The text is probably by Vadim Kaplunovsky.
The utexas.edu paper discusses the infrared divergence of the "electric" vertex correction in the scattering of an electron from a very heavy negatively charged particle X. The yellow circle depicts the electron bumping from the field of X, and also the virtual vertex correction photon.
Let us analyze the Feynman vertex diagram. The solid line is the electron.
The Green's function at the birthplace of the virtual photon k "creates" the Coulomb field of the electron.
Let the virtual photon k possess the energy E and the spatial momentum P. Let|E| and |P| very small. Then other factors in the integral are essentially constant, but the photon propagator
1 / k² = 1 / (-E² + P²)
varies a lot. The divergence has to come from photons for which -E² + P² is almost zero, that is, from almost "real", or "on-shell" photons. The paper says that the divergence is logarithmic.
Is there any reason why the integral should not diverge?
A toy model
1 90%
------------
/ \ 1% + 81% + 81%
e- ----------- 1% --------------------------------
\ /
------------
2 90%
Let us have a toy model where the electron coming close to X has a 90% probability to emit a virtual photon of the energy 1, and the same 90% probability to emit a virtual photon of the energy 2.
The probability of the electron reabsorbing the virtual photon is 90%.
We sum the probabilities of the three paths and end up in a nonsensical figure 163%.
What was wrong? We assumed that the paths 1 and 2, and their probabilities, are mutually exclusive. In the diagram, the electron never emits both 1 and 2. A more realistic model is one where it, in most cases, emits both 1 and 2.
Classically, an electron bumping into the charge X will emit and absorb a large number of real and "almost real" photons.
~~~~
/ ~~~
/ /
e- --------------------
|
X --------------------
--> t
Feynman diagrams allow the electron to emit many photons, if the diagram is more complex. Does this save the mathematical correctness?
No, not in the case of real, emitted photons. If two photons are emitted, it is a distinct end result. Its probability amplitude cannot be summed to an end result where only one photon is emitted.
What about virtual photons?
~~~~~~~
/ ~~ \
/ / \ \
e- ---------------------------------
|
X ---------------------------------
--> t
Adding a new photon means one additional photon line and two additional vertices in the integral.
Intuitively, emitting and absorbing a small virtual photon should not change the phase of the outgoing electron much. There will be no destructive interference, and adding more photon lines should not cancel the divergence of the integral with a single line. But is this true according to the Feynman rules?
Let us check what people have written about this.
C. Anastasiou and G. Sterman (2018) present a method to remove infrared divergences. They do not say that going to two loops would help and cancel divergences at one loop.
In the classical limit, an electron emits a huge number of small photons
We know that a macroscopic accelerating charge will radiate a very large number of photons whose wavelength is large. What implications does this have for Feynman diagrams?
Let k₀ be the 4-momentum of a small real photon. The correct physical model (classical) says that the probability of the electron emitting just a single photon of a 4-momentum
|k - k₀|
is essentially zero. It will always emit a huge number of small photons.
The Feynman diagram claims that the probability of such an emission is small, but it significantly differs from zero.
We conclude that the Feynman diagram calculates an incorrect result.
If an electron passes the large charge X at a relatively large distance, then we can make a wave packet to describe the electron, and the process is almost classical. Let us use the Larmor formula to calculate how many photons the electron radiates.
We assume that the electron is relativistic and passes a proton at a distance R. The acceleration is
a = 1 / (4 π ε₀) * e² / R² * 1 / me.
The power of radiation is
P = 2/3 * 1 / (4 π ε₀) * e² a² / c³
= 2/3 * 1 / (4 π ε₀)³ * e⁶ / R⁴ * 1 / c³
* 1 / me²
= 3 * 10⁻⁴⁹ * 1 / R⁴.
The radiated energy for a relativistic electron is
W = P R / c
= 10⁻⁵⁷ / R³.
One photon of the typical frequency has the energy
E = h f
= h c / (2 R)
= 10⁻²⁵ * 1 / R.
The number of photons of the typical frequency is
n = W / E
= 10⁻³² / R².
The Compton wavelength of the electron is 2.4 * 10⁻¹² m.
If the distance R = 10⁻¹⁰ m, the number of typical photons is only 10⁻¹². We conclude that the electron is solidly in the realm of microscopic particles.
The classical electromagnetic wave emitted by the electron is a "bump" which lasts for the time 2 R / c. What is the Fourier decomposition of such a bump? The Fourier transform is essentially constant for large frequencies.
The pulse is able to excite a detector which observes very-long-wavelength photons, say such that the frequency is just 1 herz. If R = 10⁻¹⁰ m, then the radiated energy is 10⁻²⁷ J, and we are able to detect a million such photons.
The "number of photons" in the pulse is not well defined. We probably can "mine" the energy in the pulse and can extract various collections of photons, depending on the detectors we are using. Nevertheless, we are able to observe a very large number of low-energy photons. This contradicts Feynman diagrams.
Feynman diagrams work reasonably well if the quanta are large?
The analysis on the infrared divergence in the utexas.edu paper
The paper at utexas.edu tries to explain away the divergence problem by resorting to the fact that any single detector can only observe photons whose energy is larger than some threshold energy
ωthr.
But the explanation is incorrect. Quantum mechanics is about what could be observed, not about what a certain real-world detector observes.
Also, as we saw above, the infrared divergence is not the only problem. Another problem is that Feynman diagrams predict a far too large probability for the output of just a single photon.
Feynman diagrams simply are a wrong way to approximate a semiclassical process which produces electromagnetic radiation. No gimmick or explanation can refute the basic problem.
*** WORK IN PROGRESS ***
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