A classical model of pair production

Let us again look at Coulomb scattering of an electron from a nucleus.

       Z+ ●

                   ^

                   |

                   |

                   e-

As the electron approaches the nucleus, its speed accelerates. The electric field lines of the electron bend and are squeezed closer to each other (just as in the Edward M. Purcell diagram of the electromagnetic wave production).

Since the field lines are now closer to each other, the energy density E^2 has increased and the tension in the field lines has grown.

In our classical model, the field lines tend to "break" if their tension grows large enough.

          field                            line

         --------------- +         - ----------------

Broken field lines end at small charges which have formed at the loose ends on the line.

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

This model is like the rubber band model in quantum chromodynamics.

How do the small charges at the ends of broken field lines clump together to form a full new electron and a positron? Our model predicts that a fuzzy spatial distribution of charges will form.

The electron is the quantum of the formed fuzzy charge distribution. A classical model cannot predict the birth of a quantum from a fuzzy distribution.

A quantum model using a Feynman diagram

A quantum model might treat the electromagnetic field as a particle, a single virtual photon, and create the electron and the positron as particles from that photon.

                                               --------- e-

                 E, p                     /

                 virtual     ~~~~

                 photon  /            \______ e+

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

                                  | virtual

                                  | photon q

           Z+ -----------------------------------

The Feynman diagram above describes a quantum process. The electron emits a virtual photon which has E >= 1.022 MeV of energy and momentum p. That photon in the classical model would be part of the increased energy and momentum of the electric field of the electron.

The increased energy escapes from the electric field to a new produced pair. We say that electromagnetism is nonlinear in this case because energy can escape to pairs.

If no pair were created, the extra energy and the momentum in the electron's electric field would return back to the electron and propel the electron out of the potential of the nucleus Z+ at the original speed of the electron. The rubber plate model helps to visualize this process. In the approach the plate lags, in the departure the plate is ahead and pulls the electron out of the potential.

But the electron lost energy and momentum in the production of the pair. The electron will exit the potential of the nucleus more slowly, and suffer an extra loss of momentum q, which is marked as the second virtual photon. Actually, q is the combined effect of several processes which transferred extra momentum from the electron to the nucleus.

Pair production in a collision: a wave-particle model

In the latest two blog posts we showed that a classical electromagnetic field might be adequate for explaining Coulomb scattering, as well as the real photons produced in it.

But classical electromagnetic theory does not explain production of electron-positron pairs in a collision.

In classical electromagnetism we have no rule about how an electromagnetic field could "induce" a wave in the Dirac field. We do have rules about how particles described by the Dirac equation produce electromagnetic fields, but no converse rule.

How an electromagnetic field A "induces" a Dirac field in the Feynman model

The Feynman diagram model does tell us how the converse rule might work: any electromagnetic field at every spacetime point constantly produces Dirac waves, using a Green's function for the Dirac field. The waves are mostly (?) virtual. In some cases, these waves can gain enough energy to become real particles.

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

The QED lagrangian contains the interaction term

       -e ψ-bar γ^μ A_μ ψ.

It is a reasonable guess that the interaction term acts as a source for the free Dirac equation. Any electromagnetic field A produces a Dirac field wave which always contains both an electron and a positron, virtual or real.

For each spacetime point, the particles are produced as an impulse response to a Dirac delta source. That is, a Green's function produces those Dirac field waves.

Dirac field waves can escape as real particles if they somehow gain the right amount of energy relative to their momentum. They must then obey the energy-momentum relation:

        E^2 = p^2 + m_e^2,

where m_e is the electron mass 511 keV.

If the waves obey that relation, they "resonate" with the free Dirac field, and can exist indefinitely.

If the waves do not obey the energy-momentum relation, they are "malformed" waves and can only exist for a very short time, or across a very short distance. The energy and the momentum that were put into malformed Dirac waves quickly return back to the electromagnetic field A. Malformed waves do not "resonate" with the Dirac field.

What original energy and momentum do the virtual pairs acquire when they are produced by an electromagnetic field? If the field A has a Fourier component whose form is something like

         exp(-i (E t - p x)),

then a reasonable guess is that the virtual pair "inherits" the energy E and the momentum p from the electromagnetic field. Another way to same the same thing is that the pair "absorbs" a photon from the Fourier component.

A time-dependent electromagnetic field A produces virtual pairs with non-zero energy

Let an electric field A produce a virtual pair. Let us treat the pair with a particle model

In an earlier blog post we noted that if A is time-independent, then the energy of its Fourier components is zero, and the electron will have some (e.g., positive) mass-energy m, and the positron a negative mass energy -m. An electric field accelerates the particles to the exact same direction, since the masses differ just by a sign. The particle with a negative mass-energy -m cannot gain any kinetic energy: its kinetic energy -m v^2 grows even more negative if it accelerates.

In a collision, the field A is time-dependent, and it has Fourier components where the energy E is positive. Then the combined energy of the virtual pair is non-zero. The electric field accelerates the particles in a different way. For example, if both particles have positive energy, then an electric field accelerates the electron and the positron to opposite directions. Both particles gain energy. They can become real particles.

A semiclassical model for virtual pair production

                         distance L

                 e-  ●                  ● e+

A classical model for a virtual pair is simply an electron and a positron which are static and close to each other. The distance L has to be > 1.4 * 10^-15 m for the system energy to be positive.

A strong external electric field can pull the electron and the positron apart. They become real particles.

We have now a semiclassical model for the process:

1. We determine the Fourier decomposition of the classical field A.

2. Each Fourier component of A can at each spacetime point (t, x) produce a virtual pair, and the pair inherits the energy E and the momentum p from the Fourier component. The probability of production at (t, x) depends on the coupling constant α as well as the strength of the Fourier component.

3. The particles in the pair move like classical point particles thereafter. But they do not observe the fields of each other until they have some distance between them? Maybe we can assume that they are born with enough energy to lift them to the distance 1.4 * 10^-15 m from each other?

If the particles cannot gain enough energy, they will collide and annihilate, producing a photon whose energy is E and momentum p.

4. How should we modify A when it lost (E, p) to the new pair? In a Feynman diagram, the new pair gets its (E, p) from a virtual photon sent by one of the colliding particles. Obviously, we must subtract (E, p) from an incoming particle.

5. How do we return (E, p) back to A if the particles annihilate? If the pair did not interact with other particles, then we probably must return the (E, p) like nothing would have happened.

Item 4 above suggests that the virtual pair is produced by a collision of an incoming particle to one or both of the components of a classical virtual pair which originally has zero energy and momentum. The Fourier component of A in this case is a Lorentz-transformed component of the Coulomb electric field of the incoming particle. The component is a longitudinal wave.

The model is probabilistic: we get probabilities for the new electron and the new positron departing with certain momenta. In that sense, the model is not classical - it is a quantum model.

