Vacuum polarization in QED

In our post on August 27, 2021 we claimed that the ultraviolet divergence of the Feynman vacuum polarization integral is canceled out by destructive interference.

Furthermore, we claimed that in the traditional analysis of vacuum polarization, in the integral there are two sign errors which cancel each other out. The Dirac sea is empty. This also solves the problem of the infinite energy density of the vacuum: the energy of the vacuum is zero, not infinite.

Let us analyze this in more detail. Our previous blog posts have taught us something about the ultraviolet divergence in the vertex correction. There, the diagrams with an ultraviolet divergence can be discarded altogether because they have a zero chance of happening. The loop will always send a real photon, which makes the loop integral to converge.

A semiclassical model of vacuum polarization

Suppose that an electron and a positron pass by each other at a very high speed. As the electric field strength grows, it "almost produces" a new electron positron pair. It is a "virtual pair" which electrically polarizes the vacuum.

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

The pair makes the vacuum to conduct electric lines better: it better "permits" electric lines of force:

                      e-

                      • -->

 

                      ° e+    virtual pair pulls on the

                      ° e-     electron and positron

      

                <-- •

                     e+

In the case an electron meeting an electron, the configuration is like this:

                        ° e-   virtual pair pulls on

                        ° e+  the upper electron

                    e- • -->

                  <--  • e-

   

                        ° e+   virtual pair pulls on

                        ° e-    the lower electron

In the above diagram, the field strength is the largest at the locations where the virtual pairs form.

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

If we have a medium where the electric polarization is superlinear in the electric field strength, then charges will behave just like above. We say that the electric susceptibility

        χ(E)

is superlinear in the field strength E.

We know that very high energy electrons and positrons will produce real pairs when they meet. It is natural to assume that the electric susceptibility is superlinear in E.

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

Classically, the extra polarization close to the meeting charges will always produce an electromagnetic wave. The virtual pair is a transient electric dipole which radiates away an electromagnetic wave.

The Feynman diagram above cannot happen. The virtual pair loop always emits a real photon.

The Feynman integral for the virtual pair loop has a logarithmic ultraviolet divergence at large 4-momenta, after using the Ward identity. Adding an emission of a real photon might make the integral to converge. But does that yield a result which matches the traditional renormalization technique? We have to check that.

In our August 27, 2021 post we suggested that destructive interference cancels out virtual pair loops with high 4-momenta. Does the emission of a real photon accomplish the same thing? The real photon makes the diagram asymmetric, which may complicate calculations greatly.

A semiclassical model for an electron-positron pair

The rubber membrane model was able to clarify bremsstrahlung and the vertex correction. We hit the membrane twice, but the second hit is a little bit displaced. What escapes from the first Green's function is the bremsstrahlung. That part is not absorbed by the second hit. The missing part causes the infrared divergence of the elastic diagram.

            #

            #======   EM field hits Dirac field

            v

      __       ___        membrane

          \__/

    e+ °     ° e-        Dirac wave

                              = created virtual pair

The electromagnetic "hit" to create the pair and the second "hit" to annihilate the pair may be somewhat similar? If there is a lot of energy available, then pair can become real and escape. That is like bremsstrahlung, this time consisting of pairs.

To create a real pair, we need at least 1.022 MeV. We get a strict upper bound on the wavelength of the escaping real "created pair bremsstrahlung". The Dirac equation does not have a solution where the pair possesses less energy than 1.022 MeV. Classically, no Dirac wave can escape if energy is missing.

                   spring            disturbance

            | /\/\/\/\/\/\/\/\/\      ----___----___----

     wall                                     <--- 

Our membrane model is not suitable for this. A better model is a spring whose resonant frequencies are high. A disturbance can squeeze the spring (virtual pair), but it can only make the spring to oscillate if the frequency of the disturbance is high enough (real pair).

The ultraviolet divergence in the vacuum polarization loop comes from the fact that we allow one of the particles to have an arbitrarily large negative energy (the absolute value is large). What is the semiclassical interpretation for this? Formally, the positron in the Dirac equation does possess a negative energy.

We observed in August 2021 that the pair, taken as a combination, is a boson: its components are fermions.

Also, we observed that a running coupling constant can break conservation of energy. If the force depends on the speed at which the particles meet, then the force may be non-conservative. This is not problem in Feynman diagrams, though, because conservation of energy is always enforced.

