How to conserve energy if the number of photons changes?

Let us analyze further the photon number change effect which we discovered in the previous blog post.

Let us have an obstacle made of a polarizable material. Let the obstacle be much smaller than the wavelength of a coherent source, e.g., a radio transmitter or a laser. The obstacle makes a very sharp area of deformation in the electromagnetic wave. An observer who moves through this deformation will in the temporal Fourier decomposition of the wave see high frequencies.

Does the effect really exist? It almost certainly does. A coherent laser beam, or a coherent beam from a radio transmitter, is a macroscopic object. It would be very surprising if the measuring device would not see high frequencies in Fourier decompositions of complex waveforms.

Thus, we must conclude that even if a laser only emits photons of energy E, a measuring device which moves at a relativistic constant velocity v may sometimes detect photons which have much higher energy E'.

Classically, energy conservation is not a problem. But how to get conservation in quantum mechanics?

Could it be that in quantum mechanics, several photons can sometimes merge into a photon with higher energy?

Maybe we should abolish the notion that the laser beam consists of photons? The beam is created by atoms decaying into a lower energy state. That is, the beam is created as kind of "photons". But maybe the beam itself does not "consist of photons"? Photons are absorbed from the beam. The beam is destroyed in units of a "photon".

In our earlier blog posts we have suggested that one should assume the minimum amount of information in the course of a quantum mechanical process. The prime example of this is that a photon does not have a definite path. It is a step further if we claim that a laser beam does not have definite "photons" either.

A "photon" would only exist when we perform a measurement. We may observe that a certain atom has decayed into a lower energy state, or has been excited to a higher energy state.

We can call this framework "minimalist quantum mechanics". Quantum mechanics moves closer to classical mechanics if we do not assume the existence of quanta in intermediate states of a physical process. Only measurements will observe definite quanta.

Conservation of energy

How is energy conserved if the number of quanta is not defined?

Maybe the quanta which are eventually absorbed from the wave will always have energy less or equal to the energy that went to create the wave?

How would nature do the bookkeeping to ensure conservation of energy?

An accelerating observer: Unruh and Hawking radiation

The derivation of hypothetical Unruh or Hawking radiation uses a Fourier decomposition of a wave as seen by an accelerating observer. Unruh and Hawking believed that negative frequencies in the decomposition represent photons popping out of nothing.

In this blog we hold the opinion that Unruh and Hawking made an error: a negative frequency is simply a classical wave which has counter-clockwise circular polarization, if we define a positive frequency wave to have clockwise polarization.

The problem of a changing number photons is prominent for an accelerating observer. What kind of quanta will he observe in his detectors if he moves within a coherent electromagnetic wave?

Classically the answer is clear: his detectors will absorb photons of various frequencies. Again we face the problem: how to conserve energy?

Does some extra energy come from the kinetic energy of the measuring device?

Suppose that we have static negative electric charges ordered in a line at equal distances. Let an antenna move along the line at a constant velocity v. The antenna will observe a fluctuating electric field and can extract energy from it.

The energy must come from the kinetic energy of the antenna. We may imagine that the antenna collides with a virtual photon sent by a charge, loses momentum, and kinetic energy is freed.

However, if the antenna moves in a coherent electromagnetic wave, then some of the energy comes from the wave, and it might be that none of the energy comes from the kinetic energy of the antenna.

The number of photons may change between inertial frames: a new "Unruh effect"

Let us work in semiclassical physics. Let us have a monochromatic laser beam which is reflected or is refracted by a small object (is scattered by the object). Let us assume that the object is smaller than the wavelength.

        laser beam

        ~~~~~~~~~~~~~~~~~~~

        ~~~~~~~~~~~~  ● object

        ~~~~~~~~~~~~~~~~~~~

                                          <o> measuring device

What kind of photons an observer may measure in the scattered light?

If the system and the measuring device are fixed in the laboratory frame, then the observer will see all oscillation of fields having the same frequency f as the laser. He will see photons whose energy matches the photons in the laser beam.

