What is the sum of many dipole transmitters?

If we sum a large number of dipoles which oscillate in a random phase, then the power output scales linearly. This is because

       σ^2(X + Y) = σ^2(X) + σ^2(Y),

where X and Y are independent random variables and σ^2 denotes the variance. The energy density of an electromagnetic wave is ~ E^2, where E is the electric field. The expectation value of E is zero. The variance tells the energy density.

The equation above explains why energy is conserved in the interference of random dipoles. Constructive and destructive interference balance each other.

In the previous posting we looked at scattering of a photon from a block of glass. In that case, we can get an almost total destructive interference of scattered waves if we let the refractive index to be 1 at the surface and gradually increase it inside the glass block.

Conclusion: there is no general rule how much destructive interference of output waves there is for a system of oscillating dipoles. We have to determine it for each case individually.

Do waves in quantum mechanics consist of particles?

In our previous blog post we asked what is the relationship of the wave and the particle interpretation of quantum mechanics.

  proton ●     

                    ^

                    |

                    |

                   e- 

                   2 MeV kinetic energy

A proton is so heavy that we may treat it as a classical particle. If an electron passes the proton at a distance, say, 10^-15 m, it can emit a bremsstrahlung photon whose energy is 1 MeV. The wavelength of such a photon is 10^-12 m.

We face the "length scale" dilemma: how can a sharp turn whose length scale is 10^-15 m produce a wave whose length scale is 1,000 times larger?

It looks like that an electromagnetic wave can behave like a "particle": all the energy in the wave is concentrated into a small volume. In our example, all the energy in the wave originates from a small area around the nucleus. In a converse process, all the energy in a wave is absorbed inside a small volume.

Classically, the Fourier decomposition of the electromagnetic wave which is emitted from a sharp turn within 10^-15 m, should contain waves whose length is ~ 2 π * 10^-15 m. If the electron turns 180 degrees, it is like half a cycle of a dipole antenna where the charge moves at almost the speed of light for a distance π * 10^-15 m.

If the charge would keep oscillating at that speed, then it would really send a wave whose wavelength is 2 π * 10^-15 m. Somehow, the incomplete cycle allows the antenna to send a wave whose wavelength is much larger.

The Fourier decomposition of a sharp "pulse" like a Dirac delta contains all frequencies. The Fourier decomposition of a nicely oscillating dipole wave contains just one frequency. This might be the reason why an abrupt turn can produce a long wave?

In an earlier blog post we introduced the "reversed time" model which claims that a nice smooth 10^-12 m sine wave can concentrate its energy into a 10^-15 m area and induce a reaction, which if time were reversed, could produce such a 10^-12 m wave.

A large momentum virtual pair corresponds to a very sharp turn or acceleration?

We have been asking if classical virtual pairs which are created very close to each other might be responsible for high momentum virtual electrons. They are under very large acceleration, and according to the nonrelativistic Larmor formula lose all their energy very quickly. Does this happen in the relativistic case, too?

Suppose that the electron and the positron are moving at opposite directions at almost the light speed. What is the radiated power?

Let us move to a comoving inertial frame for the electron. In that frame, length (and the potential of the positron) is contracted in one direction by the factor

       γ = sqrt(1 - v^2 / c^2).

In the comoving frame, the electron is nonrelativistic. 

Time t is slowed down by the factor γ in the comoving frame relative to a static frame. Energy E in the comoving frame has to be multiplied by 1 / γ to get the energy in the static frame. Thus, the power P which an observer measures in the comoving frame is equal to the power which a static observer measures.

If the electron is deep within the potential well of the positron, the comoving observer will measure enormous acceleration and radiated power from the electron. A static observer will see the same enormous power dissipation. We conclude that such electrons radiate away their energy very quickly, in classical physics. The outgoing electromagnetic wave is extremely sharp because the slowdown happens under a very short distance.

In quantum mechanics, the sharp wave mysteriously loses its sharp edges and appears as a quite smooth 10^-13 m wave, according to the energy 1 MeV it is carrying.

The analysis suggests: a collision of photons can create a virtual pair where the particles are very close to each other and have very large opposite momenta.

This, in turn, suggests that large momentum virtual pairs do contribute to photon-photon scattering.

A classical wave analogue of virtual particles

Recall the two string & rubber membrane model of two interacting fields:

1. a massless Klein-Gordon field (the electromagnetic field);

2. a massive Klein-Gordon field (the Dirac field).

Is there any classical mechanism through which nice smooth sine waves in the massless field could somehow create very sharp localized waves in the massive field?

How does an external observer detect these sharp waves? He measures some scattering which he attributes to hypothetical particles. 

Maybe there is some classical mechanism which can be interpreted as resulting from high-momentum virtual particles?

Generally, any scattering in the classical setting is a result of nonlinearity of the system. Maybe some of this nonlinearity can be explained with high-momentum virtual particles?

Suppose that we have a standing wave in the massless Klein-Gordon field, and its frequency is too low to create a particle in the massive field.

Let us assume that the standing wave disturbs a massive Klein-Gordon field. It is a source in the massive Klein-Gordon equation.

We may try to determine what kind of transient waves the source generates in the massive field. To accomplish that, let us use the Green's function for a Dirac delta source spike, and sum over all spacetime points.

There is a problem though: the Green's function determines the behavior of the free Klein-Gordon equation. Our massive Klein-Gordon field is coupled, and probably transfers all energy quickly back to the massless field.

How to decompose the movement of the massive Klein-Gordon field?

What is the steady state of the two fields? The massless field probably keeps doing the standing wave.

The massive field receives a source which oscillates like the standing wave. How does the massive field return the energy back to the massless field? How quickly?

We may try to decompose the movement of the massive field into plane wave solutions of the massive field. Those solutions correspond to real particles of the massive field.

But since no real particles can be created, we have to devise a way to return the energy in those particles quickly back to the massless field.

Another way to decompose is to allow "virtual particles", that is, plane waves

       exp(-i (E t - p x)),

where

       E^2 != p^2 + m^2.

The Green's function solution to the problem uses this kind of a decomposition. The components are not solutions of the massive Klein-Gordon equation, but their sum is a solution.

Hypotheses. 1. The particle interpretation of the particles created in a collision of quanta of the massless field is really some kind of a decomposition of the disturbance of the massive field into plane waves.

2. If no real particles in the massive field are created, then none of those plane waves satisfies the energy-momentum relation.

The hypotheses might explain why we think that virtual particles are created in a collision of quanta: the intensity of a virtual particle wave where the momentum |p| > C may be something like ~ 1 / C^2, which we may interpret that quanta of the massless field passed each other at a very short distance.

If the hypothesis is true, we can treat particle collisions purely in a classical wave model without a need to resort to a particle model.

We would still need the particle model in measurements: we observe quanta, not classical waves. Also, a state transition of a hydrogen atom is hard to explain without quanta.

But in collisions, the particles that we think are colliding are just a way to explain the nonlinear behavior of the interacting fields.

Can this interpretation work in the context of charged particles like electrons and positrons? We had hard time trying to figure out a wave interpretation for the Coulomb force. The force is easy to define for particles, but hard for waves.

Anyway, the wave model may be a way to prove that no divergence happens in a vacuum polarization loop.

