The problem with photon-photon scattering: the virtual pair has more degrees of freedom

In pair production and annihilation, the initial state and the final state have the same dimension in the degrees of freedom. The pair side has the orientation of the momentum p, and the photon side has the orientation of the momentum q. The freedom is two-dimensional.

         q            p            k

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

                   |            |  virtual

                   |            |  electron

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

       -q            -p           -k

In the diagram, two photons that come from opposite directions are scattered by a virtual pair loop.

The sum of the spatial momenta

       q - q = p - p = k - k = 0

is zero at each stage. Energy conservation requires that |q|= |k|.

Feynman rules would allow to add any 4-momentum to circulate in the virtual pair loop, which causes the Feynman integral to diverge logarithmically.

But we will now use the classical particle model of the virtual pair. Then |p| can have an arbitrarily high value which is compensated by making the electron-positron distance very short. The classical energy of the pair is set equal to the energy of the photons.

Thus, the momentum p may have an arbitrary direction, and |p| may be arbitrary. The freedom is 3-dimensional.

What are the separate "channels" of energy transmission through the diagram? If we set |p| to a constant value C, is that channel separate from the channel C' != C? If yes, then there is complete transmission because we have an infinite number of channels which carry energy to the other side.

The different number of dimensions of freedom is a fundamental problem. It is clearly connected to the divergence problem of Feynman integrals.

In a Feynman integral, the measure of the momentum space is the ordinary measure of R^3. Why should the measure be such? That is another fundamental question. Even if we through some trick get the integral to converge using the ordinary measure, it would anyway diverge if we choose some different measure over R^3.

In a deterministic universe, p would be uniquely determined by the input. The degrees of freedom of p should have the same dimension as those of q.

The universe has to be probabilistic if p has more degrees of freedom. But then we face the problem what weights should we put on various values of p, i.e., what measure to use in the space of the possible values of p. Why would p' have the same weight as p, if |p'| = 10^9 |p|?

There is a conceptual problem. This is not just about finding some trick, like regularization, which makes the Feynman integral to converge.

Restrict the cross section with the angular momentum?

If the combined energy of the photons is E << 1.022 MeV then the virtual pair has the maximum possible separation

       r_0 / 2 = 1.4 * 10^-15 m,

where r_0 is the classical electron radius. Let us assume that the particles in the pair move at the speed of light.

The photons have the same angular momentum if they pass each other at the same distance r_0 / 2.

The maximum possible cross section would be 0.06 barn = 6 * 10^-30 m^2.

M. Bregant et al. (2008) report that the cross section for 532 nm photons scattering from each other is less than 3 * 10^-60 m^2. That would mean the photons passing each other at a distance < 10^-30 m. M. Bregant et al. say that the cross section calculated from Feynman diagrams is 10^-67 m^2.

It looks like angular momentum is not the way to restrict the cross section enough.

What is the electron-positron annihilation rate at keV or eV energies?

A brief Internet search did not return any measured data for low-energy annihilation cross sections. It looks like other processes dominate over the simple annihilation process.

There is some data for positronium formation, as well as for complicated processes when a positron meets a molecule.

The formula which is derived from the Feynman diagram suggests that the cross section is

       σ = 1 / β π r_0^2,

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

The cross section using a classical particle model is very different. The pair quickly radiates away all their energy when their distance is about r_0 / 2.

Let us assume that the maximum possible angular momentum of the pair is J_a at the moment of annihilation, or just before it.

The linear momentum of a particle is initially

       p = m_e β c.

If the impact factor is b, then the initial angular momentum is

       J = b m_e β c,

and the maximum value for b is

       b = J_a / (m_e β c).

The classical cross section is proportional to 1 / β^2, not to 1 / β as in the Feynman formula.

For a 2 eV electron, β = 0.003 and for a 2 keV electron, β = 0.1.

The difference in predicted cross sections is very large.

The classical limit

Let us increase the masses of the particles by a large factor M, and their charges by a factor sqrt(M).

The particles will follow the same classical trajectory as a classical electron and a positron.

We can monitor the trajectory of the heavy particles accurately, because now they really are classical particles. If they lose energy in radiation as in the Larmor formula, they have lost all their energy when they are roughly 1.4 * 10^-15 m from each other. We may assume that they are annihilated by then.

The heavy particles send many photons whose wavelength is in the range 10^-15 m. The process is very different from the Feynman diagram for electron-positron annihilation.

The classical formula for the cross section gives the right answer in the classical limit, the Feynman formula a wrong answer.

Who is right?

