Not renormalizable theories in QFT: there is a confusion about overlapping probabilities

Wikipedia defines nonrenormalizable theories in this way:

"Not all theories lend themselves to renormalization in the manner described above, with a finite supply of counterterms and all quantities becoming cutoff-independent at the end of the calculation. If the Lagrangian contains combinations of field operators of high enough dimension in energy units, the counterterms required to cancel all divergences proliferate to infinite number"

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

Suppose that we have a process into which particles with a total energy E enter. Is it really a problem that we need an "infinite number" of counterterms?

    |                       k + q

    |                e-      ___

    |        q            /        \  q

    |    ~~~~~~              ~~~~ ● X+  massive

    |                e+  \____/                       charge

    |                          -k

    |

    |                  virtual pair

    |

    e-

   ^  t

   |

Above we have the vacuum polarization diagram. The momentum q is the input which "disturbs" the processes in a pure vacuum polarization loop with no incoming or outgoing particles.

In the previous blog post we claimed that destructive interference cancels out the entire vacuum polarization loop integral, unless q ≠ 0 disturbs it. The momentum line q is the "input" to the process.

Hypothesis. Destructive interference cancels out all loop integrals if the "input" to them from external lines goes to zero.

Does the hypothesis solve the problem of an infinite number of counterterms?

A theory where the 4-momentum k interacts: divergences really are due to the input producing many Fourier components in another field?

In gravity, the mass-energy E interacts with other particles.

A Feynman diagram loop contains arbitrary values of the 4-momentum k. Can that cause a problem if we have many loops which are connected to each other?

Let us study the diagram below.

   particle    

     • ---------------------------------------

                              |  interaction

                              |

                              |

                              |  

                             __ q + k + j

                    ___/       \___  q + k

                  /       \___/       \

               /       q + k - j        \

     q ~~                                  ~~ q

                \_______________/

                                          q - k

We assume that interactions in the diagram can depend on the 4-momentum. There are nested loops with arbitrary 4-momenta k and j.

Destructive interference almost entirely cancels any 4-momentum

       |k| > |q|,

where | | denotes the euclidean norm of the 4-momentum:

       |(E, px, py, pz)|  =  sqrt(E² + px² + py² + pz²).

If q = 0, the cancellation is perfect. If q ≠ 0, then, in very rare cases, |k| can be huge. The nested, smaller loop will in those cases have a very large |j|. Very large |j| will have a huge interaction with the particle.

The integral over j can have a very large value. Is that a problem? It should not be. If the probability of a large |k| is infinitesimal, we do not need to care much of a huge integral. In QFT there seems to be a confusion about overlapping probabilities. As if a huge integral value in some extremely rare event with a probability

       P  <  ε  <<  1

would somehow make P large, or even infinite.

Hypothesis. The huge value of the integral means that the process produces many Fourier components of a wave at the same time. The probabilities overlap.

The confusion is the same as in bremmstrahlung in electron scattering. The large value of the integral describes a complex wave which contains many Fourier components. If |k| is allowed to be large, why would it produce just one Fourier component? More likely is that it will produce a large number of Fourier components, just as happens in bremsstrahlung.

The produced wave can still be seen as "one wave", but it just happens to have many Fourier components. In classical bremsstrahlung, the complex wave certainly contains many Fourier components.

Black holes. If j has a lot of energy, then the particle may interact with a black hole. But that is extremely improbable.

What if the input q is of the Planck scale? Then we will have black holes in the diagram, and it may be that Feynman diagrams do not work at all. However, particles with a Planck scale energy are rare, or nonexistent in nature.

It looks like nonrenormalizable theories work fine. We may have a cascade of loops where the energy rises to the Planck scale, but those have an infinitesimal probability of occurring, and we can ignore them.

Hypothesis. "Nonrenormalizability" of gravity is not a problem at all.

There are other problems in the quantum field theory of gravity, though. The interaction is nonlinear and complicated.

Assaf Shomer (2007): entropy of a black hole

https://arxiv.org/abs/0709.3555

Assaf Shomer (2007) argues that the entropy of a black hole is too large for it to be describable with a renormalizable quantum field theory. He claims that a renormalizable QFT is asymptotically a conformal field theory, and the entropy in such a system must be smaller than that of a black hole.

The entropy of a black hole is roughly the same as the entropy of the wavelength

       λ  =  16 π rs

radiation which we can use to feed and grow a black hole. There rs is the Schwarzschild radius. The reverse process would be the hypothetical Hawking radiation.

Let us compare the entropy of a black hole of an energy E to the entropy of a (classical) photon gas stored into a vessel of the same size, and having the same energy E. A photon gas, presumably, is governed by a conformal field theory.

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

Wikipedia says that the energy density of a black body radiation photon gas is

       dE / dV  ~  T⁴,

where T is the absolute temperature. The entropy density is

       dS / dV  ~  T³.

Note that there is an error in Wikipedia in the table for the entropy: the table claims that one can replace the volume V with the temperature T, and derive a strange formula S = 16 σ / (3 c) T⁴, which does not depend on V.

