∫ 1 / k⁴ dV.
k ∈ ℝ⁴
The paper at the fnal.gov link uses dimensional regularization in 4 - 2 ε dimensions. At the end of the calculation, the paper states that diverging terms cancel each other.
The paper at the utexas.edu link utilizes various symmetries, and states at the end of the calculation that the integral converges.
We conclude that the integral probably is benign: if the integral is summed in the "natural" order of increasing 4-momenta in the virtual photon, then the integral will converge. There is no need for regularization or renormalization.
In the previous blog post we claimed that the mass-energy of the electric field of the electron is
α / (2 π) * me ≈ 1/861 me.
Since the integral is benign, we can claim that the result is robust: the result does not depend on dubious regularization or renormalization procedures.
The quantum imitation principle
Quantum imitation principle. 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.
We introduced the principle above in our previous blog post. Quantum mechanics tries to imitate the energy of the electric field of the electron. But the resolution is quite poor: the Compton wavelength of the electron is
2.4 * 10⁻¹² m,
which is a large value compared to the classical radius of the electron 2.8 * 10⁻¹⁵ m.
The imitation of the electron electric field only succeeds at the resolution of the Compton wavelength. Quantum mechanics believes that the mass-energy of the electric field is just 1/861 of the electron mass.
The fact that we do not need regularization or renormalization in the calculation of the Feynman integral, stresses that the quantum imitation principle is robust: it will produce finite values without dubious mathematical methods.
What if destructive interference cancels diverging integrals?
For classical waves, a wave whose frequency is
f
cannot normally produce significant waves whose frequency is > f. If we use Green's functions to construct a solution, destructive interference, in a typical case, cancels all high frequencies.
The cancellation is not absolute: if we would sum the absolute values of each wave, the integral would diverge.
We conclude that it is ok if the integral diverges in absolute values, as long as it converges when integrated in the natural order of increasing 4-momenta.
Let us calculate an example. The impulse response, i.e., the Green's function, for a static point charge is
∫ 1 / k² * Real( exp(i k • r) ) dV,
k ∈ ℝ³
where dV denotes a volume element of ℝ³, and Real takes the real part. The integral k³ / k² diverges badly. If we use Green's functions, we cannot expect absolute convergence of integrals.
If classical wave processes cancel high-frequency waves, why do high-frequency waves remain in Feynman integrals, and cause divergences?
In this blog we have long suspected that divergences are a result of a wrong way of applying Green's functions to scattering phenomena.
The reason may be that we calculate Feynman diagrams in the "momentum space", and ignore the position of the particles.
In the case of vacuum polarization, we claimed in 2021 that the divergence comes from a sign error when considering the Dirac hole theory.
The classical limit of the "electric" vertex correction
There is no classical limit for the magnetic part of the vertex correction, since the magnetic moment is a microscopic quantum phenomenon.
But for the wobbling of the electric field in a scattering experiment, there is a very natural classical limit. If we increase the mass and the charge of colliding electrons, they start to behave like classical charges. The wobbling of the electric field will remain significant, and will affect the paths of the charges.
The classical limit should have no diverging integrals. Destructive interference should cancel all high frequencies.
1/2 c
e- • --------->
| r
<---------- • e-
1/2 c
In this blog we have written about the natural "scale" of scattering in such an experiment. If electrons meet at relativistic speeds, and their minimum distance is r, then r is the natural scale. Waves shorter than r, or with a frequency higher than
f = r / c,
should get destroyed by destructive interference. The "natural frequency" of the meeting process is f = r / c.
Note about the high 4-momentum cutoff. In the case of the anomalous magnetic moment, the cutoff for high 4-momenta of the virtual photon is determined by the mass of the electron, me. But in the electric vertex correction, the cutoff is determined by the geometry of the meetup of the electron with another charge.
The diagram is from the utexas.edu link. The solid line is an electron.
In the diagram above, the virtual photon on the right is the impulse on the electron as it travels past another electron, or another charged particle.
The virtual photon on the left describes the wobbling movement of the electric field of the electron.
The far field of the electron does not have time to take part in the meeting of the electron with another charge. This means that the electron has a reduced mass as it encounters the other charge.
The Feynman diagram should calculate the reduced mass, and the effect of reducing the mass in the meetup with the other charge.
The reduction in the mass of the electron depends on how long it takes to pass the other charge. If the time is t, then we expect that the electric field at a distance
> 1/2 c t
cannot take part. The electron mass is reduced by that amount.
The reasoning in this is entirely classical. The integral should converge if analyzed as a classical process.
But is it so that the Feynman integral diverges for large 4-momenta of the vertex photon?
The incoming electron has the 4-momentum p, the outgoing p'. The vertex photon is k.
Let us denote the impulse virtual photon
q = p' - p.