The Feynman model assumes a very short interaction of particles, which otherwise fly freely. Could this be the reason for the discrepancy?

The electron and the positron according to the Schrödinger equation

The electron and the positron probably obey the Schrödinger equation. They are low-energy particles like the orbital electron of the hydrogen atom.

If we just look at the position difference vector r of the electron and the positron, we can model the system as a single particle in a central Coulomb potential.

For a free electron, the Schrödinger equation probably calculates roughly the same trajectory as classical physics.

Thus, the cross section for a close encounter really is ~ 1 / β^2, also according to the Schrödinger equation. If there is a close encounter, why would annihilation probability be proportional to 1 / β, like the Feynman formula claims?

The pair annihilation cross section once again

In the February 6, 2021 blog posting we calculated that the ratio of the Feynman integrals to opposite directions of the reaction is 1 / β^2 for a massive Klein-Gordon "pseudoelectron".

If we calculate the probability of annihilation of the "pseudoelectron" by integrating over all possible photon momenta q, and |p| is small, then the Feynman integral value is essentially independent of p.

That would mean that the annihilation probability does not depend on p at all for small p, which would be very strange.

To the other direction, the integral over possible p is ~ β^2. That sounds more sensible.

             p              q

    e-   ---------   ~~~~~~~~  photon

                     |  p - q

                     |  virtual

                     |  electron

    e+  ---------   ~~~~~~~~  photon

            -p              -q

Let us calculate the integrals again, this time using the spinor field propagator. That is, we look at the real electron instead of the Klein-Gordon "pseudoelectron".

The Feynman propagator for a spinor electron is

        i (k-slash - m I) /  (k^2 - m^2 + i ε).

There, k is the 4-momentum of the virtual electron. The energy in the virtual electron is zero in annihilation, because of symmetry. I is the unit 4 x 4 matrix.

Let |p| be small and p constant. Then |q| is a little larger than m.

We have to integrate

        i (-p-slash + q-slash + m I)

          /  (q^2 + m^2),

where we used the fact that k^2 is roughly -q^2.

Let us vary q. The denominator is constant. What do the slash terms give?

Each spatial component of q varies symmetrically around zero.

The calculation is complicated.

https://arxiv.org/abs/2003.02231

In the link, S. Bragin and A. Di Piazza (2020) perform the steps and get the same result as V. B. Beretstetskii et al. (1982) in the book Quantum Electrodynamics.

At this stage we have to believe that Feynman formulas really give the results known in literature. The cross section is ~ 1 / β, which for an unknown reason differs from the classical formula ~ 1 / β^2.

The virtual electron can be understood as the carrier of a 1 / r^2 potential "force"?

The cross section for pair annihilation is essentially constant for small β if we calculate the Feynman diagram using the propagator for the massive Klein-Gordon equation (the "pseudoelectron"). The propagator for large |p| is roughly 1 / |p|^2.

The propagator for the Dirac equation for large |p| is something like ~ 1 / |p|.

We may interpret the Dirac propagator as a carrier of a force whose potential is 1 / r^2.

When the photons scatter from this 1 / r^2 Dirac "potential", they are converted into a pair.

The Coulomb 1 / r potential has the property that if we shoot a particle horizontally from the surface of a sphere, the particle will go to infinity if its kinetic energy is larger than the negative potential energy.

But if the potential is like close to the event horizon of a black hole, then a horizontally shot particle will crash into the surface of the sphere. The particle has to be shot at an angle to make it climb up the potential.

According to the Feynman formula, the cross section for pair production is ~ 1 / β. That suggests that the potential, which the particle has to roll up, is steeper than 1 / r.

The classical cross section for annihilation in a Coulomb 1 / r potential is ~ 1 / β^2.

If the potential is steeper, the cross section may be, e.g., ~ 1 / β.

The magnetic force of the electron and positron spins may make the potential steeper than 1 / r.

The Dirac equation is kind of a "square root" of the massive Klein-Gordon equation. The propagator clearly reflects this fact. It might be that the potential 1 / r^2 and the cross section ~ 1 / β somehow are a result of being a "square root".

We need to check the form of the magnetic force between the spins, and also figure out why the classical limit suggests that the cross section should be like for the 1 / r potential.

Where do the momenta disappear in "classical" annihilation of a pair?

Let us imagine that a classical electron and a classical positron collide head-on. Let the initial velocity of the particles be small.

                   -------   electromagnetic wave

                    -----

                     ---

         e- ------>  <------ e+

                     ---

                    -----

                   -------   electromagnetic wave

We calculated using the Larmor formula that at the distance 1.4 * 10^-15 m, the pair emits all its energy, 1.022 MeV, in a very sharp "wrinkle" in the electric field of each particle.

