Also, how does the electromagnetic field give rise to a Dirac field wave in pair production? The Bhabha calculation (1935), and the Feynman diagram seem to create the Dirac field wave from "nothing". Bhabha assumes that a positron with a 4-momentum E₊, p₊ is around, and calculates its scattering first to an off-shell electron, and then to an on-shell electron. Why can Bhabha assume that such a positron is around?
A many-worlds approach might make the strange appearance of the positron understandable. Also, it could make the histories such that they are time-symmetric, and obey a typical action of interacting fields.
Bhabha (1935): positron scatters from Z+ and Z-
Z+ -------------------------------------------------
\ virtual photon
\ ______________________ e-
| E, p
| off-shell electron
|______________________ e+
/ E₊, p₊
/ virtual photon
Z- ------------------------------------------------
Bhahba calculates backward in time. He assumes a positron wave meeting a massive charge Z+, scattering into an off-shell Dirac wave, and then that off-shell wave scattering from a massive charge Z- in such a way thaf the wave becomes an on-shell electron e-.
Bhabha calculates probability amplitudes for processes which correspond to a positron E₊, p₊, and an electron E, p.
We then have a set of histories in which a positron with various E₊, p₊ scatters into an electron E, p. But something seems to be missing from a time-symmetric picture. What if the positron wave traveling back in time fails to scatter from Z- and Z+ in the described way? The same for the electron: if we calculate backward in time, most electrons should not scatter from Z+ and Z-, and should not become positrons.
Building histories backward in time: many worlds branch also backward in time
Z+ -------------------------------------------------
\ virtual photon
e+ __________\
E₊, p₊ |
| off-shell electron
|__________________ e+
/ E'₊, p'₊
/ virtual photon
Z- ------------------------------------------------
Above we have another scattering process. This time, the positron moves forward in time. It may scatter from Z+, and later from Z-. But most positrons may not scatter at all. Some might scatter backward in time.
It looks like the traditional way of understanding the diagrams cuts off scattering alternatives, which should exist in a time-symmetric interpretation of the diagram.
And time symmetry is needed to make the diagram to correspond to a unitary development. A typical action describes an unitary process.
The Bhabha approach is to build a history which looks nice. He builds it toward the future (though he lets the positron move backward in time for a while). An even nicer history can be built if we can also build backward in time. For this to succeed, and make sense, we need many worlds.
Our branch is such that the only particles entering the reaction are Z+ and Z-. But in other branches going backward in time from the reaction, e+ or e- are outgoing particles if we look backward in time.
The scattering of a Dirac wave from a heavy charge Z+ or Z- can be calculated in a nice way. It is like classical fields. We can calculate the behavior of the wave in space and time coordinates. There is no need to use the "momentum space" of Feynman integrals.
What is the phase of the Dirac waves in pair production?
Different histories produce pairs with different 4-momenta. Destructive interference cannot happen if the histories can be distinguished with a measurement.
We still have the problem: what are the phases of the produced Dirac waves of the electron and the positron? They must have a phase, or must they? Do we have an absolute way to determine what is the phase of an electron? Electrons obey the Pauli exclusion principle. There cannot be a coherent wave of electrons with the same 4-momentum, at least not in a small spatial volume.
For an electromagnetic wave, we can measure its phase, at least if we have many coherent photons.
A typical classical action determines the phase of all the waves it produces. If we have two plates A and B attached with a rubber block, then a wave in A may produce a wave in B, and the phase of the wave in B is deterministic.
| #
| # ---------- sharp hammer hits
v \/
---------------------------- metal plate
Hypothesis. If a pair is produced in a small region of spacetime, compared to the wavelength of the produced Dirac waves, then we can assume that all produced Dirac waves have the same phase at their production point. This is the result of a hit with a "sharp hammer": all the produced waves have constructive interference at the point of the hammer hit.
The hypothesis is true for classical electromagnetic waves produced, when two charges Z+ and Z- pass close to each other. It is a sharp hammer hit to the electromagnetic field.