Negative energy particles are waves which rotate to the "opposite" direction, in a classical analogue

The basic formula for a particle wave is

       exp( i (-E t  +  p • r) )

in quantum mechanics. Let us, for a moment, forget that E designates energy. Then a "negative energy" wave is something which rotates to the opposite direction. There is nothing mystical about negative energy, if we think this way.

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

                               o  spider rotates string

                             //\\

            ------------------------------  tense string

In this blog we have earlier written about a "spider" which stands on a tense string and makes the right side to rotate clockwise and the left side to rotate counterclockwise. The tense string is like children's jump rope.

What is a Green's function like? Does the spider suddenly hit the string both on the left and on the right and produce sharp rotating waves? The waves rotate to opposite directions.

This model would explain why the Green's function in a Feynman diagram always must allow also negative energy particles. The complete set of waves cannot be built from just clockwise rotation.

In the case of pair production, is it so that the negative energy waves are positrons?

Real photons have a positive energy if they are circularly polarized whichever way. The rotation in negative/positive energy must happen in an abstract (complex plane) space. It is not about polarization of light.

The Green's function which creates the pair, and the ultraviolet divergence

    |

    |                e-      ___

    |        q            /        \  q

    |    ~~~~~~             ~~~~ ● X massive charge

    |                e+  \____/

    |

    |                  virtual pair

    |

    e-

   ^  t

   |

Let us try to analyze the origin of the ultraviolet divergence in (virtual) pair production. The electron passes by a very massive charged particle X.

Let the photon q be pure spatial momentum, no energy. The Green's function hits the Dirac field with a double hammer, creating both clockwise and counterclockwise waves. The hit is very sharp: the Fourier decomposition contains waves with huge positive and negative energies E. The large energies come from the sharpness of the impulse, just like when a sharp hammer hits a rubber membrane.

The Feynman integral of the vacuum polarization diagram is infinite. But let us forget the infinity. If q = 0, then we can imagine that the second hammer strike absorbs everything from the first hammer strike.

If we let q differ from zero more, then the electric field pulls on the virtual electron and the positron, and disturbs the pair. The second hammer strike is not able to absorb "everything" from the first strike.

The escaped part is kind of "bremsstrahlung". It makes the absolute value of the integral smaller. The "bremsstrahlung" increases the cross section of the scattered electrons. The electric force appears stronger. This is the "running" of the coupling constant.

Real "bremsstrahlung" would be a real pair.

Peskin and Schroeder textbook (1995) about vacuum polarization

Peskin and Schroeder (An Introduction to Quantum Field Theory, 1995) calculate the effective coupling constant αeff for a momentum exchange q. The effective value of α is larger for large |q²|.

Dimensional regularization (ε → 0) makes the value of Π₂(0) "minus infinite".

The metric signature is (+ - - -), which means that q² < 0. The logarithm above has a negative value, and

       Π₂(q²)  -  Π₂(0)  >  0.

That is, for large |q|, Π₂(q²) is "less minus infinite" than Π₂(0).

In this blog we have claimed that if |q| ≈ 0, then the virtual pair should have very little effect on the propagation of the virtual photon q. But the Feynman integral claims that the effect is "infinite". If |q| is larger, then the integral claims that the effect is "somewhat less infinite". This sounds illogical. A large |q| means a strong electric field. Then, intuitively, the virtual pair should have a stronger effect on the virtual photon q. This is the sign error which we claim to happen in Feynman diagrams in this case. The renormalization makes the final value sensible, though.

We have claimed that for q = 0, destructive interference cancels out all virtual pairs. Then the integral is zero. If |q| is larger, then destructive interference does not cancel out all virtual pairs: they make the electric susceptibility of the vacuum larger, and make the interaction stronger.

***  WORK IN PROGESS  ***

Destructive interference fixes the ultraviolet divergence in a Feynman diagram? No, a photon fixes it

In the previous blog post we found a new idea: if we introduce "new" energy to a virtual photon, then there is nothing which would determine the phase of that energy. This implies that destructive interference cancels out any attempt to add such "new energy".