The emitted wave looks like a dipole wave and does not carry much horizontal momentum in the diagram.

What happens to the substantial horizontal momenta of the particles at the moment of annihilation?

In the Feynman diagram, a mysterious virtual electron carries the momentum between the annihilating particles.

The spins of the electron and the positron are assumed to be opposite at annihilation. Both particles are magnets whose north pole points to the same direction.

We calculated in an earlier blog posting about the Pauli exclusion pronciple that the magnetic repulsion exceeds the electric attraction already at the distance 3 * 10^-13 m if the magnets are side-by-side.

Maybe the magnets are positioned linearly? Then the magnetic force would strengthen the attraction.

There is a lot of horizontal momentum in the electric field of each particle at the moment of collision. But the electric fields almost cancel each other at that point. The neutralization of momentum may actually happen in the electromagnetic field.

The collision of static electric fields would eventually cancel the big opposite momenta of the particles. The mass-energy of the particles themselves might be already zero at that point.

In the drum skin model, we let the two pressing fingers approach each other, and at the moment of "annihilation" suddenly remove the fingers. If most of the energy escapes to the normal direction from the "collision", then the momenta in the original static fields (depressions) of the fingers has to get neutralized in some way.

Comparison of the output electromagnetic wave in a classical model versus a Feynman diagram

Let us have |p| very small in the propagator G_F(p - q) of the previous blog posting. Then the value of the propagator is almost the same for all directions of q.

The Feynman diagram predicts that the two photons will be emitted to a random direction from the annihilation (they have opposite momenta, but the direction of q is random).

The classical model gives approximately the same prediction: in most cases, the electron and the positron will orbit a little around each other. The direction of the dipole in the final annihilation may be almost random.

What about the spin? If the particles have opposite spins, their north poles point to the same direction. We suggested that they will be aligned as a line at the annihilation. The photons will be emitted mostly normal to the direction of the spin.

Let us check what literature says. A brief Internet search did not return any data about the direction of the emitted photons versus the direction of the spins of the electron and the positron.

Fermi's golden rule and "density" of states

In our previous blog post we raised the question what exactly is "density of states".

https://en.wikipedia.org/wiki/Fermi%27s_golden_rule

Fermi's golden rule relies on the concept of density of states. Wikipedia offers several derivations for the rule. For a mathematical theorem one proof is enough. Why do we need several proofs? Something is fuzzy in the theorem.

The cross section if the "hole" is at the center of a spherically symmetric potential

Let us consider a transition where a classical particle has to come to within some fixed distance r of a spherically symmetric potential in an R^3 space 1. If it gets that close, it moves to another R^3 space 2 where the potential is constant. The particle is non-relativistic.

Let the potential be larger in the space 2.

The hole appears as a sphere in both spaces. The cross section is

        σ_2 = π r^2

from the space 2 to the space 1.

Angular momentum is conserved in a spherical potential. The maximum angular momentum that the particle can possess when it falls down from the space 2 is

       J = r m|v_2|,

where v_2 is the velocity in the space 2 and m the particle mass.

Let the velocity of the particle be v_1 in the space 1. Since angular momentum is conserved, the maximum possible impact factor b for the downfalling particle in the space 1 is

        b = r v_2 / v_1.

It is obvious that if we shoot a particle in the space 1 with an impact factor smaller than b, then it will climb to a distance < r in the central potential and go to the space 2.

Any state transition path 1 -> 2 or 2 -> 1 can be traversed to the opposite direction. The laws of nature are time symmetric.

The cross section for the transition 1 -> 2 is

       σ_1 = π b^2 = π r^2 (v_2 / v_1)^2.

We see that the ratio of the cross sections to opposite directions is the same as the ratio of the "areas" that the momentum vector in each space can span. In the space 1, the area is 4 π (m v_1)^2 and in the space 2 it is 4 π (m v_2)^2.

In this case the cross section to the opposite direction can be derived from the cross section to the other direction.

Is there a general rule for obtaining the cross section to the opposite direction?

In the case of a spherically symmetric potential, we were lucky to be able to use conservation of angular momentum.

If the potential is another shape, then we cannot use the conservation law.

What about more complex processes, like pair production / annihilation? They probably are, in some sense, spherically symmetric.

What is time symmetry of natural laws in a probabilistic universe?

We need to define time symmetry of natural laws in a probabilistic universe. If a state A changes to B with a 50% probability, and otherwise to C, what does it mean to say that laws are time symmetric?