However, there is no obvious reason why the usual QED action would force the Dirac wave to satisfy this. We again encounter the insufficiency of the action.
Hypothesis of inheriting the phase from other particles. Since the produced pair gets its energy and momentum from the colliding particles, we could define that the phase of the combination e- & e+ "inherits" its phase from the particles which donate the energy and the momentum to the pair. Furthermore, we can demand that the phase of the electron and the positron must be continuous over the path of e+ -> off-shell electron -> e-.
Z+ ----------------------------------------------------
\ virtual photon
\ ___________________ e-
/ E, p
/ off-shell electron
/______________________ e+
/ E₊, p₊
/ virtual photon
Z- -----------------------------------------------------
Let us analyze the diagram above. Do we obtain reasonable phases for e-, e+?
Suppose that both e- and e+ contain a total energy E which they inherited from Z+ and Z-. The phase α of E is determined by Z+ and Z-. Does this determine the phases of e- and e+?
Imagine that the path of e+ -> off-shell electron -> e- is a metal cylinder. The cylinder is bent. Let us roll the cylinder slightly clockwise at e+. Then it will roll anti-clockwise at e-. The sum of the angles stays the same. Unfortunately, we were not able to find a unique way to determine the phase of e- and e+. The Hypothesis does not work.
Let us assume that Z+ is static. Then it cannot give any energy to the pair. We could demand that all the energy E in the pair has the phase determined by Z-.
Since Z+ is static, it cannot donate any momentum either. It can only absorb momentum. We could demand that e+ inherits its phase from Z-. This fixes also the phase of e-.
If Z- has a macroscopic mass, then its phase varies extremely rapidly. It is not clear what phase should the virtual photon sent by Z- inherit, or e+ inherit. This does not look too nice.
Suppose that we have a system which initially has a 4-momentum E, p. We ignore the spatial extent of the system. Its phase varies in time by the formula
exp( -i / ħ * (E t - p • x) ).
Then it emits particles e+ and e-, with 4-momenta E+, p+ and E-, p-. Let us define
E' = E - E+ - E-,
p' = p - p+ - p-,
Let us assume that the emission happens at the time t = 0 at the location x = 0. After that, it is natural to assume that the phase of the rest of the system obeys the formula
exp( -i / ħ * (E' t - p' • x) ),
the positron e+ obeys
exp( -i / ħ * (E+ * t - p+ • x) ),
and the electron e- obeys
exp( -i / ħ * (E- * t - p- • x) ).
The product of the phases obeys the same formula as before the reaction.
Vacuum polarization
Z+ --------------------------------------
|
| q virtual photon
|
/ \
/ \ k + q virtual electron-
-k \ / positron pair e- e+
\ /
|
| q virtual photon
|
Z- ---------------------------------------
In the phase inheritance hypothesis in the previous section, the idea is that the phase of the entire system develops according to the old formula, even if it splits into a larger set of particles.
Does the inheritance hypothesis say anything about the phase of the virtual pair particles in the vacuum polarization loop?
The virtual photon 4-momentum |q| might be very small, but |k| may be very large. Why would the phase of k be dependent on q? In the diagram above, we can assign any phase to the 4-momentum k, as long as -k has the opposite phase. Then their product is 1, and they do not affect the product of phases.
As we have noted many times in our blog, the diagram above actually summarizes what does not happen. The Feynman integral is infinite, and if |q| is infinitesimally small, then regularization removes the integral entirely. We should find reasons why the history above does not happen, even though a superficially similar pair production diagram produces pairs.
Let us apply the reasoning which is used in pair production:
1. We are allowed to assume that a flux of positrons (of electrons) exists, with a 4-momentum k.
2. The positron scatters from Z-, acquiring an additional 4-momentum q.
3. The positron scatters from Z+, donating a 4-momentum k to Z+.
4. The positron returns to its original state with a 4-momentum k. Therefore, we are allowed to assume that the history "really" happened.
