Superfluidity: time crystal model

UPDATE March 30, 2022: The time crystal model cannot explain the second sound of superfluid helium-4. Our plate model of superfluidity is better.

----

Let us study of the transition of liquid helium-4 at 2.17 K, that is, the lambda point.

Helium-4 macroscopic physical properties close to the lambda point 2.17 K

http://hyperphysics.phy-astr.gsu.edu/hbase/lhel.html

The heat of the helium I -> helium II transition is ~ 2 J/g.

https://www.sciencedirect.com/science/article/abs/pii/S0031891436802122

W.H. Keesom et al. (1935) measured as the melting heat 4 J/g at 2.5 K, at an unknown pressure less than 135 atmospheres.

https://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section2.pdf

The vaporization heat at 2 K is 23 J/g.

We may conclude that the phase transition of liquid helium-4 at 2.17 K is "substantial", compared to its melting from a solid form.

http://www.conceptualphysicstoday.com/2017/01/all-about-helium-i-ii.html

The density of liquid helium-4 grows rapidly as it is cooled from 4.2 K to 2.17 K. When it is further cooled, the density starts falling. The density curve suggests that the phase transition actually happens gradually above 2.17 K. The specific heat, on the other hand, suggests that it happens mostly under 2.17 K.

https://www.researchgate.net/publication/260597824_Modeling_and_control_of_the_String2_LHC_Prototype_at_CERN

Thermal conductivity of superfluid helium-4 below 2.17 K grows rapidly to a huge value as we cool the liquid. The conductivity according to Wikipedia is several hundred times that of copper. It is as if the liquid would contain a gas which transfers heat energy swiftly to distant locations. This may be one of the reasons why some people believe that superfluid helium contains a Bose-Einstein condensate which would behave like a gas.

Black body radiation at 2 K has a frequency ~ 2 * 10¹¹ Hz and a wavelength ~ 1.5 mm.

https://physics.stackexchange.com/questions/219921/how-does-superfluid-helium-interact-with-ultrasound

The speed of sound in superfluid helium-4 is 220 m/s at 2 K. The wavelength corresponding to 2 * 10¹¹ Hz would be 1 nanometer. The diameter of a helium atom is 0.28 nm.

Superfluid helium-4 cannot contain a Bose-Einstein gas of helium atoms

Let us have many photons in a box whose walls are perfect mirrors. There is no problem in putting many photons in the lowest energy state where the box length is 1/2 of the photon wavelength.

Low-energy photons do not interact. They do not collide and can live in a perfect harmony.

On the other hand, helium atoms do collide in a box. The assumed Bose-Einstein gas in superfluid helium-4 is so dense that it is essentially full of helium atoms. Why would the atoms not collide?

The atoms are quite heavy objects. They behave almost classically. Classically, the atoms will collide all the time. The hypothesis of a weakly interacting Bose-Einstein gas cannot hold.

                           ^

                         /

                       o   helium atom 

                     /

                   /

             __/  XXXXXX

             XXXXXXXXXXX

            uneven capillary

                       wall

Also, the atoms of a gas would collide with uneven walls of a capillary and lose their momentum if the gas is moving relative to the capillary. There would be no superfluidity.

https://www.ufn.ru/pdf/jphysussr/1947/11_1/3jphysussr19471101.pdf

Nikolai Bogoliubov (1947) suggested that a superfluid contains a Bose-Einstein gas of quasiparticles. We will look at Bogoliubov's model in a subsequent blog post.

Why sound propagates fast in a material but heat propagates much slower?

Let us have a fluid. Audible sound waves have a very long wavelength and their scattering from individual molecules in the fluid is negligible.

Short wavelength vibrations of heat scatter from individual molecules.

How could we transfer heat quickly?

Maybe adding order to the fluid would help. If the coupling is stronger, heat propagates faster. Also, scattering may be less from an ordered lattice. Thermal conductivity of ice is four times the conductivity of water.

The obvious solution to very fast heat transfer is to move matter around. If there would exist a gas of heat-carrying electrons within the fluid, thermal conductivity could be much better. Metals are good conductors of heat for this reason.