Suppose that we shoot a plane wave toward a particle. The output is a complex combination of plane waves.

We would like to know what happens to a single incoming plane wave if we reverse time. Knowing that a complex combination will produce a single plane wave is not interesting because we cannot produce such a complex combination.

We can reverse time in a Feynman diagram. That might be what we want: the probability amplitude of an incoming plane wave A producing a plane wave B is the same (or actually, the complex conjugate) as to the reverse direction. In a deterministic universe, the probability is always 1 or 0.

What about wave mechanics in classical physics? Can we calculate scattered amplitudes using the symmetry rule we sketched?

The Afshar experiment shows that there is no simple way to calculate the probabilities for a reverse reaction

We wrote about the Afshar experiment on December 23, 2020. In that experiment, two plane waves "help" each other to avoid a grid. Time reversal really requires both plane waves to be present. 

The Afshar experiment shows that there is no general rule that allows us to calculate the response to a plane wave which is sent to the reverse direction of the process. But there might be such rules in special cases?

https://en.wikipedia.org/wiki/Unitarity_(physics)

The concept of unitarity of the scattering matrix S is relevant here. The inverse of a unitary S has to be its conjugate transpose:

       S* S = S S* = I.

But if the output has more states than the input? Then S is not a square matrix and unitarity is not defined.

In the Afshar experiment, the photon interacts with external macroscopic equipment. In principle, the photon & the equipment might develop in a unitary way, but we have no means of specifying the state of the macroscopic system. Therefore, we cannot utilize the concept of unitarity.

Interaction with a macroscopic system can be seen as a "measurement". A measurement destroys unitarity.

A unitary scattering matrix S

In quantum field theory, if a collision of two particles can produce a third particle (e.g., bremsstrahlung), then the output state is 3-dimensional, though the input state was 2-dimensional. Can we apply the concept of unitarity then?

If the collision does not create new particles, then unitarity might apply? The state spaces are continuous in most cases.

Suppose that the input has a finite number N of possible states and the output the same number N of possible states. Then the scattering matrix S tells the probability amplitude P for each transition

       n -> m:    P = S(n, m)

Unitarity of S then says that the probability amplitude of the reverse transition is the complex conjugate of P.

Reversing time in a Feynman diagram

The Feynman method of keeping the probability amplitude the same (or actually, taking the complex conjugate) when reversing time looks like a good solution.

Suppose that we have particles with momenta p and -p colliding. Let the angle of p be φ relative to the x axis. 

Let us first set φ = 0.

We assume that the output is two particles with momenta q and -q. Energy conservation may fix |q|. Let θ be the angle of q relative to the x axis.

The scattering function (matrix) is then a complex-valued

       S(θ).

How to get the reverse function? The relative angular distribution might be given by the same function S. But the cross section of the reverse reaction may, for example, be a lot smaller than the reaction.

             p                     q    

    e- -----------  ~~~~~~~~~~  photon

                       | p - q

                       | virtual electron

    e+ ----------  ~~~~~~~~~~  photon

            -p                   -q

Let us look at the Feynman diagram of annihilation / pair production. Above, p and q are spatial momenta, not 4-momenta.

Let us fix p. The Feynman integral is over the possible values of q then. To the other direction, q is fixed and the integral is over possible p.

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

The Feynman propagator for a virtual massive Klein-Gordon "pseudoelectron" is

       G_F(p - q) = 1 / (-(p - q)^2 - m^2 + i ε).

In the 4-momentum of the virtual "pseudoelectron", the energy E = 0. That is why we just have the negative square of the spatial momentum in the denominator.

           p

            ^

            |

            |

             ----------------------> q

The value of |p - q| only depends on the angle θ between the vectors. If we let p or q vary, then G_F(p - q) has the exact same shape in both cases because it is a function of the angle θ. The only difference is in the area of the sphere which the vector p or q draws.

The ratio of the integrals over G_F(p - q) in each case is proportional to the integration area.

We see that the integrals are proportional to the area which the momentum vector can draw on each side of the reaction.

Annihilation at low energies is a complicated process - the spins play a role

Annihilation at low energies seems to be a complicated process. The spin of the particles affects things, and a positronium atom can form.

http://meta-phys-thoughts.blogspot.com/2018/11/a-heuristic-derivation-of-pauli.html

In a blog post about the Pauli exclusion principle we calculated that the magnetic repulsion or attraction from the spins dominates when electrons come very close to each other. That fact should show up in precision measurements of scattering processes. Feynman diagrams only take into account the Coulomb force.