Classical electromagnetism is founded on:

1. point particles, and

2. a field, the electromagnetic field.

We see that it is impossible to make a deterministic classical model for pair production. Pairs would be produced by the field, which makes the end result necessarily probabilistic. Classical electromagnetism is a mixture of quanta (electrons) and a wave (the electromagnetic field).

How can we reconcile different length scales?

The length scale of a 1 MeV collision is just 3 * 10^-15 m, while the Compton wavelength is much larger, 2 * 10^-12 m. How such a tiny length scale can produce Dirac waves whose wavelength is almost 1000-fold?

It is probably because we do not know the position of the collision better than the Compton wavelength. The Dirac waves are the combined effect of many different paths of particles. Individual paths have features whose size is only 3 * 10^-15 m. When we sum the effect of all paths, the resulting wave is of the scale 2 * 10^-12 m.

A similar thing happens with the Schrödinger equation of two electrons. The Coulomb field has a feature size 3 * 10^-15 m or smaller. The de Broglie wavelength of the electrons is larger than 2 * 10^-12 m.

Are the Feynman virtual photon waves longitudinal?

Richard Feynman uses as the photon propagator a Green's function of the massless Klein-Gordon equation.

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

An electromagnetic wave can be described by the electric field E. A real photon is a transverse wave where, for example, the electric field alternately points up and down relative to the direction to which the wave moves.

If a virtual photon carries just spatial momentum, no energy, then it does not look natural to draw it as a transverse wave. Since it pushes or pulls on another charge, a longitudinal wave looks more natural.

     Z+ ●

               ^

               |

               |

               •  e-

           

When the electron in the diagram passes the nucleus, the electron first feels a weak pull to the left, then a strong pull, and then again a weak pull. It is like the electron would have absorbed a half of a longitudinal wave.

Let the electron pass at the distance of the classical electron radius 3 * 10^-15 m.

The transferred momentum is of the order the electron mass m divided by c. The associated wavelength of the transferred momentum is the Compton wavelength 2 * 10^-12 m of the electron.

The Compton wavelength is ~ 1000X larger than the physical dimensions of the encounter. The Compton wavelength does not describe the actual, physical, length of the strong pull.

Anyway, it may be more intuitive to imagine virtual photons as longitudinal waves. That conveys the pushing or pulling aspect of such a photon.

Real photons are described as transverse waves.

Electric lines of force are tense: a new field line model of the electric field

                ________

              /               \

         +  ●-------------●  -

              \_________/

Let us assume that the energy density of an electric field is

        ε_0 / 2 * E^2.

Let us visualize the field with the lines of force. The density of lines is proportional to the strength |E| of the electric field.

We get two rules of thumb:

1. Electric field lines repel each other with a "potential" proportional to 1 / r^2, where r is the distance between the lines.

2. There is "tension" in a field line. The line wants to become shorter. The tension is proportional to E^2.

    z

    ^               E

    |    |     |     |     |

    |    |     |     |     |

    --------------------------|--->

    0                            x_0

Rule 1 follows from the energy density ~ E^2. Suppose that we have a uniform electric field E to the direction of the z axis between 0 and x_0 in the diagram above.

If we squeeze the entire field between 0 and x_0 / 2, then the electric field strength doubles, as well as the density of field lines. The energy density of the field grows 4-fold. The volume containing the field is only half after the squeezing. Thus, the total energy of the entire field doubled.

The density of field lines grows by a factor of 2 if we reduce their distance by a factor of 1 / sqrt(2). The "potential energy" stored in the repulsion of the field lines is thus proportional to 1 / r^2, where r is the distance between the lines.

Rule 2 follows immediately from the field energy density, which is ~ E^2.

Note that the tension in a field line is not constant along the line. It depends on E^2 at the location.

Let us look at the diagram at the top of this page and analyze the attractive force. The repulsion of field lines prevents them from all gathering between the charges.

If we move the charges closer, the field lines become shorter. That is the origin of the attractive force. The tension in the straight field line between the charges is much larger that the curved ones, because the field strength E is greatest between the charges.

https://phys.libretexts.org/Bookshelves/Relativity/Book%3A_Special_Relativity_(Crowell)/10%3A_Electromagnetism/10.06%3A__Stress-energy_tensor_of_the_electromagnetic_field

In the link, Benjamin Crowell derives the electromagnetic field tensor. There is negative pressure to the direction of a field line, but positive pressure to the normal directions. Our model casts light on this fact.

What are real photons in the field line model?

    ^      charge was suddenly moved up

    |               

    ● ------------

                     \              electric field line

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

                 wrinkle

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

Recall how Edward M. Purcell calculates the energy of electromagnetic radiation. Look at the diagram at the top of the linked page (by Daniel V. Schroeder, 1999).

A real photon, or an electromagnetic wave, is a "wrinkle" in field lines. The wrinkle moves at the speed of light. A bent electric field line requires a time-dependent magnetic field to be present, to bend the field line. Electric field lines are denser at the wrinkle, which implies that some extra energy E^2 is concentrated there.

The energy density of the symmetric static Coulomb field is proportional to 1 / L^4, where L is the distance. The energy density in the wrinkle is much greater than in the surrounding field. This explains how an electromagnetic wave is able to carry energy to across great distances.

A wrinkle is always transverse (the oscillation is normal to the direction of the wave movement). There exist no longitudinal electromagnetic waves.

What is a virtual (off-shell) photon, which carries only spatial momentum, in the field line model?

Suppose that we have two charges attached  to a frame so that the charges are static. The Coulomb force between them is tension in the field lines.

Thus, a virtual photon which contains just spatial momentum, no energy, is longitudinal tension of field lines.

                                        ^

      t                               /    (new frame v)

       ^

       |      |    |    |    |

       |      |    |    |    |    crests of the

       |      |    |    |    |    Fourier component

        ------------------------------> x

In a Feynman diagram, such a photon is a Fourier component of the static Coulomb field. Why is the photon a sine wave electric potential in a Feynman diagram, but the photon is longitudinal tension in the field line model? These descriptions look very different graphically. What is the connection?

What happens if we change to a different inertial frame?

If we do a Lorentz transformation to a different frame which moves up right at the speed v, in the diagram above, then the wave potential in the moving frame no longer is of the form sin(p x). It will be something like

        sin(E t - p x).

In the new frame, our virtual photon does carry some energy E, as well as spatial momentum p.

In the field line model, the a moving observer sees the field line diagram Lorentz-contracted in the direction of v. Field lines look distorted in a parallelogram fashion to a moving observer.

The moving observer sees a time-dependent magnetic field along with the electric field. He may interpret that the magnetic field has distorted the electric field lines.

How should we interpret the energy that seems to be flowing in the Feynman style diagram? If a charge pushes another, and the other charge moves to the direction of the force, then the first charge is doing work. This is the energy which flows in the Feynman diagram.

A static system of charges is easiest to study in the static frame. There is a preferred frame. Switching to a moving frame just complicates things.

