mildly relativistic
e- • --------------------------------------
|
| q virtual photon
|
/ \ k + q virtual electron-
\ / positron pair e- e+
|
| q virtual photon
|
● ---------------------------------------
e+ static
Let us analyze purely classical waves which interact. The electron and the positron are presented as waves in the Dirac field. These waves meet each other.
There is an interaction between the electromagnetic field and the Dirac field.
The waves disturb the electromagnetic field, causing waves there. The waves in the electromagnetic field, in turn, disturb the Dirac field.
The vacuum polarization loop is the disturbance in the Dirac field.
If these all are classical waves, then there cannot be any divergence. A divergence would break conservation of energy.
Deep question. Do the quantum waves in the physical world, also in a Feynman diagram, behave like classical waves, or can they behave like in the Feynman integrals, where divergences occur?
If the answer to the question is that the waves must behave like classical waves, then we have a solution to the divergence problem of QED: there is no divergence if the calculations are done in the classical way.
Photon versus a laser beam. A laser beam is a classical wave. A photon behaves just like a laser beam until we measure the photon, and the wave "collapses". That is, a photon must be described as a classical wave until the measurement. Does the same hold for any intermediate state of a Feynman diagram, until the outgoing particles are measured?
Divergences are a result of "breaking into more degrees of freedom"
If one interprets the diagram above as a Feynman diagram, then the phase of the outgoing waves is not affected by k at all. There is a constructive interference for all k. This leads to the notorious divergence problem.
The divergence of the vacuum polarization loop happens when the system "breaks into more degrees of freedom". The value of the the 4-momentum k can be chosen freely – it is a new degree of freedom.
Classical waves also can break into more degrees of freedom. But for them, this does not create any divergences. Why?
The obvious answer is that destructive interference wipes out any classical waves which have high momenta k (short wavelength). If a wave is wiped out, then it cannot pass a disturbance forward.
A classical analogue of the vacuum polarization loop
Let us have two elastic metal plates. They correspond to the electron and the positron in the loop.
Let us model the electromagnetic wave with a wave propagating in a rubber membrane. The rubber membrane is somehow loosely attached to the plates, maybe via very elastic rubber blocks. This constitutes the interaction.
metal plate (e+)
------------------
interaction
/\/\/\/\/\/\/\/\/\/\/\ rubber membrane
interaction
------------------ (e-)
metal plate
--> wave propagation direction
As the rubber wave meets the plates, it interacts with them and creates waves into the plates. Later, these created waves can be absorbed back into the rubber membrane.
Does this mean that the rubber wave "breaks into more degrees of freedom"?
We can model the interaction by assuming that each small area element of the rubber membrane hits with a "sharp hammer" both metal plates. This is the Green's function approach to analyze the process.
The impulse from the sharp hammer generates waves of varying momenta k. There is no limit on how large |k| can be.
Let us analyze different momenta. Does it make sense that the e+ plate receives k and the e- plate -k? If the rubber wave pushes the plates apart (like an electric field pushes e+ and e- apart), then this assumption is reasonable. An area element of the rubber membrane hits both plates with a sharp hammer: one upward and the other downward.
Later, the process can happen time-reversed: the rubber membrane absorbs some waves from the plates.
We seem to have a process analogous to what happens in the Feynman diagram vacuum polarization loop. We can freely choose k, and it will contribute to the waves absorbed back into the rubber membrane.
Why does this not lead to a divergence in the classical system?
One aspect is conservation of energy. Hitting with an infinitely sharp hammer would consume an infinite amount of energy? Then all hits must be done with a "blunt hammer". Destructive interference wipes away all high |k|.
Why should Feynman diagrams allow a sharp hammer? Why not require a blunt hammer?
The vertex correction: the infrared divergence with small |k|
Another loop in a Feynman diagram appears in the vertex correction of QED. Again, the system "breaks into more degrees of freedom", because of the loop. Let us try to find a classical counterpart.