The repulsion of the spins of the electrons, when the spins are parallel, is extremely strong at short distances. We suggested that this is the reason for the Pauli exclusion principle: the wave function must be essentially zero for short distances if the spins are parallel.

Photon-photon scattering

In photon-photon scattering, a virtual pair is created first and then it annihilates.

In the photograph model of quantum field theory, any classical orbit of the pair contributes to scattering. The situation is complicated, as the pair can have a hyperbolic orbit and be located anywhere in that orbit. Electromagnetic radiation quickly drains away all the energy of the pair.

We ignore the spins for now.

How to assign weights to various possible paths?

How to determine the weights of paths in a Feynman path integral?

Our previous blog post brought up this well-known problem. There is no formal definition of how the weights should be assigned to paths in a path integral.

A plane wave of electrons

            -----------------  crests

            -----------------  of

            -----------------  waves

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

Let us start from the simplest case: a plane wave.

If the flux is electrons, then it is natural to set the equal weight for each path which starts from a specific coordinate x.

But now we face a dilemma: the real axis R is infinite. The density of the probability distribution cannot be defined. It cannot be 0 nor any ε > 0.

Using nonstandard arithmetic with infinitesimal numbers would be kind of a workaround in some cases, but that does not help if we need to pick a "random" real number.

Let us just cut off the infinity in the real line at some large M. That works in most cases. But it is unfortunate if we need to start regulararization and renormalization at this early stage!

What about using a localized wave packet? That does not help either because we can never make sharp edges to it. And the Fourier decomposition of the wave packets consists of plane waves.

Maybe the mathematically most beautiful solution would be to use infinitesimal numbers, distributions, and the like. But in practice, problems do not usually arise at this stage.

Path for a "photon"?

This is a hard problem. If we treat the photon as a particle, how does it enter the stage in the deep down classical world of particles?

The electron is in classical physics a particle, but the "photon" is a wave.

In a Feynman diagram, all particles, also photons, are plane waves. They do not have a "position" at all because they exist in the momentum space. But we want to calculate path integrals in the position space.

In an earlier blog posting we suggested treating photon absorption as reversed photon emission. In classical physics electrons create complex electromagnetic waves. We may do a Fourier decomposition of the emitted wave and imagine that each plane wave component is an incarnation of a photon.

We suggested that destructive interference wipes away all but one plane wave mode from the complex emitted electromagnetic wave.

How to reverse the process of emission? Many different classical world processes can emit the exact same plane wave, i.e., the photon. If we try to add to the path integral all these infinite number of processes, how do we assign weights to them?

The classical process behind the Feynman photon-electron scattering matrix

In a Feynman diagram, an electron can emit and absorb a photon. The diagram does not specify any classical path of events. It just assumes a certain scattering matrix for the input and the output.

The scattering matrix is derived from a semiclassical calculation where an electromagnetic wave moves an electron. Thus, there is a simple classical system behind the Feynman diagram, after all. Let us analyze what exactly is assumed in the classical process.

       |    |    |    |    |    beam of laser

       |    |    | •  |    |

                   e-

We assume a classical plane wave, for example, a laser beam. We calculate how much it makes the electron oscillate, and calculate the emitted electromagnetic wave with a classical formula.

The input is a classical plane wave, and the output is a complex classical dipole wave.

The "processing system" is the electron, which is assumed to be initially static. What degrees of freedom does it have? Just its spatial location. In a path integral we simply assume that the location makes no difference. The issue of weights of paths does not come up.

The Lyman alpha absorption/emission

Let us assume that the input is a plane wave.

In this case the "processing system", the hydrogen atom, is very complex. Classically, it would be a classical electron particle orbiting a classical proton. There is an infinite number of possible paths.

We may approximate the atom as a harmonic oscillator. Or we can calculate using the known orbitals and Fermi's golden rule. In an earlier blog posting we showed that at the exact resonant frequency, the absorption cross section can be derived from classical angular momentum J conservation.

What are the weights that we put to different paths in these cases? It is not clear.

Pair production

The input in pair production is two colliding electromagnetic plane waves. The processing system is a pair which is born.

In a Feynman diagram, we assume that the photon-electron scattering process can be generalized to include scattering the electron to the past or the future. The bold generalization seems to work but why?

A more traditional calculation treats pair production as a process where a -511 keV electron is excited to a +511 keV state. This is analogous to the Lyman alpha case. A harmonic oscillator model probably gives a reasonable estimate. Why?

We in our blog have suggested using classical paths of the pair, but we have the problem how to assign weights to the various cases.