But let us then make the measuring device to move at a constant speed v relative to the laboratory frame.

The waveform close to the object has a complex form. The Fourier decomposition of the signal received by the measuring device will have many frequencies, some of them very high.

The observer may see a photon which has much higher energy than the photons emitted by the laser.

How do we interpret this? If the laser sends, say, one photon of energy E in a second, how can the observer see a photon with a much higher energy E'? Is energy conserved?

Seeing a high-energy photon has an extremely low probability, though.

We have discussed the analogous problem in the context of an accelerating laser, or an accelerating observer. There is no obvious way to match individual photons in emission to photons in absorption. We have called this problem the "real Unruh effect".

In purely classical physics there is no problem. Energy in a wave of frequency f can be transformed to energy in a wave of much higher frequency f'. It is quantization which poses the problem here.

The length scale problem is ubiquitous

We have been studying bremsstrahlung in the past weeks. The "length scale" problem is that an electron passing very close to a proton should classically emit photons of very high energy. What wipes away these photons? Our new observation in this blog post shows that a similar problem exists in very mundane scattering of laser light.

Classical bremsstrahlung

In our March 17, 2021 blog post we analyzed the Wikipedia formula for bremsstrahlung and the Gaunt factor. There we looked at processes where the electron passes the proton at a distance > the Compton wavelength of the electron.

Bremsstrahlung in a close encounter

Let us in this section analyze closer encounters with relativistic electrons.

Let us calculate the radiated energy when a relativistic electron passes a proton.

The acceleration close to the proton is

       a = k e² / (b² m_e γ),

where b is the impact parameter and γ is the Lorentz factor of the electron.

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

The radiated power is

       P = e² a² γ⁴ / (6 π ε₀ c³).

Let us determine numerical values for b = the Compton wavelength of the electron:

       b = 2.4 * 10^-12 m.

       a = 4.4 * 10^26 m/s² / γ.

       P = 1 W * γ²,

For a relativistic electron, the fly-by lasts for

       t = b / c = 8 * 10^-21 s

and the radiated energy is

       E = 8 * 10^-21 J  *  γ²,

which is 10^-7 of the rest mass-energy (511 keV) of the electron if γ is close to 1. Bremsstrahlung is very small at that distance. The total radiated energy is 0.05 eV. The acceleration happens in the time t.

The transverse end velocity of the electron is

       v = a t = 3 * 10^6 m/s / γ,

and the electron moves a transverse distance

       l = 1/2 a t² = 2 * 10^-14 m / γ

during the fly-by.

The radiated energy E scales as 

       E ~ 1 / b³.

If we set

       b = 3 * 10^-14 m

and

       γ = 1.5,

then E = 50 keV, and v is only mildly relativistic, and l < b. We see that bremsstrahlung is very large already for b which is 10X the classical radius, or a cross section of 10 barn.

Acceleration of a non-relativistic electron: kinetic energy versus Larmor radiation

Let us assume that an electron is accelerated with a force F for a time t.

The work done by the force is

       W = 1/2 F / m * t² * F,

where m is the effective mass of the electron. Part of its field does not have time to react to the acceleration. Therefore, m is smaller than the rest mass of the electron.

The energy radiated according to Larmor is

       E = e² (F / m)² / (6 π ε₀ c³)  *  t.

Obviously, W > E. If we set

       W = E,

we get

       t = 2 e² / (6 π ε₀ c³)  * 1 / m

         = 10^-23 s,

if we let m = m_e = 511 keV.

We see that

       t c = 3 * 10^-15 m,

or the classical radius of the electron. The Larmor formula does not work properly if acceleration lasts that short a time.

The bremsstrahlung spectrum

http://www.astro.utu.fi/~cflynn/astroII/l3.html

The above document (2006) in Figure 3 shows the bremsstrahlung intensity spectrum to be flat up to a cutoff angular velocity ω_cut. The cutoff angular velocity is determined by the interaction time

       t = b / v.