^
\ p + q
\
\
| \
| \ p - k + q
k | |-----------------● Z+ nucleus
| / q
| / p - k
/
/
/
e-
p
The electron e- bumps from the nucleus Z+. The momentum exchange is q. The electron sends to itself a virtual photon whose 4-momentum is k.
We can choose k freely. This causes a divergence in the Feynman integral for small |k|.
There is a direct classical analogue of this process. If we model the electric field of the electron with an elastic block of rubber attached to the electron, then the momentum exchange q makes the rubber to "wobble". The electron gives some momentum k and kinetic energy to the rubber block, and absorbs some of k back when the electron leaves the nucleus.
We already analyzed this in our post on September 24, 2025. The Feynman integral method does not understand the fact that an infinite number of "soft" real photons, or virtual photons with small |k|, are produced when the classical electron passes once by the classical nucleus. The integral thinks that these are all separate cases. The integral diverges to infinity because it thinks that the cases are separate, while they are not.
We see that the "break into more degrees of freedom" is the infrared case causes a divergence problem which is quite different from the ultraviolet case (vacuum polarization).
Further analysis of the classical vacuum polarization analogue
The metal plate analogue above is awkward. Very different from any "polarization". What about assuming a polarizable medium? When a classical electromagnetic field meets that medium, it produces a "polarization wave".
/\/\/\/\/\ ############
--->
electromagnetic polarizable medium
wave
We might have two charged fields. One would have a uniform positive charge density, and the other the canceling negative charge density. These two classical fields would correspond to the Dirac field.
The classical analysis is similar to our rubber membrane - metal plate model. Since the incoming wave is smooth, destructive interference cancels all high momenta |k|.
Quantum mechanics: the interference pattern of two created particles
We have tackled with this question many times in the past few years. Suppose that two waves are born simultaneously: e+ and e-.
For classical fields, it is enough that destructive interference cancels the waves e+ and e- separately. But in quantum mechanics, the waves are entangled. How does destructive interference act on them?
Maybe we should interpret e+ and e- as one particle which moves in 6 spatial dimensions? There is a problem: what is the 4-momentum of that single particle?
B, (E, p) + k
----------
/ \
------------- -------------
A, (E, p) \_______/ A, (E, p)
C, -k
Let us analyze this for a hypothetical particle A with a 4-momentum
(E, p).
The particle splits into two particles B and C. Above, k is an arbitrary 4-momentum. Later, the two particles are again merged back into one particle A.
Assuming that the wave function of the system B & C is the same as that of A. The typical wave function of a relativistic particle A with a 4-momentum (E, p) is of the form
ψ(t, x) = exp( -i / ħ * (E t - p • x) ).
It makes sense to demand that the phase of the "combination" B & C should advance as if they were the particle A. That is, a temporary splitting into two particles should not change the phase of A relative to the case where A moved alone.
But this assumption would mean that destructive interference cannot destroy anything. That would imply that a divergence happens, which is nonsensical.
The classical limit: the analogy with a laser pulse and electric polarization inside glass
In a laser beam we have many coherent photons. A laser beam behaves like a classical wave. In our diagram above, A is a virtual photon. Assume that we have a coherent beam of such photons. Can we argue that then the waves representing B and C must be classical waves?
Let us have a classical laser pulse entering a block of glass. There is some electric polarization in the electrons in the glass, which is manifested in the refractive index of the glass. The "polarization wave" inside the glass certainly is a classical wave. The polarization wave carries some energy of the laser pulse.
If we shoot a single photon through the block of glass, the wave associated with the photon behaves just like the classical laser pulse. Also, the associated "polarization wave" behaves like a classical wave.
What kind of a "divergence" might happen if we shoot a single photon?
Some of the photon energy may, in some cases, travel in a polarization wave. The process then is like this:
+
-------
/ \
------------- ---------
photon \______/
-
polarization
in glass
A "divergence" in this case would happen, if we would calculate a very large probability amplitude for the process above.