Classical interacting fields: a wave cannot generate waves with a higher frequency?

               wave  -->

                   ___             ___

         ____/        \____/        \____    A tense string

                        |  |  |  |  |  |       rubber bands

                        |  |  |  |  |  |

         ________________________    B tense string

Above we have two interacting fields: two tense strings. The interaction happens through rubber bands which are tense and attached to both strings.

Intuitively, the interaction cannot produce waves which have a higher frequency than the input wave.

The frequency in quantum mechanics is associated with the particle energy. In a Feynman diagram we have an analogous rule: the energy of the particles coming out of an interaction cannot exceed the input energy.

Can a transient wave in classical interacting fields contain a higher frequency Fourier component? A transient wave is analogous to a virtual, off-shell, particle.

A single rubber band in the diagram in a short time interval will impose a Green's function on both fields (= tense strings).

In the diagram above, it is obvious that for the Green's functions at various rubber strings, only those components can survive destructive interference where the component already appears in the Fourier decomposition of the wave in the string A.

In particular, a Fourier component with a higher frequency than the wave in A cannot appear in the system.

Let the wave in A be

       ψ(t, x)  =  Real( exp(i  (-E t  +  p x) ) ).

The frequency of ψ is

       f = E / (2 ψ).

At each point x and a short time t in B, a Green's function causes an impulse response in B. Can the impulse response generate a wave ψ' in B whose frequency f' differs from f? We can appeal to symmetry: there is no reason why the hypothetical wave ψ' would have a certain phase. Thus, ψ' cannot exist.

The incoming wave packet in A can be Fourier decomposed into waves like ψ above. If we assume that the disturbing process is linear, then the wave in B will have a similar decomposition. What about nonlinear disturbances? A suitably tailored nonlinear interaction certainly can create any kind of a wave in B.

For a "well-behaved" nonlinear interaction we can appeal to symmetry: for the wave ψ above, it cannot create any frequency else than f. But for a wave packet, it could create other frequencies.

https://en.wikipedia.org/wiki/Transmission_(mechanical_device)

A gearbox can transform a low-frequency wave into a high-frequency wave. A gearbox is a very complicated "field". Thus, there do exist classical interacting fields which can generate new frequencies.

Can transient classical waves contain new (high) frequencies?

In Feynman diagrams we allow virtual (transient) particles to have any energy and momentum. How is this for classical fields?

In our rubber band example, could, e.g., a rubber band move with a high frequency for a short time?

A transient wave, whatever its form, can be Fourier decomposed. We do not see how the Fourier decomposition of the rubber band movement could gain high frequencies.

High momenta

What about high momenta? If we take the Fourier decomposition of a very localized wave packet, then the decomposition will contain high momenta.

We can argue just like in the case of the frequency that if we put as an input waves with no high momenta, then interacting fields cannot create new waves with high momenta.

Note that a Lorentz transformation mixes time and space. No high frequencies implies no high momenta, and vice versa.

Why do Feynman diagrams allow unlimited energy and momenta in virtual particles?

Feynman diagrams play with individual instances of Green's functions in "momentum space".

The diagrams are not aware of destructive interference between Green's functions in adjacent spatial locations.

Sharp spatial features created by a heavy charge X

If we have a very heavy particle, like the heavy charge X in the previous blog post, then we do have very high frequencies and momenta available in the interaction of fields. Can we still claim that no high frequencies and momenta can arise?

In the particle model, the electron can pass the charge X at a very short distance, say, 2.8 * 10⁻¹⁵ m. But in the wave model, the wavelength of a mildly relativistic electron is much longer, say, 2.4 * 10⁻¹² m. Clearly, we cannot combine the wave model with an almost pointlike charge X. In a Feynman diagram this mismatch is solved by taking the Fourier decomposition of the electric field of X. If the electron wave is disturbed with a potential "grid" whose spacing is roughly 2.4 * 10⁻¹² m, the grid can bend a part of the electron wave to a large angle. This simulates a pointlike electron passing X very close.

Such a grid is like the double-slit experiment, with the spacing of the slits being somewhat larger than the wavelength.

The bent electron wave has "absorbed" the momentum q from the grid.

The Fourier decomposition of the field of X contains also grids whose spacing is much less than 2.4 * 10⁻¹² m. What do they do? If we go to a large distance from the grid, the slits will add phases which are symmetric around the circle of angles 0 ... 2 π. There will be an almost complete destructive interference. This explains why the electron cannot absorb arbitrarily large momenta from X. In the particle model, the absolute momentum of the outgoing electron is the same as that of the incoming electron.

