Why the vertex function affects the electron anomalous magnetic moment but not its charge?

SOLVED March 1, 2021! The calculation of the anomalous magnetic moment is done in the magnetic part of the Gordon decomposition. The Feynman diagram and the calculation rules in this case concern interaction with a magnetic field only. In our blog, Feynman diagrams have always been about the Coulomb interaction between charges. That caused the confusion.

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The Dirac equation predicts the ratio of the spin angular momentum and the magnetic moment of the electron, assuming that we know the charge and the mass of the electron.

If the electron were a point charge which moves along a circular path with the angular momentum equal to the spin of the real-world electron, then its magnetic moment would be exactly 1/2 of the value predicted by the Dirac equation.

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

The charge in the electron seems to rotate at 2X the radius of the mass of the electron.

Julian Schwinger in 1948 calculated the QED correction to the gyromagnetic ratio

        g_e = 2 + α / π + ...,

where α is the fine structure constant ~ 1/137.

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

The modern calculation is based on the vertex function of QED. The integral has to be regularized to remove both an infrared divergence and an ultraviolet divergence. QED predicts the measured g_e - 2 with an impressive 10 significant figures.

  virtual photon q

        ~~~~~

                     \

                       \

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

            \                   /

               ~~~~~~

            virtual photon p

If the vertex function concerns all absorptions of virtual photons by electrons, then it should affect the response of the electron in the Millikan-Fletcher experiment. The measured charge of the electron would appear larger than the "bare" charge. Let the coefficient be, for example, 1.001.

If the Dirac equation is about the bare charge of the electron, then if we use the measured charge in calculations, we already exaggerate the magnetic moment by the factor 1.001. The bare charge, which is assumed to be spinning, is a little smaller than the measured charge.

Why would we need to include the factor 1.001 the second time when we measure the electron magnetic moment in a Penning trap?

Let us check the literature. Has anyone thought about this and come up with an explanation?

How can QED predict the electron anomalous magnetic moment so precisely? Why regularization works?

There is also another mystery in the electron anomalous magnetic moment: how can the Feynman diagram approximation be so precise? The reason might be that the phenomenon is so weak that the perturbative approximation yields good results. The first term, α / π, already calculates the experimental value with a precision of 2 / 1,000. The QED prediction is obtained from the sum of third order diagrams. With good luck, the current precision of 1 / 10^9 is possible.

The prediction of QED depends on regularization, that is, an ad hoc removal of the ultraviolet and infrared divergences in the integrals. Since the prediction is correct, regularization has to be the "right" way to remove the infinities. Why?

A hypothesis is that regularization is the right way to treat an "effective field theory", where infinities arise from extrapolating the known physics to the Planck scale and beyond.

However, it is not clear to us why removing infrared divergences would be associated with the Planck scale.

Our hypothesis in this blog has been that the infinities are removed by analyzing the path integral in position space, not in the conventional momentum space.

Can an object with zero mass-energy affect any physical process? The Lamb shift

Our previous blog post mentioned the speculative infinite energy of empty space.  Some people suggest that we can renormalize the energy away by setting it zero.

In hypothetical Unruh radiation, empty space is claimed to warm an accelerating detector.

If space has zero energy, can it affect any physical process?

Let us analyze if an object with zero energy can affect anything in the physics of positive mass-energy objects.

Special relativity

We remarked in an earlier blog posting that if a zero mass object would absorb momentum for a non-zero time, that would break conservation of the speed of the center of mass (the center of mass theorem).

Of course, if there is some mechanism which restores the correct position of the center of mass, then there is no breach.

     charge                     charge

          ●  <------------------->  ●

                    Coulomb

                       force

In special relativity, a force between objects is an "object" which can transfer momentum.

In the center of mass coordinates the force transfers momentum in zero time. But if we look at the process in a moving frame, then it can appear that one object gains speed earlier than the other loses it. The concept of simultaneity is relative.

The moving observer can see a temporary breach of conservation of energy, momentum, and the speed of the center of mass. When the force no longer is in effect, conserved quantities are restored.

In our example, there is a frame where the force transfers momentum in zero time. If the separation of the events:

1. remove momentum p from the object A, and

2. add momentum p to the object B

is spacelike, then there exists a frame where the momentum transfer is in zero time.

We conclude that an "object" of zero energy can transfer momentum, as long as the transfer takes zero time in at least one frame. But if the momentum transfer is timelike, then conservation of the speed of the center of mass is broken.

Electron movement caused by zero-point fluctuations? the Lamb shift

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

The Wikipedia article contains the Lamb shift derivation by Theodore A. Welton (1948):

https://journals.aps.org/pr/abstract/10.1103/PhysRev.74.1157

Welton assumes zero-point fluctuations which "smear" the position of the electron.

http://web.ihep.su/dbserv/compas/src/bethe47/eng.PDF

Hans Bethe in his 1947 papers offers a different explanation for the Lamb shift, using electron self-energy and mass renormalization.

https://arxiv.org/abs/hep-ph/0002158

Michael I. Eides et al. (2000) have written a very detailed QED treatment of the Lamb shift. The main contribution seems to come from vertex corrections.

We do not believe in the existence of zero-point fluctuations. A QED derivation does not assume them, but could the fluctuations be hidden in the regularization and renormalization of the diagrams?

Since the electron and the proton are both present, hypothetical zero-point fluctuations might act as a temporary "force" between them. Then conservation laws would be honored.

The rubber plate model of the electron electric field and the Lamb shift

In our rubber plate model of the electron electric field, the rubber plate lags behind in the accelerating motion of the electron toward the proton. In the receding phase, the rubber plate leads the electron and pulls the electron away from the proton. The QED  vertex correction obviously is about this process.

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

In the rubber plate model, the effective electron mass is lower when it is close to the proton, because part of the rubber plate and its inertia do not have time to react. If the electron were less massive, its lowest energy according to the Sommerfeld fine structure expression would be higher.

Question. Does the rubber plate model explain the Lamb shift by making the electron effectively less massive when it is in large acceleration close to the proton?

Since the electron is "lighter" close to the proton, it moves past the proton quicker, and has less time to collect momentum. Consequently, the de Broglie wavelength close to the proton is longer than without this effect. A longer wavelength forces the orbital to become a little wider, which makes the energy level a little higher.

A non-relativistic formula for the momentum p close to the proton is

        -V(r) = p^2 / (2 m_e)

which gives

        p = sqrt(-2 V(r) m_e).

The effective mass of the electron is reduced roughly by V(r) / c^2 because the far electric field does not have time to react to the acceleration of the electron.

At the distance 3 * 10^-14 m, the effective mass is ~ 10% lower.

If the effective mass is 10% lower, then p might be ~ 5% lower. The de Broglie wavelength is ~5% larger.

We conclude that the reduction of the effective mass has a "significant" effect when r < 3 * 10^-14 m.

In Welton's calculation of the Lamb shift, the "smearing" has a significant effect on the effective Coulomb potential - and the de Broglie wavelength - when r < 3 * 10^-14 m.

This very rough calculation suggests that the rubber plate model might explain the Lamb shift.

The Darwin term

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

The Darwin term is 90 μeV while the Lamb shift is 4 μeV. The rubber plate model might be able to explain the Darwin term, too, though we doubt it.

https://en.wikipedia.org/wiki/Foldy%E2%80%93Wouthuysen_transformation

The Darwin term is explained either by zitterbewegung or by the Foldy-Wouthuysen transformation.