A Feynman diagram simplifies the process into a single hit with a "sharp hammer" into the polarization field, and the absorption later. A very crude model of the process. The Feynman integral calculates the contributions of arbitrarily short wavelengths (large |k|) in the process. This is very different from the treatment for a classical laser pulse. Destructive interference erases any large |k| waves.
Did we finally prove that the quantum waves in Feynman diagrams must behave like classical waves? So that destructive interference erases all high |k| waves?
Discussion: divergence in the classical limit versus quantum processes
If we try to model classical electric polarization in a glass block with a Feynman integral, we probably end up with a divergence. Energy is not conserved. This shows that the calculation is erroneous.
If we try to model vacuum polarization in a quantum field process with the Feynman integral, we end up with a divergence. Again, this shows that the calculation is erroneous.
The Feynman integral miscalculates both processes. It may be that the miscalculation happens for the same reason, and in the same way, in both cases.
If that is the case, the way to correct the calculations is to take into account the fact that destructive interference destroys all high 4-momentum |k| waves.
"Regularization" and "renormalization" for Feynman integrals probably work because they implement the destructive interference, in an ad hoc way.
We wrote in the fall of 2025 that quantum processes try to "imitate" classical processes, but the "resolution" of wave functions is not sharp enough. The Feynman model of one "sharp hammer" strike may be the correct one, once one removes the divergence. We cannot calculate the processes classically. Feynman integrals can calculate the electron anomalous magnetic moment to the precision 10⁻¹¹. This suggests that the Feynman model is right, once divergences are removed.
It would be wrong to say that the waves in quantum processes are strictly classical. If one tries to calculate bremsstrahlung purely classically, one obtains wrong results. The "resolution" of wave functions is not sharp enough to reproduce the classical process exactly. The classical limit is achieved only by making the resolution better: making particles heavier and charges larger.
Even though destructive interference wipes out high |k| for classical waves, it could still be that quantum waves are fundamentally different from classical ones, and that one is not allowed to use destructive interference to remove divergences for quantum waves.
However, if we approach the classical limit by making particles heavier, then the classical destructive interference must play a role in removing divergencies.
Question. What is the classical limit of various regularization and renormalization methods? Do they work in a reasonable way in the classical limit?
How "classical" were LEP collisions at 209 GeV? Not classical
The CERN LEP collider had a combined energy of 209 GeV for an electron-positron pair. The de Broglie wavelength for an energy of 104.5 GeV is λ ~ 10⁻¹⁷ m.
The energy 100 GeV corresponds to a momentum
p = E / c² * c
= E / c
= 1.6 * 10⁻⁸ J / c.
How close must e- and e- pass each other, so that the momentum exchange is on the order of p? Below, r is the shortest distance between the particles.
F = kₑ e² / r²,
t = 2 r / c,
p = F t
= 2 kₑ * e² / (c r),
r = 2 kₑ * e² / E
= 3 * 10⁻²⁰ m.
We see that the electron path was not classical for head-on collisions, where the momentum exchange was large.
The path of the electron may be roughly classical if
r > 10 λ = 10⁻¹⁶ m.
Here we assume that the uncertainty of the momentum of the electron is very large. Maybe a more realistic figure is 1,000 λ, or
r > 10⁻¹⁴ m?
This implies that the electron/positron paths in the LEP could not be considered "classical".
Wikipedia has the following electron electric potential correction. This is probably only for low-energy electrons in an atom:
The formula for r >> λ contains a coefficient
α / (4 sqrt(π)) ≈ 1/1,000
and
exp(-2 r / λ) ≈ 10⁻⁹.
We conclude that vacuum polarization has a negligible effect if the electron has a classical path, at least in the low-energy case.
