate" with the configuration. The resonant waves respect conservation of momentum and energy.
What if we let a wave packet to represent the particle? The packet can be quite small. For an electron, it could be as small as, say, a few Compton wavelengths, or 10⁻¹¹ meters.
Let us use a nonperturbative approach, in which we try to solve the differential equation for the wave packet – without resorting to a Feynman path integral.
^ V electric potential
|
e- ~~~~>
wave
packet
If we have an electron moving under a potential, we can solve its motion using the Schrödinger equation. It is nonperturbative. Let the potential be such that it will turn the electron upward in the diagram above.
The entire wave packet will steer upward. This is not a scattering experiment in which the electron might not collide with a photon. This is not a "perturbative" solution.
What about the vertex correction? The electron wave packet does disturb the electromagnetic field. A complete solution must somehow describe the electric field of the electron.
We may try to build the electric field using Green's functions. Each point in the electron wave imposes an impulse on the electromagnetic field. It is like hitting the rubber membrane with a sharp hammer. The impulse response is Green's function.
A crucial question: are we allowed to claim that destructive interference removes very short wavelengths from Green's functions?
If yes, then we can claim that the electric field of the packet essentially only contains the field which is > 1 Compton wavelength away from the center of the wave packet. Would this mean that the effective mass of the electron, me, is reduced?
The classical limit
The classical limit of the free electron is a point charge traveling at a constant velocity, having the familiar Coulomb electric field around itself.
The nonperturbative model which we suggested above, respects the classical limit.
If Feynman diagram calculations respect the classical limit (they should), then they are guaranteed to produce approximately the same result as the nonperturbative model.
Feynman diagram calculations contain regularization and renormalization procedures, which are ad hoc. The procedures must be chosen in such a way that the classical limit is respected.
Nonperturbative solutions: long waves cannot create short waves
In our blog we have stressed that wave phenomena cannot create "features" which are much smaller than the wavelength of the input.
The "wavelength" of a low-energy electron is its Compton wavelength. The wave vector is mostly to the direction of time.
That implies that virtual photons of a very high 4-momentum in the process are canceled by destructive interference.
How does the Feynman diagram achieve the cancellation?
Is the Feynman diagram perturbative, after all, in this case?
The basic idea in a Feynman diagram is that a wave in one field D (Dirac) acts as a "source" in another field E (electromagnetic). The coupling constant describes the magnitude of this source effect.
If the other field is governed by a homogeneous linear differential equation of a type
L(E) = 0,
then the source S(D) is written on the right side of the equation:
L(E) = S(D).
Does this actually calculate a nonperturbative solution for D and E?
A perturbative solution typically is a crude approximation. But Feynman diagrams give the anomalous magnetic moment of the electron correctly up to 10 decimal places. It is not an approximation!
Does it make sense to define one field E? The electron may be at any location given by the field D, and, obviously, its electric field E depends on the location of the electron. We conclude that it is not sensible to define a single field E.
What about an approach in which we assume some location x and a velocity v for the electron, and then try to construct the electric field E of that electron?
^ virtual photon
/
• ----> v
e-
In our blog we have argued that if an electron moves at a constant velocity, then destructive interference cancels almost all virtual photons emitted by the electron.
In the rubber membrane and a weight analogy, the imagined vertical impulses of the weight create static waves which form the depression under the weight. That is, the Fourier decomposition of the membrane shape only contains time-independent waves. All time-dependent waves have total destructive interference.
In electromagnetism, time-independent waves are virtual photons which only carry a momentum p, no energy E.
An excited hydrogen atom decays to the ground state
Suppose that we have an excited hydrogen atom sitting around. Let us claim that it cannot decay to a lower state and release a photon, because the photon could be emitted at any one moment, and there is a destructive interference of possible histories.
Suppose that the atom sank into the ground state at a time t, and sent a photon γ. If it would do the same at a time
t + Δt,
then the atom would be in the ground state and a photon γ' would be emitted slightly later.
There is interference between γ and γ'. Why does it not cancel the photon entirely?
Can we distinguish between the two histories with a later measurement? Let us measure the positions of the atom and the photon at some later time t'.
That will tell us the exact time when the photon was emitted?
This would resolve the paradox. The histories can be distinguished through a measurement. There is no interference.