When the electron is accelerating, its wave becomes a chirp: negative frequency waves (= the positron) are mixed with the normal positive frequency electron waves. That might explain zitterbewegung.

The Darwin term is related to the structure of the Dirac equation. We have not found a classical model for the spin and the magnetic moment of the electron.

Why the Welton method of zero-point fluctuations yields a numerically correct estimate for the Lamb shift?

If the rubber plate explains the Lamb shift, then zero-point fluctuations are not needed in the explanation. But this brings up another question: why zero-point fluctuations give a numerically correct result?

                virtual photon q

                       ~~~~~~

                     /                \

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

                             |   virtual

                             |   photon

                             |   p

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

The reason might be that the vertex correction produces an effect which imitates hypothetical zero-point fluctuations. Various possible paths store some amount q of momentum to the electric field of the electron. The momentum is later returned to the electron. These effects might mimic random pushes caused by hypothetical zero-point fluctuations.

Particle versus field interpretation of quantum field theory, and the Unruh effect

In Wikipedia, there is an example of canonical quantization. It concerns scalar Klein-Gordon field vibrations of a finite length string loop.

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

Let us assume that the string can move up and down. The classical hamiltonian contains energy from the stretching of the string:

         1/2 (δ_x φ)^2,

as well as from the vertical movement kinetic energy of the string:

       1/2 π^2.

We assume that there is a lowest energy state |0 >. Energy is added to the string by adding standing waves which have some number n of nodes.

In quantum mechanics, the string cannot stand still even in the lowest energy state. It moves up and down.

We do not know if there exists a lowest energy state if we couple the field to a detector

According to the Wikipedia article, one can prove the existence of a lowest energy state, the "vacuum state", only in a few simple cases through constructive quantum field theory.

The interacting system of a detector and the field (the string) is complex. It is probable that we cannot prove the existence of a lowest energy state.

Every vibration mode of the string contains energy, 1/2 h f, where f is the frequency. The total energy is infinite. The detector disturbs the system, and the disturbance might free an infinite amount of energy from the system.

The infinite energy of the string is normally canceled by "renormalizing" the energy of the lowest energy vibration state to zero. But that does not help if we couple a detector to the field, since the interaction might create an even lower energy state.

A particle model solves the problem of infinite energy in quantum field theory

The problem of infinite energy in a quantum field can be solved by moving to a particle model. The infinite number of vibration modes in "empty" space do not exist. Only particles exist. Their wave phenomena is caused by a path integral.

In a particle model, empty space is truly empty. There are no "zero point fluctuations" of various fields in empty space.

A crystal of atoms in the lowest energy state and a moving detector: friction

Suppose that we have a crystal of atoms in the lowest energy state. Then we move a detector close to them, or even through the electron orbits of the atoms. The detector may disturb the atoms. Kinetic energy of the detector can be converted to heat, which may excite the atoms as well as the detector itself. 

There is kind of friction between the atoms and the detector. Since atoms have mass-energy, they can absorb momentum from the moving detector, and free kinetic energy as heat.

If we assume that empty space is full of oscillators in their lowest energy state, then moving a detector at a constant speed could disturb the system. But empty space has no mass-energy (since we renormalized the infinite energy away). It cannot absorb momentum, and consequently, friction is prohibited.

Another way to say this is that Lorentz covariance prohibits such behavior, since we assume that in empty space all inertial frames are equivalent. There cannot be friction in one frame and not in another.

However, the interaction with the detector might allow the space to fall into a lower energy state. Then even a static detector might see some kind of "Unruh radiation" coming from empty space.

Also, renormalizing away the infinite energy of empty space is a dubious procedure. If we have a wildly oscillating string and let our fingers slide on it, we expect to feel violent oscillation. We cannot remove the phenomenon simply by deciding in our mind that the string has zero mass-energy and our fingers cannot feel friction because then Lorentz covariance would be violated.

Derivation of Unruh radiation

https://arxiv.org/abs/0710.5373

Luis C. B. Crispino et al. (2007) present a short derivation of Unruh radiation.

Let us have a detector on an accelerating rocket. It is sensitive to a frequency f of scalar massless Klein-Gordon field waves.

Does the detector see a frequency f wave? The transformation of an imagined frequency f wave to an inertial frame wave is a chirp. The inner Klein-Gordon product of the chirp with the vacuum state |0 > of the field is not zero. The negative frequencies in the transformed wave have a non-zero inner product with the vacuum.

This is because we can write a positive frequency mode f_i in the accelerating frame as a Fourier decomposition

        f_i =  Σ   α_Ii*  f_I  -  β_Ii  f_I*,

                  I

where f_I is a positive frequency mode in the inertial frame, and f_I* is a negative frequency mode in the inertial frame. The algebra then tells us that the annihilation operator

       â_i = Σ   α_Ii  â_I  +  β_Ii*   â_I-dagger.

                 I

That is, the annihilation operator in the accelerating frame is partly an annihilation operator in the inertial frame, but partly a creation operator (the dagger version).

Usually, the β are extremely small. Only if the acceleration is > 10^25 m/s^2 they would become significant.

The number operator of the accelerating frame, when applied to the vacuum of the inertial frame, gives a non-zero value. This is because the annihilation operator above is partly a creation operator in the inertial frame. As if there would be particles in the accelerating frame.

Negative frequencies are solutions of the Klein-Gordon equation, too. But the creation operator only creates positive frequencies. Where did the negative frequencies disappear? Apparently, they were declared "unphysical" and discarded. The negative frequency solutions may be understood as solutions of negative energy.

The above canonical quantization does not make sense in the classical limit

A detector observes a physical field. It could be a classical real scalar Klein-Gordon field.

The quantization in the previous section does not handle that case. We would need the negative frequencies, so that we can sum a mode

        exp(-i (E t - p x))

and its complex conjugate and get a real value.

The previous section does not have a sensible classical limit. It maps a classical positive frequency wave in the accelerating frame partly to an "unphysical" negative frequency wave in the inertial frame. In classical physics that cannot happen.

It looks like that the textbook example of canonical quantization is fatally flawed. All fields occurring in nature must admit negative frequencies, because those frequencies inevitably appear in accelerating motion. There cannot exist a field in nature where only positive frequencies are allowed.

The electromagnetic field, for example, definitely allows negative frequencies. We discussed them in the previous blog posting. A right-handed photon becomes partly left-handed if observed from an accelerating rocket.

The Unruh effect might be observation of the zero-point fluctuation of a field?

The derivation of Unruh radiation seems to be fatally flawed. But the phenomenon might still exist if our detector can detect energy in the hypothetical zero-point fluctuations of the assumed quantum fields filling the vacuum.

Actually, it is probable that a detector, accelerated or not, can observe such energy.

     

                             detector

                              ● 

                             /   rubber

                           /    band

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

           string in the lowest energy state

           (lowest without interaction)

Suppose that we have a string in its lowest energy state. It still vibrates, according to standard quantum mechanics.

We bring a detector close to it and attach the detector via a rubber band to the string. The rubber band represents an interaction. It would be surprising if there would be no response whatsoever in the detector. The system is no longer the same when the interaction is present.