Scattering of photons from a block of glass

Let us aim a coherent laser beam of 500 nm photons to a glass block. Roughly 10% of the light will reflect, that is, scatter from the surfaces of the block.

Glass is amorphous material. Each atom or molecule acts as a polarizable dipole which scatters some of the coherent wave. But why is there no apparent scattering inside the glass?

The reason is destructive interference. Let us analyze the light which is scattered 90 degrees to the right.

       __________

        glass

           -------               ----------> scattered wave

           -------

           -------         crests of light waves 

       __________

              ^

              |

              | laser beam

There is total destructive interference for scattered light, except for a narrow, less than wavelength thick, vertical layer in the diagram. We can imagine that scattering only happens at a thin layer at the surface of the glass block.

If we double the dimensions of the glass block, the scattered flux grows by a factor 4, while the volume of the block grows by a factor 8.

Compare this to a pure particle model of light: if each atom would scatter a small number of particles, then making the volume of the block 8-fold would make the scattered flux 8-fold.

We can reduce reflection by making the refractive index go gradually to 1 at the surface of the glass. In principle, we can cut scattering in the above experiment to zero.

Why there is no destructive interference in a particle accelerator or atmosphere?

Why there is no significant destructive interference in a collision of two particles in a particle accelerator?

Because, in principle, we can observe both particles after the collision, and distinguish between different positions of the collision. For example, if we let a nucleus collide with a photon, then we can afterwards measure the position of the nucleus very precisely.

What is the difference if we let a photon collide with a block of amorphous glass versus letting it collide with atoms and molecules in the atmosphere? 

Atoms and molecules in the atmosphere move in a random way. There is no such extensive destructive interference that is present when the photon collides with static atoms in glass.

Is there a classical analogue of the quantum interference of a system of two particles?

Let us describe two colliding particles with wave packets. We mentioned in the previous section that we can measure the position of both particles afterwards, and the interference happens between pairs of the measured final positions of the particles.

In classical physics, interference happens for a wave independently of other waves.

        

       -------------------------------- plane

                        |

                        |         coupling from large

                        |         amplitude waves

                        |

       ____/\_________/\______ string

              --->            <---

           wave X        wave Y

Let us imagine the classical system in the diagram. We have two waves X and Y colliding in a 1-dimensional string. A coupling to a plane moves energy from a large amplitude wave to the plane.

The generated wave in the plane can be understood as some kind of scattering of the waves in the string. Interference in the plane happens for a 2D point (x, y). It does not happen for x or y individually.

We may say that particles X and Y got "entangled" in the plane and one cannot treat them as independent particles.

If we measure that the particle (or two 1-dimensional particles) in the plane is located at a position (x, y), we may return the two scattered particles back to the 1-dimensional world of the string.

Conclusions

This blog posting contained analysis of the wave-particle duality from various points of view. Let us next study if a classical wave model casts light on photon-photon scattering. We have the simple model of couples massive and massless Klein-Gordon fields. We need to study it carefully.

Are there high 4-momentum virtual pairs in a collision of photons?

In our blog post on January 22, 2021, we asked if "rogue" virtual pairs are created in a collision of two photons. By rogue we mean virtual pairs where each particle has huge 4-momentum, much larger than the 4-momentum of colliding photons.

  photon      e+       

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

                    |            |

                    |            |

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

  photon      e-

A concrete example: let us assume that two vertically polarized 0.500 MeV photons collide from opposite directions.

They cannot create a real pair, but can create an "almost real" pair, which classically can move to a separation 7 * 10^-14 m from each other if all the 1.000 MeV energy is spent to the potential energy of the pair.

Feynman rules allow the electron and the positron in the diagram to have arbitrary 4-momenta, as long as the sum of energies is 1.000 MeV and the sum of momenta is 0.

For example, the electron might have 1 GeV energy and the positron -999 MeV energy.

Or the electron would have 1 GeV / c momentum to the x direction, and the positron -1 GeV / c momentum. Here c is the speed of light.

The virtual pair as the mediator particle of a "force" between the photons; a Yukawa potential?

We may interpret the diagram above that two photons scatter from each other through some kind of a force.

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

The Feynman propagator of the electron resembles the propagator of the massive Klein-Gordon equation. The Fourier decomposition of the Yukawa potential, in turn, looks like that propagator.

The Yukawa potential describes a force which has a "finite" range, in contrast to the Coulomb force which is said to have an "infinite" range:

       V_Yukawa (r) = -g^2 exp(-α m r) / r.

The Coulomb potential is recovered by setting the "mass of the mediating particle" m to zero. Alpha is a coefficient which should be set to a suitable value, so that the range of the force is as measured.

Calculations in literature claim that the cross section of photon-photon scattering is ~ 1 microbarn. We argued in a blog post that the cross section should be much larger, at least ~ 1 millibarn.

The force between photons is attractive. We can deduce that through the following reasoning. If we have two wave packets coming from opposite directions and they graze each other, then virtual pair production somehow disturbs the progress of the waves. Since waves cannot move faster than light, a disturbance probably slows them down. Then the wavelength is shorter in the grazing area, which causes the wave packets to turn toward each other.

If we assume that virtual electrons cannot travel very far, then the potential might be Yukawa.

A particle model based on classical particles

Let us assume that the electron and the positron in the virtual pair are very ordinary classical particles. It just happens that their mutual potential energy V is so much negative that the total energy of the pair is < 1.022 MeV.

If the particles in the just created pair have moderate momenta, they are at a distance whose order of magnitude is 1.4 * 10^-15 m.

If the just created particles have very high momenta, then in our classical model they must be very close to each other. The pair has very large negative potential energy V, which is compensated by very large positive kinetic energy.

Does creation of a pair which has very high momenta require that the colliding photons are very close to each other?

Let us assume that the answer is yes.

If the particles have a lot of kinetic energy they move at almost the speed of light. Then the spatial momentum p and the kinetic energy E of the particles roughly satisfies

       |p| = E / c.

The cross section for creating a pair whose particles have very large spatial momentum > |p| is then ~ 1 / |p|^2.

What about creating a pair where one particle has very large negative energy and the other very large positive energy? In our classical model, one might assign the large negative potential V in an arbitrary way to the particles. Then one would have large negative energy and the other large positive energy. But such arbitrary assignment does not change the actual physical system in any way. The most natural assignment is such that the particles have equal energies.

How to conserve angular momentum?

Let us consider the angular momentum J produced by the momenta p and -p of the colliding photons, as well as their relative spatial separation r. If the collision is very close, then J is very small. When the pair annihilates, the new photons have to leave with a very small spatial separation. 

But now there is a problem with high momenta pairs: classically, according to the Larmor formula, they radiate their energy away very quickly, because the acceleration is huge. How can we conserve J? Obviously, the created pair must possess the same J as the pair of photons. Since the electron and the positron move at a speed < c, the separation of the pair must be slightly larger than the separation of the photons was. In annihilation, how can they produce photons with the correct (small) separation?

In classical physics, conservation of angular momentum is not a problem because we can assume that the colliding laser beams possess no angular momentum, and the scattered beams have zero angular momentum, too.