How are longitudinal forces of field lines (virtual photons) converted to electromagnetic waves (real photons)?

Our rubber plate model tells us that by waving the plate up and down, we can produce transverse waves (real photons).

When we wave the rubber plate, the system is described as longitudinal forces as well as deformation of the rubber plate. How do these become converted to beautiful sine waves which escape to infinity?

It is obviously a complicated process which should be calculated in a computer simulation, or tested with a real rubber plate.

The same probably holds for the field line model. When we wave a charge, the longitudinal tension in the field lines feeds energy to form beautiful sine waves, but the process is quite complicated.

In a Feynman diagram, the production of real photons is described in a miraculously simple way. We just need to make sure that spatial momentum and energy are conserved, and we get a correct numerical prediction!

So, what is the connection between virtual photons and real photons in a Feynman diagram? These photons are quite different things. A Lorentz transformation always keeps virtual photons virtual and real photons real. There is a complicated process which can produce real photons from virtual photons.

         Z+  -------------------------

                           | virtual photon

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

                    \

                      ~~~~~~~~~~~~~

                         real photon

However, in a Feynman diagram, we draw the charge directly producing the outgoing real photon. The diagram above depicts bremsstrahlung.

What is the role of creation and annihilation operators of canonical quantization? They are supposed to create and destroy real photons.

Not much. In our example, it is fuzzy where a real photon is actually "created". The converse process would be annihilation.

The use of creation and annihilation operators confuses us, if we want to model a collision in a detailed way. These operators are only suitable for describing the measured end result of the whole process: real photons were created somewhere in the process, and observed.

What are the virtual photons in the field line model, in a dynamical encounter of two charges?

         Z+ ●                  

                             ^

                             |

                             |

                             e-

Suppose that an electron e- flies past a nucleus Z+.

The electron will first accelerate and then decelerate. Its electric field lines will "lag" in the acceleration phase and slow it down, but will "pull on the electron" to speed it up in the deceleration phase.

We may interpret this that the magnetic field of the moving electron first resists the acceleration of the electron, and then resists the deceleration.

In the approach, the electron moves spatial momentum p to its field, and gets some of p back in the receding phase. The electron moves considerable momentum to itself, via its own the field. This is the vertex correction of Feynman diagrams.

In the field line model, in the dynamic case, bending, stretching, and tension of field lines is "virtual photons". There can be large momentum p stored in these deformations of the field lines. The deformations do not obey the energy-momentum relation

       E / c = p

of real photons.

Conclusions

"Virtual photons" exist as a phenomenon of the classical electromagnetic field. There is (probably) no quantum aspect at all in them. Quanta are only observed when we measure real photons after a collision experiment.

A photon with momentum q, which is registered in the measurement, means that the classical field has "collapsed" to indicate one Fourier component of it: the component which describes an outgoing photon with the momentum q.

In the Copenhagen interpretation we speak about a collapse of the "wave function". With electromagnetic fields and macroscopic charges, we can take the wave function to be the classical electromagnetic wave. What about microscopic charges?

Does vacuum polarization exist for a static electric field?

                                virtual pair

                                        __

                                      /    \

                               e-    \__/   e+

           ● Z+ nucleus

http://w3.lnf.infn.it/the-variable-constant/?lang=en

Naive pictures on various Internet pages suggest that electron-positron pairs spontaneously pop up from empty space. If there is an electric field present, it pushes the electron and the positron to opposite directions until they annihilate. This is supposed to cause vacuum polarization, which screens part of the electric field.

We have never seen a detailed explanation how exactly this would happen.

Feynman diagrams contain vacuum polarization loops. If the incoming particles have very low kinetic energy, then we may view electric fields as almost static.

             virtual    λ / 2 phase

             photon      shift

        Z+ ~~~~~~~~~~O~~~~~~~~~~ e-

                 vacuum polarization

                                loop

In a Feynman diagram, a vacuum polarization loop is quite different from the one in the naive diagram at the top of this blog post. Feynman thought that the loop reduces the effect of the electric field, because the virtual photon is scattered (is reflected) from the loop and suffers a half a wavelength λ / 2 phase shift relative to a photon which flew directly.

The λ / 2 phase shift causes destructive interference to the loopless diagram, reducing the probability amplitude of significant scattering.

In our previous blog posts we have argued that reflection from a null-energy object, like from the vacuum polarization loop, cannot change the phase of a photon. That looked quite clear if the photon carries energy. But what about photons which carry just momentum, no energy?

A photon which carries just momentum does not proceed in the time dimension, just in the spatial dimensions. It "moves" infinitely fast.

If the electron and the positron in the loop both have zero energy, they "move" just in space, not in time. But then we have problems understanding how they end up at different distances from the nucleus Z+, and can cause polarization.

A particle model

Let us then treat the other case. We assume that the electron in the loop has positive energy E, and the positron negative energy -E.

In the naive diagram, the nucleus pulls the electron. Let it receive a momentum p toward the nucleus. The positron receives a momentum -p.

The electron moves closer to the nucleus. How does the positron move? It moves closer to the nucleus, too! Since the mass-energy of the positron is negative, it moves to the opposite direction from what one would naively expect.

                               <--- e-

                               <--- e+

         Z+ nucleus

The movement has to be this way. Otherwise, the center of mass of the system would move.

The annihilation is easy to understand: the particles are at the same position. They do not need to jump anywhere to annihilate.

Now it is obvious that the virtual pair cannot cause any polarization, because the particles are at the same position.

Thus, if we have a static setup, then virtual pairs cannot gain energy from anywhere, and there cannot be any vacuum polarization.

Note that in this analysis we apply the "Schrödinger method" to assess the effect of a pair: we take into account the electric field of both the electron and the positron. The net effect is zero.

Feynman thought that only one particle in the pair needs to interact with other particles. We coined the word "minimal quanta method" to describe the assumption that the other particle can be ignored.

Thomson scattering with a classical model

https://en.wikipedia.org/wiki/Klein–Nishina_formula

For Thomson scattering, there exists a very good classical analogue. The electric field E of the incoming wave tries to make the electron to oscillate in space. But since the electron has inertia, it lags behind and emits an electromagnetic wave which has a half a wavelength phase shift relative to the incoming wave.

                        weight

      -------------------●------------------

        tense string

It is just as with a tense string where an extra weight is attached. The weight will act as a source for the wave equation, with a 180 degree phase shift.

Suppose then that we attach two weights to the string. One has mass m, and the other has a negative mass -m.

If the weights are at the same position, then their effect obviously will be zero.

                     E     p

                     -e ---->

           < ---- +e

             -p    -E

Suppose then that we have a virtual pair e- e+, whose total energy is zero. We may assume that the electron originally has the energy E and the spatial momentum p. The positron has (-E, -p).