No, it does not resolve the paradox. The center of mass of the system moves at a constant speed. If we detect a photon γ of the energy E at a location x, then that uniquely determines the location y of the hydrogen atom. We cannot distinguish between histories in this way.
Another attempt to resolve the paradox: an excited hydrogen atom is a classical dipole transmitter. It creates a classical dipole wave according to its oscillation. Then there is no destructive interference anywhere.
The photon is a quantum of this classical wave. The photon does not exist separately from the classical wave. If the radio transmitter would be macroscopic, then this would be self-evident. For a hydrogen atom, the classicity assumption is suspicious.
An electron "decays" into a virtual electron and a virtual photon
Suppose that the electron is sitting still. Then it emits a virtual photon which contains pure energy E, no momentum. Later the electron reabsorbs the photon.
Can we use measurements to distinguish between histories where the emission happened at a time t, versus a time
t + Δt ?
Probably not. Can we claim that destructive interference cancels these histories?
What if the photon contains also a momentum p?
We really cannot measure virtual particles, or can we?
Actually, the vertex diagram above contains a "measurement" of virtual particles, since the photon q probes the structure consisting of a virtual electron and a virtual photon.
We could claim that the "measurement" cannot "see features" which are smaller than the Compton wavelength λ of the electron. This is because there is not more energy available in the process than me. Everything which the process "sees" at a resolution sharper than λ must be very fuzzy.
The measurement can then "see" a bare mass which is me minus the energy in the electric field at a distance > λ from the electron.
The bare electron still behaves like a Dirac electron, obeying the Dirac equation, but its mass is less by approximately 1/1000. The Dirac equation then gives the magnetic moment μ 1/1000 larger.
Our hypothesis suggests that the anomalous magnetic moment of the electron is larger if the probing photon has more energy and momentum.
In our blog post on March 2, 2021 we observed that the correct anomalous magnetic moment can be derived if we assume that the electric field farther than λ / 2 does not contribute to the bare mass of the electron. Why is it λ / 2 and not λ?
In our blog post we remarked that if the electron moves in the zitterbewegung loop, then the far electric field has to be constant – otherwise we would have radiation emitted. If we imagine a rope whose mass is ≈ 0, attached to the electron, the first node in oscillation of the rope is at the distance λ / 2 from the electron. But this has nothing to do with Feynman diagrams. Why does this semiclassical model work?
The Feynman integral for the vertex diagram (correction to the tree diagram) has two components:
where F₁ is goes to zero when q approaches zero. That is, when the static magnetic field B is very "flat" and does not vary much on the location. The second term F₂ does not go to zero when q goes. The term F₂ is the correction to the magnetic moment μ of the electron. It is something intrinsic to the electron and does not depend on the magnetic field B probing the electron.
Is this is the mass reduction caused by zitterbegung?
The classical limit and Feynman diagrams
\ /
|~~~~| virtual photon
/ \
e- e-
^ t
|
In our blog we have earlier studied simple electron-electron scattering, and the associated simplest Feynman diagram in which the electrons exchange a photon.
One electron "hits" the electromagnetic field. The impulse creates Green's function which contains virtual photons (Fourier components) with all kinds of 4-momenta. The coefficient of the Fourier component is the "propagator".
Surprisingly, this simple model imitates quite well the scattering calculated using a classical point charge and the Coulomb force.
That is, Feynman diagrams imitate the classical limit well. They must do so, in order to be correct.
Why does the diagram work so well in this simple case?
A possible solution to the electron mass renormalization problem
Classically, the electron is a point charge. Its electric field has an infinite energy. How is it then possible that its mass is only me? This is the classical mass renormalization problem. Is the "bare mass" of the electron minus infinite?
In quantum field theory, the mass renormalization problem comes up in various Feynman integrals which diverge.
Hypothesis. When the electron hits the electromagnetic field, it can only create Fourier components whose wavelength is > the Compton wavelength of the electron. This also holds for components which have the energy component zero and only contain momentum.
If the Fourier components of the (static) electric field are restricted in this way, how close to the electron we are able to build the electric field, and what is the energy of such an electric field?
^ V electric potential
|
___ ___
\___/ Fourier component
• e-
We see from the diagram that we are able to build the electric field roughly down to the radius which is 1/2 of the Compton wavelength of the electron.
The mass-energy of that electric field is roughly 1/1,000 of me.