The Bhabha (1935) semiclassical derivation of pair production
H. J. Bhabha (1935) calculated the pair production cross sections of a charged particle moving in the field of a nucleus. The nucleus is a heavy particle, called Z₂. A lighter charged particle Z₁ passes by Z₂. The calculation is semiclassical, since both particles have a defined position.
--------------------------- e-
/
/ \_________________ e+
/
/ virtual photon
Z₁ --------------------------------------------
| virtual photon
Z₂ --------------------------------------------
Bhabha calls the above Feynman diagram the "first-order" process.
Z₁ -------------------------------------------
| virtual photon
|-------------------------- e-
|-------------------------- e+
| virtual photon
Z₂ -------------------------------------------
The above diagram is the "second-order" process. At low velocities of the colliding particle v << c, the first process dominates.
Bhabha assumes that there is a "negative-energy electron" around. He calculates the perturbation which the electric fields of Z₁ and Z₂ impose on the Dirac wave function ψ of this negative-energy electron.
The negative-energy electron really is a positron e+ arriving from the future at a 4-momentum E₊, p₊? If yes, then Bhabha calculates the scattering of e+ from Z₁ and Z₂. If the positron is scattered into an on-shell electron, and Z₁ and Z₂ exit the process on-shell, then we have the probability amplitude for a the production of this special case of a pair, just like in a Feynman integral.
Perturbation of a zero Dirac field. This is a big problem in QED. If we have the Dirac field set to zero, ψ = 0, and use the "minimal coupling" to the electromagnetic field, then the perturbation always is zero. The function ψ remains zero. How can an electromagnetic field then produce a pair e-, e+?
We have to assume that we can freely choose a positron "arriving from the future", so that we can calculate a non-zero perturbation.
In a Feynman integral, the assumption is that the electromagnetic field somehow, in an unspecified way, acts as a source in the Dirac equation above. The source is written as a Dirac delta function on the right side of the equation above. The source is a hit with a "sharp hammer", an impulse. The impulse response, the Green's function, is the propagator of the electron.
In both the Bhabha approach and the Feynman approach, there is the problem that one has "too much freedom" to choose the response of the Dirac field to a changing electromagnetic field. The "causality" of the disturbance in the Dirac field is too weak. This leads to the divergence problem in vacuum polarization.
If the lagrangian of interacting fields A and B is correct, then in principle, the disturbance in A should causally determine the disturbance in B. Let us analyze why this does not happen in the Bhabha/Feynman approach. Is the problem in the perturbation method?
In the link we have the QED action. A disturbance in the electromagnetic field should produce a disturbance in the Dirac field, such that the action has an extremal value for that history. Note that we do not know if any solution exists at all.
Intuitively, if the Fourier decomposition of the EM field does not contain high 4-momenta |k| (short wavelengths), then it cannot happen that high |k| will appear in the Dirac field. Did we solve the mystery of divergences?
Analyzing the action of QED
Let us try to figure out what kind of a disturbance is produced in the Dirac field if the electromagnetic field changes. Does the action produce a reasonable answer?
If the energy density F² of the electromagnetic field is large, we can try to reduce it by introducing a pair e-, e+ in the Dirac equation. But now we encounter an immediate problem: how does a disturbance of the Dirac field affect the electromagnetic field? What is the electric field of a virtual pair? The action does not tell us that!
Thus, the definition of the action is badly deficient. We do not know the correct action for QED.
Hypothesis. Whatever is the correct action for QED, a disturbance in the electromagnetic field cannot produce in the Dirac field Fourier components with a very high 4-momentum |k|.
The hypothesis above is reasonable, and it solves the divergence problem of QED. The divergence problem arises from the perturbative approach of Bhabha/Feynman. Bhabha and Feynman only use conservation of energy and momentum to restrict the Fourier components in the Dirac field. For a typical lagrangian, destructive interference wipes out high |k|. Bhabha and Feynman do not use this fact.