Pair annihilation

The Feynman diagram for pair annihilation is obtained by simply turning the pair production diagram upside down. That suggests that the process really is symmetric in the real world.

The photograph model may be able handle annihilation if we assume that the pair simply disappears when it has radiated away all its mass-energy.

Note that the static electric field loses its energy as the separation of the pair grows smaller. If we assume that all the mass-energy of the electron is in its static electric field, then annihilation can be understood as the disappearance of the static electric field.

A drum skin model might illuminate annihilation. One finger presses the skin upward and another presses the skin downward some distance away. The fingers have done work to deform the skin. If we suddenly remove the fingers, all the static deformation energy in the drum skin is converted to waves.

Let the pair collide head-on. Our analysis in an earlier blog post suggests that the pair sends a very sharp wave when they are at the distance 1.4 * 10^-15 m from each other.

It is probably a dipole wave. The Fourier decomposition of the wave contains extremely high frequencies because the wave is born from a very abrupt process. Hypothesis: the components which survive destructive interference have the distribution of a dipole wave to various directions.

If the pair does not collide head-on but have an impact parameter b, what is the produced wave like?

Zero angular momentum makes the pair to move along a straight line segment. A non-zero angular momentum means a hyperbolical orbit.

The orbit must take the pair close enough, so that the distance is ~ 1.4 * 10^-15 m. The radiation must carry away all the energy and the angular momentum.

The outgoing wave is probably a mixture of a dipole wave and a circularly polarized wave. Usually, two photons are born. They have to take away the angular momentum which means that they cannot originate from the exact same point.

We need to check from literature what is known about the angular distribution of annihilation photons and their polarization states. A dipole antenna outputs linearly polarized photons and a rotating dipole outputs circularly polarized photons.

The "size" of the degrees of freedom space somehow determines the cross section

Now that we understand the photon output of annihilation, how can we calculate the cross section of the reverse process and the directions of the output particles?

Gould and Schreder (1967) give the pair production cross section as

        σ = β π r_0^2,

where the produced particle velocity is β c and r_0 is the classical electron radius.

Frank Rieger gave the annihilation cross section as

       σ = 1 / β * π r_0^2,

where the velocity of the incoming particles is β c.

The formulas are remarkably similar. The production cross section is β^2 the annihilation cross section. As if we would have particles passing through a hole whose area is π r_0^2.

                   _____________

                 |                        |

         ____|                         |

       |       |                          |

       |                                  |

       |____|                          |

                 |_____________|

 pair,

 size β^2      photons, size 1

            hole,

            size π r_0^2

The "vessel" on the pair side is β^2 the size of the vessel on the photon side. Incidentally, the momentum vector length for a pair is β times the vector for a photon, and the "area" of the available momentum space is β^2 the area of the one for the photons.

There is a rule that the "size" of the available degrees of freedom space determines the cross section in the fashion above. Obviously, this should follow from the steady state of maximum entropy. But we need to analyze this in detail.

The photograph model strongly dislikes short-lived states: this is the reason why photon-photon scattering has a small cross section?

In an the January 22, 2021 blog post we claimed that the Feynman diagram gives too low a probability amplitude for photon-photon scattering.

It turns out that Richard Feynman maybe was right and we were wrong.

Let the combined energy of the photons be < 1.022 MeV.

      photon                      photon

             ~~~~  -----------  ~~~~~

                         |            |    virtual

                         |            |    pair

             ~~~~  -----------  ~~~~~

      photon                      photon

There are four electron propagators in the loop.

We argued that once the virtual electron and positron are "born", they will annihilate at a probability 100%. Therefore, the Feynman diagram underestimates the probability amplitude of annihilation, since it keeps annihilation behind 3 extra electron propagators after the pair is born.

The annihilation is certain - that is true - but the events have too high a "resolution" because the pair is short-lived. A high resolution in the time direction requires high energy and a large palette of frequencies, which are not available.

Let us try to draw the events from the birth of the pair to its annihilation. Drawing the detailed classical paths would require extremely high frequencies and momenta, and a large palette.

They would be available if the particles were macroscopic and we would have measured their position and momentum very accurately. We then could describe the particles with sharp wave packets. Those packets contain a very wide palette of frequencies.

But such a palette is not available. We have to simplify the picture to one plane wave being born and annihilated. The simple plane wave describes the "system" of the virtual pair in the simplest possible way. The sharpest detail in the simplified picture is the short lifetime of the pair.

