Classical vertex correction reduces scattering?

Our October 22, 2025 analysis of the classical vertex correction was deficient. This deficient analysis was inherited into our August 17 and 21, 2026 blog posts.

We only considered the fact that a reduced electron mass helps it to come closer to the proton: more Coulomb focusing, resulting in more momentum exchange.

But we forgot the fact that the reduced mass also makes the electron to move faster, so that it will be a shorter time close to the proton: less momentum exchange.

Our blog post on March 13, 2021 contained a very rough calculation of the effect of a shorter time close to the proton: the effect raises the energy level of 2s by 5 - 7 μeV, while the vertex correction contribution to the Lamb shift is 4 μeV.

Which process wins in which case?

Opposing classical vertex corrections in a fast fly-by: scattering increases

                         e- • -->  v large << c

                             | 

                             v  F

                                                   1 impact factor          

                                                  

                             ● proton+

      ^ y

      |

       -----> x

We assume that the electron moves very fast and that the electron arrives along a straight line. The impact factor is 1. We assume that its mass becomes reduced during the length 2 of its path closest to the proton. We assume that the electron moves the closest to the proton in the time 1 / v.

Let us reduce the mass of the electron by a fraction

       0  <  f  <<  1

during that path of length 2. Let the electron move to its closest position to the proton:

                   1             Δx

     e-  • -------------     ---

                                    | Δy

                          1        ×  corrected position

                                                              

                          ● proton+

The reduced mass makes the acceleration in the x direction by the fraction f larger, and also the acceleration in the y direction. We have marked the corrected position of the electron after these changes.

The component of F in the x direction is, on the average,

       Fx  ≈  0.35 F.

The force downward is, on the average,

       Fy  ≈  0.85 F.

Thus,

       Δy / Δx  ≈  0.85 / 0.35  ≈  2.4.

The time to the closest position is reduced by, say 0.35%, but the force downward increases by 2 * 0.85% = 1.7%. The momentum gained downward increases by 1.7% - 0.35% = 1.35%.

In this case, the classical vertex correction increases the scattering. The effect of the electron coming closer to the proton dominates. We can almost ignore the fact that the time for the fly-by decreases.

Opposing classical vertex corrections in the hydrogen atom in the 2s orbital: scattering decreases

The electron moves very slowly when it is far from the proton. Then it dives toward the proton, and its velocity greatly increases as it gains kinetic energy from the electric potential.

                   e-

                   •     ^

                   |     |

                   |     |

                   | ● |  proton+

                    -----

The reduced mass electron is "reflected" from the proton. It first goes down, and then comes up almost the same way.

If we reduce the mass of the electron as it descends, then less momentum is "reflected" back upward. The momentum exchange q is smaller. In this case, mass reduction reduces the momentum exchange: it reduces scattering.

Our March 13, 2021 post calculated this effect very crudely. The order of magnitude agreed with the Lamb shift. We got a rise of the energy level by 5 - 7 μeV.

A summary of the many problems in the QED vertex correction

Let us recapitulate what we have learned about the QED vertex correction in the past years.

1.   The correction is ambiguous. By setting the infrared cutoff (the photon mass λ) to a suitable value, we can get almost any value for the QED correction.

2.   The vertex correction seems to be canceled somewhat by the correction caused by bremsstrahlung, but it is not clear how much. Is it canceled entirely?

3.   The QED vertex seems to produce an infinite number of infrared photons during every fly-by, but QED uses a clearly flawed infrared cutoff λ to sweep that under the rug.

4.   The correction uses a suspicious ad hoc regularization scheme for the ultraviolet divergence.

5.   The QED vertex correction claims that there is less scattering because of it, while the classical vertex correction says that in many cases, there is more scattering.

6.   The classical limit of the QED vertex correction is wrong.

7.   If we let h → 0, the QED vertex correction becomes infinite, while in quantum mechanics, the system should approach the classical limit, i.e., the classical behavior.

8.   The QED vertex correction only depends on the momentum exchange q. If we let the electron move very slowly very far away from the proton, it will gather the momentum exchange q. Intuitively, the vertex correction under such peaceful circumstances should be zero, not the value predicted by QED.

A lot of problems. As if the QED vertex correction would be an entirely useless construct.

https://en.wikipedia.org/wiki/Lamb_shift

The following can be said to be an advantage of the QED vertex correction:

- It can reproduce the Bethe term in the Lamb shift, at least if the infrared cutoff λ is tuned in a suitable way.

Hans Bethe 1947 analysis of the Lamb shift

https://harvest.aps.org/v2/journals/articles/10.1103/PhysRev.72.339/fulltext

Hans Bethe (1947) calculated the vertex correction part of the Lamb shift. He writes that the electron mass includes the "self-energy", which is formally infinite. A bound electron has a smaller self-energy, which means that its effective mass is smaller. This sounds very much like our own explanation on March 13, 2021: the far electric field does not have time to take part in the acceleration of the electron.

Let us investigate the logic used by Hans Bethe.

https://insa.nic.in/writereaddata/UpLoadedFiles/PINSA/Vol13_1947_4_Art09.pdf

R. C. Majumdar and R. C. Gupta (1947) write about the self-energy calculation. They say that the self-energy includes both the field of the electron and the "zero-point fluctuations" in the "radiation field". In our blog we do not believe that such fluctuations exist.

https://www.scribd.com/document/850438660/self-energy-of-the-electron

Jo Bovy (2006) outlines the history of self-energy calculations.

Hans Bethe writes that the self-energy of the electron in a quantum state m (2s state) depends on an integral which we take over various energies k of a hypothetical photon, up to the energy K = mₑ c², and the transition energies from the state 2s to any other state of the hydrogen atom.

The logic of Bethe sounds quite different from Feynman vertex integrals, and from our own classical vertex correction.

The self-energy of a free electron is:

The self-energy W for the bound electron in the hydrogen atom turns out to be higher than that of the free electron, W₀. The difference is easy to calculate by finding the common denominator k (En - Em + k) and subtracting the new numerators. The coefficient k cancels inside the integral:

Bethe first integrates on k:

Ry is the ionization energy of hydrogen, 13.6 eV:

https://en.wikipedia.org/wiki/Rydberg_constant

The method clearly is far from the Feynman integral method. What is common with standard QED is that the energy level shift is obtained by subtracting the self-energy of the free electron from the self-energy of the electron which is under an interaction from the proton.