The maximum energy that an electron can lose in a collision to a nucleus is the kinetic energy of the electron. The cutoff angular velocity is, of course, restricted by that value.

http://www.astro.unipd.it/franceschini/HighEnergy/2Bremsstrahlung.pdf

In the document above, the bremsstrahlung spectrum is calculated by putting cutoffs:

       b_min < b < b_max.

If the electron has low energy and comes close to the proton, then Coulomb interaction pulls it significantly closer to the proton during the fly-by. In the calculation we brutally cut off the those cases. The lowest impact parameter in the calculation is denoted b¹_min in the paper.

If the impact parameter b is smaller than the de Broglie wavelength of the electron, then the paper claims that the classical approximation does not work, and brutally cuts off those values of b. The minimum b is in this case denoted b²_min.

In most cases, b²_min is the larger one, and acts as the low-end cutoff.

If we calculate bremsstrahlung for a material where nuclei are at a distance 2 * b_max from each other, that defines the upper cutoff.

In Section 2 the Gaunt factor has an unexplained coefficient sqrt(3) / π in it.

The calculation method is to ignore what happens if the electron comes closer than the de Broglie wavelength to the proton. How can this work?

The "scale problem" once again and the inner field problem

In this spring we wrote about the problem of very close encounters of a proton and an electron, where the impact parameter b is ~ 2.8 * 10^-15 m. Classically, the electric field of the electron gains a very sharp deformation where the maximum photon energy would be around 1 GeV, regardless of the kinetic energy of the electron.

What phenomenon wipes out very sharp features from bremsstrahlung? Uncertainty of the electron position does some of the wiping. If the total energy of the electron is 1 MeV, the uncertainty is ~ 10^-12 m.

The emitted photon at most has the kinetic energy of the electron plus a few electron volts if the electron starts to orbit the proton.

We would like to know the precise bremsstrahlung spectrum, to study how the scale problem is resolved. What is the radiation from very close encounters like? But we have not found anything precise.

The texts that we linked calculate bremsstrahlung by imposing a brutal cut at b = b_min. They do not help us.

Neither were we able to find any good experimental data on Møller or Bhabha scattering, which could reveal the structure of the inner field of the electron. QED calculations seem to be very complicated, and we could not find a comprehensive analysis of all types of bremsstrahlung.

Electron-electron (Møller) scattering bremsstrahlung cross section

We want to find out what is the structure of the inner electric field of the electron, that is, the structure for a radius less than the electron classical radius of 2.8 * 10^-15 m.

High-energy Møller scattering is one way to find out.

Elastic Coulomb scattering of electrons in QED is apparently the same as for classical point charges where we only assume the Coulomb force between point masses, and do not take into account effects of the electric field of the electrons.

Bremsstrahlung: emission of a large photoni

https://www.researchgate.net/publication/234311581_Bremsstrahlung_and_Pair_Production_in_the_Field_of_Free_Electrons

E. Haug in 1975 calculated QED bremsstrahlung for Møller scattering.

Figure 7 in the paper plots the differential cross section for a single energetic photon emission per MeV for various collision energies. The plot is actually

       cross section × energy of the photon in MeV.

It is kind of a power spectrum.

Let us look at a 10 + 10 MeV collision. A 5 MeV photon can be called a "large" photon. The differential cross section at 5 MeV is 6 mb / MeV. The total cross section for an emission of a > 5 MeV photon is roughly 20 mb.

For a 50 + 50 MeV collision, the total cross section for a > 25 MeV photon is roughly 40 mb.

Elastic scattering to a large angle without bremsstrahlung

What about elastic scattering to a large angle? Large angle scattering requires that the potential at the impact parameter b is comparable to half of the total energy of the colliding particle.

For a 10 MeV electron, b should be less than 2.8 * 10^-16 m, which implies a cross section of 3 * 10^-31 m^2, or 3 mb.