Maybe we can solve the angular momentum problem in the following way: we calculate the sum of the waves produced by various possible collisions and spacetime points. The sum is essentially the classical picture. Then we require that the observed photons conserve angular momentum.

A position space analysis of the Feynman model

In our classical model, rogue pairs are no problem. They contribute little to the cross section. Why are they a problem in the Feynman integral? Why does the integral diverge? What is the difference of the classical model to the Feynman model?

   photon

       ~~~~~  -------------------- e+ 4-momentum q

                       | virtual

                       | electron

       ~~~~~  -------------------- e-  4-momentum p

   photon

Let us consider the above diagram where we allow the outgoing particles be virtual, that is, have any 4-momentum, even off-shell.

          laser                       laser

      t    ----->                      <-----

      ^

      |        /   \               /   \          

      |      /       \           /       \

      |    /           \       /           \

      |  /               \   /               \

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

                       2 * 10^-14 m

         oblique lines are the crests of the

         waves of two colliding laser beams

Let us try to visualize the situation in position space where we have a cube whose side is, say, 4 * 10^-14 m, and two laser beams of energy 0.500 MeV collide. We may assume that both beams have one photon in our cube.

Let δ > 0 be a fixed small real number. In the Feynman diagram, for any 4-momentum p_0, we are allowed to assume the existence of a "standard flux" of virtual or real electrons whose 4-momentum p satisfies

       |p - p_0| < δ,

where | | denotes the euclidean length of the 4-momentum vector. We may assume that the cube contains one such electron.

We should calculate the flux of electrons which scatter from both photons, and become virtual positrons.

The Feynman formula reduces the flux of scattered electrons with the electron propagator, which is ~ 1 / |p| for very large |p|.

What is the contribution of electrons whose

      r < |p| < r + dr?

It is ~ r^3 dr / r, or

       ~ r^2 dr.

If we try to integrate over r, it diverges very badly.

What is the problem? The input of photons to the cube is a finite flux. How can they cause an infinite flux of scattered electrons?

The problem is the approximation which pretends that the flux of photons is not reduced at all by scattering. A single photon can cause an infinite number of electrons to scatter.

A "first order perturbation" approximation makes the assumption: the input flux is not affected by a small scattered flux. But in this case, the assumption is very much wrong. The scattered flux is not small - it is infinite.

The Feynman formula reduces the scattered flux by a coefficient 1 / |p|. Is that sensible? 

We have argued that the "lifelength" of a virtual particle is ~ 1 / |p|. If that is the case, the virtual electron has to meet the other photon in a volume ~ 1 / |p|^3. That coefficient would cut off high momenta very efficiently. But that does not sound right. The coefficient 1 / |p|^3 corresponds to a meetup of three particles. We believe that the pair is created by a collision of two photons.

The classical model which we introduced in a section above suggests that the coefficient should be ~ 1 / |p|^2. It is like the propagator for the photon. The Yukawa potential model for photon interaction, which we introduced in a section above, suggests a similar coefficient.

The coefficient 1 / |p|^2 applies to the collision of the two photons. The virtual pair will meet and annihilate with a 100% probability if the combined energy of the photons is < 1.022 MeV. What is the behavior if the combined energy is > 1.022 MeV? If the created particles are very close, they classically radiate away all their energy very quickly, and must annihilate. It might be that the probability of annihilation is slowly reduced when we increase the energy to several MeV.

Destructive interference in the position space model

Let us again return to the difficult problem of destructive interference. What is the phase of the virtual particles which are produced in a photon collision? Should we calculate the effect of destructive interference on them?

In Feynman rules, destructive interference is ignored, except in the final result.

In the double-slit experiment, we certainly have to take into account the interference pattern.

If the phase of the created virtual particles depends on the phases of the colliding photons in some simple and natural way, then we expect that the phase of the created high-momentum virtual particles is almost constant in a spatial area whose size is

       λ = h / |p|,

where h is the Planck constant and p is the spatial momentum of the virtual particle. There λ is the de Broglie wavelength of the virtual particle.

The phase is almost constant also during a time interval λ / c at a constant spatial location x.

The phase of the real photons changes quite slowly relative to our high-momentum virtual particles. Therefore, we expect the virtual particles in such a small area be born with a relatively constant phase.

If the lifelength of the virtual particles is just 0.1 λ, then we can ignore interference effects. Waves born far away never reach the spatial position x we are interested in.

Note that this is a general rule with oscillating physical systems: if we have a field F which disturbs another field G, and the disturbance does not "resonate" with G, then the disturbance in G moves a very short distance and is retained in G for a very short time. If the "wavelength" of the disturbance in G is short relative to waves in F, then we can ignore interference effects.

A virtual particle is a disturbance which does not "resonate" with the field.

We conclude that we can ignore interference effects if the virtual particles are a lot off-shell. If |p| is large, then they are a lot off-shell.

Why do we need to care about very small features at all, like high |p|? Classically, an oscillating system where wavelengths are long does not care much about small detail. The answer: quanta can be understood as particles, and particles care about small detail; for instance, about other particles.

We will look at this fundamental question of the wave-particle duality in the next blog post.

Conclusions

Our classical model suggests that high-4-momentum virtual pairs do contribute to scattering of photons, but their contribution is not large, something like ~ 1 / |p|^2.

The Feynman loop integral diverges for the following reasons:

1. To work, a first order perturbation approximation requires that the scattered flux is small compared to the input flux. In the case of a Feynman loop, the scattered flux is infinite, which makes the approximation nonsensical.

2. The electron propagator, which is ~ 1 / |p| for large momenta, overestimates the chance that photons come very close. The coefficient should rather be  ~ 1 / |p|^2. The process is not about an electron bouncing from two photons, but about two photons colliding.

3. Feynman allows arbitrary energy E to circulate in the loop. The classical model suggests that energy must be divided evenly between the virtual positron and the virtual electron.

Pair production from collision of gamma rays and a nucleus

In the previous blog post we mentioned that Frahm et al. (2009) have measured pair production when a photon hits a gadolinium nucleus. The energy of the photon in their measurements is moderately over the threshold energy 1.022 MeV.

We asked why the cross section at 1.1 MeV is only 1 / 100 of the cross section at 2 MeV incident photon energy.

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

Pair production is the dominant process when > 10 MeV photons hit atoms.

  gamma ray photon

                              _______  e-

                            /

       ~~~~~~~

                            \________ e+

                                  |

                                  | virtual photon

      Z+  --------------------------

  nucleus

The photon gives up energy and some momentum to create the real electron. The photon transfers the rest of its momentum and energy to the created positron which then has too much momentum to be on-shell.

The positron bounces from the nucleus and the excess momentum is absorbed by the nucleus. The positron becomes on-shell, that is, real.

Our various classical models claim that the potential energy of the combined field of the photon and the nucleus can be reduced by creating a pair. Electric field lines "break" and the loose ends are the particles being created.

Our classical models do not make any numerical prediction for the cross section. Intuitively, creating a pair from a photon whose energy is only a little over 1.022 MeV should be much harder than from a 2 MeV photon. The result of Frahm et al. makes sense from the classical point of view.

We remarked in a blog post about classical virtual particles that the momentum and energy of the pair can in interaction with the nucleus be equal to the momentum and energy of a single photon. In principle, we could calculate the probability of pair production in this classical model, if we assume that the photon particle has a certain probability to split into a pair under a time interval.