Items 1 and 4 differ drastically from the corresponding principle in pair production. In pair production we have the principles:
1'. We are allowed to assume a flux of real positrons with a 4-momentum E+, p+ exists.
4'. The positron becomes a real electron, and escapes. Therefore, we are are allowed to assume that the history "really" happened.
In Feynman integrals, anything that conserves 4-momentum is a history that "really" happens. Divergences prove that that rule is wrong.
Long-lived virtual photons in the vertex correction
k virtual photon
~~~~
/ \
e- -------------------------------------
1 |
| q virtual photon
proton+ -------------------------------------
In our blog post on October 22, 2025 we suggested that the subtracted probability amplitude of the electric form factor
F₁(0) - F₁(q²)
represents long-lived virtual photons which reduce the effective inertial mass of the electron which scatters from the proton. We have a very good classical analogy for this: the inner field of the electron follows the electron closely, and does not reduce its inertial mass. But the far fields "falls off" for a while, and does not contribute to the inertial mass.
Interpretation of the vertex Feynman diagram. We can interpret the diagram in this way: the "hammer hit" (impulse) at the position 1 always produces the same full spectrum of virtual photons, regardless of q. That is, F₁(0) is the "response" to the hammer hit.
The Feynman integral calculates which of these are absorbed so quickly that they have no effect on the process. The missing part of the integral is those virtual photons which live long and are only later absorbed back to the electron. The long-lived photons reduce the inertial mass of the electron, and contribute to the scattering amplitude.
Let us model the electric field with an elastic rubber plate attached to the electron. When the electron is under an acceleration, energy flows to the rubber plate, to distort and bend the plate. Less mass energy is left in the electron. This is another way to explain the reduced inertial mass.
If the electron is static, the Fourier decomposition of its electric field only contains time-independent waves, whose energy E is zero:
exp( -i / ħ * (E t - p • x) ).
We could say that the electric field of a static electron has "zero energy".
Long-lived off-shell pairs in vacuum polarization
e- --------------------------------------
| q
/ \
/ \ vacuum
-k \ / k + q polarization
\ /
| q virtual photon
proton+ ---------------------------------------
On November 4, 2025 we suggested that in vacuum polarization, the subtracted probability amplitude
Π₂(0) - Π₂(q²)
represents long-lived off-shell pairs which exist for a long enough time, so that they will reduce the energy in the electric field between e- and the proton, and effectively make the attractive force between e- and the proton stronger.
Interpretation of the vacuum polarization diagram. Let us assume that the electron passes the proton at a distance r such that the momentum transfer is q. The value q measures the "disturbance" of the electric field between the electron and the proton. The disturbance q hits the Dirac field "with a hammer". An off-shell pair may be born. The hammer hit is represented by Π₂(0).
The Feynman integral Π₂(q²) calculates those off-shell pairs which annihilate quickly and have no effect, except relaying the momentum q to the proton. That is, those pairs do not affect the scattering amplitude at all.
The remaining off-shell pairs are long-lived, and they add to the scattering amplitude. The difference
Π₂(0) - Π₂(q²)
tells us how much they add. Why do they add to the scattering amplitude? Because they can relay the momentum exchange q between the electron and the proton, even if the particles pass at a distance > r.
A standard principle in interpreting Feynman diagrams is that each different diagram with the same input and the same output adds to the scattering amplitude of a reaction. Why should it be so? If the second diagram "eats the market share" of the first diagram, then that principle does not need to hold.
Another way to explain why Π₂(q²) has no effect on anything: if the electron is moving at a constant velocity, then Π₂(0) represents a "hammer strike" which destructive interference from earlier hammer strikes completely wipes out (except maybe Fourier components with E = 0). As the electron passes the proton, then Π₂(q²) is that part which is destroyed by interference from earlier strikes. That is, Π₂(q²) has no effect on anything.
The Feynman diagram in the case of vacuum polarization describes the complement of the relevant process.
The action of QED for the vertex correction
Above, we wrote about hammer strikes. Let us try to interpret these in the framework of the (unknown) QED action.