The electron moves to the direction of p. Since the positron has negative energy, it moves to a surprising direction - it moves to the same direction as the electron! They stick together.

If there is an electric field present, it will give the electron some extra momentum q, and the positron the opposite momentum -q. But again, the particles will move to the same direction. They stick together.

The electron and the positron act as sources of electromagnetic waves as they oscillate. Since the particles are at the same position, their waves totally cancel each other. There is no Thomson scattering.

It is not a surprise: how could a real wave bounce back from a zero-mass object, the virtual pair? Zero mass means that the pair is essentially "nothing". One cannot bounce back from nothing.

This means that vacuum polarization loops cannot occur for a free photon. Vacuum polarization has no effect at all for a photon which flies freely.

What about virtual photons for which the momentum is off-shell? The classical analogue of a virtual photon is a malformed electromagnetic wave, whose E and p do not match. It is hard to see how the system of an electron and a positron at the same position could have any effect on the malformed wave.

Thomson scattering with a malformed electromagnetic wave

                  ^

                  |    rubber plate

     -----------●-----------------•-------

             charge              weight m

Our rubber plate model can be used to illustrate a malformed wave. Let us move upward the charge in the diagram.

The rubber stretches, but the stretching is not in the right form to leave as a sine wave (= real photons). The stretching is a malformed wave.

The weight m in the diagram has the role of a positive energy electron. But at the very same position, there is a weight with a negative mass -m. Together, these weights have a zero effect.

A static electric field in the rubber model is modeled as charges pushing or pulling horizontally on the (very stiff) rubber plate. What is the effect of an extra weight attached to the plate then? Zero. This suggests that virtual pairs have no effect whatsoever on static electric fields.

Collisions with energy > 0

If a static electric field induces no vacuum polarization, what happens in a dynamic setup where particles collide?

Collisions do produce real pairs of electrons and positrons. It is very likely that collisions cause vacuum polarization, too. If the collision "almost succceeds" in creating a real pair, the pair probably exists for a while as virtual, and will cause vacuum polarization.

We need to study the collision process with the rubber plate model as well as with a particle model.

How to interpret the Afshar experiment? Is complementarity violated?

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

The optical experiment by Shahriar Afshar was a cover story in New Scientist in 2004.

light --->                

                           photon detector B

                  |              /     /

                  |            /     /

hole A

                  |         •  grid

                  |         •          interference of the

                  |         •          two waves

hole B          

                  |             \     \

                  |               \     \

                            photon detector A

Plane waves of light enter from the left. They are diffracted by the two holes. They form an interference pattern in the spatial volume right of the holes. We put obstacles to the dark spots of the interference pattern, the "grid" in the diagram. The grid might block, say, 20% of the routes (without the interference pattern).

We have two photon detectors at the bottom of two tubes. From the geometry it is obvious that the detector A will detect almost exclusive the wave which went through the hole A, and the same for B.

Diffraction at the grid will scatter some photons to the wrong detector, but not many, because the grid is located at the dark spots.

Afshar's setup contains a lens which focuses the light from the holes. The lens can, in principle, collect 100% of the light that goes through each hole. The lens offers a better "distinguishability" D of the paths, as defined by B.-G. Englert. If a photon does not make it from the hole A to the detector A, it is not "distinguished".

The standard way to analyze the experiment is to calculate the diffraction of the waves at the holes, and the interference of the classical plane waves. We get the intensities at each detector A and B.

Suppose that the intensity of light is so low that we send only one photon per second.

When we measure a photon at a detector, we may interpret that the wave which described the process "collapses" to a single photon at the detector.

Suppose that the detector A measured a photon. Is there sense in saying that the photon passed through the hole A?

The wave at the detector A is able to avoid the grid because it interferes with the wave that went through the hole B. Without the help of the B wave, it might bump into the grid.

The interference of the two waves A and B near the grid is a linear phenomenon. In a linear system we can sensibly talk about the origin of some wave.

In that sense, we may say that the wave at the detector A comes through the hole A. But the wave gets "help" from the B wave in avoiding the grid.

In quantum mechanics, we are really not allowed to talk about the "paths" of photons before the measurement. Everything has to the treated as a wave. In the strict sense, the photon at the detector A did not have any path.

But informally we can say that the wave at the detector A really comes through the hole A.

The experiment is in harmony with all the interpretations of quantum mechanics.

The interpretation by various authors

Shahriar Afshar claimed that the experiment clashes with "complementarity". We do not think so. The particle nature of light is only present when the photon is measured. Everything else is a wave.

William Unruh (2004) claims that a setup with half-silvered mirrors is equivalent to Afshar's experiment, and one cannot deduce which path the photon took:

http://www.theory.physics.ubc.ca/rebel.html

The setup of Unruh is not very much equivalent. If we calculate a plane wave backward in time in Unruh's diagram from the detector 5, the backward wave will be divided evenly between the paths 1 and 2.

In our diagram, calculating back from the detector A, the wave will almost exclusively go through the hole A.

Lubos Motl calculates the effect of the wire thickness in the grid.

If the wires are thick, there is a big difference in the wave intensity at the detector A when 1) both holes are open, and 2) only one hole is open. But thick wires scatter some light to the "wrong" detector.

The distinguishability value D defined by B.-G. Englert is bad if the wires are thick: for the reflected and absorbed photons we do not know the path.

Motl correctly states that the experiment is a classical one and can be calculated using Maxwell's equations. There is nothing in the experiment which clashes with complementarity.

Extracting energy from a static electric field with capacitors

Suppose that we have a pointlike charge static in space.

                         _____________

                  ___|___

                  _______    capacitor

                       |_____________

          

                       ● negative charge

We can extract energy from the static field of the charge with a capacitor. In the diagram, the charge will repel electrons to the upper plate of the capacitor. If we put a resistor to the wire connecting the plates, we get energy in the form of heat.

Does this show that energy is contained locally in the electric field of the charge?

The charge might be far away and would not even know that we connected the capacitor plates and started extracting energy.

If we explain the process with Coulomb forces, then the charge will not know about the extraction after some time, because of the finite speed of light. This is called retardation.

Let us make a thought experiment. We put so many capacitors so close to the charge that we can extract more energy than is the mass-energy of the charge itself. That would break conservation of energy.

We do not believe that conservation can be broken. The capacitors have to "negotiate" with the electron if there is still mass-energy available which they could take.

If there is no more mass-energy available, Nature solves the problematic situation with annihilation.

We see that the huge energy density close to the charge cannot be utilized without asking the charge first - in this sense we may claim that the huge energy really is "located" inside the pointlike charge.

Transactional interpretation of quantum mechanics

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

In a Feynman diagram, the time order of close encounters, emissions, and absorptions is insignificant. As if Nature would perform transactions. Inside a single transaction, the time order of various operations does not matter. The crucial thing is that either all operations succeed, or none at all.