Hypothetical Unruh radiation is just the tip of an iceberg of the strange phenomena which are implied by zero-point fluctuations.

Another one is the hypothetical infinite energy of the vacuum.

If we believe in zero-point fluctuations of the electromagnetic field, then a single electron interacts with infinite energy in those fluctuations. Why the electron sits still if it is bombarded by infinite energy?

Also, a single electron couples the modes, containing infinite energy, all together. How does the infinite energy react?

It looks like we have to abandon the concept of quantum fields filling the entire space. Classical fields are ok because they do not have zero-point energy. A particle model with a path integral might be a way to avoid the most serious problems of infinite energy.

How to quantize the wave sent by a freely falling electric dipole?

Let us have a very heavy rotating electric dipole which falls freely in the gravitational field of Earth. We assume that the charge in the dipole is very small, so that it only emits one photon per second. The dipole sends right-handed photons to space.

          \        <----

           |          ●                                     ●

           |           |                                       |

           |          ●                                     ●

          /

 Earth      rotating                          dipole

                 dipole in                         antenna

                 free fall                          in space

We have a dipole antenna far away in space. It receives some of the energy, momentum, and angular momentum sent by the free falling dipole.

The question is: how do we quantize the emission / absorption process? We believe that the absorber receives energy, momentum, and angular momentum in "packets". When are the packets formed and how do they behave in the gravitational field?

The length scale problem again

Distortion of electromagnetic waves is clearly related to the length scale problem. We have been blogging about the length scale problem in bremsstrahlung. How can a sharp classical wave whose form has features of size 10^-15 m appear as a nice sine wave of a wavelength 2 * 10^-12 m to the absorber of the photon?

We suggested that the uncertainty of the position of the electron causes destructive interference which wipes away all frequencies except the sine wave 2 * 10^-12 m frequency, and constructive interference moves the energy to the 2 * 10^-12 m frequency.

But now we realize that by making the electron very heavy and forcing it to make a turn of size 10^-15 m, we can eliminate the destructive interference. Does the spectrum of bremsstrahlung then contain waves whose wavelength is ~ 10^-15 m? Probably yes. A very heavy particle possesses a lot of kinetic energy it can afford to lose. We assumed that the electron in bremsstrahlung radiates away most of its kinetic energy. Since the electron did not have much kinetic energy, it could only produce a wavelength ~ 2 * 10^-12 m.

Are there different kinds of photons of the same energy and polarization?

In quantum mechanics we believe that photons of the same energy and polarization are identical. They behave in the same way regardless of their birth process. A 2 eV photon which is born in bremsstrahlung is identical to a 2 eV photon which is born in a state transition of an atom.

In the case of the falling dipole, naively the photon gets "stretched", and becomes a chirp. Could the stretching somehow show up at the absorber?

Classical aspects

Since the rotating dipole is very heavy, we can almost exactly know the phase of the wave sent by it. Also, we can know the location of the dipole very precisely.

The classical wave received by the absorber is a chirp where the frequency falls with time. Its Fourier decomposition contains all kinds of frequencies, and even "negative frequencies" which have left-handed angular momentum.

An accelerating observer meets an electromagnetic right hand polarized plane wave

The observer in a rocket will see the wave a as chirp. Its Fourier decomposition contains various "positive frequencies" which correspond to right hand polarized waves, and a small portion of "negative frequencies" which are left hand polarized.

Let us assume that the incoming plane wave has a very low intensity and contains just one photon per second.

How is it possible that an antenna held by the accelerating observer absorbs a left-handed photon, while the incoming wave only contains right-handed angular momentum?

Let us look at the classical limit. If an antenna somehow absorbs left-handed angular momentum, it either:

1. has to transfer right-handed angular momentum to the rocket, or

2. scatter the incoming wave in such a way that the scattered flux contains right-handed angular momentum (linear or in its spin).

The antenna might be a wind turbine which uses the radiation pressure to turn the turbine left. The right-handed momentum flows through its axis to the rocket, and right-handed momentum is also sent away in the reflected wave.

We suggest that one can calculate the response using classical wave equations. Though, of course, one has to use quantum mechanics to describe the absorption of a photon to an antenna.

The output wave can be decomposed into Fourier components. Each component may contain one or more output quanta. The quantization should be done in such a way that energy, momentum, and angular momentum are conserved.

How do we determine the probabilities for various collections of output quanta? We do not know. There might be a Feynman diagram like method for the calculation.

The wavelength of left-handed components is typically much larger than the main frequency of the right-handed components. If an antenna somehow is able to absorb a left-handed photon, then a lot of excess energy has to be scattered away. The outgoing wave may easily contain enough right-handed angular momentum, so that angular momentum is conserved.

In the previous blog posting we mentioned the "real Unruh effect", which is a paradoxical absorption of a left-handed photon from a beam of right-handed photons. It looks like the real Unruh effect occurs in nature. It happens in the classical limit and does probably happen also at the quantum level. The effect is extremely weak and we do not expect an experimental confirmation.

An accelerated emitter of a right hand polarized electromagnetic wave and an inertial observer

The Fourier decomposition of the emitted wave contains various frequencies and a small portion of negative frequencies.

If the observer absorbs a left-handed photon, it is typically of a much larger wavelength than the right-handed main component of the wave. We may assume that the emitting process somehow split a photon in two or more splinters, one of them having the left-handed angular momentum.

We suggest that one can calculate the response using classical wave equations, like in the previous section.

Deformation of an arbitrary electromagnetic planar wave under a gravitational field

Suppose that we aim a beam of laser light toward a large mass. The waveform is originally planar. The waveform gets distorted in a complex way by the gravitational field. Let the frequency of the laser be f.

The Fourier decomposition of the wave becomes very complex in the distortion process. The frequency f components have almost all energy, but they are mixed with longer wavelengths.

How do we interpret the longer wavelengths? Did photons collide with gravitons and split into smaller energy packets? Those smaller packets would have a longer wavelength.

How do we interpret the redshift of a laser beam sent up from the surface of a planet? A photon collides with a graviton, the graviton steals some energy and goes to build up the energy in the static gravitational field of the system?

Redshift is not a perturbational effect because every photon gets redshifted with a 100% probability. Maybe a non-perturbative process is better handled as a classical wave? 

We might assume the following:

The appearance of a longer wavelength is associated with splitting of a photon in a "collision" with a "graviton". The rest of the energy in the original photon is contained in the other splinters of the photon.

We are not sure if there is a sensible way to explain a non-perturbative process in terms of collisions of quanta. That is why we put the quotation marks "" above. We suggest that the correct method is to calculate the distortion of the wave using classical wave equations like in the two previous sections.

The error of William Unruh and Stephen Hawking

We believe that the correct way to transform electromagnetic waves to an accelerating frame is to use classical field equations. Then "negative frequencies" is quite a mundane phenomenon of flipped handedness of a photon.

Unruh, Hawking, and several other researchers have thought that the quantum field, with its creation and annihilation operators, should be transformed to an accelerating frame. The algebra of those operators then leads to strange phenomena like Unruh and Hawking radiation.

In a Feynman diagram, "creation" of a photon happens by using the Green's function of the massless Klein-Gordon field. Annihilation is the time reverse of creation.