Theodore Welton's calculation from "zero-point fluctuations" (1948) – fluctuations are actually bremsstrahlung sent by the electron itself?

https://en.wikipedia.org/wiki/Lamb_shift

https://journals.aps.org/pr/abstract/10.1103/PhysRev.74.1157

Wikipedia contains a heuristic calculation, due to Theodore A. Welton (1948). Welton assumes that zero-point fluctuations push the electron around, making its path "fuzzy", so that the average energy held by the electron is higher than for an electron without this disturbance. The logic of Welton is similar to the calculation of the Darwin term, but in the Darwin term the fuzziness comes from zitterbewegung.

In our blog we do not believe that zero-point fluctuations exist. Let us compare Welton's approach to the classical vertex correction.

Welton assumes a large box in which there is a standing wave of the minimum possible energy (1/2 h f) for each mode k such that the standing wave has its field zero at the borders of the box.

https://en.wikipedia.org/wiki/Zero-point_energy

Let us calculate the numerical value of

       ⟨(δr)²⟩vac  =  1/200 * 10⁻²⁵ * ln(50) m²

                        = 2 * 10⁻²⁷ m²,

       δr  ≈  5 * 10⁻¹⁴ m.

The value above is 700. The rise in the energy of the 2s orbital is 2.5 * 10⁻²⁵ J, while the correct value is 7 * 10⁻²⁵ J. The Welton model gives a qualitatively correct answer. Why?

Do the Bethe and Welton formulae determine the value of the Planck constant h? No. The power of h in the Bethe formula is -6, plus the the logarithm also depends on h. In Welton's formula, the power of h is only -5, plus the logarithm depends on h. If the formulae are supposed to calculate the same figure, we get a formula for h in terms of other constants of nature.

https://arxiv.org/abs/2306.01000

The paper by G. Jordan Maclay (2023) gives a different formula for the Bethe calculation. Both formulae depend on α⁵, and the exponent for h is -5. We probably misinterpreted the Bethe (1947) paper formula.

What is the dependence of the classic vertex correction on h? In our March 13, 2021 calculation, we used the probability density of the 2s orbital.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html#c3

The probability density for small r for 2s is

       ~  1 / a₀³  ~  1 / h⁶.

The volume of a sphere of size λₑ is

       ~  h³.

If we double h, then the potential in the March 13, 2021 calculation drops to a half:

       ~  1 / h.

The velocity of the electron drops by a factor 1/sqrt(2), and the mass reduction is then

       ~  1 / h¹⋅⁵.

The classical vertex correction depends on 

       -5.5th power of h.

The power is different from the Bethe/Welton calculation. We may have found a way to calculate the Planck constant from the mass and the charge of the electron?

Why would the Welton formula calculate anything like the classical vertex correction?

The Welton "zero-point fluctuation" is actually bremsstrahlung sent by the electron itself, which cannot escape? If the electron is on a stationary orbit, it cannot radiate in quantum mechanics. But if the electron still sends bremsstrahlung, then it has to absorb it at some point. Theodore Welton's random fluctuations could actually be the electron absorbing this radiation?

We can imagine that the sphere of the hydrogen atom, 1.1 * 10⁻¹⁰ m in diameter, is the "box" Welton is assuming to exists. Electromagnetic waves bounce around in that box, but cannot exit the atom, because the electron is on a stationary orbit and cannot lose energy. The electron moves in an accelerated fashion inside the box and excites some modes of electromagnetic waves, and absorbs some modes.

The largest photon the electron can send has the wavelength λₑ. This explains the ultraviolet cutoff used by Bethe and Welton. The longest wavelength that can fit in the box is 2 π a₀. This explains the infrared cutoff.

The Casimir effect is explained through a hypothesis that the vacuum between two metal planes can only accommodate "zero-point fluctuations" of certain wavelengths, and therefore contains less energy than free space. If we move the plates closer, then the energy drops even more. This creates a force pulling the plates together. An alternative explanation is that a fluctuation of electron density in one plate create an opposite density to the other plate. There is an attractive force between opposite density deviations.

Analysis of the probability density "smearing" caused by the Welton fluctuations. The Schrödinger probability density at distances r < 5 * 10⁻¹² m from the proton is almost a constant for 2s. The smearing does not change the probability density close to the proton at all.

Since ∇² 1 / r = 0 if r ≠ 0, the smearing of the probability at r > 5 * 10⁻¹² m does not change the average potential for that probability.

              probability            ------>  smearing

              density                 /

             ●===============|                  

             proton+                5 * 10⁻¹² m

              ----> r

We conclude that it is the smearing of the probability at the edge of the sphere r < 5 * 10⁻¹² m raises some of the edge probability to a higher potential V(r).

The electron does not react to "vacuum fluctuations", but sends the photons itself? This makes a lot of sense. In Welton's thinking, a vacuum fluctuation accelerates the electron. But we can as well imagine the process time-reversed: the electron itself sends a photon and reacts to the sending process by moving. This sounds like the sharp hammer model, in which the electron hits the electromagnetic field. We do not need to assume "vacuum fluctuations". A "vacuum fluctuation" actually is the photon sent by the electron, looked at in a time-reversed manner.

In the QED vacuum correction, the electron sends a photon with an arbitrary 4-momentum k.

This may explain why Welton's approach produces the same result as Bethe's and also the same result as QED calculations.

This still does not explain why Welton's very simple "smearing" approach works. It is not at all self-evident that the position of the electron is randomized in the fashion Welton assumes.

Another coincidence: orbital 1s and 2s probability density at the distance λₑ from the proton is roughly the same as for a classical elliptical orbit

https://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html

The probability density is obtained by taking the square of the radial wave function.

If one calculates the probability of the electron being within λₑ from the proton, the numerical value is roughly the same for a classical orbit with an angular momentum very close to zero, and for Schrödinger orbitals 1s, 2s, 3s.

The electron in scattering experiments behaves much like a classical charged particle. The dive of the electron toward the proton is like a scattering experiment.

The probability densities above are

       ~  1 / a₀³  ~  1 / h⁶.

The volume of the sphere of a radius λₑ is

       ~  h³.

The probability of being in that sphere for the Schrödinger solutions thus is

       ~  1 / h³.