Another option is to transform the fluid into a time crystal where molecules move large distances around without bumping into each other. They can carry thermal vibrations at a large speed to distant locations.

A pot of water boiling on the stove may have a fairly stable convection pattern. It is an approximate time crystal with a very efficient heat transfer rate.

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

Superfluid helium seems to have a third method of a very fast heat transfer: heat can propagate in waves which do not scatter much. The speed is 20 m/s at 1.8 K. We have to analyze this.

The time crystal model of superfluid helium-4

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

Huge thermal conductivity suggests that superfluid helium-4 is a time crystal where atoms can move in rings over large distances.

                    ring of atoms

                           ------>

                         o o o o o

         ^   o o o o o o o o o o o o o   |

         |   o o o o XXXXXX o o o o   v

   XXXXXXXXXXXXXXXXXXXXXXXXX

                  uneven capillary wall

  

The model may also explain why the liquid is superfluid. A ring can move over uneven walls like the continuous track of a bulldozer moves over an uneven terrain.

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

             chain of helium atoms

                     o o o o o o o  --->

      o o o o o o o o o o o o o   static helium atoms

      XXXXX o o o XXXX o XX

      XXXXXXXXXXXXXXXXX

        uneven capillary wall

Besides the continuous track, there is an alternative explanation for superfluidity: static helium atoms form a layer over the uneven capillary wall and make the wall smooth. After that, chains of helium atoms can slide within the capillary without friction.

The time crystal model requires that chains of helium atoms can slide past each other with zero friction. Classically, the atoms will radiate electromagnetic radiation or phonons as they slip past each other. It might be that the highly ordered structure of the time crystal causes destructive interference in the created waves, such that the radiation to the outside is essentially zero.

Superfluid helium-4 at 0 K has to be a time crystal

Helium-4 can only become a solid crystal at pressures larger than 25 atmospheres. At the normal 1 atm pressure it stays as a superfluid even at 0 K. 

Matter at 0 K must, in some sense, be highly ordered at 0 K. Otherwise, it would lose energy through radiation. But a superfluid does not form a static crystal lattice.

If a superfluid is highly ordered but does not form a static lattice, then it probably is a "moving crystal", that is, a time crystal.

Conclusions

The time crystal model explains superfluidity and huge thermal conductivity.

We need to check if the time crystal model fits all the facts that we know about superfluid helium.

Can we explain the second sound with the time crystal model?

https://www.physics.umd.edu/courses/Phys404/Anlage_Spring10/Donnelly%2520Two%2520Fluid%2520LHe.pdf

Russell J. Donnelly (2009) has a nice paper about the second sound. Let us look at it.

Landau's criteria of superfluidity: the model is wrong?

UPDATE March 30, 2022: The "phonon" and the "roton" of Lev Landau seem to be something like an atom, or a lighter or a heavier particle moving and carrying the energy p² / (2 m) and the momentum p. There, m is the mass of the particle. Landau's phonon is not a vibration of a lattice of atoms. A high-energy phonon may be a single helium atom. A roton is a group of ~ 5 helium atoms.

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UPDATE March 28, 2022: In this blog post we treat an "excitation" as a phenomenon of classical mechanics. Helium atoms are relatively heavy objects, and groups of them even heavier. Therefore, a classical treatment makes sense. Friction between solid bodies is usually modeled as a classical process.

https://en.wikipedia.org/wiki/Newton%27s_cradle

How would we quantize a classical excitation? For example, in Newton's cradle, the shock wave carries very large momentum compared to its energy. The simplest example of a shock wave is a single atom flying and carrying momentum. The momentum and energy in such a setup is not quantized, or, the "quantum" is a single atom.

The problem of quantization of excitations in a lattice is a fundamental one. We will look at this problem.

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Let us analyze the superfluidity model of Lev Landau in the following manuscript by Yoshihisa Yamamoto:

https://www.nii.ac.jp/qis/first-quantum/e/forStudents/lecture/pdf/qis385/QIS385_chap5.pdf

https://www.nii.ac.jp/qis/first-quantum/e/forStudents/lecture/index.html

It is Chapter 5 in the above link.