The Wheeler-Feynman absorber theory tries to extend the transactional paradigm to cover an emission of a photon today to the absorption of the same photon a trillion years from now (in a Big Crunch, for example). That is counter-intuitive, and the absorber theory has gained little acceptance.

The conjecture that the energy of a static electric field is zero solves the infamous 4/3 problem of classical electrodynamics

This Physics Stack Exchange post (2013) describes the 4/3 problem:

https://physics.stackexchange.com/questions/80856/does-the-frac43-problem-of-classical-electromagnetism-remain-in-quantum-m

Richard Feynman in his Lectures on Physics (1964) defines the energy density of the electromagnetic field as

      u = ε_0 / 2  E^2  + ε_0 c^2 / 2   B^2.

https://www.feynmanlectures.caltech.edu/II_27.html

One can derive from the definition of u that the Poynting vector of energy flow is:

       S = ε_0 c^2 E × B.

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

The energy flow agrees with what we empirically observe in electromagnetic radiation.

                    |  -

       ----------------------

         |    |    |    |      E

       ----------------------

                    |  +

                          ---->

                          v

However, for a static electric field, S gives nonsensical results. Imagine a capacitor where the electric field lines are as indicated.

If we move the capacitor to the right at a speed v, the magnetic field vector B stands up from the diagram toward the reader of this blog. The Poynting vector S correctly says that there is energy flow to the right in the space between the capacitors. 

Near the capacitor plates or the edges, the electric field is more complicated and we do not claim anything about the energy flow there.

But if we move the capacitor upward, then E and v are parallel, B is zero, and S is zero. The Poynting vector claims that there is no energy flow in the space between the capacitors! The energy flow is different for an upward movement than a sideways movement. That defies our intuition about how mass-energy moves in space.

The energy flow happens in a complex way close to the plates and at the edges of the capacitor, and probably that flow transfers the energy up when we move the capacitor upward.

The 4/3 problem concerns the electric field of a charged sphere when it is moved. Also in that case, we get a nonsensical result. The Poynting vector gives a larger momentum for the field than what we get from the expression of u above.

A static field has zero energy

The problems disappear if we define the energy density of a static electric field as zero. The energy density E and B of electromagnetic radiation we define in the old way.

Do we always know what field is static and what field is radiation? If a charge is accelerated, then part of the field is static and part is radiation.

The "sharp hammer" model, which we introduced a couple of days ago, helps us to distinguish what is the static field component and what is radiation. The point charge repeatedly applies the Green's function to create the electric field. If the charge moves at a constant velocity, then there is a total destructive interference of energy-carrying photons. But if the charge accelerates, then the destructive interference is not complete. There is energy in the field.

Thus, if a charge moves at a constant velocity then its field has zero energy. If it accelerates, then part of its field does contain energy.

Electromagnetic waves have no static field. Their whole field contains energy.

Our rubber plate model of the electric field helps to understand all this. If the charge is moving at a constant velocity, then there is no stretching in the plate and zero energy. If the charge accelerates, then the charge makes a "hill" in the rubber plate. There is now energy in the field because stretching requires energy.

If the charge moves back and forth, it creates waves in the rubber plate. Those waves carry energy far away - they are electromagnetic waves.

The choice of u and S is not unique. Does there exist a choice of u which would be non-zero for a static electric field, but for which S would give sensible results?

Divergence in the vertex correction in QED

NOTE December 22, 2020: 

https://www.mpi-hd.mpg.de/personalhomes/harman/HCI-WS2013-2014/Lecture-12.pdf

Zoltan Harman (2014) writes that the ultraviolet divergence is only in the case where the momentum q of the incoming photon is zero.

There is no ultraviolet divergence if q != 0.

The case q = 0, of course, can be reduced to the electron self-energy case, which we handled by banning self-energy loops altogether (the center of mass argument).

There is an infrared divergence. That is handled by setting a little mass m to the photon carrying the momentum k.

-----

In the past two weeks we have argued that the Feynman integral in QED for vacuum polarization converges (but not absolutely), and the electron self-energy loop can be removed altogether.

The third divergence in QED is in the vertex correction:

                     photon

                     4-momentum k

                     ~~~~~

                   /               \

          e- ------------------------

                        /

             ~~~~

          photon

          4-momentum q

The divergence is logarithmic.

For example, the electron may scatter from the field of a nucleus Z+. Then the incoming photon carries spatial momentum q.

Our rubber plate model of the electric field of the electron says that the plate will bend when the electron accelerates toward the nucleus. The plate resists the acceleration.

When the electron recedes from the nucleus, the rubber plate tries to keep it going. The electron emitted a photon which contains spatial momentum k, to the rubber plate, and then absorbed the same photon.

Thus, in the classical world, the electron does send an off-shell photon to itself in the encounter with the nucleus.

If the electron exchanges momentum q with the nucleus, we can calculate the rough geometry of the encounter. 

We have the principle that virtual photons only live for about 0.1 times their (Compton) wavelength. From the geometry of the rendezvous we can then set a cutoff for possible momenta k. The divergence is removed.

Alternatively, we may use destructive interference. If the approaching electron would send a short-wavelenth off-shell photon (momentum k) to itself, there would be lots of destructive interference at the point where it absorbs the photon. We would get an extra coefficient 1 / |k| to the Feynman integrand, which would remove the divergence.

The famous anomalous magnetic moment of the electron is calculated using many diagrams with the vertex correction. We need to check if our cutoff procedure would somehow alter the calculation. The calculation agrees with the experimental value at a precision 10^-10. We do not want to change that.

Note that in the vertex correction, we could have very heavy particles with large charges doing the rendezvous. The calculation has to agree with the classical limit. We need to verify that is the case.

In classical electrodynamics, the magnetic field resists changes in the current. A single electron can be regarded as a current. The rubber plate in this case seems to do the work of the magnetic field: it resists changes in the velocity vector of the electron.

Our rubber plate model tries to solve the long-standing problem of the interaction of a charge with its own field. The vertex correction is a prime example of such self-interaction.

A running coupling constant breaks the classical limit

There is nothing which prevents the particles in a Feynman diagram from being extremely heavy, making them essentially classical particles. We could have 1 kg particles doing Coulomb scattering.

https://canvas.harvard.edu/files/936398/download%3Fdownload_frd%3D1%26verifier%3DPAHa18X0trEAYjwcsmQXrhusjECF3ct6JNxDQ41E

In formula (63), Matthew Schwartz (2012) presents the vacuum polarization effect on a spatial momentum transfer. The effective coupling constant becomes smaller when -p^2 is increased:

       α_eff = 1 / 137 (1 +  0.00077 ln (-p^2 / m^2),

where m is a fixed energy, for example, m might be 511 keV.

Let us then imagine that we have very heavy particles with very large electric charges. They are the classical limit of Coulomb scattering. We can monitor the paths of the particles very accurately.