Let us look at the classical orbit for 1s. It is a very narrow ellipse which extends to

       2 a₀

The electron moves at λₑ = 0.045 a₀ roughly

       sqrt(44) ≈ 7

times faster than at a₀ because its kinetic energy is 44-fold.

Thus, the electron spends within a₀ roughly

       1 / 0.045 * 7  =  154

times more time than within λₑ.

The electron "falls" from the distance 2 a₀. It may spend between 2 a₀ and a₀ some 3 times more time than below a₀. We get an extremely crude estimate. The electron spends

       1/600

of the time below λₑ. This value is 1 / h¹⋅⁵.

The Schrödinger 1s (R₁₀) gives 3.6 as the probability density at r = λₑ, if we set a₀ = 1. The probability of being within λₑ is roughly

       3.6  *  4/3 π * 0.045³

       ≈  1/700.

Is it a coincidence that the orbital 1s so closely imitates the classical elliptical orbit down to the Compton wavelength of the electron?

If it is not a coincidence, then this determines what the value of h must be.

If we double the radius, to 2 λₑ, then the classical probability grows 2.8X, but the Schrödinger probability by 7X. Could this be just a coincidence that the probability for the electron being in the Lamb shift zone is the same classically and for Schrödinger?

The orbit for 2p: is it close to the classical? Yes     

The energy of the 2p orbital is the same as on the second lowest Bohr orbit, whose radius is 4 a₀. The total angular momentum is sqrt(2) ħ.

In the Bohr model, the lowest orbit has a total angular momentum of ħ. If we make the velocity there sqrt(2)-fold, then the electron escapes. The lowest point of the classical elliptical orbit of 2p must be slightly higher than a₀. In the diagram above, we the probability of 2p to be closer than a₀ is very small. The average distance is ≈ 5 a₀, which is roughly the same as for a classical elliptical orbit.

We conclude that the 2p Schrödinger orbital is quite close to the corresponding classical elliptical orbit. This is expected because the de Broglie wavelength of the electron on tge corresponding Bohr orbit is only 1/2 of the orbit length. The electron will behave somewhat like a classical particle. Changing the value of h does not change anything in the above analysis. Thus, this analysis does not restrict h in any way.

Arnold Sommerfeld model (1919)

Arnold Sommerfeld developed a model in which the electron in the hydrogen atom tracks an elliptical classical orbit, quantized by the angular momentum. Did he prove that the probability density is close to the Schrödinger equation solution?

https://en.wikipedia.org/wiki/Bohr%E2%80%93Sommerfeld_model

Wikipedia states that the Sommerfeld relativistic model fails to predict the Lamb shift. In our blog we have suggested that the Lamb shift is caused by the classical vertex correction. Then the Sommerfeld model might predict the Lamb shift, too.

Does the Sommerfeld model imply that the value of h is determined by the mass and the charge of the electron? Probably not. Someone would have noticed.

Reanalysis of the far field reabsorption in the QED vertex correction

On August 21, 2026 we outlined why the QED vertex correction might calculate the same thing as the classical vertex correction, in a special setting. But we had the wrong impression that the vertex correction increases the scattering probability, while it decreases that. Let us write a new analysis. We study the hydrogen 2s orbital mentioned above.

            e-

              •  p                                     ^  p + q

                \                                      /

                  \                k                /

                    \~~~~~~~~~~~ / virtual photon

                      \                          /

                        \          ●          / proton+

                          \        | q     /

                            --------------

The classical vertex correction suggests that k represents the mass of the far field of the electron. The far field does not have time to take part in the "reflection", or the bounce-back of the electron from the proton.

In the classical case, some of the kinetic energy of the far field would escape as radiation. But we assume that energy does not escape.

Semiclassical scattering assumption. If a fraction f of the mass of the electron does not take part in the scattering, then q is reduced by the same fraction. That is, the scattering probability (cross section) is less for each value of q. It is as if the charge of the proton would have been reduced by the same fraction f.

Let us compare this to the idea we used on March 13, 2021 to calculate the Lamb shift. There, we noted that if the kinetic energy of the electron is approximately

       1/2 mₑ v²  ≈  -V,

where V is the electric potential, and mₑ is reduced by a fraction f, then the momentum of the electron is reduced by f / 2. It is as if the potential would have been reduced by the fraction f, or that the charge of the proton would have been reduced by the fraction f.

We conclude that the semiclassical scattering assumption is equivalent to our Lamb shift idea from 2021.

https://en.wikipedia.org/wiki/Quantum_electrodynamics

The Feynman rule for a vertex is:

The factor -i suggests that adding the k photon line with two vertices flips the phase of the outgoing electron by 180 degrees. The Feynman diagram above probably will reduce the scattering amplitude: the correct behavior.

***  WORK IN PROGRESS  ***

QED vacuum polarization – a new analysis

On October 22, 2025 we left the analysis of the QED vertex correction incomplete. In our previous blog post we (maybe) were able to complete the analysis.

Let us next attack QED vacuum polarization, based on the insights we got from the vertex correction. On November 4, 2025 we wrote a previous detailed analysis of vacuum polarization.

Is the Feynman rule, that a closed fermion loop adds -1, correct?

In this blog we have criticized that rule. Let us have the following Feynman diagram:

 

                                         ___

                               e-   /         \

   photon      ~~~~~                 ~~~~~

                               e+  \_____/

The diagram, with the -1 rule, claims that the phase of the photon changes 180 degrees, even though it does not interact with any outside object. That cannot happen with classical waves. We do not believe it is possible with quantum waves, either.

https://www.bohrium.com/en/sciencepedia/feynman/keyword/fermion_loops

The text in the link says that a fermion loop involves an odd number of fermion field swaps, and that the factor -1 comes from the fact that fermion fields anticommute.

https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem

https://physics.stackexchange.com/questions/18342/why-is-there-extra-minus-sign-in-feynmans-rules-for-every-closed-fermionic-loop

In the link, Lubos Motl explains that an operator ψ₁↑ has to be transported over 3 other operators, anticommuting, and yielding -1³.

The subscripts 1 and 2 denote the two vertices of the vacuum polarization diagram. A propagator is the line between two vertices. But why should we transport ψ₁↑ to the end? This might suffice:

       ψ₁ ψ₂↑  ψ₁↑ ψ₂.