Superfluid in a capillary hits the wall

                 capillary

            -----------------------

            ooooooooooo   ----> v   superfluid M

            -----------------------

                     <---- p'  momentum from the hit

A body of a superfluid is moving in a capillary at the velocity v.

Let us first work in the frame where the superfluid is static. The fluid may be hit by the wall of the capillary and receive the momentum p' to the left.

The momentum p' is the total momentum received by the whole body of the fluid. Let p be the momentum of the created excitation in the fluid body. There is no obvious reason why p and p' should be equal.

Landau assumes that the hit produces a single excitation of the fluid, such that the vibrational energy of the excitation can only take a single value determined by a function

       ε(p).

Is the assumption of a single value for the energy function realistic?

We may imagine that we have harmonic oscillators attached to the body of the fluid. The oscillators may have various masses m. Let an oscillator absorb the momentum p. The vibrational energy after that is

       p² / (2 m).

Besides p, the energy depends on m. The mass m can be at most the total mass M of the superfluid. There is no obvious lower limit for m.

An "oscillator attached to the fluid" may be just some volume of the fluid body, e.g., 1/10 of the total fluid. We may divide the fluid into 10 such oscillators. The oscillators are coupled to each other.

We conclude that ε(p) is not a function. It may have many values for the same p and the values may be very large.

Let us then work in the laboratory frame. We assume that P = M v  >> | p |. If

       E = P² / (2 M),

then

       dE / dP = 2 P / (2 M) = P / M = v.

If v is small, then the body of the fluid, when it hits the wall, loses little energy compared to the momentum (p') that it loses.

On the other hand, an excitation typically contains a lot of energy compared to the momentum it holds.

If the excitation would be required to hold the entire momentum p' lost by the body of the fluid, then the excitation would not be able to happen. There would not be enough energy available to create the excitation.

This is the error of Landau: he assumed that the momentum p of the excitation has to be the same as the momentum p' lost by the body of the superfluid. When the body of the superfluid hits the wall, there is an interaction between the different parts of the fluid. The entire momentum p' is not required to go to the excitation.

Conclusions

Lev Landau considered a body of a superfluid that is hit by the wall of a capillary. His analysis produced a counter-intuitive result: the kinetic energy lost by the body cannot go to vibrations of the fluid body because there is not enough energy available.

The claim is strange. We know that if we have a moving rod of a material, and the rod hits an obstacle, then the lost kinetic energy does go to vibrations.

The paradox is solved through the realization that the vibrations do not hold the entire momentum p' lost by the rod. Landau did not consider this.

The other error made by Landau is that he assumed that the energy of an excitation ε(p) is uniquely determined by its momentum p.

The third error is to assume that the hit only produces a single excitation. If a rod hits an obstacle, many different vibrations are born at the same time.

Superfluidity is plates of helium atoms?

A peculiar phenomenon is that the superfluid phase of helium-4 coexists with the normal fluid phase at temperatures 0 K < T < 2.17 K.

https://en.wikipedia.org/wiki/Superfluid_helium-4

The share of the normal fluid phase is 100% at 2.17 K, and drops to almost 0% at 1 K as we cool the liquid.

This behavior is very different from, say, ice and liquid water. The phase transition happens to the whole body of water at a fixed temperature 0 C.

How can a phase change happen gradually as we lower the temperature?

If the different phases would exist as large 3-dimensional volumes of matter, and be separate from each other, then we should be able to separate the two phases and make 100% superfluid helium-4 at, e.g., 1.5 K.

Since that is not possible, the "structure" of helium-4 at 1.5 K has to contain matter from both phases. We cannot separate the phases if we keep the temperature constant.

Atoms in superfluid helium-4 probably are ordered in some way. That would explain the peak in the specific heat just below 2.17 K.

The plate-layered model of superfluid helium-4

      o   o   o   o   o   o   o   o   plate of superfluid

              o   o     o   o   o       

          o    o      o      o    o       ordinary fluid

             o   o  o     o    o  

      o   o   o   o   o   o   o   o   plate of superfluid

Let us introduce the plate model of superfluid helium-4. The superfluid component arranges itself into 2-dimensional plates of ordered helium atoms. Between the plates there is helium-4 in the normal fluid phase. Without the normal fluid layers, the plates would be broken apart by thermal vibrations.