A wave description of the Feynman diagram

The Feynman diagram can be understood with the following plane wave model:

        t

        ^ 

        |                   ----------  on-shell

        |                   ----------  positron wave e+

        |                                                           ●  Z+ 

        |       |          |          | Fourier component

        |       |          |          | of static Coulomb

        |       |          |          | field of nucleus

        |

        |          |       |      |  off-shell

        |         |       |      |   electron wave e+-

        |          /  ---  /  ---  / ---------   

        |        /  ---  /  ---  / ----------   on-shell

        |      /  ---  /  ---  / -----------   electron

        |    /  ---  /  ---  / ------------   wave e-

        |  gamma laser

        |  beam wave

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

The nucleus Z+ is not really located anywhere in the diagram. We work with the Fourier components of its Coulomb field. The components are plane waves which span the entire diagram at a constant amplitude, regardless of the position of the nucleus.

1. We assume a coherent laser beam of gamma photons. The oblique lines are the crests of the electromagnetic waves.

2. We assume a coherent flux of electrons with little momentum. The electron wave crest lines are almost horizontal because the momentum is small.

3. The strange thing in this model is that the electrons arrive from the future, or are assumed "just to exist there." Alternatively, we may assume that the electron wave is created "when needed".

4. The interaction term between the electromagnetic field and the Dirac field makes a source to the Dirac field. The source produces off-shell (virtual) electron waves. Another way to say this is that electrons absorb photons from the laser beam. The wave crest lines are almost vertical, because the off-shell electron has too much momentum relative to its energy to be real.

5. The interaction term between the off-shell electron wave and a Fourier component of the Coulomb field of the nucleus Z+ makes another source to the Dirac field. That source can produce on-shell positrons. We may say that the off-shell electron absorbed a virtual photon from the Coulomb field of the nucleus and became an on-shell positron.

For clarity, we did not superimpose all waves on each other in the diagram above. They should be superimposed in the accurate diagram.

We may imagine that the input plane waves are standardized, e.g., one particle per square meter per second.

How far can the virtual electron fly?

The diagram above is a momentum space description of the process. Points (t, x) are not genuine spacetime locations but just "substance" on which we can visualize plane waves. Particles in momentum space must be described as plane waves.

What would a position space diagram look like? We can draw the nucleus Z+ as a classical particle because it is very heavy.

The positron in the diagram must come very close to the nucleus to lose the extra momentum. If the nucleus is gadolinium (Z = 64), then the minimum distance should be around 64 * 1.4 * 10^-15 m = 10^-13 m, which corresponds to potential energy 1 MeV for the positron and the same negative potential for the electron.

A. The Compton wavelength of a real electron is 2 * 10^-12 m, which is quite a long distance. In an earlier blog posting we calculated from the time-energy uncertainty relation that the "lifelength" of a virtual particle might be 0.1 Compton wavelengths. That would be 2 * 10^-13 m.

B. Our classical model of virtual particles says that they must exist quite deep in the Coulomb field. That suggests that the virtual electron is ~ 10^-13 m from the nucleus.

C. The pair is born from some kind of a source in the Dirac equation. The source presumably is a result from some kind of interaction between the Coulomb field of the nucleus and the photon. That would suggest that the pair is born "close" to the nucleus.

D. We had a classical model where electric field lines break and the loose ends are the created particles. That suggests a "close" birh to the nucleus.

E. The inverse reaction is pair annihilation close to the nucleus, so that only one photon is produced, instead of two. Our classical model of annihilation suggests that it happens in a volume of size ~ 1.4 * 10^-15 m. This is evidence for a short lifelength for the virtual electron.

F. If we turn and twist the diagram, the positron seems to absorb a virtual photon from the nucleus, and emit a real photon. Our classical model of Thomson/Compton scattering suggests that the absorption and the emission are simultaneous. This is evidence for a short lifelength of the virtual electron.

G. Bethe and Heitler in their 1934 paper, in the case of bremsstrahlung (page 84), write that the transition occurs under the simultaneous action of the perturbations by the nucleus Coulomb field and the photon field of the produced photon. This suggests that the lifelength of the virtual electron is short.

F. If we try to measure experimentally where the electron and the positron are born, we have to face an uncertainty relation about the position and momentum. We may observe the created positron annihilating with the orbital electron of the atom around the nucleus. That way we might be able to localize the positron with an accuracy 10^-10 m. This, of course, requires that we assume that an orbital electron is a particle and has a definite location.

Conclusions: our classical models suggest that the virtual electron only flies 10^-15 m ... 10^-13 m. The uncertainty principle model suggests a lifelength up to 2 * 10^-13 m. Experimentally, we can hope to restrict the lifelength to < 10^-10 m.

In the plane wave diagram, the virtual electron seems to fly an arbitrarily long "distance", but that is just an illusion of the momentum space model.

Cross section for bremsstrahlung versus pair production from a photon

Feynman integral formulas must take into account the small volume of the actual position space diagram of the process. The virtual electron probably flies only a short distance.

The probability amplitude of pair creation is curtailed by the propagators of the virtual electron and the virtual photon, as well as the small coupling constant ~ sqrt(α) at the vertices.

The cross section for an on-shell positron flying within 10^-13 m from a gadolinium nucleus (potential > 1 MeV) is 300 barn.

Frahm et al. measured that the cross section for pair production with a 2 MeV photon is 0.2 barn.

Thus, the cross section for pair production is just 1 / 1,500 of the cross section of a "close hit".

If we turn and twist the Feynman diagram, we get one for bremsstrahlung of a relativistic electron:

                                       ~~~~~ 1 MeV photon

                                     /

 1.7 MeV e- ------------------------------  0.7 MeV e-

                             |  virtual

                             |  photon

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

Let us check what is a typical cross section in bremsstrahlung in the MeV range.

https://ntrs.nasa.gov/citations/19710001524

S. H. Morgan, Jr. (1970) in Figure 9 has

       dσ / dE = 10 barn / MeV

for Z = 50 and incident electron kinetic energy 1.7 MeV, where we require that the photon energy is around 1 MeV.

The cross section for the electron coming to a potential < -1 MeV is 200 barn.

The cross section for a 1 MeV photon production is 1 / 20 of the cross section of a close hit.

What is the cross section for inverse reactions?

A classical calculation as well as a Feynman diagram calculation tell us that the cross section for a particle to make a "close" hit (1 MeV Coulomb potential) to a typical nucleus is ~ 250 barn.

Empirical data as well as calculations by Bethe and Heitler (1934) show that when a 2 MeV photon hits a typical nucleus, the cross section for pair production is ~ 0.2 barn.

For bremsstrahlung producing a ~ 1 MeV photon, the cross section is larger, ~ 10 barn.

Why is a pair harder to produce? In pair production, "kinetic" energy of the photon is converted to "potential" energy of the pair: their masses. Physical systems generally tend to increase kinetic energy and reduce potential energy.

What about the inverse reactions? An inverse reaction requires three particles to meet. Or, at least, first 2 particles have to meet and after that, 2 particles have to meet again. The probability for an inverse reaction is extremely small unless the density of particles is huge.