Creation is like hitting a drum skin with a sharp hammer. It is the impulse response. We need to analyze in detail if anything strange, like the Unruh effect, can happen under acceleration in the drum skin model. That is unlikely, since the drum skin should behave in the classical way.

Quantum field theory in curved spacetime

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

Daniel Kastner and Rudolf Haag developed a local quantum field theory approach since the 1980s. We have to compare our ideas to their framework. 

The real Unruh effect: absorption of photons by an accelerated observer

In this blog we have been critical of the claimed Unruh effect, where an accelerating observer would see empty space as "warm".

https://arxiv.org/abs/1212.2409

https://arxiv.org/abs/1412.5892

Detlev Buchholz et al. (2012, 2014) have analyzed Unruh radiation in two papers quite recently. They criticize the interpretation that the detector would see "heat".

https://www.sciencedirect.com/science/article/abs/pii/0375960195007857

Vladimir Belinski (1995) holds the opinion that Unruh or Hawking radiation do not exist.

Unruh radiation does not exist

Let us repeat our arguments against Unruh radiation.

1. We do not see a mechanism through which the accelerating machinery of a rocket could excite the radiation detector in the rocket.

2. If empty space would somehow heat up the detector on its own, energy conservation would be violated as heat would be generated from nothing.

3. Turning kinetic energy of the rocket into radiation is not possible because light-speed radiation cannot carry away all the momentum which the rocket loses if kinetic energy is turned into heat.

4. Unruh radiation would be a macroscopic effect: under very great acceleration, the rocket would heat up to, say, a billion degrees Kelvin. There is no such effect in classical physics. Thus, the classical limit of quantum physics would not hold.

5. A static radiation detector on Earth is under constant acceleration in the gravitational field of Earth. If the detector on its own gets excited, from where does the energy come? The Unruh effect would violate the equivalence principle of gravity and acceleration, or alternatively, energy conservation.

Quantum field theoretic derivation of the Unruh effect

             wave W of frequency f

             in the frame of the rocket

         ~~~~~~~~~~~~~~~~

                      ● detector

                       |

                     <==

                     rocket

The derivation of Unruh radiation goes as follows.

Let us assume that a detector responds to photons of frequency f. Imagine a wave W of frequency f observed by a detector in the rocket in its accelerating frame.

1. The wave W would be a chirp in an inertial frame. That is, its frequency changes with time in an inertial frame.

2. The Fourier decomposition of a chirp contains a small amount of negative frequencies. We get the Fourier transform in an inertial frame with a Bogoliubov transformation of the wave in the accelerating frame.

3. Canonical quantization of a quantum field only contains positive frequencies.

4. The algebraic machinery of quantum field theory then claims that the negative frequencies in the inertial frame make a non-zero positive contribution to the expectation value of observing the wave W in the accelerating frame.

The error in the above derivation

The correct way to convert the wave W to another (accelerating) frame is to treat W as a classical electromagnetic wave. If W is circularly polarized, then the negative frequencies correspond to the opposite circular polarization. Thus, negative frequencies are very mundane, ordinary physics. There is nothing mysterious in them.

The algebraic machinery of quantum field theory, where only positive frequencies are assigned the status of real quanta, is a human invention. Why would Nature transform waves between different frames using the human-invented machinery? There is a more natural transformation using classical waves.

Another way to say this is that there is no way to quantize waves in an accelerating frame. The machinery of quantization was developed in inertial frames. There is no a priori reason why the machinery would work in an accelerating frame.

Our previous blog posting touched this issue. What is the ontology of quantum field theory? Is W a classical wave, or a quantum field object which transforms into an accelerating frame using some algebraic machinery?

The real Unruh effect: what happens to a laser beam if the laser or the observer is accelerated?

Suppose that a laser sends a right-handed circularly polarized wave. The laser is floating static in a Minkowski space.

An observer is in an accelerating rocket which speeds up toward the laser.

In the static frame, all the photons are right- handed.

But in the accelerating rocket, the observer sees a chirp, and if he has a detector which is only sensitive for left-handed photons, he may observe some.

If the observer absorbs a left-handed photon, he receives some left-handed angular momentum. But the laser beam only contained right-handed momentum. How is this possible?

Let the "laser" be a miniature rotating electric dipole. The dipole loses its angular momentum and radiates the angular momentum away in one or more right-handed photons.

If an observer absorbs a left-handed photon, how can angular momentum be conserved?

Acceleration of a frame is not a perturbative process: we have to quantize using a base of chirps, not plane waves?

           ● +

            |                               ●  antenna

            |                                |

            |                             <==

            |                            rocket

           ● -

        dipole

        rotates

Let us make the rotating dipole macroscopic and analyze the classical electromagnetic wave. That is, we analyze the classical limit of the process. The dipole is floating freely in the Minkowski space. It rotates and produces right-handed polarized radiation toward the direction of a rocket.

On the rocket, it is very well possible to absorb left-handed angular momentum from right-handed photons, for example, by using a turbine which turns under the radiation pressure from the incoming radiation. Angular momentum is conserved because the turbine reflects, or scatters, incoming photons.

Let us then on the rocket have an antenna which can absorb a certain frequency. If we accelerate the antenna toward the rotating dipole, then the antenna sees a chirp, and absorbs some left-handed polarized radiation.

How can angular momentum be conserved?

We have a feeling that using the Fourier decomposition to "quantize" the chirp is not the right way to describe the process.

The blueshift of the wave when the rocket accelerates is not a perturbative process, since the wave gets shifted at a 100% probability.

In the classical limit, one can build a dipole antenna whose axis contains a ratchet, and the antenna might be able to extract left-handed angular momentum from the wave. In this case, the right-handed momentum would be absorbed by the rocket through ratchet.

If we have a dipole which can rotate either way, the classical wave will always make it to absorb right-handed angular momentum.

The right way to decompose the chirp to quanta might be to use a base of chirps.

If we have a free electron, then a plane wave is the natural way to decompose a wave packet.

But if the electron is flying under a static electric field, then a natural decomposition consists of bent orbits. Every electron will at a 100% probability fly along a bent orbit. It is not a perturbative phenomenon.

We need to think this in more detail. What is the relation between perturbative processes and a Fourier decomposition with plane waves? Should we decompose non-perturbative processes into chirps?

A path integral is about the orbits of particles. There is no problem if there is a static electric field which bends every orbit. It may be best to abandon plane wave Fourier decompositions in non-perturbative contexts and use a particle model and a path integral.

This realization may be a fatal blow to Steven Hawking's 1975 derivation of Hawking radiation. The derivation fully depends on decomposing a chirp to plane waves and finding negative frequencies in the decomposition. But the process of light passing through a collapsing star is non-perturbative. Using plane waves may be a wrong approach.

Steven Weinberg and Meinard Kuhlmann: opinions about particles versus fields in quantum field theory

In our blog we have tried to analyze collision experiments from the particle viewpoint. Our photograph model of quantum field theory claims that deep down the processes happen in classical mechanics. It is the path integral which makes the processes to look like wave phenomena.

https://plato.stanford.edu/entries/quantum-field-theory/

Meinard Kuhlmann (2020) has written a detailed account about various standpoints in quantum field theory. Paul Dirac and Richard Feynman supported particle interpretations, Wolfgang Pauli a field interpretation. The string model (some people call it a theory) has a string interpretation.