We showed that even though the charge X in the particle model is sharply localized, it cannot add high momenta to the electron in the wave model.

Virtual photons emitted and absorbed by an electron

A real or a virtual photon forms a potential "grid" in spacetime. A part of an electron wave can be bent, or scattered, by the grid. We say that the bent wave "absorbed" the photon.

An emission of a virtual photon is a time reverse of this process.

An electron can, of course, absorb a real photon of any phase.

What determines the phase of an emitted virtual photon? Classically, the phase of a real photon is determined by the acceleration of the charge emitting the photon. In the rubber membrane and the sharp hammer model, the phase of a wave in the Green's function is determined by the hit of the hammer at a certain location and time. If we describe the electron as a wave, then the electron has a phase – but a sharp hammer does not have a phase. What to do?

In Huygens's principle, the wave itself "hits with a hammer", and the new wave inherits its phase from the old wave.

The virtual photon "inherits" the phase from the electron? This is the spirit of Feynman diagrams. The action is of the form

       ∫  exp(-i E t) dt

       t

If the energy E is converted to another form, the formula stays the same. If a free electron emits and reabsorbs a virtual photon, the phase of the electron should not change. It would be very unnatural if a particle can manipulate its own phase by releasing and recapturing parts of it.

                     ___

                  /       \

         •  ----           -------   satellite

        ●  --------------------   central particle

Above we have a possible model for a virtual photon emission and reabsorption. We have a composite particle which temporarily releases a satellite part of it. The part may be attached with a rubber band which the central particle uses to pull back the satellite.

Could it be that a real photon can have a particle phase which is different from the electromagnetic phase of the wave? Yes, the particle phase rotates in the space of complex numbers, while the electromagnetic phase is in 3D space. If the central particle in the diagram is a small spaceship, and the crew sends a photon inside the ship, there is no reason to require that the electromagnetic phase (a 3D vector) somehow "matches" the particle phase (a complex number).

Inheriting the phase. Suppose that we have a static particle of the energy E, the phase of its wave function is exp(i α), and it sends a virtual photon with the energy E / 2, and a zero momentum. "Inheriting the phase" can be defined as meaning that the virtual photon starts its journey at the phase exp(i α). Note that the product of the phases of the particle and the virtual photon will from now on progress just like the photon would not have been emitted. The product can be seen as representing the full system the particle & the virtual photon.

If an electron sends a virtual photon with more energy than the electron, what is the phase of the photon?

Let a virtual photon have an energy

      E'  >  E,

where E is the energy of the electron. In this case, it is not natural to say that the photon "inherited" its phase from the electron. The electron will have a negative energy

       E  -  E'.

The large photon gained the energy E' - E which the electron did not possess in the first place. What phase should this extra energy carry?

We could claim that destructive interference cancels any such extra energy. There is no reason to assign a certain phase to the extra energy. This is like the symmetry argument which we used above for classical fields.

Hypothesis. A virtual particle cannot have negative energy.

The hypothesis, if true, assigns an ultraviolet cut-off for all Feynman diagrams.

The hypothesis is a way of saying that the energy of a process determines how small features can the process describe. It is like a generalized Heisenberg uncertainty principle. If virtual particles were allowed to have an arbitrary energy, then they could probe features of an arbitrarily small size. That would be against the spirit of the uncertainty principle.

Tunneling seems to contradict our hypothesis, or does it? A particle can have a negative kinetic energy when it tunnels through a potential wall. The momentum of the particle is imaginary.

In Feynman diagrams, the momentum cannot be imaginary. This differs from tunneling.

The hypothesis probably is too strict. In the calculation of the anomalous magnetic moment, the integral allows arbitrary energies for the virtual photon. If we impose a sharp cutoff at me c², the integral probably yields a wrong result.

Can we detect the physical state of an oversized photon and its parent electron?

Generally, we can only sum probability amplitudes of two states if the physical states cannot be discerned between with experiments. The double-slit experiment is spoiled if one of the slits contains floating particles which the photon will nudge if it goes through that slit.

    photon

                        |

       ~~~>            • •  floating particles

                        |

                        |

We cannot sum the probability amplitudes of:

1.   the photon went through the lower slit, and

2.   the photon went through the upper slit and nudged a floating particle.