Let the Schwarzschild radius rs vary. The energy of a black hole is

       Ebh  =  C rs,

where C is a constant of nature. The entropy, according to Bekenstein and Hawking, is

       Sbh  ~  rs².

Let us then put photon gas worth Ebh = C rs to a vessel whose volume is

       V  ~  rs³.

The total energy inside the vessel is then

       ~ T⁴ rs³  ~  C rs,

which implies that the temperature

       T  ~  1 / sqrt(rs).

The entropy inside the vessel is

       S  ~  rs³  *  T³

           ~ rs^3/2.

We see that as rs grows, the entropy of a black hole grows quadratically, while the entropy of a photon gas vessel of the same energy only grows by an exponent 1.5.

Assaf Shomer has a calculation error in the paper. He claims:

Since d = 4, that would mean that S ~ rs^¾ Shomer confused the density of energy and entropy to the total energy and entropy in the vessel.

Anyway, Shomer's main argument still seems to stand: the entropy of a black hole grows faster than the entropy of a equivalent photon gas vessel, if we increase the total stored energy.

In this blog we have claimed that the ingoing matter "freezes" at the horizon of a black hole. The calculation above assumed that the vessel containing the photon gas has no freezing effect: time flows at the same rate everywhere in the vessel.

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

Assaf Shomer's argument suggests that a black hole cannot globally, in static spatial coordinates around a black hole, behave asymptotically like a conformal field theory.

But we are interested in local phenomena in freely falling coordinates, where particle energies are much less than the Planck energy. Thus, Shomer's argument does not prevent us from having a fruitful quantum field theory of gravity.

Destructive interference cancels large frequencies exponentially well

When an electron e- passes by a massive charge X+, the time variation of the electric field in comoving coordinates of the electron, or in the comoving coordinates of X+ is something like

       ΔE(t)  =  1 /  (1  +  (v t)²).

Let us assume that v = 1.

https://math.stackexchange.com/questions/1047977/find-the-fourier-transform-of-frac11t2

The "disturbance" ΔE(t) then has a Fourier transform

For large frequencies, or 4-momenta, the Fourier component is absolutely negligible. If f = 100, the component is ~ 10⁻⁶³⁰.

However, is this cancellation even too strong? Let us compare this to the renormalized vacuum polarization value. Is the contribution of |k| > 100 |q| absolutely negligible?

https://www.scribd.com/document/259722615/Quantum-Field-Theory-by-Hagen-Kleinert#page=892

Hagen Kleinert (2013) calculates the effect of q ≠ 0 by using the q² derivative of Π₂(q²):

The arbitrary 4-momentum is in his nomenclature p, not k as in our blog text. The contribution of large |q| seems to be a decreasing geometric series. It is not exponentially decreasing.

This is not paradoxical. Even if the "disturbance" is very much free of high frequencies, its "impact" may be less free.

Is it possible that for some loop, the impact of a disturbance diverges, too? What would an ultraviolet divergence mean in such a case? If there are overlapping probabilities, can it mean an infinite energy for the generated wave?

Quantum gravity

https://link.springer.com/article/10.12942/lrr-2002-5

Zvi Bern (2002) writes about divergences in quantum gravity. In Section 2.2 he states that gravity with matter typically diverges badly at one loop, and pure gravity at two loops.

Let us check if we can find a way to alleviate the problem.

There is another problem, too. The Feynman integral for just five loops of gravitons contains 10³⁰ terms! We have to find a simpler way to calculate the interaction. In our blog we hold the view that gravity is a combination of an attractive force and an increase in the inertia of a particle in a gravitational field. Could it be that calculations with inertia would be simpler than calculations with the metric tensor?

Ultraviolet divergence in a loop

In an infrared divergence, on September 24, 2025 we were able to explain the problem away by claiming that the produced wave contains an infinite number of low-energy bremsstrahlung photons. Does the same principle work for an ultraviolet divergence?

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

If we hit a rubber membrane with an infinitely sharp hammer, then it, presumably, creates the Coulomb field of a point charge, which has an infinite energy. That is why the hammer is never allowed to be infinitely sharp: destructive interference must cancel the infinite energy. The hammer must be blunt.

   --> t

Above we have a Feynman diagram. Let the lines represent gravitons and the vertices their gravity interaction. Two gravitons enter from the left, create new virtual gravitons, interact, and exit on the right.

The Feynman integral calculates the "4-point function", or the probability amplitude for the process to happen, assuming that the input flux from the left has a certain value.

The input gravitons coming from the left have some modest probability amplitudes. If the Feynman integral is infinite, that would mean that the output flux of the gravitons on the right is infinite. That is, we have created an infinite energy. This is clearly nonsensical. What is the problem?

The process has a classical limit. The input gravitons could be wave packets, and the output gravitons are wave packets. General relativity is supposed to conserve energy. We conclude that the ultraviolet divergent Feynman integral miscalculates the process, and badly.

Let us imagine that the graviton q interacts with the electron and the positron gravitationally. In that case, the interaction is proportional to 

       |k|.