The dagger version of the operator ψ can be interpreted either to annihilate an electron or to create a positron? Then the first two operators above create and annihilate an electron (= first propagator), and the two last create and annihilate a positron (= second propagator). The operator ψ₁↑ was only transported over two other operators. There is no sign change.

Classical processes suggest that there cannot be a factor -1 in a fermion loop 

Classical radio antenna. Let us have a radio wave which meets a radio antenna.

         

                      reflected waves, 180º phase shift

                                /                                \

                                        /                \

    radio wave   ~~~~~~~   

                                                #

                                                #

                                                #

                                          antenna

Radio waves are first absorbed by the antenna, and then emitted again (= reflected). The emitted wave has destructive interference with the wave which would continue directly through the antenna. The missing energy goes to the wave emitted to other directions. The emitted wave has a 180 degree phase shift relative to the original wave.

The antenna keeps some of the momentum of the incoming wave. Therefore, the antenna is able to re-emit the wave with a 180 degree phase shift. In vacuum polarization, the lonely electron-positron loop cannot absorb momentum. It is not plausible that vacuum polarization could do the same trick.

Classical metal body between the electron and the proton. 

                   e- • --->

                      ++   metal body is polarized and

                       --    increases the attraction

                               

                       ● proton+

A metal body acts as a radio antenna. What if we put that body between the electron abd the proton? The body is polarized and increases the attraction.

If we interpret the attraction as a virtual photon carrying only spatial momentum, the antenna does not cause any "destructive interference" to the virtual photon. Instead, it makes the virtual photon stronger. There is no phase shift.

Classical polarization:

1.   Classical polarization makes a real photon to be reflected with a 180 degree phase shift. The polarized body must keep some spatial momentum.

2.   Classical polarization increases the attraction between opposite charges. Here we assume that the charges are passing by each other, and that polarization increases in the strong electric field between them.

Classical analogy between a real photon and a photon only carrying spatial momentum. In the radio antenna example, the antenna absorbed some of the energy of the incoming radio wave. The antenna could not keep that energy. It had to send the energy to other directions. Let us investigate an equivalent process for the momentum exchange between the electron and the proton.

                              Fd 

                           ^

                           |        • e+  

                       e- •          dipole

                           |        • e-

                           |       |

                           v       v F' weak

                             F

                          ● proton+

The proton pulls on the electron with a force F.  We try to reduce the pull by putting a dipole near the electron. The dipole exerts an upward force Fd on the electron, and reduces the pull of the proton. The electron pushes the dipole down with the force -Fd.

We were able to divert some momentum transfer to the dipole, so that the electron does not receive it. The missing momentum transfer was absorbed by the dipole. Bu we now face a dilemma: to which body can the dipole "emit" that downward momentum it stole from the electron? The dipole should pull something else downward. It cannot pull the proton downward.

The only way for the dipole to get rid of its momentum is to collide with the proton. The dipole is a transient process created by the passing electron. It is highly inlikely that it could collide with the proton.

We found an analogy between a real photon and a momentum-only photon versus an "antenna". The difference is that the "antenna" cannot work in the momentum transfer case.

A real photon temporarily absorbed by an electron does change its phase by 180 degrees – what about the electron?

  photon   ~~~~~                 ~~~~~~

                               \         /

             e-  ----------------------------------

This is just like the classical radio wave plus an antenna case. The electron is the antenna. The photon undergoes a 180 degree phase shift.

Does the electron undergo a 180 degree phase shift? It is, in a sense, reflected from the photon.

In the wave model, a photon can be understood as a "grid" which systematically disturbs the Dirac wave representing the electron. An electron wave absorbing the photon is scattered from the grid. 

Under the Schrödinger equation, an electron scattered from the electric field of a proton does not undergo a phase shift. Light scattered from a grid is not phase-shifted.

Hypothesis: no phase shift unless a real particle is "absorbed". A phase shift is a major thing in the life of a real particle. If the real particle is, in some sense, absorbed, then there can be a conceptual "delay" in its re-emission, and a phase shift of 180 degrees is possible. The emitted particle is "created again" from scratch. A phase shift is not possible if a particle is not absorbed.

A real particle travels in time. A static observer sees the wave undulating up and down.

The hypothesis leaves open what happens with a virtual particle. Our reasoning in the previous section suggests that no phase shift can happen with virtual particles which only transport spatial momentum.

In the vacuum polarization diagram, we have a virtual photon which creates the pair into which it is "absorbed". If it were a real photon, it would be a miracle if this Baron Munchausen trick could change its phase. Also, the photon mostly transfers spatial momentum, which suggests that its phase cannot change.

Furry's theorem is almost correct

https://en.wikipedia.org/wiki/Furry%27s_theorem

The classical limit of this theorem is very suspicious. It claims that if an electron passes a proton, and vacuum polarization occurs, then that vacuum polarization cannot interact with a second electron.

Why should that be? Vacuum polarization means that there are charges present in previously empty space. Those charges can be an "almost" on-shell electron-positron pair. That pair can, of course, interact with a foreign electron.

The pair is very short-lived. Therefore the probability of an interaction with a foreign electron is extremely low. Furry's theorem is almost correct.

Does vacuum polarization classically affect a foreign electron coming around?

                          • e-            

                                                              • e- foreign

                        + +   polarization

                         -  -

                          ● proton+

The dipole field of the electron and the proton is reduced because of polarization. Polarization "conducts" field lines. Thus, vacuum polarization reduces the interaction of a foreign electron with the dipole.

The proof of Furry's theorem in Wikipedia goes like this: let us do a charge conjugation to the diagram, i.e., switch the signs of charges. Each of the three photon fields Aμ (lines) changes its sign. The method to calculate the diagram probability amplitude multiplies those photon fields. The calculation yields a factor (-1)³ = -1.

Most calculations in QED are unchanged under charge conjugation. If the Furry diagram changes its sign, and its value is nonzero, then the sum of the diagrams (= scattering amplitude) would change under charge conjugation. But charge conjugation must preserve the physics unchanged.

The conclusion in the proof is that the Furry diagram must have a zero probability amplitude.

Why does the calculation method cause a sign change in charge conjugation? A sign change means a 180 degree phase shift.