If we assume that the electron is very heavy, and its electric field is very strong, then the vertex correction is a classical process. It should obey a classical action of electrodynamics.
Note that we argued on February 25, 2025 that Maxwell's equations fail for accelerated systems. Thus, we do not know the correct action for classical electromagnetism, either.
In the classical action, high frequencies, or large 4-momenta |k|, probably play no role. Destructive interference cancels them.
Why would the Feynman integral correctly calculate the effect of the reduced mass of the electron in the vertex correction?
The Quantum Magnification Principle of November 4, 2025 may explain that. The scattering of the electron happens in a very small volume of spacetime. If we try to model the mass-energy of the far field of the electron with a wave, that wave will have a very short wavelength, and a very large energy. To work around that, quantum mechanics assigns a large mass reduction in a small number of cases. Other cases have no mass reduction.
e- -------------------------------------
|
| q virtual photon
proton+ -------------------------------------
The bulk of the scattering amplitude comes from the plain Coulomb scattering diagram, above. There is no mass reduction.
In a small number of cases, the Feynman integral for the vertex correction diagram has the mass of the electron greatly reduced. Does the vertex correction diagram (integral) then calculate the added scattering amplitude right? The integral should calculate the difference which the mass reduction makes to the ordinary scattering amplitude. Does it?
The formula
F₁(0) - F₁(q²)
calculates a difference. Could it be that F₁(0) represents the elastic scattering behavior? The subtraction then calculates the impact of adding the disturbance q to the process?
If that is the case, then if we add the integral of the elastic scattering diagram, and the integral of the vertex correction diagram, then we do not count overlapping probabilities twice.
On October 22, 2025 we calculated that the QED vertex correction in a hydrogen atom roughly agrees with the classical vertex correction. But we do not understand the details: how do the Feynman integrals manage to calculate an estimate for the classical vertex correction? In that post, we just presented a hypothesis that it calculates the classical effect.
-k virtual photon
~~~~~~~~
/ \
/ A B \
e- -----------------------------------------
p p + k | p + k + q
|
|
| q virtual photon
proton+ -----------------------------------------
When we add the virtual photon k to the elastic scattering diagram, the Feynman integral also gain the electron propagators for the lines A and B as coefficients.
If q is infinitesimal, then A and B are symmetric.
If q is larger, it "displaces" the Feynman integral more.
Let us assume that k contains a large amount of positive energy. Adding the contribution of the Feynman integral is equivalent to saying that then the electron comes closer to the proton, and adds to the scattering amplitude. It is an additional history which contributes to the total amplitude.
The classical vertex correction is increased "Coulomb focusing" from a lower inertia of the electron
e- • ------------------
\
b \
●
proton+
Let us assume that the electron is mildly relativistic, and scatters from the proton to a large angle. Then, using natural units,
|mₑ|, |q|, |k|
can be assumed to be of roughly the same order of magnitude.
Let
r
be the classical impact factor b which gives the electron a pull q as it passes the proton.
When a classical electron approaches the proton, the attraction to the proton will make the impact factor b less when the electron is close to the proton. This increases the scattering amplitude, and is called Coulomb focusing. The elastic scattering Feynman diagram does not understand Coulomb focusing.
Classically, the inertia of the electron is less than mₑ close to the proton, because the far field of the electron does not have time to react.
Suppose that k takes a large portion of the energy of the electron away. Then, Coulomb focusing will increase the scattering cross section greatly, possibly twice or more. This proves that it may be reasonable to add the Feynman vertex correction integral to the elastic scattering cross section.
For very small |k|, the integral is small. Presumably, the main contribution to the integral comes from values |k| ~ |q|. In the classical analogy, most of the contribution to the scattering amplitude will come from electrons whose impact factor is > r. That is, they will increase the scattering amplitude.
We are interested in the number of electrons which scatter to > |q|. The elastic diagram gives us the first estimate. The vertex correction diagram gives an estimate of the extra amplitude we should add. We can add it because most of that extra amplitude comes from electrons with the impact factor b > r.