Maybe the Feynman integral diverges, even after renormalization? What does that mean, physically?

The contribution to the probability amplitude of the process might be such that each interval

         n  ≤  |k|  <  n + 1

contributes an equal amount. The sum is infinite. Then the classical probabilities must overlap. That would mean that the process at the same time launches many pairs with various momenta k. We have suggested that the pair, as a whole, is a boson. Many bosons with various k are launched at the same time. The right side of the diagram then would describe a simultaneous absorption of all these bosons. In this interpretation, the various k are not separate classical probabilities. They overlap. Then there is no divergence in the classical probability.

An analogy: a sharp hammer hit to a rubber membrane produces many Fourier components with a large |k|. The next hit absorbs them all.

Suppose that q produces a wave packet of a pair which carries the momentum q. The Fourier decomposition of the wave packet contains many different momenta k. Why would the sum of the classical probabilities for each k be less than 1, or even finite?

Destructive interference should almost completely cancel high 4-momenta. For classical waves, it probably establishes a cutoff.

Why does a Feynman integral diverge? It may be due to the following:

1.   the integral does not take into account destructive interference, or

2.   the integral does not understand that classical probabilities for various 4-momenta k in a loop overlap.

Classically, the infrared divergence in bremsstrahlung is easy to understand. But having a large amount of very high 4-momenta k would be strange. When an electron passes a charge X+ inside a material, the polarization of the material behaves varies smoothly. The Fourier decomposition of the polarization will not contain high frequencies.

Classically, a hit with an infinitely sharp hammer will contain very high frequencies. The energy of the hit probably is infinite. If the Green's function is not completely absorbed, then it may "leak" lots of very high frequencies, and we might get a classical divergence. An infinite amount of energy is created.

We already discussed one type of a classical ultraviolet divergence. If an electron receives an instantaneous impulse, and its acceleration is infinite, it will radiate an infinite energy.

Why do Feynman diagrams operate with a single hit of the hammer? The method does work in some cases, but in the general setting, it is prone to create an infinite energy – fail miserably.

Question. Is destructive interference the correct method to regularize and renormalize all Feynman integrals? It is suspicious that integrals of the type

        ∫ d⁴k  * 1 / kⁿ

only converge slowly, or not at all. Exponential convergence might be more realistic?

In quantum mechanics, one natural "convergence mechanism" is the uncertainty relation which says that a particle can "borrow" an energy E at most for a time

       t  <  ħ / (2 E).

The "convergence" in it is quite slow.

Conclusions

Let us close this long post. We will continue the study of ultraviolet divergences in an upcoming post.

We have to analyze more thoroughly what does an ultraviolet divergent Feynman integral really mean. It can remain divergent also after the "renormalization", if the input (like q ≠ 0 in vacuum polarization) "disturbs" the integral value enough.

Classically, an ultraviolet divergent approximation formula is a very bad approximation, since it generates an infinite energy from a finite input. Feynman integrals may in some cases calculate the processes wrong, regardless of regularization and renormalization procedures.

In quantum gravity, ultraviolet divergences seem to happen inevitably at two loops. Furthermore Einstein formulae are too complicated, so that a simple Feynman integral may contain 10³⁰ terms. We have to find a simpler method to calculate interaction processes.

Vacuum polarization in QED

UPDATE October 20, 2025: When two electrons approach each other, superlinear polarization reduces the energy of the electric field? Then the repulsion is weaker. We have to check if someone has measured the running of the coupling constant for Möller scattering. 

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

Wikipedia only mentions the LEP experiment, where the running was observed in Bhabha scattering e- e+ → e- e+. Also, Wikipedia states that vacuum polarization is not relevant in low-energy processes.

----

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 between the electron and the positron 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.

Toy model of superlinear polarization: flat plates of opposite charge

     E = 0        E = 2         E = 0

                       -->

                 |              |

                 |              |

                 |              |

                 |              |

                 |              |

                 +              -

We assume that the polarization of the air between the plates is linear in E when the electric field strength is

       |E|  ≤  1,

and the polarization is much stronger when

       |E|  >  1.

The plates, when alone, have the electric field |E| close to the plates. We bring the plates parallel, close to each other.

Then the field between the plates is E = 2, and there is a lot of extra polarization there. The energy of the electric field between the plates is reduced between the plates. We conclude that when we bring the negative plate close to the positive plate, the superlinear polarization increases the attractive force felt by the negative plate. The field has less energy => we were able to harvest more energy when we lowered the negative plate close to the positive plate.

Question. Is the electric attraction "asymptotically free"? If the polarization at a very strong electric field entirely erases the energy of the electric field, then opposite charges will not feel any force. This might happen when the distance of an electron and a positive charge e+ is the classical radius re = 2.8 * 10⁻¹⁵ m.

Since the Compton wavelength of the electron is much longer, 2.4 * 10⁻¹² m, the electron can only enjoy its asymptotic freedom for very short times. It is not a permanent.

The Uehling potential around an electron is steeper than the Coulomb potential. If we have a symmetric spherical shell of positive charge around an electron, and lower it close to the electron, then the entire energy of the electron electric field is exhausted earlier than for a Coulomb potential. The force felt by the shell is stronger than Coulomb, but the force should become zero earlier.