Does the sign change show that the calculation method gets the phase wrong? the previous sections of this blog post we showed that determining the phase is not trivial. QED has suspicious rules about the sign of the probability amplitude.

In the case of the Furry diagram, if the photons are pure momentum exchange, there should be no phase change, ever. Not after a charge conjugation, either.

We have to check the 1937 paper by W. H. Furry and determine what is the error there. Furry seems to have shown that the two diagrams in which the fermion loop flows to opposite directions, cancel each other. There is some sense in this, also classically. Since vacuum polarization is a dipole field, its opposite charges almost cancel any interaction with a foreign electron which is some distance away.

https://rohankulkarni.me/posts/notes/furrys-theorem-part-1/

At the link we have two proofs of Furry's theorem. The link says that if we reverse the direction of the electron arrows in the Furry diagram, then the probability amplitude flips its sign. For a real physical process, we always have to sum both diagrams because we have no way of knowing which one happened. This sounds sensible. But why should the amplitude flip its sign if we change the direction of the electron arrow? Is it because in that case we interpret that the particle interacting with the foreign electron flips from an electron to a positron or the other way around?

Vacuum polarization should be symmetric between the electron and the proton. How we formally draw the diagram should not matter.

Furry's theorem is true for real photons. A simple energy-momentum conservation argument proves that. Assume that two photons are not traveling to the same direction.

                     \

                       \

                        O ~~~~

                       /

                     /

Let the photons collide. A third photon cannot take away the entire energy of the two photons because it would have too little momentum compared to its energy.

In the next section we argue that energy-momentum conservation also bans a diagram in which the two upper lines are a real photon and the down line is pure momentum exchange (Delbrück scattering).

If the photons represent momentum exchanges, then a single photon can, of course, take away the contribution of the two other. Also, if the photons are virtual, they are allowed to break the energy-momentum relation.

Classically, the interaction with a foreign electron is very weak for the following reasons:

1.   The transient vacuum polarization field between an electron and a proton exists for a very short time. The foreign electron should be close to the proton at that moment. The probability of that is much smaller than, e.g., the probability of the foreign electron meeting the electron and the proton on separate occasions. The interaction between vacuum polarization and the foreign electron is a three particle interaction. 

2.   Vacuum polarization creates a dipole field which is very weak far away.

We conclude that Furry's theorem is true for practical purposes. However, its proof suggests that the sign rules in QED are incorrect. Charge conjugation should not flip the sign of a probability amplitude.

Delbrück scattering has a classical limit?

      E

       ^       __            __

       |     /      \___/      \___/      laser beam

                          ^ E'

                          |

                         ● 

                          nucleus+

Let us have a nucleus. The electric field close to the nucleus is strong. Let us have a laser beam in the plane of the nucleus, such that the electric field E of the laser beam is in the same plane. Let E' be the electric field of the nucleus.

When the electric fields E and E' point to the same direction, a virtual electron-positron pair may appear, and reduce the energy of the electric field. The medium carrying the electromagnetic wave becomes "looser", and the electromagnetic wave slows down. We expect to see the wave to turn toward the nucleus.

An analogous process happens when light propagates in a medium, for example, glass. If there is a lot of polarization of electrons in the medium, then light travels slower – the refractive index of the medium is larger.

Light travels slower if it interacts with the medium. In the case of Delbrück scattering, the interaction is with the strong electric field of the nucleus.

      real

      photon   ~~~~  --------   ~~~~~~

                                 |          | virtual e- e+

                                 |          | pair

                                   --------

                               /          /  electric field

                             /          /    of nucleus

                                   ●

                            nucleus+

 

Note that Furry's theorem bans the diagram in which there is only one leg pointing toward the nucleus. We must have two legs. What is the classical reason for this?

Let us imagine that the nucleus is holding a block of glass which makes the photon to move slower. Since the photon moves slower, it must pass some of its momentum to the nucleus as it enters the glass. When the photon exists the glass, it must get the momentum back. There must be two momentum exchanges between the photon and the nucleus. This may explain why the diagram must have two legs pointing to the nucleus.

  real photon ~~~  --------   ~~~~~~

                                 |          | virtual e- e+

                                 |          | pair

                                   --------

                               /  electric field

                             /    of nucleus

                           ●

                    nucleus+

The one-legged diagram above cannot happen because the virtual pair is moving at a speed less than light. It cannot be converted to a real photon at the end of the diagram.

Is there a classical limit to this? If one real photon (gamma radiation) passes the nucleus, it is a quantum process. But if there is a laser beam, do we arrive at a classical process?

Magnetars may have a magnetic field which exceeds the Schwinger limit 4.4 * 10⁹ tesla. Pair production there may be a "classical", macroscopic process. Vacuum polarization might be classical, too.

Why is it important if vacuum polarization has a classical limit?

The vertex correction has an obvious classical limit, as we showed in our previous blog post. That allows us to argue that, at least with large charges, high |k| are wiped out by destructive interference. We do not need regularization or renormalization.

Our goal is to get rid of regularization and renormalization in vacuum polarization, too.

https://en.wikipedia.org/wiki/Born%E2%80%93Infeld_model

Max Born and Leopold Infeld in 1934 introduced a model in which the electric field has a maximum value. They attempted to remove the infinite energy in the field of a point charge. Their model is an example of nonlinear electrodynamics. Nonlinear electrodynamics models might be able to act as semiclassical models of vacuum polarization.

The Uehling potential

https://en.wikipedia.org/wiki/Uehling_potential

https://www.scribd.com/document/204531248/An-Introduction-to-Quantum-Field-Theory-Peskin

Peskin and Schroeder (1995) give these formulae:

 

The above formulae are in "atomic units". In them:

       e² / (4 π)  =  α  ≈  1/137.

The constant A = exp(5/3).

Why does the Uehling potential correction drop off exponentially when r is larger than the electron reduced Compton wavelength

       λₑ / (2 π)  =  ħ / (mₑ c) ?

Is the potential dependent on the speed of the electron? Then it would be wrong to call it a potential.

Peskin and Schroeder (1995) take the Fourier decomposition of the Coulomb potential, and modify it with the vacuum polarization correction Π₂(-q²). This sounds very logical. But how does vacuum polarization behave for a very slow electron? Let us have an electron extremely far from the proton, moving very slowly. Then the momentum exchange |q| can be quite large. But it is unlikely that vacuum polarization affects the Coulomb force much. Peskin and Schroeder calculate the integral of various q to determine the formula of the potential V(r). The integral drops off exponentially when r > the reduced Compton wavelength of the electron.