How can the Feynman integral "know" the mass-energy of the electric field far away?
p'
\ |
\ |
| \ |
k | | ~~~ q
| / |
/ |
/ |
p Z
e-
^
| t
In the diagram, an electron meets a heavy nucleus Z. The nucleus give a pure momentum (no energy) q to the electron. The virtual photon k reduces the mass of the electron in the encounter.
It has to be that the virtual photon q somehow "modulates" the Feynman integral, so that the electron mass me is suitably reduced when the electron meets the nucleus Z. Since the mass of the electron is reduced, the nucleus Z will pull it a little closer, and the electron will be scattered a little more.
If p = p', then the Feynman integral is quite simple. Does it diverge? If yes, we could "renormalize" the integral value to zero. If p ≠ p', the value of the integral will change somewhat. We can use the change as the correction to the scattering probability amplitude. Why does the change when q is made nonzero, tell us the corrections to the probability amplitudes of the tree-level diagram?
Classically, destructive interference cancels any high-frequency phenomena (large |k|) in the process. If the Feynman integral is infinite, that means that the integral does not describe the classical process adequately.
Missing information. Let us analyze classically: in the diagram, we know the velocity of the electron, since we know p. If we know the charge of the nucleus Z, we can calculate what is the minimum distance of the electron from the nucleus for a given value of q:
R(q).
But if we do not know the charge Z, we cannot know R(q). Then we cannot calculate the reduced mass of the electron. The vertex correction is not well defined then.
Let the charge of Z be +e and velocities in the encounter close to c.
If |q| is very large, close to me c, then the reduced mass of the electron should be similar to the case of the anomalous magnetic moment. A reduction of ~ 1/2 me would make scattering to large angles much more probable.
What if q contains lots of energy, not just momentum? That corresponds to a slow electron meeting a very fast nucleus Z. In that case we might be able to identify q with m in the Feynman integral. By switching to a different frame, we can make the electron to receive a lot of energy in the encounter.
The fnal.gov link has the Feynman integral in a clearer form:
If the the nucleus Z passes the electron close to the distance of the classical electron radius 2.8 * 10⁻¹⁵ m, the electron will receive a momentum which is close to me c, and an energy which may be ~ 1/2 me c².
Classically, the encounter happens very quickly, in just ~ 2.8 * 10⁻¹⁵ m. Most of the electric field of the electron does not have time to react. Classically, the electron mass is reduced a lot, maybe to ~ 1/2 me.
But the quantum imitation principle says that quantum mechanics cannot imitate the energy of the electric field of the electron closer than its Compton wavelength 2.4 * 10⁻¹² m. Thus, the mass reduction seen by quantum mechanics is ~ 1/861 me.
Let us set q to zero for a while. The integral above, probably, diverges. We regularize and renormalize the integral to zero. We kind of pretend that the integral has a fixed value. We want to calculate the difference that setting q ≠ 0 makes to the value of the integral.
Let q then be very large, close to me c. Its energy "reduces" the electron mass m in the factor
(l-slash - q-slash + m)
/ ( (l - q)² - m²)
in a major way.
The factor 1 / (2 π)⁴ at the start of the integral tones this down, so that the probability amplitudes are much smaller (by a factor ~ 1/861) than for the tree-level diagram.
Feynman miscalculates the classical limit? If q is small, then the quantum imitation principle allows quantum mechanics to imitate the classical wobbling of the electron electric field very well. Quantum mechanics should then approach the classical behavior. But the factor 1 / (2 π)⁴ tones the probability amplitudes down too much? Our note about the Missing information above suggests that, indeed, the Feynman diagram is incorrect if we study a macroscopic system.
Regularization and renormalization work because their ad hoc trick implements the destructive interference seen in the classical treatment? In the classical treatment of the process, destructive interference cancels all high-frequency waves. In the Feynman integral taken in "momentum space", this classical mechanism is missing. Regularization and renormalization are ad hoc tricks which are mathematically dubious. But they happen to implement the effect of destructive interference. That is why they produce empirically correct results?
The electric form factor F1 and the classical limit: infrared divergences
The paper at the utexas.edu link calculates the "electric" form factor F₁(q²). Let us check if the result is reasonable for the classical limit.
The calculation of F₁(q²) contains both an "infrared" and an "ultraviolet" divergence.
Let us try to analyze what infrared divergences are in the context of a classical system. We assume that the electron and the nucleus are classical particles with a macroscopic mass and a macroscopic charge. What problem could be involved with very long wavelengths?
If we try to solve a classical wave problem with Green's functions, could it be that the low frequency spectrum requires some destructive interference, too?
Let us, once again, look at the rubber membrane model, which is hit with a sharp hammer.
Ultraviolet problems arise from the sharpness: the hammer must input an infinite energy to create the analogue of a point charge electric field.
A single hammer hit will output energy in the outgoing waves in the undulating membrane. The energy is not infinite, though.
An infrared divergence seems to be associated with the zero mass of the photon.