Feynman diagrams describe events in the momentum space. There are no spatial locations in the diagrams, nor points in time. In this sense, the diagrams are a pure field interpretation.

Our view in this blog has been that we must work in the position space and use a particle interpretation to gain more insight on events, and possibly remove divergences from Feynman integrals.

Ontology of quantum field theory is unclear

Meinard Kuhlmann writes about the unclarity of the "existing things" in quantum field theory, that is, the ontology.

In this blog we have been perplexed by the ontology. Does quantum field theory study:

1. classical particles,

2. classical fields,

3. quantum fields with the creation and the annihilation operators and a Fock space,

4. virtual particles,

5. wave functions,

6. path integrals, or

7. measurements?

Or maybe Hilbert spaces, operators, or algebras?

In addition, is it just a perturbative theory, or a general physical theory?

Many confusions arise from the ontology. For example, the Unruh effect definitely does not exist in classical fields. Could it exist in quantum fields?

Hawking radiation definitely does not exist in classical fields. Could it exist in quantum fields?

What is the vacuum energy per cubic meter? Classically it is zero. Is it infinite in quantum field theory?

The ontology problem exists in ordinary quantum mechanics, too: it is the wave-particle duality.

The electron clearly is a "particle"

Charge conservation requires that the number of electrons does not change spontaneously. Quantization of charge requires that electrons do not break up.

Dirac formulated his equation in the way that there is a probability density for the location of the electron, and probability is conserved.

The electron is very much a particle. We can look at the path of an electron in a cloud chamber and see it drawing a line there.

The Schrödinger equation has had great success in explaining natural phenomena. The electron is a particle in the equation, moving under a potential.

A rotating electric dipole falling in a gravitational field; is the photon a "particle"?

An observer in free fall measures the dipole emitting right-handed photons of a fixed frequency.

An observer far out in space measures red-shifted photons of varying frequencies. Some of the photons are left-handed, because the dipole sends a "chirp" and the Bogoliubov transformation turns the handedness of a small number of photons in a chirp.

It is the general spirit of quantum mechanics to make a minimum number of assumptions about a system before a measurement is made. It would be wrong to say that the photons "existed" before any measurement. The "charge" to make the photons, that is, the energy, did exist. Before the measurement, there was a superposition of states of the various locations of the "charge".

If an excited hydrogen atom sends a photon, and we measure afterwards the recoil of the atom, then after the measurement it is safe to say that a photon particle is flying somewhere in space.

If we have an electron in our hand and wave the hand, the electron classically sends a very complex electromagnetic waveform with various frequencies. Should we say that the electron sent a set of photons which was completely defined immediately after the hand waving?

Again, the spirit of quantum mechanics is to make the least assumptions before a measurement is made. It sounds wrong to claim that the set of photons is defined before a measurement. Usage of a creation operator sounds wrong because it seems to assume that the set of photons is definite and determined before they are measured.

Creation operators are a form of a hidden variable model - assuming that some property is defined before a measurement.

The photon may be a particle, but in most cases we do not know how many and what frequency photons exist in an experiment.

The photon clearly is a particle in a digital camera or Compton scattering

If the photon is not a particle, how does it then appear as a single pixel in the image sensor of a digital camera?

In Compton scattering, a classical particle seems to collide with the electron. If the photon were a wave, why would it imitate a classical particle in a collision?

Is the virtual photon in a Coulomb scattering Feynman diagram a "particle"?

In his 1949 paper, Richard Feynman derives the virtual photon from the Fourier decomposition of the 1 / r Coulomb potential.

The QED lagrangian talks about the electromagnetic field A interacting with the Dirac field, but nowhere it is defined where we get A from. How does a Dirac electron wave generate A? If the electron wave packet is far away, we can calculate A classically. How to calculate A when the electron wave packet is all around?

The particle status of a virtual photon is much less clear than the status for a real photon. The alternative is just to use the Coulomb classical potential.

In an earlier blog post we wrote about an electric field bending a beam of electrons. The phenomenon is easy to understand with a potential and the Schrödinger equation. But it is hard to use the concept of a virtual photon to explain the electron paths. The reason probably is that the bending is not a perturbative phenomenon. The path of every electron gets bent with a 100% probability. Then it is not a "perturbation". Incidentally, a particle moving in the gravitational field of a macroscopic object, that is, curved spacetime, is not under a "perturbation" either. Has this been overlooked by researchers?

Virtual photons seem to exist in "perturbations" only. They are a useful concept in Feynman diagrams, but probably just fiction as "particles".

Steven Weinberg's view

https://arxiv.org/abs/hep-th/9702027

Steven Weinberg (1996) writes about the history and philosophy of quantum field theory.

Weinberg tells us that in the 1960s he did not like the path integral approach of Feynman, because an objective of quantum field theory was to work with fields only - no particles.

Weinberg strongly supports the regularization / renormalization machinery of quantum field theory.

Weinberg writes that the standard quantum field theory is an inevitable consequence of Lorentz covariance, quantum mechanics, and the cluster decomposition principle, which requires that distant experiments give uncorrelated results. But then he admits that the string model (which some people call a theory) does satisfy the above principles without being a quantum field theory!

Weinberg says that a weakness of quantum field theory is that it is just a perturbative theory.

Our blog studies various problems in mainstream quantum field theory: the need for regularization / renormalization, the obscure status of Unruh and Hawking radiation, non-renormalizability of gravitation, and failure to handle a simple non-perturbative situation like an electron beam in a static electric field. Steven Weinberg seems to have a practical attitude: if the theory agrees with the empirical data, then it is ok.

The power output of an oscillating charge in the rubber plate model agrees with Larmor formula

UPDATE March 16, 2021: there are calculation errors in this blog post. See the March 15, 2021 post for a correct calculation of the lagging mass.

---

Let us finally calculate a value for the power output, if we assume that the mass of the far electric field of an electron is lagging behind in the oscillation, and causing a drag on the oscillation.

This is the rubber plate model. The "plate" is the static electric field of the electron. The inertia of the plate is caused by the mass of the field.

Let us calculate a concrete example. Let an electron orbit a proton at a distance r = 0.5 * 10^-10 m. The speed of the electron is

       v = c / 137 = 2 * 10^6 m/s.

The acceleration of the electron is

       a = v^2 / r = 10^23 m/s^2.

The electron makes 10^16 cycles per second.

The Larmor formula gives as the power output

        P = 5 * 10^-8 W.

That is 3 * 10^-5 eV per cycle. An aside: a 10 eV state transition in a hydrogen atom may require a million cycles of the electron.

The mass of the electric field of the electron outside the distance 1/4 * 3 * 10^8 m / 10^16 = 15 nm is

       m = 2 * 10^-7 m_e.

The field may cause a dragging force of 

       F = m a = 2 * 10^-14 N

on the electron.

We get a power drain of

       P = F v = 4 * 10^-8 W

on the electron. Quite a good agreement with the Larmor formula.

If we double the speed of the electron, the acceleration (a = v^2 / r) is 4-fold. The Larmor formula says that the power is then 16-fold.

In our own calculation, the mass of the field dragging on the electron is double, the acceleration 4-fold, and the speed double. The power is thus 16-fold. We get an agreement with the Larmor formula.