We may say that the entropy of a system grows in a reaction if the inverse reaction is less probable. A random soup of particles tries to increase its entropy.

 

                 --------------------

                |                         |

          -----|                         |

         |  •       hole              |

          -----|                         |

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

    small       large vessel

   vessel 

Let us think of a small vessel which is connected to a large vessel through a narrow hole. A particle bouncing randomly in the small vessel tends to move to the large one. We may say that the space of states in the large vessel is larger. The cross section for a reaction is a measure of how large is the hole with respect to the volume of the states in the vessel.

Photon-photon scattering: what is the role of the virtual pair?

UPDATE February 4, 2021: the analysis in this blog post has been improved. See our new blog post.

---

Photon-photon scattering is sometimes grouped under the name Delbrück scattering. Max Delbrück in the 1930s studied single photon scattering in the static Coulomb field of a nucleus. But in this blog post we are interested in a collision of two real photons.

A collision of photons can produce a real pair. Real pair production is a tree-level Feynman diagram, and consequently, does not involve diverging integrals.

                         loop

          ~~~~    ----------    ~~~~

                       |           |

                       |           |

          ~~~~    ----------    ~~~~

 2 photons                    2 photons

 total energy < 1.022 MeV

We may interpret the diagram in this way: in the middle is a pair whose total energy is less than 1.022 MeV. On the right the pair annihilates and produces two photons. On the left is the reverse reaction to annihilation: pair production - but the produced pair never gains enough energy so that the electron and the positron could escape from each other.

https://arxiv.org/abs/1111.6126

Yi Liang and Andrzej Czarnecki (2011) have written a tutorial on photon-photon scattering.

They write that there are three essentially different diagrams for scattering. Each diagram individually diverges, but when all are added together, divergences cancel. However, the sum is different for dimensional regularization and for Pauli-Villars regularization.

Pair production or scattering from the collision of two photons has not been observed experimentally.

 V

  ^

  |

  |                        ______ ------------  1.022 MeV

  |                   /

  | ------------------------------------------ 0 MeV

  |             /

  |         /

  |      |

  |    |

   ---------------------------------------------------> r

                   if V = 0 then r = 1.4 * 10^-15 m

   high

   momenta?

In the diagram, V is the total energy of a static electron-positron pair. The separation between the particles is r.

The diagram is in very poor character graphics, but the reader of this blog probably gets the idea.

Our previous blog posting suggested that the arbitrarily high 4-momenta in a diverging Feynman loop integral may be (classical) electron-positron pairs where the separation is r < 1.4 * 10^-15.

Annihilation of a pair

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

How fast does an electron radiate energy if it starts static at a distance r_0 = 1.4 * 10^-15 m from the positron = half of the electron classical radius.

The Coulomb force is

       F = k e^2 / r_0^2 = 120 N.

The acceleration of the electron is huge, 1.3 * 10^32 m/s^2.

The Larmor formula for the radiated power is

       P = e^2 a^2 / (6 π ε_0 c^3)

           = 10^11 W.

The mass-energy of the electron is 10^-13 J.

The electron would lose all its mass-energy in 10^-24 s, which at the speed of light corresponds to r = 3 * 10^-16 m. That is just 1 / 5 of r_0.

Let us calculate r symbolically, using the formulas for P, a = F / m (where m is the electron mass), and

       r_0 = 1 / 2 *  1 / (4 π ε_0)  *  e^2 / (m c^2),

       k = 1 / (4 π ε_0).

We get

       r = c * (m c^2 / P)

          = m c^3 * 6 π ε_0 c^3

                         * m^2

                         * e^8

                         * 4^2 π^2 ε_0^2

                / (e^2

                   e^4

                    8^4 π^4 ε_0^4 m^4 c^8)

          = 6 * 16 e^2

             / (8^3 m c^2 8 π ε_0)

   

          = 96 / 512 r_0.

If we cut the distance to half, the acceleration is 4X and the power 16X.

If we double the distance, the power is just 1 / 16 X.

Classically, the speed of the electron will be significantly lower than c until it is close to the distance 1.4 * 10^-15 m. This is simply because the negative potential goes as 1 / r. Thus, the Larmor formula may be roughly right at larger distances.

Hypothesis: classically, in annihilation, the electron will radiate its mass-energy away at a distance 1.4 * 10^-15 m from the positron, in a short burst.

It is interesting that the classical analysis agrees this well with the classical electron radius. Suppose that the particles would radiate their energy away much slower. Then, in principle, we could have a pair which would still possess positive energy even though the particles are static and closer than 1.4 * 10^-15 m. But such a pair has negative total energy. This would be a contradiction.

We have the length scale problem here. The energy is radiated in a very sharp electromagnetic wave whose length scale is 1.4 * 10^-15 m. But the wavelength of the photons is 2 * 10^-12 m. Our earlier blog posting suggested that we must sum the electromagnetic waves over many spacetime points, over a volume 2 * 10^-12 m (uncertainty of the position of the pair), to get the wave observed by an external observer. Then the electromagnetic wave's wavelength is of the right order of magnitude.

Creation of a real pair from 1.022 MeV photons versus a virtual pair

      ~~~~   ---------------------- e+

                  | virtual

                  | electron

      ~~~~   ---------------------- e-

 2 photons

Production of a real pair from a photon-photon collision is called the Breit-Wheeler process.

Let the photons have a combined energy of just over two times the electron mass.

The Feynman integral has one propagator, the virtual electron in the middle.

But if we reduce the combined energy of the photons just under the mass of two electrons, the produced pair will inevitably annihilate. The Feynman integral then has four propagators. If we assume that the electron and the positron are almost on-shell, then the propagators for them have very large absolute values around the pole which corresponds to an on-shell electron. 

We need to check what happens if we increase the mass of the electron so that it becomes a classical particle. Do the large propagator values around the pole make sense then?

What is the correct Feynman diagram for photon-photon scattering?

If the photons have combined energy < 1.022 MeV, then the produced virtual pair will certainly annihilate.

The diagram at the top of this blog post contains three "unnecessary" propagators for the electron and the positron, the ones which do not appear in the real pair production diagram. If we know that annihilation is inevitable, why should we use those three propagators?

It might be better to use the real pair production diagram, with the electron mass adjusted lower?

We know that Feynman diagrams do not handle bound states correctly. The pair with energy slightly below 1.022 MeV is kind of a bound state. The pair might even form a positronium "atom", whose lifetime is 0.12 ns for antiparallel spins, and 142 ns for parallel spins.

Numerical cross sections for pair production and scattering in a real photon-photon collision: the values differ too much?

Let us check what people have calculated for the pair production cross section at combined energies close to 1.022 MeV.

The production of virtual pairs of moderate momenta should have roughly the same cross section.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.155.1404

Gould and Schreder (1967) give the pair production cross section in the center of mass frame as

        σ = β π r_0^2,

where the velocity of the produced electron and positron is β c, and r_0 is the classical electron radius.

If β = 0.1 c, then the cross section is 2.5 * 10^-30 m^2, or 25 millibarns.

https://link.springer.com/article/10.1140/epjc/s10052-018-6364-1

T. Takahashi et al. (2018) have the scattering cross section in their Figure 1 as less than 10 microbarns, or less than 10^-33 m^2.