Suppose that we then double the mass and the electric charge of our particles. The formula above claims that the coupling constant changes, because the momentum transfer p is double. The change is not small, about 0.1 %.

That does not make sense, since we are working at the classical limit. The Coulomb force should have the same formula for all classical objects.

What is wrong?

Our new way of handling vacuum polarization restores the classical limit.

Recall that a virtual pair typically lives just for 0.1 Compton wavelengths. Here, we define the generalized Compton wavelength for a general 4-momentum as

        λ = h c / |k|,

where k is a 4-momentum (E, p), and we define

      |k| = sqrt(E^2 + p^2).

The norm || is the euclidean length of the 4-momentum vector. It is not the Minkowski metric length sqrt(-E^2 + p^2).

Let the heavy particles pass each other at a distance L. We add to the usual Feynman vacuum polarization integral an extra coefficient

         C(|k|) = 0.1 λ / L = 0.1 h c / (|k| L),

if C(|k|) < 1. If it is bigger than 1, we do not add the coefficient.

That is, we only take into account those pairs which are born at a distance < 0.1 λ from the passing particle.

The coefficient C(|k|) makes the Feynman integral to converge. Or does it? We have to find online a brute force calculation of the Feynman integral with a cutoff |k| < Λ and check that it only diverges logarithmically on Λ.

https://cds.cern.ch/record/893077/files/1-94%2520Alvarez-Gaume.pdf

L. Alvarez-Gaume and M. A. Vazquez-Mozo (2010) have done the calculation (though they did not include it in their paper).

Their formula (372) states:

        Π(q^2) = e^2 / (12 π^2)  log(q^2 / Λ^2)

                         + finite terms.

There, the integral has been taken over all |k| < Λ. The integral diverges only logarithmically when we increase the (large) cutoff  Λ.

Our corrective term C(|k|) makes the integral to converge, because it has 1 / |k| in it. But we are also interested in a numerical value.

Suppose that heavy particles pass each other at the distance of 1 meter and exchange momentum q. How does our corrective term C(|k|) affect the value of the integral?

The value of |q| is macroscopic. Its Compton wavelength is microscopic, of the order 10^-34 m. If |k| is of the order |q|, then our corrective term is roughly 10^-34.

In the SI unit system, the 4-momentum of the electron at rest is ~ 10^-21. Our corrective term for it would be ~ 10^-13.

The integrand in formula (365) of the paper for |k| << 10^-21 is roughly

        1 / |q|^2,

which is integrated over a 4-volume 10^-84. The value is negligible.

We conclude that the integral is essentially zero when we correct the integrand with our term C(|k|).

The effect of vacuum polarization is essentially zero when macroscopic charges pass each other at a distance 1 meter.

We have restored the classical limit. Heavy particles behave as in the simplest Feynman diagram for Coulomb scattering. Vacuum polarization, which would depend on q, has essentially a zero effect.

Apparently, prior to this blog entry, no one has noticed the classical limit problem in Coulomb scattering vacuum polarization. Breaking the classical limit is a serious error in quantum mechanics.

The rubber plate model of the electromagnetic field explains momentum conservation

         ^

         |

         |

        ●                     ●

        +                      -

Assume that we have opposite charges initially static, like in the diagram. They pull on each other.

Let us then use a rocket to accelerate the positive charge upward very fast, so that it attains almost the speed of light.

The negative charge on the right still sees the positive charge at its original location for a while. The negative charge gains a momentum p to the left during that time.

We assume retardation. Signals do not travel faster than light. The negative charge does not know that the positive charge has moved.

Meanwhile, the positive charge gains a momentum q right, but |q| < |p| because the positive charge rapidly moves away.

Where did the difference of those momenta go? Could it be that the negative charge somehow emitted photons to the right, and those photons took the difference? That looks very implausible. The accelerating positive charge does emit energy, but that happens symmetrically and cannot take the extra momentum.

How can we ensure that momentum is conserved?

The rubber plate model comes to the rescue. The negative charge on the right pulls on the rubber plate of the positive charge. The negative charge gains a momentum p left, and the rubber plate gains a momentum p right.

The rubber plate delivers the momentum p later to the positive charge.

How can a massless rubber plate store momentum? Where is its inertia? Again, it is the finite speed of light which mimics inertia. When the negative charge pulls, the elastic  rubber plate stretches a little bit. The longitudinal stretch wave travels at the speed of light to the positive charge and tugs on it.

We need to check if anyone has proved conservation of momentum under retarded forces. Gravity and the Coulomb force should honor conservation.

https://www-liphy.univ-grenoble-alpes.fr/pagesperso/bahram/Relativite/Biblio/wheeler_feynman_1949.pdf

The Wheeler-Feynman (1949) absorber theory tries to tackle this issue.

http://cds.cern.ch/record/401893/files/9909087.pdf

S. Carlip (1999) writes that in General Relativity, a binary system conserves angular momentum (ignoring gravitational radiation). Retardation of forces would violate angular momentum conservation, but the effect is canceled by other factors.

Laplace in 1805 calculated that the "speed of newtonian gravity" has to be at least 7 million times the speed of light, so that retardation effects do not make the orbit of the Moon unstable.

Apparently, momentum conservation in classical electromagnetism is an open problem.

The energy of the static electric field of the electron is zero, not infinite?

Our "sharp hammer" hypothesis from last week suggests that the energy of a static Coulomb field is zero! The field consists exclusively of virtual photons which contain spatial momentum but no energy. Why would the field then contain any energy of its own at all?

The sharp hammer hypothesis is that the static Coulomb electric field consists of the Green's function of the electromagnetic field applied at the location of the point charge at infinitesimal time intervals.

                  | knock knock knock

                  |

                  v

               #      

               #=======   sharp hammer

               v

  _______________ drum skin

               ●

              "point charge"

The classical analogue is that we hammer a drum skin at the same location at very short time intervals. A permanent, time-independent, depression is created in the drum skin. It is like a static Coulomb field.

In the classical case, there is energy stored in the static deformation of drum skin.

Classically, in newtonian mechanics, we can explain the movements of charges without assuming any energy stored in the electric field. It is just the particles that exert forces on each other.

Maybe the right quantum field model for the static electric field of an electron is that it is a pure spatial momentum field that does not contain any energy at all?

Energy is only present in Fourier components where the 4-momentum k contains energy. Since for a static field, those components have a total destructive interference, there is no energy in the static electric field.

This would solve the paradox which has been present since the discovery of the electron at the start of the 20th century: how can a point charge have a finite mass even though the energy of its static electric field is infinite?

An early attempt to "renormalize" the electron mass

A simple attempt to resolve the paradox was to assume that the electric field is cut off at some very short distance dr from the electron, and the mass of the "bare" electron has a large negative value -M. When we add -M to the very large energy E of the electric field, we get:

       E - M = 511 keV,

that is, the well-known electron mass visible to the outside world.