For a 50 MeV electron, the cross section for large angle elastic scattering is 0.1 mb.

We see that the cross section for a large photon emission is much bigger than the cross section of elastic scattering to a large angle.

Strange behavior: in elastic scattering the electron acts like it would have no electric field

Classically, according to the Larmor formula, if relativistic electrons pass at a distance < 1.4 * 10^-15 m, they will lose much of their energy in radiation. They cannot scatter elastically at all at short distances.

But in QED, electrons can also behave like point charges with no electric field: they can scatter elastically at very short distances, down to at least 10^-18 m. The elastic scattering is the simplest Feynman diagram.

What kind of a classical model could explain this peculiar behavior?

A real photon self-energy Feynman diagram

On May 12, 2021 we analyzed the electron self-energy and concluded self-energy has no effect on a free (real) electron.

Let us consider a real ("free") photon flying in a vacuum. The photon self-energy means the hypothetical effect of photon conversion into a virtual electron-positron pair mid-flight.

                        e- virtual

                     _____

                   /           \

      ~~~~~                ~~~~~     real photon

                   \______/

                        e+ virtual

Above we have a Feynman diagram of photon self-energy.

If the virtual pair would possess positive rest mass, then the flight of the photon would be slowed down below the speed of light. We conclude that the entire energy of the pair must be kinetic. Then the flight of the pair is probably described by the massless Klein-Gordon equation, just like the flight of the photon.

We do not believe that "empty space" can change the phase of the photon. Conclusion: self-energy has no effect on real photon propagation.

Just as in the case of electron self-energy, it is unclear what the Feynman diagram above means. If the photon proceeds for 1 meter, how many times does it get converted into a virtual pair?

There may exist some kind of polarization of vacuum around the photon. We have suggested in this blog that the photon itself might be a rotating virtual electron-positron pair.

Photon-electron (Compton) scattering: a particle model of the photon

Let us analyze scattering of a high-energy photon from an electron. In previous blog posts we have stressed that we do not know the structure of the inner field of the electron, at a distance considerably less than the classical radius of the electron.

The wavelength of a 1 a MeV photon is 10^-12 m, much more than the classical radius of the electron. To probe the inner field, we should use 1 GeV photons. But then a collision produces a jet of particles. It is hard to deduce the structure of the inner field from a complicated jet.

Collision of a 511 keV photon and an electron: is this an analogue of pair production?

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

                      /          \

          ~~~~              ~~~~

      511 keV photon

Suppose that a slow electron absorbs a 511 keV photon and subsequently emits a photon of roughly the same energy. If we switch the time axis t and a spatial axis x, the process somewhat resembles pair production or annihilation. A major difference is that after the switch, the electron/positron move superluminally.

In a Feynman diagram, one is allowed to rotate the diagram through 90 degrees. That suggests that there might be some connection between mundane Compton scattering and mysterious pair production/annihilation.

Compton scattering of a 511 keV photon produces a photon whose wavelength is 2.4 * 10^-12 m, while annihilation is a process where the electron and the positron come within 2.8 * 10^-15 m from each other. The length scales are very different, or are they? The uncertainty in the position of the electron and the positron is > 2.4 * 10^-12 m.

In our February 1, 2021 blog post we discussed the problem of different length scales.

A particle model of the photon

Let us introduce an alternative model for Compton scattering: the electron moves very sharply and abruptly when it absorbs a (pointlike?) particle photon. In the path integral of all histories, destructive interference wipes out sharp features of the wave produced by the abrupt movement. The remaining smooth emitted wave contains the emitted 511 keV photon.

Classically, we think of the electron as a particle and the photon as a wave. The new model may partially explain how the photon can be a particle and still appear as a wave.

Collisions of particles produce abrupt and sharp movements. Destructive interference in the path integral evens out sharp features in the produced waves.

The electron self-energy Feynman diagram

In our December 19, 2020 blog post we remarked that the Feynman diagram for the electron self-energy seems to break conservation of the speed of the center of mass. If the electron would send a virtual photon to itself, it would change the phase of the electron wave function.