Normally, in Feynman diagrams, we assume that the electron moves at almost the speed of light relative to the proton. The Peskin and Schroeder method of correcting the Coulomb potential aims to determine the potential which would cause the same scattering behavior as the one calculated with the vacuum polarization correction.

Fourier decomposition of a potential generally does NOT replicate the scattering behavior. Would the Peskin and Schroeder potential replicate the scattering behavior also if the path of the electron is calculated nonperturbatively from the potential?

Probably not. For the Coulomb potential, its Fourier decomposition produces the same scattering behavior as the potential itself. But the same is not true for an arbitrary potential. 

         ---------    --------   V = 0

                    ||                 

          potential well

If we have a very narrow potential well in a plane of a zero potential, the Fourier decomposition of the potential contains many long wavelengths, meaning a lot of "mild" scattering – but the potential itself may have infinitesimal "mild" scattering.

Question: what does the Uehling potential actually mean?

The velocity of the electron affects q, and the vacuum polarization correction in Feynman diagrams. The momentum exchange q depends on the velocity of the electron. If we calculate the vacuum polarization correction based solely on q, then a slow electron far away would experience the same vacuum polarization as a fast electron close to the proton. This sounds wrong. Feynman digrams probably miscalculate the correction for electrons whose velocity is << c.

Electron velocity in a hydrogen atom. The average velocity is only 0.008 c. At the distance of the reduced Compton wavelength 4 * 10⁻¹³ m, the velocity is 0.13 c. At the classical electron radius 2.8 * 10⁻¹⁵ m, the velocity is close to c. Thus, the electron is "mildly relativistic" in the zone where vacuum polarization is assumed to be relevant.

The electron in the hydrogen atom can be conceived to make "dives" toward the proton. Thus, the electron performs "scattering experiments" in the hydrogen atom. The potential calculated by Peskin and Schroeder (1995) may approximate these scattering processes relatively well, if not exactly.

The electron moves significantly slower in the hydrogen atom than in a typical collision experiment. The fact that the Uehling potential calculates the contribution to the Lamb shift correctly, suggests that vacuum polarization depends on the effect that it has on a relativistic electron, and that slower electrons obey the potential calculated for relativistic electrons. That is, vacuum polarization is not "dynamic" in the sense that it would depend a lot on the velocity of the electron.

However, the Feynman diagram for vacuum polarization very much describes a dynamic process: a scattering experiment. This suggests that, after all, vacuum polarization is a dynamic phenomenon.

Electron-proton scattering vacuum polarization is also Delbrück scattering? No

When an electron passes a proton at a large speed, then the electric field of the electron looks somewhat like a photon electric field from the viewpoint of the proton. Delbrück scattering slows down the photon close to the proton. The electron turns more toward the proton.

Could it be that the familiar electron-proton vacuum polarization diagram calculates this Delbrück effect? Probably not.

https://arxiv.org/html/2307.08380v1

According to Sommerfeldt et al. (2023),  Delbrück scattering for a 300 keV photon has a cross section ~ 0.1 microbarn/steradian from a tin nucleus. The scattering of an electron to a large angle from such a nucleus is ~ 1 barn/steradian. The Delbrück effect for a photon is much smaller than the vacuum polarization correction for an electron.

The electric field of the electron and the proton is highly concentrated into a small volume between them. It is quite different from the electric field of a 300 keV photon, which can be produced by an oscillating dipole which is some 10⁻¹¹ m long.

Real photons as vacuum polarization?

We have tossed this idea before in our blog: could it be that real photons are vacuum polarization propagating at the speed of light? The propagator of an electron is, in a sense, a "square root" of the photon propagator. Maybe the propagator of a virtual electron-positron pair is equivalent to a photon propagator?

Since electric field lines are source-free in empty space, this hypothesis does not work, or does it?

Let us have a light wave packet propagating inside a glass block. It polarizes the electron distribution inside glass. If the electric field lines are not broken anywhere, there cannot be an extra charge density anywhere? In that case, it would be the movement of electrons which affects the propagation of the wave packet.

Let us forget about the wave packet for a while, and concentrate on the movement of electrons. That produces a magnetic field. A changing magnetic field produces an electric field. Polarization is then a magnetic phenomenon.

According to this hypothesis, vacuum polarization between an electron and a proton might be viewed as a transient "photon".

Quantitative analysis of vacuum polarization

Peskin and Schroeder (1995) give the following formula for the vacuum polarization correction:

If |q²| << |m²|, then

       ln(m²  /  ( m²  -  x (1 - x) q²) )

       ≈  x (1 - x) q² / m².

We see that the correction is ~ q² / m².

 

                       e-  • --> v ≈ c

                             r

                             ● proton+

Let a mildly relativistic electron pass a proton at a distance r. Then

       q  ~  1 / r.

The vacuum polarization correction (where 1 corresponds to the simple elastic scattering diagram) is

       ~  1 / r².

That is, the correction depends linearly on the strength of the electric field between the electron and the proton. Why?

Wave description as a magnified picture of the classical particle process?

            p     --> v ≈ c 

            e- --------------------------------------

                                     |          rₑ

                                     | 

                                     | q,  |q| ~ |p / 2|

                                  /   \

                               /         \ k + q virtual electron-

                          -k  \         /  positron pair e- e+

                                  \   /

                                     |  

                                     | q virtual photon

                                     |

proton+ ---------------------------------------

Let us have a mildly relativistic electron which passes close to a proton, and is scattered a lot, say, 90 degrees. Classically, the electron passes the proton extremely close, at a distance

        ~  rₑ  =  2.8 * 10⁻¹⁵ m.

But the wave description of the process uses an electron wave whose wavelength is much longer:

       ~  λₑ  =  2.4 * 10⁻¹² m.

The wave description "magnifies" the classical process by a factor 861X.

This suggests that the wave description of vacuum polarization is a magnified picture of a classical process which happens in that very small space ~ rₑ.

The creation of the virtual pair can be understood as a scattering process in which a positron traveling backward in time scatters from q forward in time as an electron. It is natural to assume that, classically, this process happens at a distance < rₑ from the proton.