We assume that our photons are very precisely tuned to a certain energy. The wave packet has to be at least 10 million cycles long to ensure that the uncertainty in energy is less than one millionth.

  input ------> | virtual pair| -----> output

  precisely                                   scattered

  1,000,000 eV                            photons

  of energy

The input to the virtual pair picture essentially has just a single frequency. We cannot really draw the virtual pair in the photograph in a meaningful way.

But probably the "channel" in that precise frequency (1,000,000 eV) still relays some energy to the output.

Drawing a short-lived virtual pair is like making a wave packet whose time dimension is short. The packet has a spectrum of frequencies (= energies). Of that spectrum, the narrow band which matches the input frequency can relay energy to the output.

The lifetime of a virtual pair as classical particles versus the oscillator model

The classical model of a virtual pair tells us that it is very short-lived if the energy is even 1 eV below 2 m_e.

The "lifetime" of forced oscillation in a harmonic oscillator is

        E / (E - E') cycles,

if E is the resonance frequency of the oscillator and E' is the frequency of the driving sinusoidal force.

If h f = 1 MeV, then one cycle is

        1 / f = h / 1.6 * 10^-13 J

                 = 4  * 10^-21 s.

Let E' < E. What is the lifetime of a virtual pair whose energy is 

       2 m_e E' / E?

Let the distance of the pair be r, and largish. The classical acceleration is

       a = k e^2 / (m_e r^2).

We get a rough approximation for the fall-down time t by setting

       r = 1/2 a t^2.

Then

       t = sqrt(2 m_e r^3 / (k e^2)).

Let us calculate some examples. If r = 2 * 10^-12 m, then the acceleration is

       a = 5 * 10^25 m/s^2,

and

       t = 3 * 10^-19 s,

or 75 cycles. The oscillator formula gives a longer lifetime, 1300 cycles.

What if we let the virtual pair orbit each other and calculate the lifetime with the Larmor formula?

The dissipated power is

       P = e^2 a^2 / (6 π ε_0 c^3)

          =  10^-2 W.

The Larmor lifetime is

        t = 1.6 * 10^-13 J / 10^-2 W

           = 1.6 * 10^-11 s.

Very long! But if we assume that the virtual pair is created from a close collision of photons, maybe the orbiting pair has way too much angular momentum? Also, the orbiting pair would send very long wave radiation. We assume that the pair was created from two large photons.

If the distance of the virtual pair is very short, say, r = 2 * 10^-15 m, then our classical model gives an extremely short lifetime t ~ r / c = 10^-23 s for the pair. The oscillator model gives a lifetime ~ 10^-20 s. Which one is right?

The oscillator model would simplify all temporary states to behave in the same way: the lifetime of the state would only depend on how many percent of energy we are lacking. That sounds implausible. The classical model allows temporary states to have versatile behavior. We conclude that the oscillator model is too simplified. But we do not know yet if the classical model is the right way.

If the particles are very heavy, then we are working in the classical limit, and then classical physics is certainly the right way to calculate the lifetime of a temporary state. Annihilation of macroscopic particles will send a huge number of photons. The converse reaction is practically impossible then.

How does the path integral cause virtual pair paths to have an almost complete destructive interference?

We have claimed that the wave-like behavior in the photograph model of quantum field theory is a result of path integrals.

What are the "paths" in our example and how does the destructive interference occur? We need to study this.

What is the effect of high-momentum virtual pairs?

We have shown that classical virtual pairs have an extremely short lifetime if they have high momentum. An extremely short lifetime means that drawing the picture would require a huge palette of energies. That, in turn, implies an almost complete destructive interference for the process.

But there is an uncountable number of high-momentum states of virtual pairs. How do we assign "weights" to them?

The Sommerfeld atom model

Let us consider classical orbits of the electron and the positron. The energy E and the angular momentum J determine the shape of the ellipse.

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

http://www.thestargarden.co.uk/Sommerfelds-atom.html

This sounds a lot like the Sommerfeld atom model (1915) with elliptical orbits. His model explains most of the spectrum of hydrogen - this fact supports our claim that deep down particles are classical, and wave behavior is the result of path integrals.

We may model the pair as a single particle which orbits under the mutual potential.

The particle is under huge acceleration and radiates its energy away very quickly. The polarization of the produced photons reveals the current direction of the velocity vector v of the particle.

But what determines |v|? It affects the sharpness of the wrinkle in the electromagnetic wave produced by the particle. However, destructive interference wipes away the information about the sharpness in the final wave output.

Should we assume that the virtual pair decides |v| at "random"? Random according to what measure?

The problem is: if the "channels", which drain energy from a system, have an extra degree of freedom, what happens?