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

It is a "lyotropic liquid crystal" where ordered structures of a superfluid are immersed into a normal fluid.

Thus, the microscopic structure of helium-4 under 2.17 K is a layered structure. Layers of the normal fluid phase may act as a some kind of a lubricant or a cushion between the more ordered structures of the superfluid phase.

When we lower the temperature to 0 K, the entropy of helium-4 drops to a very low level. There no longer can exist layers of the normal fluid phase. Apparently, the ordered structure of the superfluid phase cannot be broken apart by the zero-point fluctuation.

Why can the superfluid phase flow independent of the normal fluid phase? Why is the viscosity zero below 2.17 K?

If the superfluid phase exists as plates, those structures are able to move even though the normal fluid phase stays static. The two phases can move independently.

Why is the viscosity zero at a low temperature? Our hypothesis is that an ordered structure of helium atoms can slide through the normal fluid phase without friction. The atom lattice of the superfluid phase does not contain potential wells which would be able to capture atoms of the normal fluid phase, so that there would be an exchange of momentum.

Conclusions

The specific heat of helium-4 at less than 2.17 K suggests that the superfluid phase must be highly ordered.

The gradual phase change requires that the structure of helium-4 at less than 2.17 K must be layered at the atomary level. There must be layers of normal fluid between the highly ordered superfluid structures. The thickness of the superfluid structures is probably just one atom, while the layers of normal fluid may be up to tens of atoms thick.

Superfluidity requires that superfluid structures must be able to slide through the normal fluid without friction.

Could it be that the structure of helium-4 under 2.17 K is not as "classical" as we have presented? The disorder, or the normal fluid component, might be excitations, for example, phonons, which move in a highly ordered material. In semiconductors we have electrons and holes: it is like a two-fluid model.

Superconductivity is two "crystals" sliding within each other?

The Wikipedia page states that it is imperfections in the lattice of atoms which causes resistance in a normal metal.

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

Phonons are a time-dependent distortion of the lattice.

In our previous, March 14, 2022 blog post we remarked that scattering of electrons in a perfectly ordered crystal might be negligible.

Thus, the problem of superconductivity is why phonons or zero-point fluctuations do not cause scattering of electrons at a low temperature.

Copper, silver, and gold are not superconductors even at zero kelvin. There is some residual resistivity. Could it be that zero-point fluctuations in those metals are able to scatter electrons?

Two crystals sliding within each other

Ordinary resistance comes from the interaction of electrons with phonons. An electron is an individual particle, but a phonon is a collective motion of the lattice of atoms.

Could it be a general property of lattices that they tend to interact through their defects or distortions (usually phonons) and not through their individual particles? Let us assume that.

Let us assume that the conducting electrons at a very low temperature form a highly ordered structure, some kind of a "crystal". As the cloud of the conducting electrons slides past the lattice of nuclei, then only the "defects" of the electron crystal and the lattice interact. That is, only the phonons in each interact. The interaction of phonons probably does not cause much friction between the electron crystal and the lattice?

A phonon of the lattice can be converted to a phonon in the electron crystal. Can that process take momentum away from the sliding electron crystal? How does a phonon "jump" to the other material?

An analogous setting is a block of glass moving in a photon gas. If the glass would reflect the photons, it would certainly lose momentum. But if it is transparent, then it does not lose.

We conclude that the electron crystal can slide through the lattice of atoms without friction. This may explain superconductivity. We must analyze this in more detail.

Copper, silver, and gold

Could it be that the lattice of these metals is very stiff? A stiff lattice reduces the amplitude of phonons, and reduces resistivity at high temperatures.

We speculated about a "crystal" of conducting electrons in the previous section. The crystal apparently needs phonons of the lattice to form - the isotope effect strongly suggests that. The phonons in a stiff lattice may have too small an effect to bind electrons into an ordered structure even at zero kelvin.