Is it reasonable to assume that if the inertial mass of the electron is halved, then Coulomb focusing will double the scattering cross section?
When the electron passes the proton, its angular momentum is conserved relative to the proton. A large scattering angle requires that the total energy of the (relativistic) electron is very crudely double close to the proton. Then its distance to the proton is only b / 2. Coulomb focusing is a very prominent phenomenon.
If the inertial mass of the electron is halved close to the proton, then it will accelerate faster toward the proton. It will pass the proton much closer. It is reasonable to assume that the cross section is double in such a case.
We conclude that for mildly relativistic electrons, adding the Feynman vertex correction integral to the elastic scattering integral makes sense.
Earlier in our blog we have shown that the electron anomalous magnetic moment correction, which Julian Schwinger in 1948 calculated from the vertex correction, agrees with the classical vertex correction if we assume that the electron in zitterbewegung makes a loop whose length is the Compton wavelength.
An mildly relativistic electron is very far from the classical limit. The bulk of the mass-energy in the electric field of the electron is close to its classical radius
rₑ = 2.8 * 10⁻¹⁵ m,
but its Compton wavelength is much larger,
λₑ = 2.4 * 10⁻¹² m.
We cannot "build" the electric field from waves which are 861 times longer than the most prominent detail in the field. Thus, we cannot expect that the classical vertex correction in this case is anything close to the quantum vertex correction. Classically, an electron scattering to a large angle passes the proton at a distance rₑ. Then the mass-energy of the far field is close to mₑ, and the classical vertex correction could easily be 100% to the scattering amplitude. Looking at the literature, the QED vertex correction seems to be much smaller, on the order of 1/861.
Comparing the classical vertex correction and the QED vertex correction in a hydrogen atom
On October 22, 2025 we calculated that if an electron passes a proton at the speed (0.008 c) and the distance (5 * 10⁻¹¹ m) as in a hydrogen atom, then the classical vertex correction and the QED vertex correction roughly agree. In this case, the electron and its field can, maybe, be considered an almost classical object. But how does the Feynman integral manage to calculate the classical phenomenon?
-k virtual photon
~~~~~~~~~~
/ \
/ A B \
e- -----------------------------------------
p k + p | k + p + q
|
|
| q virtual photon
proton+ -----------------------------------------
p
e- • -----------------
\
● v
proton+
| p + q
v
The 4-momentum q is sufficient to turn the path of the electron by 90 degrees.
The Feynman integral for vertex correction mostly consists of contributions in which
|k| ~ |q|.
If the electron emits such a virtual photon, and the photon survives for a long time, then the emission substantially affects the path of the electron. But why would all such emissions contribute positively to the scattering amplitude?
If q is infinitesimal, then, obviously, the vertex correction integral cannot affect the scattering amplitude. The integral describes a symmetric process in which the electron sends a photon k to an arbitrary direction. The difference
F₁(0) - F₁(q²)
describes how much does the integral change from the q infinitesimal case. Could it be that the entire change in the integral adds to the scattering amplitude?
The Feynman integral contains coefficients from the electron propagator:
1 / (E² - s² - mₑ²),
where E is the total energy of the electron, s is its spatial momentum, and mₑ is its mass. The denominator essentially calculates how much the electron is off-shell.
The 4-momentum p is on-shell, as well as p + q.
The 4-momentum k takes the electron off-shell. Then the absolute value of the integral grows smaller, since with p or p + q it is infinite.
Let us denote
k = (E, s).
If s is parallel to the spatial momentum of p + q, then the propagator for k + p + q is significantly smaller than the propagator for k + p. The velocity in k + p + q is significantly larger => it is more off-shell.
If s is antiparallel to the spatial p + q, then the velocity in k + p + q is smaller. It is off-shell. But about as much off-shell as k + p. The integral value does not change much.
We see that the integral grows smaller in the case that k is rougly parallel to spatial p + q. And in this case, the electron moves closer to the proton, so that the impact factor b becomes smaller. This makes sense.
*** WORK IN PROGRESS ***