Conclusion. We can explain the Larmor formula with a model where the mass of the far field of the electron lags behind the movement of the electron and uses its inertia to drain the maximum possible power from the movement of the electron.

How does the electron "move" in the Compton scattering Feynman diagram?

The Feynman diagram for Thomson/Compton scattering seems to depict an electron "absorbing" a photon, then "moving" a short distance, and then "emitting" a photon.

                                                      photon

                                                   ~~~~~~

                                                /

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

                        /      virtual 

             ~~~~        electron

      photon

The traditional classical description of the events is quite different: the field of the incoming photon shakes the electron.

   

    |      |      |

    |      |      |              ^             \  scattered wave

    |      |      |              |       \      

    |      |      |             ●  e-

    |      |      |              |

    |      |      |              v

     incoming            electron

     vertically            moves up

     polarized            and down

     wave

When the electron is shaking, it is in an accelerating movement, and produces a wave. The processes are simultaneous.

The rubber plate model of photon absorption / emission

                         ^             waves are created

                         |              in rubber plate

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

                         |   electron

                         v  moves up and down

The field of the incoming photon exerts a force on the electron. In this blog we have written several times about momentum conservation in emission of electromagnetic waves. Namely, the electron must give up a lot of spatial momentum to provide the energy for the photon. But the photon can only take away a small fraction of that momentum. Where does the excess momentum go?

The rubber plate model solves the mystery. The static electric field of the electron is like an elastic rubber plate attached to the electron. For abrupt movements of the electron, the inertial mass of the electron appears to be small, since the field does not have time to react to the movement of the electron. Some, or all, of the mass of the electron is in its static electric field.

The electron in this temporary state appears as off-shell: it has absorbed more energy relative to the absorbed momentum than what would be allowed for a perfectly rigid object of mass m_e.

The excess energy is subsequently emitted as waves in the rubber plate.

Momentum is, of course, conserved in the rubber plate model. The rubber plate has a backreaction on the electron. The backreaction solves the mystery of the momentum flow.

The rubber plate model is close to the Feynman model.

We notice the following thing: to produce new waves, the electron has to be off-shell. A perfectly rigid system of the electron & the field would not produce any waves.

Special relativity makes a perfectly rigid field impossible. There is always retardation.

What is the classical analogue of the electron propagator?

http://www.mso.anu.edu.au/~geoff/HEA/5_Compton_Scattering_I.pdf

Geoffrey V. Bicknell gives the total cross section for low-energy photons (Thomson scattering) as

       σ_T =  8/3 * π r_0^2,

where r_0 is the classical electron radius. That is, the cross section is 8/3 times the area of the classical silhouette of the electron.

The total cross section for high-energy photons (Compton scattering) is

       σ = π r_0^2 / x * ( ln(2 x) + 1/2 ),

where x = E / (m_e c^2) >> 1 and E is the energy of the photons.

http://web.hep.uiuc.edu/home/serrede/P436/Lecture_Notes/P436_Lect_14p5.pdf

The Thomson scattering formula follows from the classical calculation for a small dipole, as can be seen from the lecture notes of Steven Errede (2015).

The propagator controls the amount of power which flows through the scattering system to the scattered flux. Since the power classically is constant regardless of the photon frequency, there is no classical analogue for the electron propagator.

The 1 / r^2 potential of the "electron force"

We can qualitatively explain the Compton cross section if we assume that the photon is a rotating small electric dipole which is not traveling - it only rotates.

The electron moves in the dipole potential 1 / r^2 and is scattered. If we double the energy of the electron, then for "significant" scattering it has to come closer, within a 1 / sqrt(2) distance from the dipole.

The analysis is as for a pair annihilation, since the probability amplitude for the Feynman diagram is the same as in pair annihilation.

Alternatively, we may imagine that the photon scatters from a 1 / r^2 potential of the electron.

Are the interactions in photon-photon (Rayleigh?) scattering all simultaneous?

A Feynman diagram is strictly about plane waves, which, in principle, span the whole spatial universe, as well as an infinite time.

The diagram itself looks like that things would happen in some specific order, but that is probably just an illusion.

            q                p               k

        ~~~~~   ---------------   ~~~~~~

                      |                 |  virtual

                      |                 |  electron

                      |                 |  p - k

        ~~~~~   ---------------   ~~~~~~

           -q                -p             -k

Classically, in Thomson or Compton scattering, the incoming electromagnetic wave shakes the electron, and the shaking electron simultaneously produces the scattered wave.

In the diagram above, we have photon-photon scattering. Each internal line describes an instance of Compton scattering.

In an earlier blog posting, we noted that "off-shell" electrons can classically exist if they are interacting. A free electron is classically on-shell.

In the diagram, all the electron lines are off-shell. We suggest that all the electrons and positrons are interacting with other particles in the diagram.

A photon on the right side is being born at the same time that the virtual electron is still absorbing the photon on the left side.

The electron and the positron are interacting with the photons, but not with each other? If the mutual interaction of the pair is negligible, then the 1 / r^2 potential which governs pair production / annihilation has to be produced by the incoming photons.

How would the incoming photons interact with a dipole?

They might start to pull the dipole apart, but at the same time, the dipole would transmit an electromagnetic wave because the charges in the dipole would be accelerating.

Rayleigh scattering

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

If the dipole is much shorter (< 1 / 10) than the incoming wave, we speak about Rayleigh scattering.

Rayleigh scattering is much larger at short wavelengths. That is why the sky appears blue.

The virtual pair in photon-photon scattering can be understood as a tiny dipole which the much longer electromagnetic waves are disturbing.

The cross section of Rayleigh scattering from a droplet is

       σ = 2/3 π^5 d^6 / λ^4

              * (n^2 - 1)^2 / (n^2 + 2)^2,

where d is the droplet diameter, λ is the wavelength of light, and n is the refractive index of the droplet.

https://arxiv.org/abs/1111.6126

Liang and Czarnecki (2011) quote the classic Euler and Kockel (1935) cross section for low-energy photon-photon scattering:

       dσ / dΩ = 139 α^4 / (180 π)^2

                        * E^6 / m_e^8

                        * (3 + cos^2 θ)^2,

where α = 1 / 137 and E is the photon energy.

Let us think about a photon coming from, e.g., the left. It meets another photon that comes from the opposite direction. Let us imagine that the other photon is a Rayleigh type "droplet". What is the size of the droplet?

Let us use green light, λ = 532 nm. The cross section for photon-photon scattering is assumed to be 10^-67 m^2, and it has been measured to be smaller than 10^-59 m^2.

The corresponding Rayleigh droplet would be 0.3 * 10^-15 m in diameter. We assume that the coefficient at the end (n^2 ...) has a value ~ 1.

If the wavelength of the photons is λ = 2 * 10^-12 m, then photon-photon scattering has a roughly microbarn cross section, that is, 10^-34 m^2.

The corresponding Rayleigh droplet would have a diameter of 30 * 10^-15 m.

The "force" model of the virtual electron and the associated potential: the potential has to be a "black hole" potential?

In earlier blog posts we suggested that the virtual electron line in pair production or annihilation can be interpreted as a "force" which has a 1 / r^2 potential. It turns out that the force must have an even steeper potential than that. It looks like the pair falls into a "black hole".