A phase change in the wave function of the Schrödinger equation corresponds to a temporary slowdown or speedup in the movement of the particle. That is, the speed of the center of mass is not constant. We proposed that such phase changes are forbidden for a free particle.

Classical self-energy

Let us assume that the classical electron is static. In the sharp hammer model, it hits a drum skin with a hammer and later absorbs the produced wave. The hammer hit produces a symmetric wave. The electron will not move anywhere. There is no conceptual problem in the classical process.

The Dirac equation versus the self-energy Feynman diagram

                            photon E, q

                               ~~~~~

                             /                \ 

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

         p

In the Feynman diagram, the electron emits a photon (real or virtual) and reabsorbs it.

We cannot measure the photon. There is no wave function collapse. To determine the behavior of the system, we have to sum the probability amplitudes of all possible paths it may take.

Let us prepare a free electron as a wave packet. Conventionally, it is assumed that the packet will spread obeying the free Dirac equation. The Dirac equation describes a relativistic particle.

The Feynman diagram depicts the electron as emitting a photon and then absorbing it back. What does the sum of all possible paths look like? The Feynman integral diverges, which indicates that the sum is not well-defined.

Another question is if the process of emission and reabsorption affects the behavior of the electron wave packet. Does the process make the wave packet to spread faster or slower than what the Dirac equation says?

The success of the Dirac and Schrödinger equations in calculating the hydrogen atom shows that they describe the electron wave function very accurately.

If a wave packet of the electron would spread faster than what the Dirac and Schrödinger equations say, then the uncertainty relation for the electron position and momentum would differ from the conventional one which we know.

The Dirac equation is derived from the principles:

1. the equation has to be Lorentz covariant;

2. the number of particles has to be positive or zero and must be conserved.

It looks natural that whatever the free electron is, it should obey principles 1 and 2.

The Feynman diagram does not make much sense for a free electron which moves, say, 1 meter. How many photons per millimeter is the electron supposed to emit and absorb?

In conventional quantum field theory, the Feynman integral of the self-energy diagram is first regularized and after that, the mass of the electron is renormalized, so that the propagator with the renormalized mass looks like the propagator for the Dirac equation. What is the meaning of this procedure? Why we do not simply define that the electron has the measured mass and obeys the Dirac equation?

The spin and the anomalous magnetic moment

In this blog we believe that the spin and the magnetic moment of the electron are a result of genuine circular physical movement. We have called this movement zitterbewegung because we believe that the zitterbewegung which Erwin Schrödinger in 1930 found in solutions of the Dirac equation somehow reflects this circular movement.

Our March 2, 2021 blog post was able to explain the anomalous magnetic moment of the electron by assuming that the inertia of the far field of the electron does not have time to take part in the circular movement: the electron has a reduced mass.

Thus, the static electric field has a measurable effect on zitterbewegung, even if it would have no effect on linear movement of a free electron.

Are there degrees of freedom in the static field of the electron?

In quantum mechanics, if we attach two particles together through some force, there will be zero-point oscillation even in the lowest energy state. The electron is, in some sense, attached to its static electric field. Are there zero-point oscillations in the field? Could zitterbewegung be zero-point oscillation?

No. The electric field of the electron is 137 times too weak to keep it on a light-speed circular orbit whose length is the Compton wavelength. Moreover, zitterbewegung is a property of the Dirac equation, and the equation does not assume any electric field for the particle.

The static electric field of the electron seems to be mostly a classical thing.

Conclusions

It is not clear to us what the self-energy Feynman diagram for a free electron is supposed to describe. We believe that in this context it is best to treat the static field of the electron as a classical object. For a free electron, the field moves with the electron.

In the classical model the regularization means that we cut off the energy density of the static field somewhere around the classical radius of the electron. The renormalization in the classical model is that we set the system mass at the observed mass of the electron.