The Bhabha method for solving the (unknown) QED action: vacuum polarization from a produced pair
1. If we have an electron, and its associated Dirac field wave, then a good guess is that the behavior approximates the trajectory of a classical charged particle. This is why Paul Dirac in his 1928 paper added the "minimal coupling" to his equation.
2. When an electron wave packet approaches a massive charged particle, the wave is distorted. It gains Fourier components which do not correspond to an on-shell electron. Bhabha guesses that these off-shell, virtual Fourier components approximate the correct Dirac field wave for pair production.
3. Bhabha requires that outgoing waves represent on-shell particles. This follows from the action for a free Dirac field outside an external electromagnetic field.
4. Bhabha requires that energy and momentum are conserved. This is implied by any reasonable action.
The most uncertain guess is number 2. Is there any reason why the unknown action of QED should produce the disturbance of the Dirac field in this way?
Since various Bhabha/Feynman calculations correctly predict pair production, it is an empirical fact that 2 is approximately true.
Z+ ● ------>
| E strong electric field
v
<------- ● Z-
A physical system typically "tries" to minimize its potential energy. When opposite charges Z+ and Z- pass close to each other, the energy density of the field between them grows. Pair production is a way to reduce the energy of the field. Some of the energy goes to the produced pair, and some to the kinetic energy of the pair. This is the natural semiclassical analogue of pair production.
Let us play the semiclassical process backward in time. We assume that we cannot know the exact path of e- and e+.
Z+ ● <----- • e-
Z- ● <----- • e+
The electron e- tends to approach Z+, and the positron e+ approaches Z-. They will "screen" a part of the electric field of Z+ and Z-.
The electron and the positron will reduce the energy of the fields of Z+ and Z-. In particular, they will reduce the field energy density between Z+ and Z-. Generally, these new particles popping up should reduce the total energy of the electric field, which should show up as an increased attraction between Z+ and Z-. Here we assume that moving Z+ and Z- close to each other causally produced the pair, which reduces the total energy of the electric field.
This is "vacuum polarization" with a real pair e-, e+. Has it been observed?
In our November 4, 2025 post, we speculated that vacuum polarization in QED is caused by "long-lived" virtual pairs, which reduce the energy of the electric field when Z+ and Z- are close.
In quantum mechanics, a system can "tunnel" into a lower potential. When the electron and the positron are "almost" on-shell particles, then we may expect them to have the ordinary electric field of such a particle. Then we can approximately calculate how much energy they freed from the electric field.
In the Bhabha paper, in the more important second-order process, both the particles Z+ and Z- "nudge" the forming pair. After the nudge, the pair is almost on-shell.
In a semiclassical model, after the nudge, e- is still close to Z+, and e+ close to Z-. We see that the pair really did reduce the total energy of the electric field. There should be vacuum polarization from an on-shell pair.
The classical limit and the large Compton wavelenght of the electron
In Bhabha's approach, the particles Z- and Z+ can be very massive. They have classical paths. The pair production typically happens at very short distances. These classical particles might pass at a distance
r = 10⁻¹⁵ m
from each other. This is much shorter than the Compton wavelength 2.4 * 10⁻¹² m of the electron. If a pair is produced, then we know the location of the production much more precisely than the Compton wavelength of the electron. How is this possible?
The explanation may be that even a short antenna can receive a radio wave of a much larger wavelength. The disturbance in the Dirac field has a wavelength which is much larger than 10⁻¹⁵ m. The disturbance in the electromagnetic field is an "antenna" which produces a wave in the Dirac field.
This analysis suggests that classical electric fields do have pair production, and vacuum polarization, as long as pairs are able to "tunnel" into existence.
Suppose that Z- and Z+ have very large charges. We see no reason why their collision could not produce many pairs. If the charges are only one unit e, then a single pair e-, e+ can "saturate" the process, and only one pair can be produced, in most cases.
If many pairs are produced, then thr resulting Dirac wave would be classical, consisting of a large number of coherent electrons and positrons. Do we finally have the classical limit?
*** WORK IN PROGRESS ***