Semiclassically, the creation of the pair might be a process in which q "excites" an electron, so that a "hole" is left. The hole is the positron. The strongest electric field is between the electron and the proton. It is natural to assume that the electron gets excited there.

If we have a polarizable material, like glass, then an electron can be kicked up from its valence band and freed at the location where the electric field is the strongest.

What is the role of the arbitrary loop 4-momentum k in this? 

Let us assume that a positron with a huge 4-momentum k moves backward in time. Why would it scatter from the proton into an electron with a huge 4-momentum -k of an opposite sign? This cannot happen to an electron or a positron moving forward in time. The proton is unable to change the 4-momentum of a colliding electron by a huge amount.

Hypothesis: scattering from a particle cannot change the 4-momentum of another particle by an arbitrarily large amount. It is not possible even in the case in which a positron traveling backward in time scatters from a photon into an electron traveling forward in time.

The "scale" of a possible 4-momentum exchange in our example is q. The hypothesis above would give us a natural cutoff for |k|. That would free us from the ultraviolet divergence in vacuum polarization.

Is our analysis of the vertex correction somehow a "dual" of this hypothesis? In the vertex correction, the electric field of the electron receives a "hit" whose scale is q. In the vertex correction, we appealed to the smoothness of the electron orbit around the proton. Destructive interference cancels large |k|.

In the vertex correction, the smooth path of the electron prevents it from creating high |k| in its electric field. In vacuum polarization, the electric field creating polarization is smooth. Does this imply that large |k| are canceled by destructive interference?

4-momenta q and k in the description of (virtual) electron paths. The momentum exchange q is a summary description of the path of the electron close to the proton, in a very small space ~ rₑ. Maybe the 4-momentum k describes the path of the virtual electron in the same very small space? Then it makes sense that |q| ~ |k|.

In the vertex correction, the classical process is present in all fly-bys of the electron, but the Feynman integral formula assumes that it only happens in a small fraction of the fly-bys, at a large magnitude |k|.

What about vacuum polarization? If it is a classical process, it must happen during every fly-by. There are no random processes in a classical world. Could it be that the Feynman integral formula "concentrates" a small classical effect into a prominent effect in a small fraction of fly-bys?

Suppose that a e- e+ pair appears between the electron and the proton. If the distance of the particles is ~ rₑ, they will affect the orbit of the electron in a major way. Since vacuum polarization is a minor correction, this major disturbance can only happen in a small fraction of cases.

What is a classical "minor" vacuum polarization? Classically, the vacuum polarization correction must happen during every fly-by. Is classical polarization such that, say, there is 1/1000 of a pair disturbing the fly-by?

The Dirac wave may have any amplitude. If we consider it a classical wave, then there must exist any fraction of an electron. Similarly, for a weak classical electromagnetic wave, there must exist any fraction of the photon. In quantum mechanics this is avoided by assuming that the wave is a probability amplitude wave.

Can a classical description explain the birth of an electron and a positron as particles?

Probably not, because a classical description cannot explain the existence of the quantum of the electromagnetic field, the photon, either.

The electron and the positron are quanta of the Dirac field.

Once an electron exists, then a classical description of it as a "point charge" is relatively good in many setups.

The Dirac field can be interpreted as a classical field, though. Vacuum polarization should be some kind of a wave in that classical field.

A further complication is the fact we mentioned in the previous section: the Dirac field presents a "magnified" version of the events in the particle realm.

Could it be that high |k| do not represent separate processes, but they are overlapping probabilities?

The resolution to the infrared divergence in the vertex correction is to realize that a single fly-by of the electron by the proton involves an infinite number of low energy photons. The probabilities overlap. The infrared divergence is an illusion.

In a similar fashion, many large |k| in the vacuum polarization loop could happen during every fly-by. They would have overlapping probabilities. The ultraviolet divergence would be an illusion.

The infrared problem is due to the fact that we have to process an electric field which extends to infinity.

Let us imagine that we have a classical model which puts a localized Dirac field wave between the electron and the proton during the fly-by. Since the wave is localized in a very small space, in our example 3 * 10⁻¹⁵ m, we must have very large |k| to build the wave from Fourier components.

In this classical model, the probabilities for various k overlap. We could claim that the Feynman integral divergence is just an illusion caused by the assumption that the probabilities for various k do not overlap.

Let us recapitulate:

1.  A Dirac delta impulse (a hit by a sharp hammer) creates the full spectrum of 4-momenta k.

2.   Classically, the probabilities of various k greatly overlap. Assuming that they do not overlap leads to the wrong conclusion that there is an infrared or ultraviolet divergence.

3.   In bremsstrahlung, and in the vertex correction, an infinite number of infrared photons are always created when the electron passes by the proton. This is because the electric field extends to infinity. This is true both classically, and in quantum field theory. High |k| are eliminated by destructive interference in the classical model.

4.   If we model vacuum polarization classically, we must create a sharply localized Dirac wave between the proton and the electron passing it. The Fourier decomposition of the wave contains very high |k|. There probably is a cutoff for extremely large |k|, since the wave probably is smooth. Large |k| have overlapping probabilities. An ultraviolet divergence is an illusion.

Why do |k| ~ |q| determine the probability of vacuum polarization? Probabilities overlap for various k. Why is it ok to cut off |k| which are much smaller than |q|, and also to cut off |k| which are much larger than |q|? Why do these cutoffs produce the right result for the probability amplitude, and remove overlapping probabilities?

Standard QED calculates the complement of the Feynman integral (which is an incorrect way, as we have explained). Since q affects the Feynman integral value mostly for |k| ~ |q|, standard QED considers those values of k the most important.

For the vertex correction, we showed that the Feynman integral "concentrates" the small effect in every fly-by to a very large effect (~ 1,000X) in a small fraction of fly-bys. In that small fraction, |k| ~ |q| are the most important, which numerically yields the classical vertex correction.

                     ---> v ≈ c

             e-  • -------------

                                      \

                               rₑ      \

                         ●               |     q momentum

           proton+         b     v        exchange

The fly-by in our mildly relativistic example can be defined to take the distance 2 rₑ. The momentum exchange

          |q|  ~  mₑ,

where we have set c = 1.