This is because we can observe that a particle was nudged. The states 1 and 2 can be discerned between with an observation.

If we claim that destructive interference wipes out something, we must be sure that the states cannot be differentiated between through an observation.

Do Feynman loop diagram integrals with an emitted photon diverge? No, they do not diverge

The simplest Feynman calculation of the anomalous magnetic moment does not diverge. On September 24, 2025 we showed that we do not need the elastic (loopy) diagram with loops for bremsstrahlung. Could it be that a Feynman integral does not diverge if the loop emits a real photon? We have to check the literature.

                    ~~~    k virtual photon

                  /          \

          e- ---------------------

                      |    \

                      |      ~~~ real bremsstrahlung

                      |

          X ---------------------

In the diagram above, if the mass of the electron is reduced by the virtual photon, then the Larmor formula indicates that there will be more bremsstrahlung.

Classically, the mass of the electron is reduced somewhat because its field does not have time to follow it. The effect is very small, though.

https://dspace.library.uu.nl/bitstream/handle/1874/4847/14006.pdf

The paper by t'Hooft and Veltman (1979) contains general calculations for one-loop diagrams up to four legs. Unfortunately, bremsstrahlung has five legs: e-, X, and the photon.

https://dspace.library.uu.nl/bitstream/handle/1874/4828/14007.pdf

The paper by Passarino and Veltman (1979) does not help us.

Actually, adding the photon emission to the loop, as above, probably removes the ultravioet divergence because there is another electron propagator in the loop. We get an additional factor

        1 / k

into the integral. In the elastic vertex correction diagram, the ultraviolet divergence is logarithmic. Adding 1 / k should fix the problem. This may solve our ultraviolet divergence problem. An elastic loop is never possible, because the electron will always send a photon in the loop.

But now we face another problem: the photon emission probably introduces an infrared divergence to the diagram. We already have an infrared divergence in the tree level bremsstrahlung diagram. Does this change anything?

Classically, the photon sent from inside the loop describes the process where the far field of the electron does not have time to react, and the mass of the electron is reduced. The reduced mass electron creates a somewhat different electromagnetic wave than the full mass electron. Thus, the diagram does make sense classically. Maybe it is correct to add the bremsstrahlung from the reduced mass electron to the bremsstrahlung from the full mass electron.

Elastic scattering plus a real photon

                    ~~   k virtual photon

                  /       \

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

                     |                \

                     |                 ~  real bremsstrahlung

                     |

          X --------------------------

What about a diagram where there first is an elastic diagram, and later the electron sends a real bremsstrahlung photon?

One can argue that the partial diagram with just the loop is not possible. The electron will always send a real photon, even in that loop.

Conclusions

There are many reasons to believe that an ultraviolet divergence is not possible in quantum field theory. Such a divergence cannot happen with well-behaved interacting classical fields. Classical fields cannot produce waves which have a higher frequency than the input waves, except in pathological cases, like a gearbox.

Also, the spirit of uncertainty principles is that one cannot describe or probe arbitrarily small features with limited energy. This rules out the possibility that a huge (borrowed) energy in a virtual particle could have observable consequences. A divergence is a prominently observable consequence.

In the vertex correction, the ultraviolet divergence arises from the simplest one-loop elastic scattering Feynman diagram with no bremsstrahlung. We argued on September 24, 2025 that elastic scattering can never happen. There are always real photons created. Thus, we can ignore the simplest one-loop elastic diagram.

We also considered a diagram where the virtual electron in the loop sends a real photon. Then the loop integral converges, and there is no ultraviolet divergence.

We may have eliminated all ultraviolet divergences in vertex correction, simply by banning diagrams where divergences happen.

Note that Feynman diagrams probably miscalculate the vertex correction. It is unlikely that they could reproduce the complicated behavior in the classical limit. A more accurate method in many cases is to calculate the electron path and the produced radiation with classical methods. Even if we proved that we do not need diverging Feynman diagrams, we did not present the "correct" quantum field theory, and did not prove anything about its divergences.

We still have to look at the ultraviolet divergence in the vacuum polarization loop. During the fall of 2021 we claimed that we had solved that problem. But let us look at it again.