The solution to the paradox is often viewed as an early example of renormalization, where we assume that the "bare" mass (or some other property of a particle) has a wildly different value from the "dressed" mass visible to the outside world.

The notion of the electron having a negative bare mass is strange. Negative masses behave in a strange way in classical physics. 

Suppose that the renormalization scheme is correct. Then we could collide the electron with another at a very high energy. The electric field of the electron cannot react to the collision instantaneously because the speed of light is finite. The collision might reveal that the bare mass of the electron is -M, a negative value. That would be a strange observation.

The self-energy diagram of the electron

                            photon which the electron

                            sends to itself,

                            4-momentum k

                                ~~~~~

                             /                \

 electron e-   -------------------------

 4-momentum p

                            A                B

The "self-energy" diagram of the electron is depicted above. The Feynman integral for the diagram contains the product of:

1.  the electron propagator (4-momentum p - k) from A to B and

2. the photon propagator (4-momentum k) from A to B.

The integral diverges when we integrate over all 4-momenta k.

The standard way to solve the divergence problem is to put a cutoff to k and assume some "bare mass" -M for the electron, such that the electron mass visible to outside world becomes m = 511 keV.

However, our considerations about the conservation of the speed of the center of mass suggest that the diagram above is forbidden, or that the photon must be absorbed in zero time, and the the effect of the diagram is then actually null.

               

              real photon, energy E

               ~~~~~~~~>

             /

       e- ● 

        <-

Let us analyze the diagram using the particle model. We may assume that the electron is initially static.

Suppose that the pointlike electron emits a real photon to the right, but after some finite time interval dt, reabsorbs it. During the time interval dt, the electron moves left.

The real photon carries some energy E to the right at the speed of light and is partially responsible for keeping the center of mass of the system static.

But then a miracle happens: the electron magically captures back the energy E carried by the photon. The electron is again static, but it has moved a little bit to the left. The center of mass has moved.

Could it be that at the time of absorption, the electron somehow jumps to the right to its original position? It might be, but in the momentum phase wave description of the process there is no such sudden jump in anything. When an electron absorbs a photon in the wave model, the electron does not suddenly jump to another place. How could the electron even know where it has to jump when it absorbs a photon?

What is going on here? If we think of classical particles, they only absorb other particles which are at their current location. They do not magically absorb particles which are far away. Working in momentum space in Feynman diagrams obfuscates this simple fact and people start to think that, in the plane wave representation of the momentum space, particles can actually absorb other particles which are located anywhere in the experiment area.

The misunderstanding comes from the fact that if we have independent particles at random locations of the experiment area,  then they can meet at any location and one can absorb the other.

Is newtonian mechanics right about static forces between charges?

In newtonian mechanics, we can get rid of the energy of the electric field by assuming that signals travel infinitely fast. Then the positions of the charges explain everything, and there is no need to assume the existence of an electric field at all, let alone a field which would contain energy.

When we calculate a Feynman diagram where a photon only moves spatial momentum, no energy, then we do assume that the force acts instantaneously. We can drop the time dimension from the model. We are calculating using newtonian mechanics.

What about dynamically changing electromagnetic fields? They do contain energy in on-shell photons, or in virtual photons where the 4-momentum has the energy component non-zero.

An improved rubber plate model of the electromagnetic field

Two years ago, we were able to explain the flow of spatial momentum and energy in a radio transmitter by introducing the model where the electric field of a charge is an elastic "rubber plate".

                             ^

                             |  rubber plate

        ------------------●------------------

                             |  charge

                             v  waved up and down

When an external force waves an electric charge, the rubber plate attached to the charge oscillates and waves in the plate propagate towards infinity.  The waves carry energy but little spatial momentum. The charge exchanges quite a lot of spatial momentum with the rubber plate, as the charge is waved. But the net spatial momentum flow from the charge is zero.

The large gross momentum flow between the plate and the charge explains how the kinetic energy of the charge can be transformed into on-shell photons which carry very little spatial momentum. The charge loses a lot of momentum when it loses its kinetic energy. Where does the extra momentum go? It cannot go to photons.

The answer is that the momentum is temporarily stored in the rubber plate and returned back to the charge when it swings back. The net flow of momentum is zero.

Two years ago we thought that the mass-energy of the static electric field of the charge temporarily absorbs the extra spatial momentum.

But our new hypothesis is that the mass-energy of a static Coulomb field is zero. What to do?

Suppose that the charge is initially static. Then we suddenly start to accelerate it up. Since the speed of light is finite, the plate  has to stretch a little to accommodate the sudden movement of the charge. The rims of the plate are not moving yet but the center is moving up.

We do not need to assume that the plate has any rest mass at all. The stretching will create a force which resists the movement of the charge. The person who is pushing the charge up must give spatial momentum p upward to the plate.

Thus, the finite speed of light can be used to mimic inertial mass, even though the plate is massless.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

The birth of electromagnetic waves, or real photons, is understood using Edward M. Purcell's diagram of the electric lines of force bending. See the top of the linked page for a picture how the field lines get wrinkled.

The wrinkle in the rubber plate moves outward. The wrinkle contains both an electric and a magnetic field. That is, the wrinkle is a dynamic electromagnetic field and contains energy.

In Purcell's diagram, the static electric fields can be regarded as containing zero energy. What matters is the energy in the wrinkle.

Scattering of colliding classical electromagnetic waves

A couple of weeks ago we tried to introduce a model of the electromagnetic field where a massless tense string is waved. The waves contain some mass-energy, and will cause inertia in the string. The problem in the model is that then energetic waves should propagate slower than low-energy waves because energetic waves contain a lot of mass-energy and inertia.

What about the rubber plate model? It looks plausible that waving a plate is harder if the plate contains a lot of energy in wrinkles. If there is little energy, then waves propagate precisely at the signal speed of special relativity, that is, c.

But if a wave contains a lot of energy, it should progress slower. Does this sound right?

If two energetic waves C and D collide, then there should probably be scattering of waves because inertia of the wave D disturbs the medium where the wave C propagates. This might be the root cause for the scattering of photons by photons? There would be scattering in classical electromagnetism, after all.

Electromagnetism is nonlinear in QED because energy can escape from the electromagnetic field into the Dirac field. Can we somehow harmonize this fact to the classical scattering of electromagnetic waves?

In Feynman diagrams, photon-photon scattering is modeled as a collision to a virtual pair.

  photon

       _____     _____

                \  /

                 O virtual pair loop

       _____/  \_____

              

   photon

In our classical model, scattering of colliding waves is calculated from the electromagnetic coupling constant (~ tension of a unit electric field line).

In Feynman diagrams, scattering is probably determined by the electron charge / mass ratio and the coupling constant.

If these figures happen to agree, then we have found a formula which determines the electron mass from its charge and the electromagnetic coupling constant.

https://arxiv.org/abs/1111.6126

Yi Liang and Andrzej Czarnecki (2011) write that for low-energy photons, the cross section is proportional to the sixth power of the photon energy. That can hardly happen in our classical model. Conclusion: the classical model cannot explain quantum effects like the electron mass and photon-photon scattering.

