The text is probably written 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 is incorrect
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.
The paper says that the divergence in the photon emission has the opposite sign to the divergence in the vertex correction, and that the divergences "cancel each other out". This is not possible. If the electron loses kinetic energy to a real photon, then the momentum of the outgoing electron differs from every vertex correction electron, because a vertex correction electron does not lose any energy. Electron waves with different momenta cannot cancel each other out.
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.
The Peskin-Schroeder textbook An Introduction to Quantum Field Theory (1995), contains the same strange claim as the utexas.edu paper, that we can sum probability amplitudes for processes which have a different end result:
Jonathan Gleason (2012) noticed the same thing as we: one cannot add probability amplitudes for bremsstrahlung and an elastic scattering. Lubos Motl answers and states our other observation: bremsstrahlung is always emitted.
The Feynman probability amplitude for a real photon emission: we can allow divergence
There are two diagrams. Above is one of them. The probability amplitude is
Let |k| be very small. If we double the charge of the electron e, the mass of the electron m, and the momenta q and p, then the the probability amplitude grows twofold, and the probability flux fourfold . This agrees with the Larmor formula. The classical limit is ok, in this sense.
Now we realize an important thing:
- The Feynman probability amplitude is the PRODUCT of the electron flux and the photon flux. We can ALLOW the integral to diverge, if an infinite number of photons are produced!
Classically, the process will generate an infinite number of photons, if we look at ever longer wavelengths. The Feynman formula may be a fairly good approximation?
But Feynman diagrams miscalculate the effect of the emitted photons in the scattering of the electron from the large charge X. The electron loses its kinetic energy to the radiated photons. That affects its scattering from X. Feynman diagrams only consider one emitted photon, while in the classical limit, the electron will always emit a large number of photons.
How do Feynman diagrams work at all? We calculated above that the probability to emit a photon of a "typical size" is only 10⁻¹² in a typical case of scattering. Corrections which come from photon emissions are so small that they do not spoil the accuracy.
Also, the correction mainly comes from the very rare case (probability 10⁻¹²) when the electron does emit a photon of the typical size. The Feynman diagram is correct to reduce the mass of the electron by that one photon. We can ignore small photons.
In particle accelerators, the margin of uncertainty for a scattering probability is typically on the order of 1%.
The "electric" vertex correction
● X
e- • ----------- R = minimum distance
\
\
Let us look at the divergence in the vertex correction. Let us guess that the vertex correction really is about how much the electron mass must be reduced, because its far electric field does not have time to follow the electron in the abrupt scattering from X.
"Far" means something like > 2 R, if the electron is relativistic. There R is the minimum distance between the electron and X.
If we think of the far electric field of the electron built from virtual photons of various wavelengths, then, obviously, an infinite number of virtual photons are needed. This explains why the Feynman vertex integral diverges.
As we noted above, then the Feynman integral calculates the product of the electron flux and the virtual photon flux.
In the classical limit, a correct approximation method would calculate the expectation value of the combined energy of the produced virtual photons, and reduce the mass of the electron by 1/2 of that value when the electron meets X. The Feynman method is erroneous in the classical limit.
Let us then analyze the process when the electron is microscopic.
The paper at utexas.edu states that the infrared divergence is logarithmic when we go to smaller |k|. Making the dimension larger than 4 makes the integral to converge. This suggests that the divergence really comes from building the electric field at the distance > 2 R. We can build an approximation "exponentially" by using waves of length 2ⁿ, where n is a positive integer.
After a lot of calculations, the integral becomes logarithmically diverging:
where D = 4 is the dimension.
A strange detail: the divergence has a negative sign (marked in red) in the vertex correction, while it had a positive sign in the real photon emission. Why is this?
In the Schrödinger equation, when an electron wave bounces from another negative charge, there is a 180 degree phase shift.
The vertex correction diagram has one vertex more than the real photon emission diagram. This might explain the sign difference. But it does not explain what the phase of the two-particle combination e- and γ should be. We only considered the electron wave.
How to get rid of infrared divergences?
In the case of the real photon emission, we argued above that the electron always emits an infinite number of real photons. We argued that the Feynman probability amplitude for the combination e- and γ is the product of the e- and γ fluxes. Therefore, the integral must diverge. It is the correct behavior for the integral.
We did not check if the integral agrees with a classical calculation of the emitted wave, though.
Also, we remarked that the classical electromagnetic wave cannot be divided into a fixed combination of real photons. We can "mine" energy away from a classical wave packet in many ways.
In the case of the electric vertex correction, we claimed that the electron always emits an infinite number of almost "on-shell" virtual photons, and reabsorbs them later. These virtual photons will reduce the effective mass of the electron. The scattering should be calculated with the reduced mass. In the classical limit, this is self-evident: the far field of the electron does not have time to react to the scattering of the electron e- when the electron comes close to the charge X.
The Feynman diagram calculations in the classical limit are definitely wrong. Feynman cannot handle a case when many quanta are always produced in a process.
Actually, we could say that when many quanta are created, we come to the "non-perturbative" regime. With many quanta, we can model classical processes accurately. Feynman diagrams only work in a "perturbative" setting where a single quantum is created at a time.
Conclusions
We found the reason why there is an infrared divergence in the emission of a real photon in electron scattering.
It is the expected result, and not incorrect: the probability amplitude is the product of the photon amplitude and the electron amplitude. An infinite number of photons are always emitted in the process.
However, we did not check if the Feynman diagram correctly reproduces the classical emitted electromagnetic wave. There must be lots of empirical measurements of bremsstrahlung. Feynman diagrams must reproduce the measured results. Thus, Feynman diagrams do calculate the emitted wave approximately right for microscopic processes.
In the case of the electric vertex correction, many, almost on-shell, photons are always emitted and absorbed. This explains the infrared divergence. How to fix it? Putting a suitable cut-off will remove annoying small photons, and should yield approximately correct results. Feynman diagrams can handle one photon which is rarely created. The divergence problem shows up with many simultaneously created photons.
In the utexas.edu paper it is claimed that the divergences in the real photon emission and in the electric vertex correction "cancel out" each other. That is definitely a wrong claim. A state which has a different number of particles, and a different electron 4-momentum, can never cancel another state.
We introduced a new hypothesis:
- When many quanta are always produced in a process, then the process approaches the classical limit, and the "perturbative" method of Feynman diagrams will fail.
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