If the electron has a large kinetic energy, then LEP collider experiments show that the electric field does possess a force down to distances of 10⁻¹⁸ m or less. The "asymptotic freedom" would only hold for low-energy electrons.

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 is used: we integrate in 4 - ε dimensions, where ε is a small positive real number. This makes the integral finite.

We then let ε approach 0. That makes the value of Π₂(0) "minus infinite". Peskin and Schroeder renormalize the infinite integral by declaring Π₂(0) to be the reference point, to which the integral with other values of q have to be compared:

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.

Can we somehow interpret that the vacuum polarization diagram makes the interaction stronger? Yes

If the renormalization calculates the interaction correctly, then, for some reason, the disturbance caused by q on the virtual pair makes the interaction stronger, so that the cross section of a q momentum exchange is larger. Can we find an intuitive explanation why this should be so?

      

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

                           | q

          X+ --------------------------

We assume that X+ is very heavy and has a positive charge. The tree level diagram makes most of the cross section.

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

                           |  q

                          O       vacuum polarization

                           |  q

          X+ ------------------------

The vacuum polarization loop maybe is an independent phenomenon from the tree level diagram? The loop denotes polarization between e- and X+. The polarization pulls both on e- and X+. Then the loop diagram would come on top of the tree level diagram, and increase the cross section. Classically, the loop would be an object which is polarized in the field of e- and X+.

We can interpret the vacuum polarization diagram this way: the field of X+ disturbs a virtual electron wave and the virtual electron wave absorbs the momentum q from the field of X+. That is, X+ pulled the virtual electron. The virtual positron pulls the real electron and absorbs the momentum -q from the real electron. Then the virtual pair annihilates. Since the pair was able to get rid of the extra momentum (q - q = 0) this process respects conservation laws and is allowed.

This all sounds very logical. But why does the Feynman method with renormalization calculate this right? We can imagine that if q = 0, then the Feynman integral calculates the "life" of a virtual pair in empty space. Having q ≠ 0 disturbs this peaceful life.

New kind of "bremsstrahlung": the virtual pair is a boson with a mass

Hypothesis of why a disturbed virtual pair contributes to the interaction.

1.   When the life of a virtual pair is peaceful, destructive interference, actually, cancels out the pair completely. It never existed and did not have any effect.

2.   If q ≠ 0, then not all virtual pairs are completely absorbed in the second "hammer strike". Non-absorbed pairs did not suffer a complete destructive interference. They came into existence.

3.   A virtual pair cannot escape as bremsstrahlung because it does not have enough energy.

4.   The pair must eventually annihilate and disappear. The pair did contribute to the interaction, transferring momentum between X+ and the real electron e-.

The hypothesis above explains why the "missing part" of the integral tells us the extra interaction between X+ and the real e-. In Feynman rules, this is implemented by flipping the sign of the integral, because of the fermion loop, and declaring the integral value for q = 0 the "benchmark", to which the integral with q ≠ 0 has to be compared.

The extra interaction between X+ and the real e- is kind of "bremsstrahlung", which has a real-world effect on the process.

Since the pair, if made real, has a mass 1.022 MeV, there is no infrared divergence in this new kind of bremsstrahlung. Recall that the infrared divergence of bremsstrahlung was removed by giving a small mass to the photon.

Note that the virtual pair is a boson. The virtual pair is analogous to a boson with a mass.

How does the new concept of bremsstrahlung affect our view of the anomalous magnetic moment?

Supersymmetry

We realized that the virtual pair can be seen as a massive photon. Does this have anything to do with supersymmetry?

A fundamental idea of supersymmetry is that the fermionic field and the bosonic field form one whole, similar to time and place being one whole in special relativity. If we have a polarizable material, then a photon can be defined as polarization in the material. Does this have anything to do with supersymmetry?

In this blog we have presented a hypothesis that a photon really is polarization of the vacuum. We will look at supersymmetry later.

Conclusions

We argued that the renormalization, or the counterterm, in the vacuum polarization integral really removes what "does not exist". The scale of things in the vacuum polarization process is determined by the momentum exchange q. There is a dynamic process as the electron passes close to the charge X+, and |q| determines the "scale", or the resolution, of the process. Momenta much larger than |q| are canceled out by destructive interference.

The Feynman vacuum polarization integral calculates how much of the Green's function is canceled out by destructive interference. The remaining part,

        Π₂(0)  -  Π₂(q²)

is the one which is not canceled, and pulls the charge X+ and the electron e- toward each other with the momentum exchange q.

In the correct interpretation, we do not need any "renormalization", meaning an obscure cancelation of an infinity. Rather, we simply remove the nonexistent part of the integral.

This blog post ends our analysis of QED divergences. We showed that they are not problematic:

1.   The infrared divergence of bremsstrahlung is the correct result: it describes the fact that an infinite number of low-energy real photons are always emitted.

2.   We can discard the vertex correction elastic scattering Feynman diagram which contains the other infrared divergence: the history can never happen because photons are always emitted.