High-energy annihilation

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

Frank Rieger gives the high-energy cross section as

       σ_e+e- = π r_0^2 / γ  *  (ln (2γ) - 1),

where

        γ = 1 / sqrt(1 - v^2 / c^2)

is the Lorentz factor of the electron and the positron in the center of mass frame, and r_0 is the classical radius of the electron.

Imagine that there would be a "potential"

       V = -B e^2 / r^2

between the electron and the 

positron, where

        B = 2.5 * 10^-5 J m^2 / C^2.

We set the numerical value of B as 2.8 * 10^-15 times the Coulomb constant, so that V is equal to the Coulomb potential at the distance of the electron classical radius r_0.

To achieve scattering to a "significant" angle, a fast electron of total energy E must go deep into the potential well, so that

       E = -V.

What is the cross section for such a process?

The total energy of the electron is

        E = γ m_e c^2.

Let us solve r from -V = E, that is,

       B e^2 / r^2 = γ m_e c^2.

We get

        r = sqrt(B e^2 / (γ m_e c^2))

           = 2.8 * 10^-15 m * 1 / sqrt(γ)

as the radius of the closest approach.

The impact parameter b is probably 2 r for a 1 / r^2 potential. The cross section is

       σ = 4 π r^2

          = 4 π r_0^2 / γ.

If γ = 100, the formula of Frank Rieger gives the cross section

          (π r_0^2 / 100) * (ln(200) - 1),

where ln(200) - 1 = 4.3.

Our formula agrees quite well with Rieger at the 50 MeV scale, but at other energies, the factor

        ln(2 γ) - 1

causes some difference.

From where does the logarithm term come in the Rieger formula? It makes the cross section larger when we increase the energy E a lot. The potential must be steeper than 1 / r^2. Maybe something like

        ~ ln(r) / r^2

for r << 1?

Low-energy annihilation of a pair

Frank Rieger gives the low-energy cross section as

       σ_e+e- = 1 / β * π r_0^2,

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

https://arxiv.org/abs/quant-ph/0202091

Sidney A. Coon et al. (2002) in their Section 2.4 in formula (51) state the classical scattering angle for the 1 / r^2 potential as a function of the angular momentum L.

If we in the formula cut the initial velocity v of the particle to a half, we can keep L constant by doubling the impact parameter b. This means that the cross section grows 4-fold if we halve the initial velocity v. Thus, the 1 / r^2 potential is classically not steep enough for the Rieger formula.

The potential in pair annihilation is a black hole potential?

The 1 / r^2 potential has circular orbits at every r (incidentally, the energy of every orbit is the same). To decrease the cross section, we need a potential which does not have circular orbits arbitrarily deep.

Such a potential exists around a Schwarzschild black hole. Close to the event horizon, a ray of light has to be sent almost vertically for it to reach infinity. Otherwise, the potential makes the light to curve back to the horizon.

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

The lowest circular orbit is at 3/2 times the Schwarzschild radius.

The annihilation of a pair is somehow related to falling into a black hole. The infinitely deep Coulomb potential might form a singularity if the energy were not sent away as photons.

Let us check what are the cross sections associated with a Schwarzschild black hole.

https://arxiv.org/abs/gr-qc/0503019

Chris Doran et al. (2005) in their formula (3) have the black hole capture cross section in terms of the initial velocity v of the particle. The cross section is ~ 1 / v^2. Thus, it is not like the annihilation cross section of a pair.

Making the potential flat at large distances cuts the dependence of the cross section on a small initial velocity

We can cut the dependence of the cross section on the initial velocity v simply by making the potential flat at larger distances. The Yukawa potential is very flat far away, and for small v, the cross section is essentially constant.

Thus, there probably exists a potential for which the classical cross section is linear in v. We did not yet find that potential.

Scattering in the 1 / r^2 potential differs in classical mechanics from quantum mechanics

https://aapt.scitation.org/doi/10.1119/1.1987905

Boris Kayser (1974) writes that the cross section for scattering from a 1 / r^2 potential is quite different from classical mechanics if calculated with the Born approximation.

The classical cross section is linear with the potential strength while the Born approximation claims it is quadratic.

Kayser explains that classical scattering happens in a "strong coupling regime", and therefore, the Born approximation does not apply.

Which is the right way to treat annihilation? Feynman diagrams probably use the Born approximation, and empirical data seems to support the Feynman way.

The fight against degrees of freedom

In photon-photon scattering, individual Feynman integrals diverge. Their sum does not diverge, but the value of the sum depends on how we regularize (remove the infinities in an ad hoc way) individual integrals.

The problem seems to be the extra degree of freedom which the virtual pair loop introduces to the reaction.

The Dirac equation describes strictly one particle and conserves charge

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

Paul Dirac tried the massive Klein-Gordon equation to describe the electron. A problem was that the equation is about the second derivatives. To specify initial values, we would need to give the initial function and its first derivative. There is too much freedom in the equation.

Defining a probability density function for one particle does not succeed, nor defining the probability current.

Charge conservation requires that the electron is strictly one particle which travels around. It cannot be a fuzzy ensemble like a classical electromagnetic wave.

Dirac took a "square root" of the massive Klein-Gordon equation, and was able to define a Lorentz covariant equation where there is a probability density and a conserved current.

 

Pair production spoils the strictness of the Dirac equation

The Schrödinger equation is a "deterministic" equation in the sense that the number of particles does not change, and we can use path integrals or other methods to develop the wave function forward in time in a deterministic way.

Creating new pairs can spoil this determinism if the initial conditions do not dictate all the parameters of the new particles. Creating a virtual pair in photon-photon scattering is an example: besides an arbitrary direction of the spatial momentum vector p, also the length of |p| can be set arbitrarily large.

The space of directions in R^3 is finite and we can define a constant probability density function on it.

But if we allow |p| to be arbitrary, then we face the problem of defining a constant probability density function on the real line R. That is not possible.

Photon-photon scattering

        q                 p                k

     ~~~~~    --------------    ~~~~~

                    |               |  p - k

                    |               |  virtual

                    |               |  electron

     ~~~~~    --------------    ~~~~~

     -q                  -p                -k

Photon-photon scattering is a prime example of too much freedom. One can choose the 4-momentum |p| as large as one wants, and is only punished by the product of the four propagators, which is something like

          ~  1 / |p|^4

for large |p|. The space of 4-momenta is R^4, which makes the Feynman integral to diverge.

The way forward may be to describe the virtual electron line as some kind of scattering from a 1 / r^2 potential. That way get a handle on what is happening at the deep down level, and can make the behavior more "deterministic".

The photon-photon scattering Feynman integral converges if we assume a maximum lifetime for a virtual electron which "borrowed" energy

If the euclidean length |p| is large in the diagram, then the virtual electron has to "borrow" a lot of energy and its lifetime t is at most

        t ~ h / (4 π|p|).

The maximum "lifelength" of the virtual electron is

       ~ t c.

Why? Because if the virtual electron would move a large spatial distance x, then in another (moving) inertial frame the electron would appear to live a long time (as a tachyon).

If |p| is large, then the virtual electrons move almost at the light speed (or even faster). The emission of the photons at the vertices has to be very quick, because of the great speed of the electrons, which means that the spatial separation of the created photons is at most the "lifelength" apart.

This means that the cross section seen from the left, from the incoming photon side, is at most

       ~ 1 / |p_0|^2

for |p| > |p_0|.