Conclusions

In BCS theory, electrons form Cooper pairs through an unknown attractive force which is mediated by phonons. BCS theory does not try to explain what is this force and why the pairs can move within the lattice of atoms without scattering from phonons and losing their momentum.

Our crystal-within-lattice model may explain these things, at least qualitatively. Instead of pairs of conducting electrons, phonons at a low temperature bind whole ordered structures, "crystals" of electrons. 

The Cooper pair may describe the interaction between two "adjacent" electrons in the crystal.

What are these ordered structures of electrons? They cannot consist of a static lattice of electrons. Electrons move at large speeds.

A crystal is a collective phenomenon where a large number of particles are tightly bound to each other. Then the interaction of a single particle with an outside disturbance has to be replaced with the interaction with a large collection of particles. The excitation of a crystal structure is often a phonon.

What are Cooper pairs? Time crystals?

The Physics Stackexchange contains varying descriptions of Cooper pairs. Do they consist of electrons, or of dressed electrons, or are they denser zones of electrons?

https://physics.stackexchange.com/questions/28699/intuitive-reasons-for-superconductivity/28705

The energy of electrons at the top of the Fermi sea is typically 7 eV, which corresponds to the energy of a particle at the temperature 81,000 K. The energy which binds a Cooper pair is 1 meV, which corresponds to 11.6 K.

Conduction of electricity

Suppose that we have a block of metal at 0 K. The electrons in the Fermi sea have the lowest possible energy, and they cannot lose any of their kinetic energy to the nuclei. The electrons are, in a sense, in a highly ordered state. It is some kind of a "crystal".

Then we heat up the metal to the room temperature. The electrons bump into each other and into nuclei and radiate photons. The state of the electrons is no longer highly ordered. Outgoing radiation is a symptom of high entropy.

Then we apply an electric voltage over the block. Electrons get accelerated by the voltage. Electrons bump into nuclei and radiate away some of their kinetic energy. That is, resistance is producing heat from an electric current.

Superconductivity occurs at a low temperature. Presumably, the ordered state of electrons can somehow make electrons avoid bumping into nuclei when we apply a voltage over the block.

In BCS theory it is Cooper pairs which bring order to electrons.

Time crystals

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

A time crystal is a system where the lowest energy state contains a "repetitive motion" of particles. Frank Wilczek proposed time crystals in 2012.

Every atom or a molecule is, in a sense, a time crystal. There is a repetitive motion of electrons. A permanent magnet is a time crystal which produces a macroscopic magnetic field.

If a superconductor can make electrons to move without radiating anything, then, if we have a current flowing through the superconductor, it is a time crystal where new crystallization happens on that side where electrons enter, and melting happens on the other side.

We may define an ordinary crystal as an ordered structure which is static in some frame (static except of zero-point oscillations of particles). If electrons are flowing in a superconductor, then it is not an ordinary crystal. Nuclei are static but electrons flow.

What do we know empirically about superconductors?

We know at least the following things:

1. A superconductor stops being a superconductor above a transition temperature T which may vary between 4 K and 120 K.

2. A superconductor can move a surplus of electrons without radiating heat, or it radiates very little heat.

3. The specific heat of the material is high around the transition temperature. As if we were melting a "crystal" at that temperature.

4. The transition temperature is lower with heavier isotopes of the nuclei. To form, the "crystal" apparently requires nuclei which can oscillate wide enough.

5. The energy of a particle at the temperature 4 K ... 120 K is 0.3 meV ... 13 meV.

The facts above do not necessarily mean that electrons are ordered into pairs, or that they would form a Bose-Einstein condensate. Is there evidence that such a condensate really is formed?

The "crystal" of electrons at 0 K

Since a block of metal, or any material, does not radiate at 0 K, it has very low entropy, and its electrons and nuclei form some kind of a "crystal". We do not know much about what that crystal is like. Apparently, the Fermi sea is filled up to some surface. But what are the wave functions of the electrons or nuclei like?

If the block is a superconductor, it can move a surplus of electrons from one side to the other with zero or very small radiation. The system acts as a time crystal up to some transition temperature T, or a current density.