The electromagnetic force grows larger at short distances because of van der Waals effects?

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

The van der Waals force is believed to exist because transient dipoles in molecules attract each other. It is a polarization force and always attractive, while ordinary polarization is always repulsive.

Suppose that polarization of vacuum is superlinear on the electric field E. We have a good reason to believe that it is superlinear. Very strong electromagnetic forces in particle collisions "tear out" electrons and positrons and they become real particles. Virtual pairs are bound together by the Coulomb force, and that force grows weaker at a greater distance.

                  vacuum

                  polarization

       e- ●     +++   - - -       ● e+

                                      +    -

                                      vacuum

                                      polarization

Suppose that we have an electron and a positron close to each other. The electric field E is double between the charges, and produces some additional vacuum polarization between them because of superlinearity.

The additional dipole of the vacuum polarization strengthens the attraction between the charges.

We need to figure out what is the effect of superlinearity on ordinary, repulsive, vacuum polarization. Does it grow in strength when the charges come closer?

The electron mass is finite because of extreme vacuum polarization close to it?

The classical radius of the electron is 3 * 10^-15 m. The mass-energy of the electric field outside that radius is the electron mass, 511 keV.

Why is the 1 / r^2 electric field massless inside that radius? The reason might be that vacuum polarization almost totally cancels the electric field close to the electron.

There has to be an electric field E inside the classical radius, to maintain vacuum polarization, but the strength of the field E might be of the same order as at, say, the Coulomb force at two classical radii from the electron.

What about the energy which is required to stretch the vacuum dipoles close to the electron? That energy might be very small compared to the infinite energy that is required to create a 1 / r^2 electric field.

A Feynman diagram does not work if an electron passes a nucleus from just one side

This blog post returns to the problem of  "bending" an electron beam under an electric field.

Suppose that we have a static nucleus and a uniform Dirac wave passes symmetrically around it on all sides.

         ------>

      |    |    |    |

      |    |    |    |     ●  Z+ nucleus

      |    |    |    |

      e- electron wave

We can calculate the scattering using the simple Feynman diagram.

    e-  -------------------------

                      |

                      | photon with

                      | momentum p

    Z+ -------------------------

The Fourier decomposition of the Coulomb electric potential 1 / r of Z+ consists of sine wave potentials for each p. The potential undulates in spatial dimensions and is constant in time.

        potential for p component

        ----------------

 e- ----->

        ----------------

        ----------------

In the diagram above, the lines mark the "crests" of the potential of the Fourier component

       1/|p^2|  sin(p • x)

of the Coulomb potential. In the formula, p is the momentum vector of the Fourier component and x is the spatial position vector.

The potential acts as a "diffraction grating" which scatters electrons up and down, giving them the momentum p up or down. The potential crests act as sources for the Dirac equation. These sources have constructive interference for electrons which have "absorbed" the momentum p up or down.

Note that the scattering is both up and down. That is ok because the electron wave arrives symmetric relative to Z+.

        ---->

       |     |     |    ● Z+ nucleus

       |     |     |

       |     |     |

       e- electron wave

If we want to model the scattering for an electron wave which is not symmetric around Z+, then the Feynman diagram method does not work. It predicts scattering both up and down, but the real scattering is just up.

We can calculate the correct scattering pattern with the Schrödinger equation, or even with classical physics.

The Schrödinger calculation can be viewed as a non-perturbative calculation where the passing electron absorbs very many photons from the various Fourier components of the 1 / r potential. The sum of all these photons guides the electron to the correct path up.

The Feynman diagram method is a vastly simplified model where just one photon is exchanged between Z+ and e-. We could call it a minimal quanta method. If the method gives correct answers, it is, of course, much simpler to calculate than the Schrödinger method.

A big question is: why does the Feynman diagram work at all? Why does it work for a symmetric flyby? Is it a coincidence? We have not yet found an answer for that question.

An electron flies past a group of opposite charges

Suppose that we have a bound state of opposite charges. It could be a hydrogen atom, or just some ions attached together somehow. We have the charges static in space. An electron flies by.

The "minimal quanta" method of Feynman diagrams would assume that an electron normally absorbs just one photon, from one of the static charges.

    

                    ● • Z+ e- hydrogen atom

 

      e ----->

The one photon method would give a lot of scattering up and down. But we know that the atom appears electrically neutral to the outside world. There is no scattering.

The experiment has to be calculated using the Schrödinger method. We may imagine that the electron receives many photons from both of the opposite charges. The effects of these photons cancel each other out.

Feynman diagrams do not handle bound states. We have presented one example of that.

We could introduce a tentative rule:

Feynman diagrams are not the correct way to calculate a flyby of charges when the charges do not come "very close". In some (symmetric) cases, Feynman diagrams do work, though. Feynman diagrams do not work for bound states of opposite charges.

                         vacuum

                         polarization

                         loop

       Z+ ~~~~~~~O~~~~~~~ e-

             virtual

             photon

A vacuum polarization loop very much looks like a bound state of opposite charges. Furthermore, the loop is usually "far away" from the electron flying by. There is no reason why a Feynman diagram would model the setup correctly. The divergence of the vacuum polarization integral is a symptom of an incorrect method.

Feynman diagrams do not work if we know something about the relative positions of the particles

The two examples of Feynman diagrams failing have one thing in common: 

The particles in the experiment area are not at random positions. We know something about their relative positions.

In the first example we knew that the electrons pass just on one side of the nucleus.

If there is a bound state, then we know the relative positions of its component particles.

Feynman diagrams assume that free particles collide or fly by at random places in the experiment area. In that case, the single photon formula of Feynman diagrams produces (approximately) correct results.

If there is a collision of particles, say, two photons collide to produce a real pair, then we do know that the initial positions of the electron and positron are close to each other. Apparently, that is not a problem if the produced particles will be at random positions when they interact with other particles the next time.

A vacuum polarization loop contains two particles which are very close to each other for their whole lifetime. We cannot model the process as random collisions of free particles. That is why Feynman diagrams do not work for them.

The same problem as with vacuum polarization, exists for any diagram containing a loop. When the participants of the loop interact again, in most cases they are not at random positions relative to each other.

A correct way to handle bound states in a Feynman diagram might be to invoke the Schrödinger method at certain places. For example, if an electron interacts with a neutral hydrogen atom, we have to assume that the electron always interacts with both of the components of the atom. Or if we have a vacuum polarization loop, then a real electron always interacts with both the electron and the positron in the loop.

The Schrödinger equation calculates the electron wave function under all the potentials which are present. In a Feynman diagram, this is simplified to the one photon method in a special case.

It may be that the divergence in Feynman vacuum polarization is the result of applying the simplified method to a bound state, which is an error.