The Feynman integral for the loop does converge if we use this additional constraint. The contribution of very large |p| is negligible because they require the photons to pass very close to each other.

Note that slow electrons can emit photons in such a way that the photons are "created" very far away from each other. A simple example is a slowly rotating dipole. If the speed of the electrons in the dipole is, say, 1 / 100 c, then the emitted photons are "created" at a separation which is 200 times the dipole length. This is seen by calculating the emitted angular momentum J.

What is the fundamental reason why we can get the Feynman integral to converge?

In the diagram, large |p| represent an infinite number of "channels" which can take energy from the photons on the left, and transmit the energy to the right. Our argumentation shows that those channels have to compete for a very small flux of energy which is provided to the channels when the photons pass very close to each other. Even if the channels can transmit all the energy in that flux, the effect is negligible.

Without our restriction, the integral thinks that it can tap the energy of photons which pass very far from each other, and use that energy to create a virtual electron with huge 4-momentum |p|. That sounds counter-intuitive.

The interaction of a pair is either through the Coulomb force or "virtual electron force"

In an earlier blog post this week we observed that in annihilation, the virtual electron can be understood as a carrier of a "force".

         e+ ---------------------------

                          | virtual

                          | photon

                          | p

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

In Coulomb scattering, the pair exchanges a virtual photon. The propagator of the photon is

        ~ 1 / |p|^2,

which corresponds to the Coulomb potential 1 / r.

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

                            | virtual

                            | electron

                            | p

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

Annihilation of a pair can be understood as scattering where the pair exchanges a virtual electron and changes to a pair of photons. The propagator of the virtual electron for large |p| is something like

        ~ 1 / |p|,

which would correspond to a 1 / r^2 potential. The force carrier is charged, in contrast to the Coulomb force.

There is a dilemma: why does the pair in the first reaction obey the Coulomb potential, but in the second reaction an unknown 1 / r^2 potential?

The pair certainly is under the Coulomb force in the second reaction, too.

It is as if there would be a 1 / r^3 force acting in the second reaction. We speculated that it might be the magnetic force between the spins, but that force between two dipoles is 1 / r^4. Also, the force between magnets is electromagnetic. It does not carry charge.

The "virtual electron force" acts between photons as well as pairs. That explains why the force affects the cross section from both directions in the second diagram.

The Coulomb force only acts between pairs. It would affect the reaction cross section only on the left side.

The photon as a rotating dipole would explain the 1 / r^3 force

Recall the "teleportation model" of the photon which we introduced in an earlier blog posting.

Let us assume that a circularly polarized photon is a rotating electron-positron dipole.

The strength of a dipole field on a single charge is ~ 1 / r^3. The potential is ~ 1 / r^2.

photon

         e+   --------------------- e+

         e-    ------ 

                          \    

                             ---------- e-   zero energy

                             ---------- e+  photon

                          /

         e+   -------

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

photon

In pair production, the dipole field of a photon succeeds in breaking apart the dipole of another photon. More precisely, both dipoles break apart. An electron and a positron fly away, and the remaining two particles form a zero energy photon.

Thus, pair production is a process where two photons collide and are broken apart into their constituent particles: the electron and the positron. Two particles fly away, the rest form a new photon of zero energy.

Pair annihilation is a process where an electron and a positron break apart a zero-energy photon and form two new photons.

If we have a lattice which contains charged particles, there photons can propagate as vibrations of the lattice. There pair production might really look like in the diagram above. The photon is a vibration which causes a surplus density of electrons at one location, and lack of electrons at another location. The collision of photons may be able to free one electron, leaving behind a hole.

Pair production is governed by a 1 / r^2 potential in the lattice and the cross section is described by the ~ 1 / |p| propagator.

To the other direction, the electron and the positron "roll down" the 1 / r^2 potential.

This would explain the β and 1 / β in the cross sections calculated from Feynman diagrams.

If the description is correct, an electron and a positron do not truly annihilate each other. They just regroup to form new photons.

Annihilation is not governed directly by the Coulomb force between the particles, but indirectly through the force that a dipole exerts on a single charge.

The Coulomb force does guide the opposite charges closer, but the annihilation reaction itself is between dipoles.

The "Coulomb correction", which is mentioned in literature about Delbrück scattering, probably tries to account for the focusing effect of the Coulomb attraction.

Does the new model solve the problem of extra degrees of freedom?

In the previous blog posting we discussed photon-photon scattering and the problem that the virtual pair stage has an extra degree of freedom, that is, |p| can have arbitrarily high values.

The problem arises from the assumption that the virtual pair can be born arbitrarily close to each other. If the particles are born very close, they must have huge kinetic energies and huge momenta.

In the diagram above, the e- and e+ which free themselves are not born at an arbitrary place. They are not born at all.

Rather, the electron and the positron are continuations of the two photons.

It may be possible to model the pair as a single particle which moves under the 1 / r^2 potential. Then there are no degrees of freedom at all. It is a classical deterministic path.

If an electron moves under the field of a nucleus and exchanges two virtual photons with the nucleus, then there is a loop in the Feynman diagram, and the rules would allow us to circulate any momentum p in the loop. But common sense says that such arbitrary momentum p cannot just pop up and join the electron. A single particle under a field does not have any degrees of freedom.

Yet another model: the semiconductor model of annihilation

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

Let us have a lattice of atoms. Let an electron escape from the lattice. What does the hole look like?

It probably polarizes the neighboring atoms. The electric field might look a bit like a dipole field?

That might explain the 1 / r^2 potential which governs annihilation.

When the hole and the electron are at a larger distance, polarization has no effect on the field of the hole far away.

Let us look at what models people have developed about semiconductor annihilation. LED lights work by doing this annihilation.

If a semiconductor is a perfect analogy of real electrons and positrons, then we can experimentally study pair production and photon-photon scattering without the problems of generating gamma quanta.

A brief Internet search does not reveal any potential associated with the electron-hole recombination in semiconductors.

Actually, polarization is always present in the lattice, whatever the distance of the electron and the hole.

The potential might be 1 / r far away from the hole, but steeper close to the hole if polarization for some reason is less effective close to the hole. In the case of atoms, once the distance of the electron is less than the grid spacing of the lattice, polarization might be zero. This would fit the picture that there is a cloud of polarization around an electron. The bare charge is larger but partially hidden by polarization.

The running of the coupling constant requires collisions of 100 GeV electrons and positrons to make a 8% difference. How could that explain the 1 / r^2 potential of annihilation at 1 MeV energies? It cannot.

Why the electron in the hydrogen atom does not "annihilate"?

Our classical calculation of the electromagnetic radiation says that when the electron approaches the proton, the acceleration is so large that the entire energy of the electron should radiate away when the distance is ~ 1.4 * 10^-15 m.

This is the classical problem about why electron orbits are stable in an atom.

In the s orbital, the angular momentum of the electron is zero. It swings through the proton every 5 * 10^-17 s.

The photograph model (or the uncertainty relation) says that to "draw" the electron at a position 10^-15 m from the proton, we would need spatial momentum > 1 GeV / c. Such an electron has energy > 1 GeV. The energy is not available, not even if we take into account the negative potential energy 1.5 MeV at that distance from the center of the proton.

Destructive interference in the photograph model wipes out all histories where energy is not conserved.