In the update today to our previous block entry (March 12, 2022) we remarked that in a highly ordered system, scattering from collisions of particles may have very low power because there may be almost total destructive interference in the scattered wave.

Classically, a crystal of electrons might slide past the lattice of the nuclei without touching. There would be no energy loss.

Another way to avoid scattering is to use a material which cannot be excited at the available energies. Could it be that the system below the transition temperature somehow becomes unable to be excited? We can warm up the block of metal to, say, 1 mK. The system can absorb very low-energy photons. Thus, there are excited states which could be produced by bumping electrons.

But could it be that a bumping electron cannot find a free state for itself?

Let us assume that the cloud of electrons stays static, and the nuclei are slowly dragged through the cloud. In this frame, it is a nucleus which bumps into an electron. If the electron cloud is a kind of a "crystal", then it probably can start to vibrate, and be excited. The lattice of the nuclei can do the same.

Conclusions

The (almost) zero resistance of superconductors may be due to:

1. A band gap: there is no free state for the system, such that kinetic energy of the flowing electrons could be converted to vibration. Is that really possible? Crystals tend to have very many excited states, and the states can have very small energies.

2. Electrons and nuclei form a time crystal. There is no scattering, that is, electrons and nuclei do not bump at all. The time crystal probably is bound by vibrations of the lattice of nuclei. Heavier isotopes damp vibrations, and the time crystal melts at a lower temperature if the nuclei are of a heavier isotope.

We need to investigate what evidence do we have of a Bose-Einstein condensate in a superconductor.

The Fermi gas: why we can ignore the repulsion of electrons in a superconductor

UPDATE March 14, 2022: Our reasoning about destructive interference of scattered waves from random collisions was erroneous.

If we sum random variables Xⱼ, then the variance of the sum is the sum of variances of each Xⱼ.

Let Eⱼ be the electric field strength of an electromagnetic wave at a certain time t at a location x. The expected value of Eⱼ is 0. The variance is defined as the expected value of Eⱼ². The power density of the wave is linearly proportional to the variance.

If we have random waves Eⱼ, then the power density of the sum of all Eⱼ is the sum of the power densities of each Eⱼ. Destructive interference does not reduce the power density of the scattered wave at all.

On the other hand, if scattering happens from an ordered structure, then destructive interference can wipe out almost all the power from the scattered wave. We will write more about this.

----

In a superconductor electrons can carry an electric current without losing energy in resistance. It is easy to design classical models where this is true: simply order the movement of electrons in a way where they are able to avoid collisions with nuclei.

But what is the classical model which best describes the phenomenon? What is the relation of Cooper pairs to this?

If electrons move in a random way, like molecules in a gas, then the electrons certainly will collide with nuclei and lose energy. Electrons in a superconductor presumably must have low energy and must be ordered into some kind of a "flowing crystal" to avoid collisions with nuclei.

When we raise the temperature of a superconductor, the "flowing crystal" melts. The specific heat of the superconductor must be high at this transition temperature, because it is a phase transition of the free electron system.

The BCS theory and Cooper pairs

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

The Bardeen-Cooper-Schrieffer theory (1957) explains superconductivity.

http://hyperphysics.phy-astr.gsu.edu/hbase/Solids/coop.html

Electrons are supposed to form Cooper pairs which can "condense into the same state" like bosons can.

The electron speed and the electron drift speed in a metal

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html

The link contains an example of a copper wire, where the speed of electrons at the top of the Fermi sea is 1,570 km/s, and the drift speed of electrons is only 4 mm/s.

The drift speed is minuscule. It is not that individual electrons flow in the wire, but the "cloud" of electrons flows.

The Fermi surface and Cooper pairs

https://arxiv.org/abs/hep-th/9210046

Joe Polchinski (1992) considers the surface of the Fermi sea under the weak interaction (~ 1 meV) which is assumed to form Cooper pairs.

How can Cooper pairs survive if the sea of electrons has much larger energies (~ 7 eV)?

An electron should do a Brownian motion under the bombardment of other electrons. How can a weak ~ 1 meV interaction make any difference?