Cross section of photon absorption: hydrogen Lyman alpha

In a blog post last week we asked what restricts the absorption cross section of a hydrogen atom. If the atom A outputs a plane wave in the 2p -> 1s transition, then in principle, the atom B could absorb that plane wave with a 100% probability.

https://arxiv.org/abs/1704.03416

Mark Dijkstra (2017) in Figure 16 has calculated the cross section with classical methods as 7 * 10^-15 m^2.

We suggested in a blog post that the angular momentum, which the electron in the hydrogen atom can absorb, restricts the absorption rate. Let the atom A emit a 122 nm Lyman alpha photon. The atom A recoils to a random direction.

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

The speed of the 1s electron in the hydrogen atom is 1 / 137 c. Its momentum is 2 * 10^-24 kg m/s.

The distance of the electron from the proton is the Bohr radius r_0, 5 * 10^-11 m. We conclude that the electron may absorb an angular momentum

       J = 2 p r_0 = 2 * 10^-34 kg m^2 / s.

The coefficient 2 comes from the fact that the electron can make a U-turn.

The momentum of a 122 nm photon is

       p = c * h f / c^2 = h / λ

          = 5 * 10^-27 kg m/s.

The maximum distance at which the absorbed photon may pass is

       r = J / p = 4 * 10^-8 m = 40 nm.

The maximum possible cross section is

       π r^2 = 5 * 10^-15 m^2.

The figure given by Dijkstra was 7 * 10^-15 m^2. Why is the Dijkstra figure larger than ours? We assumed that the electron would stay at the Bohr radius, but it climbs to the second orbital.

The radius of the second Bohr orbital is 4 r_0 and the electron speed there is 1/2 of the speed of the first orbital. The angular momentum is 2X of the first orbital. We must raise our figure for J by 50%. Then we get a maximum cross section

       π (60 nm)^2 = 11 * 10^-15 m^2,

which agrees with Dijkstra.

The Dijkstra cross section is surprisingly large. The silhouette of a hydrogen atom only has an area ~ 10^-20 m^2. The cross section is 700,000 times larger.

If the photon passes the hydrogen atom at a distance < 0.4 λ, it gets absorbed.

Cross section for photon-photon scattering

We can get an upper limit for photon-photon scattering cross section through calculating the maximum angular momentum that a (virtual) pair can possess if it has the energy of the colliding photons.

If the pair has 0.5 MeV of energy, then the maximum possible distance is 2.8 * 10^-15 m, but then the particles stand still.

The dipole electric field energy is concentrated between the particles.

Let us calculate with a particle-force model and ignore the effect of the electric field energy distribution.

If we put the particles at a distance 1.4 * 10^-15 m, they have 1.0 MeV mass-energy, -1.0 MeV potential energy, and 0.5 MeV of kinetic energy. Their speed is 0.85 c.

Their angular momentum J is the same as the photons passing each other at a distance 3.6 * 10^-15 m.

We conclude that the cross section is less than 3 * 10^-29 m^2, or 0.3 barn.

Our blog post on January 22, 2021 contained figures that have been calculated from Feynman diagrams. The diagrams have four propagators and the cross section has to be very small then. Literature says that the Feynman diagram cross section is less than 10 microbarn.

Individual Feynman diagrams diverge but their sum converges. Regularization (removing the divergence in an ad hoc way) has to be used.

There is a huge difference from our estimate. Let us check what are the known bounds for scattering.

https://arxiv.org/abs/0805.3036

M. Bregant et al. (2008) report that the measured cross section for 532 nm photons is less than 2.7 * 10^-60 m^2. The upper bound is still 20 million times higher than the calculated QED value.

Our maximum angular momentum argument removes the infinity in a Feynman integral, because the cross section gets an upper bound. The bound is very big, though.

We need to find a way to curtail the probability that a virtual pair absorbs long wavelength photons. Intuitively, it is extremely improbable that a pair whose separation is very small absorbs a 532 nm photon.

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

Rayleigh scattering cross section for nitrogen for 532 nm light is only 5 * 10^-31 m^2.

A virtual pair lives an extremely short time. That might be the reason why the probability of photon-photon scattering is very small.

Conclusions

The angular momentum argument makes individual Feynman integrals (or, more precisely, path integrals) to converge in the position space in photon-photon scattering, because there is a maximum angular momentum which can be held by a virtual pair. The colliding photons have to come quite close - otherwise the angular momentum of their relative motion is too large.

For the hydrogen atom Lyman alpha absorption, the angular momentum argument gives quite a precise correct value.

We need to understand better how a virtual pair absorbs the colliding photons. The measured cross section is very small for 532 nm photons.