3.   If we add a real photon emission in the vertex correction loop, then the Feynman integral converges.

4.   The "renormalization" in the vacuum polarization diagram just removes the part of the integral which is canceled by destructive interference. We, of course, should not integrate over something which does not happen at all. We do the logical, right thing, and there is no need to devise complicated explanations about why we can remove the infinity.

We will next look at theories which are not renormalizable. Our goal is to understand what problems exist in quantum theories of gravity.

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.

Ultraviolet divergence in QED

UPDATE October 21, 2025: Below we claim that we can discard the elastic scattering loop diagram altogether. This may be a wrong conclusion. In vacuum polarization, the diagram does alter the the scattering probability, even though no real photon is emitted.

The question is if the reduced mass (or reduced momentum) in the elastic loop has any effect on the probability of scattering by q. Classically, the mass of the electron really is a little reduced, but that has an extremely small effect on scattering amplitudes. We probably can ignore the elastic loop diagram.

The loop, presumably, always emits an infinite number of real photons, though. Is it likely that the Feynman integral correctly calculates the probability amplitude for this? This cannot be separate from the photons emitted in the tree level diagram.

Classically, tree level diagrams tells us a rough approximation of the process. If we add a loop, that refines the approximation, but the probabilities may overlap. The mass of the electron is somewhat reduced because its far field does not have time to react. That slightly affects the electromagnetic wave emitted by the electron.

We will write a new blog post about this.

----

In this blog we have for many years claimed that destructive interference removes ultraviolet divergences in QED. Our argument is based on the classical limit.

Regularization or renormalization are not needed if one uses a mathematically correct approximation method. Ultraviolet divergences are a result of a Feynman diagram only "hitting" the electromagnetic field with one Green's function – which is a poor approximation of the process.

In the diagram above, we see an electron passing close to a very heavy negative charge X.

Let us switch to the classical limit. The electron is then a macroscopic particle with a very large charge, 1.8 * 10¹¹ coulombs per kilogram.

Feynman diagrams have no restrictions about the mass of the particles. The particles are allowed to be macroscopic.

As the large electron passes X, it emits a classical electromagnetic wave which has a huge number (actually, infinite) number of real photons.

The electric vertex correction is about the wobbling of the electric field relative to electron as it passes X. In particular, the far field of the electron does not have time to take part in the process. The electron appears to have a reduced mass as it passes X.

Classically, it is obvious that the inner electric field of the electron tracks the movement of the electron very accurately. The inner field is "rigid", and does not affect the movement of the electron much.

The rubber membrane model

                     #

                     #=========   sharp hammer

                     v                      keeps hitting

        ______       _____ tense rubber membrane

                   \__/

                     • e-         weight makes a pit

In the rubber membrane model of the electron electric field, we can imagine that the weight of the electron is implemented with a sharp hammer hitting the membrane at very short intervals.

                              |

                              |

                              |

                              |

                              • e-

        ^ t

        |

         -----> x

Let us analyze the Green's functions of the hammer hits if the electron stays static in space.

We see that if E ≠ 0, then there is a complete destructive interference for any

       exp(i (-E t  + p • r) / ħ).

That is expected, since the electric field is static.

Let us then assume that the charge X passes by the electron e-. The electron is accelerated, and gains some final velocity v.

For large |E|, the destructive interference still is almost complete. For what values of E is the destructive interference incomplete?

Let a be the acceleration of the electron. Let Δt be the cycle time of a wave with E ≠ 0.

During the cycle time, the electron moves a distance

       R  =  1/2 a Δt².

The wavelength is

       λ  = c Δt.

We see that if Δt is very short, then the electron moves negligibly during a cycle, compared to the wavelength λ. Intuitively, the destructive interference is strong then.

Let t be the time when the electron is accelerated. Intuitively, destructive interference is spoiled the most if the cycle time is t. That is, the wavelength is

       c t.

In this blog we have claimed that the electric field "does not have time to follow the electron", if it is at a distance c t from the electron. Destructive interference matches this.

The ultraviolet divergence is due to the fact that a Feynman diagram only hits the electromagnetic field once with a Green's function. In reality, the electron keeps hitting all the time.

Regularization and renormalization of the ultraviolet divergence in the electric vertex corrention

https://web2.ph.utexas.edu/~vadim/Classes/2024f-qft/notes.html

https://web2.ph.utexas.edu/~vadim/Classes/2024f-qft/vertex.pdf

Let us look at how Vadim Kaplunovsky handles the ultraviolet divergence in the vertex correction.

If q² = 0, then the vertex function should be 1. We decide that the "counterterm" δ₁ must have the value:

With that value, the vertex function F₁^net(q²) has the right value 1 when q² = 0.

What is the logic in this? The idea is that the infinite value of the integral F₁^loops(q²) is "renormalized" to zero when q² = 0. We calculate a difference of the integral value when q² ≠ 0, compared to the integral value when q² = 0. The difference, defined in a reasonable way, is finite, even though the integral is infinite.