The wavelength associated with q is the electron Compton wavelength λₑ ≈ 861 rₑ. The wavelength is much too long to describe a process which happens in a space rₑ.

Suppose that we want to build a photon wave packet which fits in rₑ. The Fourier decomposition of a "bump" of such a size such contain roughly "an equal amount" of each wavelength down to the size of the bump. Does the Feynman integral for vacuum polarization have an ultraviolet divergence which fits this description? Peskin and Schroeder (1995) and other sources say that, after applying the Ward identity, the divergence of the vacuum polarization integral is logarithmic. That sounds reasonable.

For |q| << |m|, the "missing part" of the integral,

       |Π₂(0)  -  Π₂(q²)|  ~  q².

That is, the integral Π₂(0) for small |k| probably grows quadratically. For very large |k| > |m|, the integral grows only logarithmically.

The overlapping hypothesis does not work in the ultraviolet divergence. Consider an infinitesimally small |q|. The electron passes the proton at a distance of, say, 1 kilometer. Let the semiclassical vacuum polarization be a transient Dirac field wave. Since the wave spans a large spatial distance over a long time, there should be no Fourier components in the wave with |k| ~ |m|. This implies that high |k| must "not exist". High |k| cannot be overlapping probabilities (which do exist).

The magnification in the vacuum polarization integral?

Classically, vacuum polarization in our mildly relativistic example should happen in a space whose size is rₑ, but in the Feynman integral, the interesting values are |k| ~ |q|, that is, a scale of 861 rₑ.

The simplest elastic scattering Feynman diagram has this same 861X magnification property.

Could it be that the vacuum polarization integral magnifies the virtual particle process by 861X? The virtual pair consists of particles, and they can be treated like elastic scattering? Since the virtual pair happens at the midway between the electron and the proton, the "scattering" or creation of the pair, happens at the same scale ~ |q|?

Let us make h much smaller. Then the wavelength of q can easily fit into rₑ. We can then build wave packets whose spatial dimension is small enough. What happens? Then the value of the fine structure constant α grows as it is inversely proportional to h. Pair production becomes easier, and the vacuum polarization correction grows. As if the fact that rₑ << λₑ would act as a "pinhole" which prevents the pair from forming in most cases.

https://en.wikipedia.org/wiki/Bremsstrahlung

The Bethe-Heitler formula says that the bremsstrahlung differential cross section is proportional to 1 / h.

Bremsstrahlung, pair production, and vacuum polarization depend on 1 / h. They are high |k| phenomena.

Making h smaller means that less energy is required to create a quantum. It is logical that then the cross section becomes larger for phenomena which need to create new quanta. The simplest elastic scattering involves a momentum exchange q which does not contain any energy at all. That may explain why its cross section is not affected by h.

Since vacuum polarization depends on 1 / h, we conclude that it must temporarily create quanta which contain energy. Peskin and Schroeder (1995) formula 7.91 says that the relative vacuum polarization correction to the simplest elastic scattering is

       ~  1 / h  *  q² / mₑ².

If we would shrink h by a factor 1/861, then the electron Compton wavelength λₑ would fit in rₑ, and the vacuum polarization correction would be very large, on the order of 100% for a mildly relativistic electron passing a proton at a distance rₑ. In that case, we can "build" a Dirac wave packet which fits between the electron and the proton. Intuitively, that should help in creating a lot of vacuum polarization.

The classical vertex correction does not depend on h, but it agrees (?) with the QED vertex correction which does depend on h: did we find a way to derive the value of h? No

Our analysis of the vertex correction on October 22, 2025 and our August 2026 blog posts showed that the classical vertex correction very roughly agrees with the QED one in the case where the electron speed is the same as it is on the lowest orbit in the hydrogen atom. It does not agree at other speeds.

But is that really possible? The classical correction comes from the reduced mass of the electron, which depends on the speed of light, not h. The QED vertex correction depends on the fine structure constant

       α  =  1 / (4 π ε₀)  *  e² / (ħ c)

            ≈ 1/137.

Did we find a way to derive the value of the Planck constant h?

https://www.southampton.ac.uk/~doug/qft/aqft3.pdf

The value of F₁(q²) in the link is ambiguous, though. It depends on the "photon mass" λ used as an infrared cutoff.

https://en.wikipedia.org/wiki/Natural_units

In particle physics, natural units have

       c  =  ħ  =  ε₀  =  1.

In the formula for F₁(q²), q = m v, where v is the change in the velocity vector of the electron. Thus,

       q² / m²  =  v².

The value of v tells the scattering angle. It cannot depend on h. The value of α does depend on h.

Assuming that the QED vertex correction calculates the same thing as the classical vertex correction, we found a dependecy between h and the mass and the charge of the electron?

The answer to the mystery probably is this: the speed of the electron in hydrogen depends on h. The classical vertex correction at that speed depends on h. Thus, the classical calculation is not independent of h.

The QED vertex correction formula may be wrong, since it only depends on q. Let us fix some value of q. We can obtain the momentum exchange q by letting the electron move very far from the proton at a very slow speed. The electron is under a very small acceleration. It is strange if the vertex correction under such peaceful conditions would stay the same, as it does with QED, since the formula only depends on q. The classical vertex correction would be much smaller, which intuitively is a better hypothesis.

Does a vertex correction which only depends on q break conservation of energy?

Let us multiply the charge of the electron by some large number N, and the mass of the electron by N². The electron becomes a classical charged body. The vertex correction numerical value stays the same, according to the formula above. The correction can be substantial, say, 0.1% for q. If we let the "big electron" move very slowly past another classical charge at a large distance, so that the momentum exchange is q, then the formula claims that the Coulomb force is altered by 0.1%. If we double the speed, then the formula claims that the Coulomb force is only altered by 0.025%. Classical charges do not behave like that. We conclude that the QED vertex correction breaks the classical limit.

How can the vertex correction and vacuum polarization depend on 1 / h?

In quantum mechanics, letting h to approach zero should make the system classical. But in classical physics there are no infinite corrections. Are the QED formulae correct?

Conclusions

We cannot draw any conclusions yet. The analysis was still left incomplete.

In the next blog post we will again analyze the vertex correction. We realized that the classical vertex correction does not agree with the QED vertex correction if the speed of the electron is not the one (0.008 c) on the lowest orbit of hydrogen. What does that imply?