Could the problem be that the hammer can never create the entire Coulomb field, since the field spans the entire space? The far parts of the field must be "inherited" from earlier hammer strikes? Having a partial electric field is a sure way to end up in problems with Maxwell's equations.
The photon propagator in the Feynman integral with p = q = 0 is
1 / l²,
where
-E² + P²,
if E is the energy and P is the photon spatial momentum. If the photon is real, "on-shell", the denominator is zero. This probably is the source of the divergence. What does this correspond to, classically?
Low-frequency waves DO escape as real photons. Suppose that the electron consumes the time ~ t to pass the nucleus Z. Then a part of its Coulomb field farther than c t from the electron "breaks free" and escapes as electromagnetic waves. Those long waves cannot be reabsorbed by the electron. This imposes a cutoff to an infrared divergence. The Feynman integral miscalculates the probability amplitude of the electron reabsorbing the waves which broke free.
The paper at the utexas.edu link mentions that the infrared divergence is canceled if we consider the "soft photons" which the electron sends. This suggests that the Feynman integral error really is about the waves which break free. However, this does not yet explain why the integral result diverges.
Destructive interference inside a limited volume: Feynman diagrams ignore it
A Feynman diagram is in "momentum space". We imagine a wave arriving to, say, a cubic meter of volume, and interacting with another field there. Waves then leave that cubic meter. We calculate the Fourier decomposition of the departing waves.
The process should conserve energy. If the arriving wave denotes an electron, and leaves 100X stronger, energy was not conserved. Even worse if the departing wave is infinitely strong.
Is there any reason why using the Green's function method should conserve energy? If the method ignores destructive interference, then it will certainly break conservation of energy.
Suppose that wave components with different momenta p,
exp(i (p • r - E t))
leave the cubic meter. Can we extract the energy in the wave components separately? If yes, then we can sum the energies of the components.
In principle, we can tune an antenna which absorbs the component p, but not a nearby component p' where
|p' - p| > ε.
If ε is small, the antenna, presumably has to be very large. It cannot fit inside a cubic meter.
Inside that cubic meter, we have to take into account destructive interference of Fourier components of waves.
We uncovered a major error in the Feynman diagram calculations. They ignore destructive interference inside a limited volume. The calculation happens in "momentum space", and assumes that the space is infinite and that Fourier components can be handled separately.
All physical experiments happen in a limited volume.
Switching to "momentum space" is a valid approximation in many situations. A cubic meter is a vast volume for microscopic processes. But the approximation, in many cases, will cause wave energy to grow infinitely. This probably is the fundamental reason for diverging Feynman integrals.
Diverging integrals mean that energy (the number of particles) is not conserved. Energy is not conserved because the integral ignores destructive interference in a finited spatial volume.
The problem is not that we "do not know the physics at the Planck scale". The problem is a simple mathematical error in the treatment of waves.
The author of this blog never understood how calculations in "momentum space" can work. The answer is that they do not work. However, the approximation in many cases can be saved with ad hoc tricks like cutoffs, regularization, and renormalization.
The Feynman path integral method is inaccurate
The path integral method assumes that we can calculate all the distinct paths of "particles", and sum the probability amplitudes of the paths, to obtain the probability amplitudes of the end results.
The method assumes that the paths are independent and other paths can be ignored. But this is a wrong assumption. Destructive interference in the volume where the process happens, plays a crucial role in conservation of energy and particle numbers. We cannot ignore other paths.
If we look at a physical process with classical waves, no one would assume that we can ignore destructive interference when we do calculations. If we hit a tense rubber membrane with a sharp hammer, the energy of the impulse response, or the Green's function, depends in a complex way on the interference of the Fourier components. If we try to sum the energies of the components separately, we certainly will obtain strange results.
Why was the mathematical error ignored in quantum field theory?
1. Feynman diagrams produce correct results for many processes, like the tree-level diagram for electron-electron scattering. It is somewhat surprising that the diagram works in this case.
2. Subatomic particles are viewed as mysterious and exotic. They do not necessarily obey classical physics. People speculate about new physics at the "Planck scale".
3. The idea that a virtual photon is a "particle" is nice. A Feynman diagram is aesthetic.
4. People confuse a probability amplitude with a classical probability. Classical probabilities often are separate and can be summed without much thinking.
5. People did not realize that Feynman integrals should also work in the classical limit. They must be able to calculate classical wave processes, or they are erroneous.
6. People thought that a "momentum space" exists. But all physical processes take place in a finite spatial volume. There is no momentum space.
7. Ad hoc methods, like renormalization, remove most of the problems caused by the mathematical error.
Conclusions
We believe that we found the fundamental reason why Feynman integrals diverge in many cases. The problem is that the Feynman path integral method ignores destructive interference of waves, which takes place in a finite spatial volume.
We have to analyze the electric vertex correction in more detail. Does it miscalculate the classical limit?
Freeman Dyson presented an argument which claims that the sum of Feynman diagrams will always diverge. We will next look at that.
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