What is the relationship of this to our own idea in which destructive interference is used to make the integral to converge?

If q² = 0, then we claim that destructive interference cancels, for the Green's function, every Fourier component for which E ≠ 0. This is equivalent to the "renormalization" in the utexas.edu paper, where a "counterterm" δ₁ erases the entire Feynman integral.

What is the meaning of the difference

      F₁^loops(q²)  -  F₁^loops(0)

for q² ≠ 0?

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

In the link, for q² << m²,

where λ is the "photon mass" used to regularize the infrared divergence. There is no rule for how we should choose λ. The formula is vague.

Is the electric form factor F1(q²) a microscopic quantum effect?

We are struggling to find the analogue of the electric form factor F₁(q²) in the classical limit. If the electron is a macroscopic charge, then the wobbling of its electric field will reduce the mass of the electron, since the far electric field of the electron does not have time to react.

If the mass of the electron is reduced, and it passes a negative charge X, then X will push the electron away a little bit more: the momentum exchange is reduced, and the cross section is less.

But if X is positive, then the reduced mass of the electron allows it to come a little bit closer: the momentum exchange is larger and the cross section is more.

In the literature, the form factor F₁(q²) only depends on the square q² – it does not differentiate between X being positive or negative.

The tree level diagram of e- X scattering only depends on q, not on the electron mass. Thus, the mass reduction would not even show in the Feynman integral cross section.

The quantum imitation principle, which we introduced on September 19, 2025, may solve the problem. When the electron meets X, the electron tries to "build" its electric field with a photon. But the resources of the electron only suffice to send one large photon (mass-energy ~ me) at a time. The electric form factor F₁(q²) would be a result of this shortage of resources.

In the classical limit, the electron is able to send many large photons simultaneously, and build its electric field at a high precision.

The Feynman integral may work correctly if the electron passes very close to X. Then the resources of the electron are severely limited. It may send a single large photon, attempting to build its electric field.

In the classical limit, the form factor F₁(q²) clearly is wrong. It does not describe the wobbling of the classical electric field.

We have to check if any empirical experiments have verified the factor F₁(q²). Does the anomalous magnetic moment depend significantly on F₁(q²)?

A practical calculation when e- is relativistic and meets a massive charge of size e-

Let us assume that the electron e- is relativistic and is deflected by X into a large angle. Let us try to estimate the magnitude of the electric vertex correction.

The southampton.ac.uk link above suggests that 

     F₁(q²)  ~  1  +  α / (3 π).

That is, the cross section increases by ~ 1 / 1,300.

                               ^  v ≈ c

                              /

                            /

           e- • --------    

                                ● X

The mass-energy of the electric field of the electron at least a Compton wavelength

       λe  =  2.4 * 10⁻¹² m

away is α / (2 π) = 1/861 of the electron mass me.

Since the relativistic electron is scattered to a very large angle, its closest distance to X must be

       ~ re  =  2.8 * 10⁻¹⁵ m.

If we reduce the mass of the electron by 1/861, then as it passes X, it will come closer to X. Let the time that the electron spends close to X be

       t  =  2 re / c.

The acceleration of the electron toward X is something like

       a  =  c / r

            = c / (2 re / c)

            = c² / (2 re).

The acceleration takes the electron closer to X, very crudely:

       Δr  =  1/2 a t²

             = 1/2 c² / (2 re)  *  (2 re)²  / c²

 

             = re.

If we reduce the mass of the electron by 1/861, the impulse that the electron receives is ~ 1/861 larger. We expect the scattering cross sections to grow something like 1/861.

However, we do not understand how the Feynman diagram could be able to calculate this. The Feynman diagram calculates the probability amplitudes for various scattering momenta p' ≠ p, if p is the initial momentum of the electron.

Could it be that the two electron propagators in the integral become larger if we reduce the mass of the electron:

No. The value of the propagator measures "how much" the electron is off-shell. Reducing the mass of the electron does not bring the electron closer to on-shell.

              far field

              • ---> v

             |

             | rubber band

             |

         e- • ---> v    

                                       ● X

Classically, the electric field of the electron becomes distorted when the electron bounces from X. If we treat the electron and its inner field as a single particle, that single particle (reduced electron) is "on-shell" after the bounce.

After the bounce, the electron must supply the missing momentum to the far field of the electron, and must get rid of the excess kinetic energy that the electron has. In this sense, the electron is "off-shell" after the bounce, if we treat the electron and its entire electric field as a single particle. The excess kinetic energy escapes as electromagnetic radiation.

The bounce puts the electron and its far field into an "excited state". The process can be understood like this:

-  The bounce from X converts kinetic energy of the electron into an excitation of its electric field. The excitation decays by emitting electromagnetic radiation.

The process will always radiate real photons which have large wavelengths. Elastic scattering is a process which does not happen at all. If Feynman diagrams claim that it is possible, then they miscalculate the process.

Feynman diagrams calculate the energy radiated in an inelastic collision, and subtract that energy from the elastic path

Now we figured out what the inelastic and elastic Feynman diagrams calculate. (Inelastic = a real photon is radiated. Elastic = no photon is radiated.) This interpretation is inspired by the rubber membrane model.