Suppose that the Feynman tree diagram and the vertex correction diagram are able to calculate the mass-energy of such an electric field, and are able to determine the magnetic moment μ of the reduced-mass electron, where the energy of the field is subtracted from me.
Then the Feynman diagrams calculate the magnetic moment correctly.
This gives us the answer to the problem: what is the "bare mass" of an electron where the energy of its static electric field is subtracted? For an electron sitting still, it is the reduced mass which gives the right anomalous magnetic moment. That is,
mbare = me (1 - α / (2 π)),
where α ≈ 1/137 is the fine structure constant.
The electron is not able to "build" more of its electric field. The reduced mass is the mass which the Dirac equation receives as its input, and the reduced mass determines the magnetic moment of the electron.
Electrons moving fast in a particle collider possess more energy, and are able to build a larger portion of their electric field.
A more detailed analysis of the Coulomb field versus a Feynman diagram
\ /
|~~~~| virtual photon
/ \
e- e-
^ t
|
Let us try to figure out how the simple Feynman diagram accomplishes the same as the Coulomb field.
The electron on the left hits the electromagnetic field with a Green's function. Let us study a Fourier component of the Green's function.
^ V electric potential
|
<-- • e-
__ __ __ Fourier component
\___/ \___/
• e-
The component may pull the rightmost electron to the left or to the right, depending on the position of the rightmost electron.
__ ____________
\___/
• e-
If we sum many such components with varying wavelengths, say, 100 of them, we can build a potential pit close to the leftmost electron. Far away, the components mostly cancel each other out. The potential is roughly constant far away.
Let the leftmost electron to pass through this potential. It has some probability, say, 1/10 to scatter in the potential pit.
Let us multiply the Fourier components by 1/1,000. The factor 1/1,000 is the "propagator" here.
The leftmost electron has a probability 1/1,000 to scatter in each Fourier component. Summing over the components yields 1/10 as the scattering probability.
But in what way can we interpret this that the rightmost electron "absorbed" a virtual photon from the leftmost electron?
We can imagine that each component is "attached" to the leftmost electron. If the component pushes the rightmost electron, then the momentum, indeed, is passed between the electrons. It is a "virtual photon" which transferred momentum.
How do we arrive at the energy of the Coulomb field? The leftmost electron must use energy to build the component? If yes, how is the Feynman diagram aware of this energy?
Or is it so that energy is only present when the Coulomb field will do work on some charge? In the vertex correction, the virtual photon probably contains some energy, besides the momentum. The field receives energy from the electron, and returns the energy back later.
Classically, the electric field does work through giving up some of its field energy to the charge which the field is accelerating. How is this related to Feynman diagrams?
====== hammer keeps hitting
|
v
_____ ____ membrane
\_/
pit
If we keep hitting a rubber membrane with a hammer at short intervals, the membrane assumes a time-independent pit form.
The hammer outputs momentum into the rubber, but does not output energy. The energy was spent when the experiment started, and the hammer made the pit.
The energy of the electric field is easily found by looking at the potential between an electron and a positron: this is the connection to scattering experiments
Imagine that we have a spherical shell, whose radius is R, and whose total charge is the charge of the positron. At the center of the shell is an electron.
The shell erases the electric field outside it. The potential of the shell reveals us the energy of the electric field of the electron at radii > R.
In a scattering experiments, the potential between two electrons, or an electron and a positron is important.
This is how scattering experiments are connected to the energy of the electric field.
A wild guess about the anomalous magnetic moment
A wild guess about the anomalous magnetic moment vertex diagram: the incoming electron "throws" energy to its later incarnation, which is the outgoing electron. That is why the electron which meets the magnetic field B, has a lower mass by approximately 1/1,000 than me, which in turn makes the magnetic moment μ 1/1,000 larger.
Semiclassically, we can interpret that the electric field farther than 1/2 the Compton wavelength from the electron does not "take part" in the Dirac equation. The "bare mass" of the electron is 1/1,000 lower.
Why is it 1/2 the Compton wavelength?
In the Feynman diagram integral, let the mass-energy E of the virtual photon be close to me. If
E / c² = me - δ,
where δ > 0 is small, then the reduced-mass electron (mass only δ) will have a huge magnetic moment μ as it meets B.
But that is compensated by the virtual photon with
E / c² = me + δ,
*** WORK IN PROGRESS ***