This is strange. How can the probability for a virtual pair production be less than 1 / 1,000 of the probability for real pair production at just a little higher energy?

We need to check the figures from other sources.

http://adsabs.harvard.edu/full/1990ApJ...349..415S

R. Svensson and A. Zdziarski (1990) in their introduction give a value ~ 4 * 10^-34 m^2 for the cross section of scattering of ~ 1 MeV photons. They write that scattering is a fourth order process - therefore the cross section is small. Their value is consistent with Takahashi et al.

A concrete example: combined photon energy 2 m_e +- 0.5 eV

Let us calculate a concrete example. Let β = 0.001 in the pair production formula. The combined kinetic energy of the pair is then 0.5 eV. The cross section for pair production is 10^-32 m^2.

If the pair would have the energy 2 m_e c^2 - 0.5 eV, the virtual particles could climb to a separation of 3 nanometers until falling back and annihilating. Literature above claims that the cross section is < 10^-33 m^2.

A separation of 3 nanometers is a huge distance in particle physics. The diameter of a hydrogen atom is 0.1 nm.

How can the colliding photons, that are lacking 0.5 eV, know beforehand that their energy just fails to produce a real pair, and the photons then "decide" that the cross section should be much smaller?

Or is it so that the annihilation will send the photons to the exact same directions that they arrived to the collision? Why would that be?

Feynman formulas fail to take into account the attraction between the created particles. One of the basic assumptions in Feynman diagrams is that particles fly between vertices without interaction.

What about scattering at substantially higher energies than 2 m_e? Classically, the created electron and the positron will move to opposite directions, and will never annihilate. In quantum mechanics, the positions of the particles have uncertainty. In the MeV scale, the uncertainty about the relative position is ~ 10^-12 m.

The cross section for annihilation should be roughly the same as for pair production, that is, ~ 10^-29 m^2. If the electron and the positron move through an area ~ 10^-24 m^2, the probability for annihilation is ~ 10^-5.

We conclude that the cross section for scattering if the combined energy is substantially larger than 1.022 MeV, is ~ 10^-34 m^2, or one microbarn. This figure agrees with the ones that T. Takahashi et al., and R. Svensson et al. give.

Conjecture: cross sections for ~ 1.022 MeV combined photon energy are larger both for pair production and scattering

How could the colliding photons, who have a combined energy ~ 1.022 MeV, know what to do? The formula for pair production gives a very small cross section for real pair production if the energy is only slightly over the limit 2 m_e.

The formula for scattering gives an extremely small cross section regardless of the energy.

Suppose that a real or virtual pair is produced in a photon collision.

Classically, the eventual fate of the pair would depend on its energy: if the energy is < 2 m_e, then the pair falls back and annihilates. Then it is a case of scattering.

In wave mechanics, a wave which is born from a source in a small spatial volume must go through a "hole" whose area is 4 π r^2 where r is an arbitrary distance from the source. The hole can be considered a circle whose radius is 2 r. If the de Broglie wavelength of the particle is long, it has hard time going through the hole, and is reflected back. That prevents real pair creation.

Let us calculate an example. The de Broglie wavelength of an electron with 50 keV kinetic energy is 2.3X the Compton wavelength. The de Broglie wavelength of a 500 keV kinetic energy electron is 0.5X the Compton wavelength. 

We also need to check if the very small cross sections in literature can be explained by destructive interference. Maybe destructive interference is very important if the output is slow electrons?

              __          __

       __/      \___/      \___

     photon wavelength 2 * 10^-12 m

Let us assume that we collide two linearly polarized laser beams whose wavelength is the Compton wavelength of the electron.

They form a standing electromagnetic wave which has the same wavelength.

We assume that there exists a nonlinear process which causes the standing wave to induce waves in the Dirac field. It is reasonable to assume that the created waves in the Dirac field inherit their phase from the standing wave.

If we replace the Dirac field with a massive Klein-Gordon field, we observe that almost all the energy that we take from the standing wave is spent to make the harmonic oscillators to oscillate at each spatial point x. Only a very small amount of energy goes to the kinetic term d^2 ψ / dx^2 of the massive Klein-Gordon equation.

This means that the energy flow from the electromagnetic field to the Dirac field is almost unharmed by destructive interference. We conclude that destructive interference is insignificant in the process.

Pair production when a 1.1 MeV photon meets a nucleus

There exists measured data for a different process, namely, a single photon hitting the nucleus of an atom.

https://www.researchgate.net/publication/235335360_Positron-Electron_Pair_Creation_Near_Threshold

R. Frahm et al. (2009) measured pair production cross section at low gamma ray energies > 1.022 MeV when photons hit atoms. They calculated with the program XCOM 3.1 that the cross section for 2 MeV should be 0.2 barns per gadolinium atom and 0.0007 barns for 1.1 MeV.

Their Figure 4 shows that the measured cross section at 1.1 MeV is roughly 0.002 barns, or 3X times the calculated value.

Why is the cross section so much lower for 1.1 MeV than for 2 MeV? Bethe and Heitler apparently use Fermi's golden rule in their 1934 paper. If 1.022 MeV is spent to create the electron and the positron, then the "number of states" in the momentum universe of the final outcome is less with just 78 keV of kinetic energy.

https://en.wikipedia.org/wiki/Solution_of_Schr%C3%B6dinger_equation_for_a_step_potential

When a nonrelativistic electron hits a sharp potential step, the transmission is roughly the ratio of kinetic energies.

Cross section for pair annihilation

The Feynman diagram formula for the probability amplitude has the same value if we turn the diagram upside down.

We know that if we put a static electron and a static positron close to each other, the probability of annihilation is 100%.

If we model the electron and the positron as de Broglie wave packets, we cannot really make them static.

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

       σ_x σ_p >= h-bar / 2.

Let us assume that the uncertainty of the electron position is 1 meter. Then the uncertainty in its momentum is 5 * 10^-35 kg m / s, which corresponds to a velocity 5 * 10^-5 m/s.

The kinetic energy of the electron is

       E = 1/2 m v^2 = 10^-39 J.

The Coulomb potential of an electron and a positron at the distance 1 m is

       V = -k e^2 / r = 2 * 10^-28 J.

We see that the probability of annihilation is 100% if the electron and the positron move at a speed < 15 m/s.

A thermal electron at 300 K moves at 1.2 * 10^5 m/s.

Let us check what literature says about annihilation cross section.

https://www.mpi-hd.mpg.de/personalhomes/frieger/HEA10.pdf

Frank Rieger gives the low-energy cross section as

      σ = 1 / β * π r_0^2,

where the velocity of the particles is β c and r_0 is the classical electron radius.

The annihilation cross section for a thermal 300 K electron is

       2,500 * π r_0^2,

or a circle whose radius is 50 r_0 = 1.4 * 10^-13 m.

Gould and Schreder give a cross section

       σ = β π r_0^2

for the inverse reaction, i.e., pair production (300 K) from the photons. It corresponds to a circle whose radius is 1/50 r_0.

The kinetic energy of a thermal 300 K electron is 0.04 eV. If the energy of a pair is 2 m_e - 0.08 eV, what is their separation? It is 2 * 10^-8 m, a huge distance in particle physics.