1.   If the electron is not under an acceleration, it will always reabsorb all the real photons which it emitted when it hit the electromagnetic field with a Green's function.

2.  The diagram which concerns a real photon emission, calculates the probability amplitude of that photon being emitted. It does not calculate the electron flux. We cannot expect the sum of these amplitudes be 1 – rather it is infinite, since an infinite number of photons are always emitted. The diagram, loosely, calculates radiated energy.

3.  The elastic Feynman diagram simply reflects the fact that if energy is radiated out, then the electron cannot reabsorb that energy.

The above interpretation explains why the infrared divergent parts of the tree level radiation diagram and the elastic diagram cancel each other exactly: they both calculate how much energy is radiated out!

Since the electron will always radiate real photons when it passes by a charge X, the elastic Feynman diagram is useless? It does not describe any process in nature. It does not calculate anything useful.

Radiated photons reduce the energy and change the momentum of the electron. The photons, all of them, must be taken into account when we calculate the momenta of the electrons coming out from the experiment.

What does this mean concerning the ultraviolet divergence? We discard the purely inelastic Feynman diagram, but there are Feynman diagrams which contain both the emission and reabsorption of a virtual photon and an emission of a real photon. The ultraviolet divergence can dive up there.

                    ~~~~~~

                  /                \

          e-  ---------------------

                         |      \

                         |        ~~~

                         |

          X   ---------------------

Does the elastic collision Feynman diagram calculate anything useful?

Let an electron pass by a charge X. We pretend that there is no radiation out. Then the elastic Feynman diagram is relevant. If q² varies, what does the integral calculate?

                             k

                         ~~~~

                       /            \

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

The diagram for the free electron is above.

                             k

                          ~~~~

                        /            \

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

                              | 2

                              | q

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

The diagram for the colliding electron differs from the free electron, because we have the vertex and the electron propagator marked with 2 in the diagram, as well as the photon propagator for q. The photon q propagator is factored out from F₁(q²).

The electron propagator measures how much the electron is off-shell: being more off-shell means that the absolute value of the propagator is smaller.

It is obvious that q² affects the Feynman integral, but does it have any intuitive meaning?

In the rubber membrane and the sharp hammer model, if we assume that there is no energy loss, then the hammer must hit at the same place as it did earlier. The process is the same as for the free electron. If we make the (unrealistic) assumption that there is no radiation loss, then the Feynman integral should have the same value as for the free electron.

Classically, we can think like this: we strip the far electric field of the electron off, to remove radiation losses. We keep the rigid inner field. Since the inner field is rigid, we can assume that the electron has no electric field at all, and the mass-energy of the field is in the mass of the pointlike electron. The elastic Feynman diagram is useless: we can just look at the tree-level diagram.

But could there be some microscopic effect which comes from the quantization?

Let X be static. The spatial momentum p of the electron changes its direction as the electron absorbs q, but the energy and p² do not change.

The ultraviolet divergence in the classical limit would come from the abrupt momentum change q?

Suppose that we have a classical charge which makes an instant turn. We ignore the fact that it will slow down as it radiates electromagnetic energy.

The power of the radiation is, according to Larmor,

       P  ~  a².

If we let the charge to accelerate to a speed v in half a time, the power is fourfold and the the time a half: the radiated energy is double.

In this way, we get an "ultraviolet divergence" in the classical limit. In the classical picture, this could show up as an ultraviolet divergence of the opposite sign in the absorption of waves sent by a Green's function.

                     ~~~~  "cut-off" for a real photon

                   /               in tree level diagram

         e- ----------------

 

                    ~~~

                  /         \     no cut-off in elastic

         e- ----------------   diagram

The requirement of energy conservation implements an ultraviolet "cut-off" to the tree level Feynman diagram. But there is no cut-off in the elastic Feynman diagram. This explains why there is no matching ultraviolet divergence in the tree level diagram.

The classical model suggests that the electron rarely radiates a large real photon

Let us have a mildly relativistic electron and let X have a charge -e. We assume that the electron is scattered to a large angle. We calculated in the previous blog post from the Larmor formula that the energy radiated by the electron is

       W  =  10⁻⁵⁷ / R³,

where R is the minimum distance, e.g., 10⁻¹⁴ m. The radiated energy is then 

       W  =  10⁻¹⁵ J

             =  6 keV.

The electron can, in principle, radiate away most of its kinetic energy, which for a mildly relativistic electron is ~ 500 keV.

This suggests that at most in ~ 1/100 of cases does the electron radiate a large photon. In most cases, the encounter is "mostly" elastic.

If the tree level Feynman diagram correctly calculates the probability of radiating a large photon, then that (small) probability should be deducted from the corresponding "mostly" elastic encounter probability.

This result is hard to map to the traditional interpretation of Feynman diagrams. There are no totally elastic encounters at all: an infinite number of small real photons are always radiated.

Most encounters are "mostly" elastic: there is no large photon radiated.

A small portion of encounters are very much inelastic: a large photon is radiated.