The Rieger formula gives a cross section in the 1 m^2 range if β = 10^-30, while our own potential calculation found that β = 5 * 10^-8 is enough to ensure a 1 m^2 cross section. The Rieger formula probably is correct for 300 K electrons, but fails for electrons whose temperature is in the microkelvin range.

Divergence of individual Feynman diagrams

Liang and Czarnecki mention that each of the scattering diagrams has a diverging Feynman integral.

The divergence is probably logarithmic, because there are four electron propagators in a diagram.

Above we argued that the diagrams are not the right way to calculate because photons cannot know in advance what exactly is the value 2 m_e.

Anyway, whatever method we use, we have to face the problem that virtual pairs with arbitrarily high 4-momenta contribute to scattering. How to get rid of divergences?

We will look next at those rogue virtual pairs.

A classical particle model of QED vacuum polarization loop divergence

                <---- e-     e+ ---->

                    -p             p

Let us put a classical particle electron and a positron very close to each other, less than 10^-15 m apart. If the pair is static, its negative potential energy is < -1.022 MeV. The pair has negative total energy.

We can raise the energy by giving the pair large opposite momenta. Then the total energy of the pair may be zero or positive.

This may be a classical analogue of the vacuum polarization loop divergence in a Feynman diagram. In principle, we can give the particles arbitrarily high momenta, if we place them close enough.

Can we give the particles arbitrary energy? Let us work in a frame where the sum of momenta is zero. Then the system is symmetric. The natural way to assign energies is that both particles have the same energy. But if we allow ourselves to divide the large negative potential energy arbitrarily between the particles, then we can also give them arbitrary energies.

This has a lot of similarity to the strange Feynman diagram vacuum polarization loop, where e- and e+ can have huge opposite momenta and "opposite" energies.

The magnetic field of fast moving particles complicates the calculation of their potential energy, but that probably does not spoil the setup.

Electromagnetic radiation from the fast pair

Our pair can be understood as a (time-reversed) final stage of a classical pair crashing into each other. The negative potential energy as well as the kinetic energies of the particles grow without limit.

But our setup is not realistic. The pair would radiate away electromagnetic radiation at a fast rate. It would quickly lose, for example, 1.022 MeV in radiation.

What about Delbrück scattering? How much energy in radiation does the virtual pair lose, if we calculate the radiation with the Larmor formula?

Actually, annihilation of a pair might be that the pair radiates away all its energy in electromagnetic radiation which is produced by the acceleration.

In the Larmor formula, the power of radiation is proportional to the acceleration squared. The acceleration in the static case is 1 / r^2, where r is the distance. The power is 1 / r^4 if the speed is not relativistic.

The negative potential is 1 / r. In the nonrelativistic case, putting the particles at half a distance (double the negative potential) makes the power of radiation out 16-fold. This classical model suggests that the lifetime of a virtual pair with high opposite momenta is extremely short.

Delbrück scattering by a nucleus produces a virtual pair which then annihilates

             

                         ____

photon     e+ /         \

            ~~~~                ~~~~

                  e-  \_____/

                           | virtual photon

                           | which acts on e+, too,

                           | because of Furry's theorem

  Z+ ----------------------------

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

In Delbrück scattering, a photon collides with a nucleus, forms a pair, and the pair soon annihilates. The annihilation produces just a single photon, not two.

The momentum of the new photon differs from the incoming one because the Coulomb field of the nucleus interacts with the pair. Furry's theorem requires that there must be an even number of photon vertices in a fermion loop. We have only drawn one photon line, to avoid cluttering the diagram.

http://14.139.211.59/bitstream/123456789/1500/6/06_chapter_01.pdf

B. Kunwar (2011) writes about Delbrück scattering.

This scattering forms a good setting for studying vacuum polarization. Earlier, we have looked into pair production. Here we have a pair which never gains enough energy to become real. Nevertheless, the pair produces a measurable scattering effect on the photon.

Besides Delbrück scattering, there is Rayleigh scattering from the electrons of an atom, and scattering caused by the nucleus. Empirical tests must take into account these effects.

Experimental tests of scattering have been performed in the photon energy range 0.89 MeV - 1 GeV. According to B. Kunwar, the agreement of theoretical calculations and measurements is good, often within a few percent.

We should check if various authors allow an arbitrary momentum k to circulate in the virtual pair loop. If yes, then the integrals easily diverge. Maybe they use some cutoff for k?

A "reversed time" model for a photon or a wave function collapse

When a pair annihilates, it typically emits two photons. The converse reaction would convert the entire energy of the two photons into a pair.

How can we use a particle model for the produced pair and erase the entire electromagnetic field of the two photons?

The only way to make such a process seems to be running the annihilation in reverse.

We have the same problem when a hydrogen atom is excited by a photon. How can the electric dipole formed by mixed 1s and 2p states absorb the entire electromagnetic field of the photon?

We need to change our "teleportation model" of the photon. The new model is a reversed time model. A photon (or photons) is magically able to run in reverse a process which could emit such a photon.

Classically, it would be a miracle that a complex process runs in reverse because it would dramatically reduce the entropy of a complicated electromagnetic wave.

In quantum mechanics, such a miracle is the norm.

The miracle enables life to exist in our universe. Classical electromagnetic waves become soon too diluted to excite anything enough. Entropy grows too fast in classical physics. Photosynthesis would be difficult in a classical universe.

The reversed time model has similarity to the absorber theory of Wheeler and Feynman. The electromagnetic field of a photon is just a mechanism for relaying a "message" to the absorbing system that it is allowed to make a certain state transition, that is, absorb a certain photon.

Our teleportation model brought the emitting and the absorbing system very close, to explain why the electromagnetic field is not diluted. Our reversed time model augments this by saying that a transition in the emitting system can induce a transition in the absorbing system, such that the entire emitted energy is absorbed.

A classical model where the created pair absorbs the whole momentum of the photon

In a blog posting a couple of years ago we noted that an electron which is close to a positive charge may have the inertial mass large, even though m + V is zero. Here m is the electron mass and V is its negative potential close to the positive charge.

That is because the electron makes a "hole" in the electric field of the positive charge. If we move the electron, the energy content of the electric field moves to the opposite direction.

If m + V is negative, the inertial mass of the electron might be m + |m + V| = -V.

A positron may have an inertial mass m + V', where V' is the positive potential it has close to the nucleus.

                4 * 10^-14 m      3 * 10^-15 m

       Z+ ●                             • e-            • e+

                                                      ^

                                                      |  photon E

Let us put the particles in a line, such that the total energy of the pair equals the energy of an incoming photon, say, E = 511 keV. The separation of the pair is about 3 * 10^-15 m.

The inertial mass of the electron in the diagram may be, for example, 1.022 MeV. If Z+ is iron, its charge is 26 e, and the distance to the electron 4 * 10^-14 m, so that the potential V of the electron in the field of the nucleus is -1.022 MeV. The potential in the field of the positron affects the inertial mass of the electron, too, probably increasing it somewhat.

If we let the electron move downward in the diagram at the speed ~ c / 2, the electron can absorb the whole momentum of the incoming photon. Then the electron will have some kinetic energy, too. We need to readjust the distance between the electron and the positron so that the total energy

      2 m + potential + Mv^2 / 2= E,

where M is the inertial mass of the electron and v is its (nonrelativistic) speed.