Suggested solution: we do not need the elastic Feynman diagram at all – use only tree level diagrams

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

                             | q

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

The simplest Feynman diagram approximates the scattering of the electron quite well. Bremsstrahlung typically contains very little energy and momentum.

                         ~~~~~~~~

                       /

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

                            | q

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

The simple tree level bremsstrahlung diagram can be used to calculate real photons which are emitted. In rare cases, a very large real photon is emitted, and that significantly alters the 4-momentum of the outgoing electron. Here we assume that the diagram fairly well imitates the classical spectrum. That should be checked.

If no large photon is emitted, then the process is almost classical. An infinite number of very small real photons are emitted. A classical approximation probably is much more accurate than a Feynman diagram. That also holds for the scattering amplitudes of the electron: they should obey the Schrödinger equation, which in turn must obey the classical limit.

There is no need to do anything about the infrared divergence. The divergence simply is a sign that an infinite number of small photons are emitted.

What about the ultraviolet divergence? There is no ultraviolet divergence in tree level diagrams. No need to do anything.

Analysis of the ultraviolet renormalization in textbooks

Why would the counterterm

       δ₁  =   -F₁^loops(0)

work at all in the elastic Feynman diagram?

Let us once again refer to the rubber membrane and the sharp hammer model. If q² = 0, then then the entire Green's function from a hammer hit will be absorbed. But if q² ≠ 0, that means that the hammer position is accelerated. Not every wave produced by the previous hammer hit will be absorbed. The difference

       F₁^loops(q²)  -  F₁^loops(0)

will then contain the escaped waves, the sign flipped.

Now we see a problem here. If a real photon contains more energy than the kinetic energy of the incoming electron, then the photon cannot escape.

But the elastic Feynman diagram still thinks that such a photon produced by the Green's function can avoid being absorbed? That would be illogical. That photon would contribute negatively to the difference above.

The textbook renormalization may have an error here. If a virtual photon is such that it cannot escape, then we have to assume a 100% absorption for it. This differs from the textbook treatment.

We suggested above that we can avoid renormalization altogether by using tree level Feynman diagrams. Thus, it may be unnecessary to worry about the renormalization of the elastic Feynman diagram.

More complicated diagrams with loops

Our analysis above suggests that tree level diagrams are sufficient for the analysis and we can skip the simplest Feynman diagram with a loop: the elastic scattering diagram. What about more complex diagrams?

                          k virtual photon

                    ~~~~~~

                  /                \

          e-  ---------------------

                         |      \

                         |        ~~~

                         | q

          X   ---------------------

Above we have a complicated diagram. Classically, some of the field of the electron does not have time to take part in the movement of the electron e-, as it passes by the charge X. If the virtual photon contains a lot of mass-energy E, then the effective mass of the electron is greatly reduced. The reduced mass classically should increase the radiation of the electron a lot as it passes by X.

This effect should be significant and measurable in experiments. In a few cases, the electron sheds a lot of mass to the virtual photon.

Now we have to deal with the ultraviolet divergence. What prevents the electron from losing temporarily all of its mass, or even more?

If q is small, then we can appeal to the classical limit: virtual photons cannot affect the path of the electron much. We simply can ignore the diagram with a loop.

How to prevent huge virtual photons in a loop?

Suppose then that q is large and k carries more energy than the electron possesses in the first place. Can we then extract that energy and break conservation of energy?

A Feynman diagram enforces conservation of energy for outgoing particles. Is this enough? It seems very odd if we probe the electron with another particle, and that particle scatters from a photon which has much more energy than the electron.

Suppose that we have a particle Y which can jump over a high potential wall and keep the energy if it absorbs a very large photon. If the virtual photon k "really can exist", then it can push Y over the potential wall, and conservation of energy is broken. This suggests that large virtual photons cannot exist. Virtual photons have to obey conservation of energy.

Let us then look at the wave model of a free electron whose 4-momentum is p. If the electron wave ψ(p) meets an electromagnetic wave k, the k wave will disturb the electron wave, according to the interaction in the lagrangian. A part of the k wave is "absorbed" by the electron, producing a

       ψ(p + k)

wave.

The emission of a virtual photon by the electron wave is the reverse of this process.

                ___________   ψ(E, 0)

                ___________

                ___________

      ^ t

      |

       ------> x

Why would the electron wave ψ(p) emit a photon wave k? In the diagram we have the electron wave for a static electron with the energy E = me c².

If the electron emits a virtual photon k = (E, 0), then the photon, obviously, will inherit its phase from the electron.

If the photon has a larger energy than E, is there anything which could determine the phase of the "extra energy" in the photon?

If not, then we found a cut-off which is like the conservation of energy cutoff for the outgoing particles in a Feynman diagram.

Conclusions

If an electron scatters from a charge X, we showed that we do not need the elastic scattering Feynman diagram at all. There is no need to handle the ultraviolet divergence in the simplest elastic diagram.

However, for more complex Feynman diagrams, the ultraviolet divergence still appears. In the next blog post we will study if destructive interference imposes a cut-off which removes the divergence.