There we have a classical model where we have a "virtual" pair absorbing the whole energy and the whole momentum of the photon.

The pair is not very "virtual" in any mysterious sense, but quite normal particles which interact. We have stressed that interacting particles do not need to obey the energy-momentum relation. They may be off-shell.

Our classical model suggests that Delbrück scattering would be possible quite far away from the nucleus. We need to look at calculations of Delbrück scattering.

The relevance of the classical model in real pair production

                       ____________ e-

                     /

          ~~~~

                      \____________ e+

                                | virtual photon p

                                |

     Z+ -----------------------------

Above is a typical diagram for real pair production from a photon hitting atomic matter.

The virtual photon gives the excess momentum p to the nucleus.

The positron is in the mysterious virtual state until it gets rid of the extra momentum. Our classical model shows that the virtual state might not be so mysterious, after all.

Does the Feynman diagram plane wave assumption produce incorrect results?

Let us assume that two electrons pass a nucleus. What is the probability that one of the electrons gains momentum p in the experiment?

We shoot the electrons from opposite directions toward the nucleus. We model the electron fluxes as plane waves with a definite momentum q or -q.

    q  e-  ------>      ● Z+      <-----  e-  -q

  

Let us draw some Feynman diagrams.

 -q  e- -----------------------

  q  e- -----------------------

                    |  p  virtual photon

      Z+ -----------------------

Above we have the basic case.

    q e- -------------------------

                       | p

   -q e- -------------------------

                   | p

     Z+ ------------------------

Above is another case.

The end result is the same in both diagrams: the nucleus and one of the electrons exchange momentum p. We measure the momenta of both electrons after the experiment.

Feynman diagrams calculate probability amplitudes in the momentum space. The amplitude is a function of the momenta of the output particles.

Feynman rules say that we must add the probability amplitudes (which are complex numbers) of the diagrams to get an approximation of the scattering probability. Typically, the probability amplitudes have a different phase. There is some destructive interference.

Adding probability amplitudes is correct if the output is a product of plane waves.

Note that the nucleus is an almost classical object. After the experiment, we can measure its path. Let us assume that the nucleus had very small momentum - it essentially stayed static in the experiment.

Electrons which gained the momentum p directly from the nucleus came from a relatively small spatial volume. They do not form a plane wave.

Electrons which gained the momentum via the other electron come from a much larger spatial volume.

Let us next do a thought experiment.

Let us assume that that we send two arbitrary waves which have a 180 degree phase shift. The wave A comes from a small spatial volume v. A weaker wave B comes from a large spatial volume V.

Why would there be a lot of destructive interference? There is not. We cannot destroy the energy that we used to create the wave A by putting energy to the wave B.

Feynman rules assume that the output of a collision is a product of plane waves. But often it is not. If the electron receives the momentum p in the first diagram, then it flew close to the nucleus. We gained knowledge about the relative position of the nucleus and the electron. Then we cannot model the output with plane waves.

We could calculate the experiment result using the Schrödinger equation because no new particles are created in it. Who thinks that a very crude plane wave model can replace the Schrödinger equation calculation? That would be a miracle.

How to fix Feynman rules? Obviously we have to be very careful if we try to add the probability amplitudes of two diagrams. If the output cannot be approximated with plane waves, then we cannot calculate any interference of the diagrams.

What about vacuum polarization? There Feynman did add the interference of two diagrams, the simplest one and the one-loop diagram. Is that an error?

What is the role of virtual particles or malformed waves?

Let us look at Coulomb scattering again.

    proton ●

                        ^

                        |

                        e-

url=https://www.unige.ch/isdc/ferrigno/index.php/download_file/50/149/

Andrii Neronov (2017)

We were not able to find from the Internet figures for the total cross section for pair production, but let us guess it is of the same order of magnitude as for 5 MeV gamma rays, if the electron is in the MeV range. That is, around 0.1 barn = 10^-29 m^2.

That suggests that the electron has to come within 1.5 * 10^-15 m from the center of the proton, to produce pair(s).

We could satisfy energy and momentum conservation even if the electron would pass the proton at a very large distance, say 2 * 10^-12 m, or the Compton wavelength of the electron. Any output from a collision satisfies conservation principles - there is no need for the particles to be born close to each other.

Tunneling, with 1 MeV of energy borrowed, is easy through a barrier of 0.1 Compton wavelengths, or 2 * 10^-11 m. The tunneling barrier cannot explain why the electron and the proton must come very close.

Let us imagine that the process creates a positron close to the electron, and an electron close to the proton. Both particles have to get their energy from the kinetic energy of the electron. To give up that kinetic energy, the electron must lose a lot of momentum. To lose that momentum, it must come very close to the proton. The Compton wavelength is way too far away. Here we have a semiclassical explanation why the cross section is ~ 0.1 barn.

The process is described using a particle model of the electron, and almost classical physics.

What are virtual electrons?

To analyze vacuum polarization we should understand what exactly are the virtual electrons in various processes. There is a virtual electron line in the Feynman diagram for pair annihilation or the converse process. There is a virtual electron in Thomson/Compton scattering.

The virtual electron in Thomson/Compton scattering is a "very real" electron that moves in interaction with the photon field(s). There is nothing unreal or mysterious in that electron.

The virtual electron in pair production

The virtual electron in pair annihilation, or the converse process, is mysterious. When we analyzed pair production from colliding photons using a the drum skin wave model, we suggested that the virtual electron is some kind of a malformed wave which transfers momentum in the Dirac field.

But later we used a particle model, and tunneling to potential wells, to explain pair production. How can we reconcile a particle model with a malformed wave?

As we have noted, the wave equation source, which creates an electron or a positron, is quite small, ~ 10^-15 m (in the MeV range). The source is as close as we get to a pointlike "particle" if we work in a wave model.

A source so small will produce also electrons in the GeV range, unless we use a cutoff, or sum over many different source locations, to eliminate waves which have too high a 4-momentum.

A possible model:

1. The virtual electron in pair production is just a symbol for the (complex) process which happens, in a particle model, in a volume of a size 10^-15 m.

This is analogous to the fact that the virtual photon which transfers momentum in Coulomb scattering is just a symbol for Coulomb interaction.

2. We external observers cannot discern details which are much smaller than the Compton wavelength 2 * 10^-12 m. For us, the electron and the positron just pop up in a volume ~ 2 * 10^-12 m.

An external observer may interpret the 2 * 10^-14 m scale process in the Feynman way: a positron coming from the future was scattered by a photon into a virtual electron, and that virtual electron was scattered by a second photon to a real electron.

This model reconciles the almost pointlike character of the photons and the pair, with the larger scale view of waves scattering from each other.

Collapse from a wave to a particle: the teleportation model

The "collapse" of a wave to a particle can be explained by the teleportation model which we introduced in an earlier posting. The original particle system which produced the wave is magically teleported close to the system which observes the particle. The original system is (in the MeV range) ~ 10^-15 m in size. This explains why the particle appears quite pointlike.

The Wheeler-Feynman absorber theory had some similarity to the above idea. In the absorber theory, a mysterious mechanism simultaneously decides the emitter and the absorber of a photon. It is like the teleportation process.