Joe Polchinski suggests that Cooper pairs form in an effective field theory where we can ignore individual particles of the Fermi sea. It is like a wave in water: we can ignore water molecules and study waves in a higher-level description of the liquid.

A wave in a material with lots of particles

Let us consider a wave of visible light which propagates inside a block of glass. Each atom of glass individually scatters the wave. Is there destructive interference which removes almost all of the scattering?

Question. What is the long-wavelength electron wave? Is it an individual electron or a collective phenomenon?

The water wave analogy suggests that the wave is not an individual electron. A hypothesis is that it is a "dressed" electron, with interactions included in the quasi-particle.

The double-slit experiment

In the standard double-slit experiment there is the following problem: the photon passing through a slit interacts with atoms in the edges of the slit. Can we from this interaction somehow find out the the slit through which the photon passed?

If we can determine the slit through which the photon passed, then there should be no interference pattern on the screen, according to rules of quantum mechanics.

A possible solution: look at the behavior of a classical wave of light. If the classical wave forms an interference pattern, then the quantum wave must do the same because otherwise the classical limit of quantum mechanics is broken.

Conclusions

In a material where particles are dense, there can be almost total destructive interference of scattered waves. This may explain why glass is transparent and why we can ignore the repulsion of individual electrons in a Fermi gas.

We have to think more about this. Does the behavior of classical waves determine quantum mechanics, just like in the double-slit experiminent? What is the precise role of the quantum then?

A classical Pauli exclusion principle for electrons

Let us consider the following model. We have a 2-dimensional plate with a uniform positive electric charge density. We put electrons close to the plate, so that the positive and negative electric charges cancel out.

                          electrons

             ● e-     ● e-   ● e- ● e- ● e-

          +          +          +          +          +

                               plate

What happens if we let the electrons have zero or very little kinetic energy?

If the kinetic energy is zero, then the electrons will form a "crystal" where they minimize their repulsive potential energy. We could say that each of the electrons is in a different "state" from others, since the electrons are static and at different positions. A Pauli exclusion principle holds in this case. The electron system is in a "solid" state.

If we then add a little bit of kinetic energy, the electrons probably will vibrate around their initial positions. We can still say that that the electrons have different states.

Adding still more energy, the electrons can slide past each other. The electron system is in a "liquid" state. The electrons are in different states, but can exchange their states easily.

If we add still more energy, we will have an electron gas where the electrons bounce around. A gas has a certain distribution D of energies. We could say that the distribution D is the "Fermi sea" of the classical electrons.

Can we squeeze a very large number of electrons on the plate, such that the energy distribution D stays unchanged?

Probably not. The repulsive force between the electrons grows stronger as we add more electrons. The system will become a liquid or a solid again.

We showed that there may exist a classical Pauli exclusion principle: the distribution D can only fit a certain number of electrons before the electrons start to crystallize into separate (almost) local states. As if the distribution D would only contain a finite number of "separate states" for electrons.

We have a classical analogue of the quantum mechanical Pauli exclusion principle. The classical principle follows from the simple fact that electrons repel each other.

https://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem

The Pauli exclusion principle in quantum mechanics is a consequence of the spin-statistics theorem. The dubious "proof" of the theorem in Wikipedia uses the algebra of spin rotations. A non-integer spin implies the Pauli exclusion principle while an integer spin allows many particles to have the same state.

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

In a superconductor, electrons are believed to form bound Cooper pairs. A pair has an integer spin, and many pairs can exist in the same "state". 

What is the relation of these two things: having a non-integer spin and having a repulsive or attractive force between particles? We need to investigate this.

How to solve fundamental open problems in physics

In this blog we have made progress in solving the black hole information paradox, the non-existence of a singularity in a classical black hole, the non-existence of Unruh and Hawking radiation, renormalization and regularization of quantum electrodynamics, we have predicted the existence of dark energy, and so on.

How are we able to solve problems which earlier researchers have struggled with but have not found a solution? Is this magic?

Our method is simple:

A. Check if a proposed physical model obeys principles which people generally feel that a physical model should satisfy.

B. Consider also second order effects and the "backreaction" of the system.

C. Also, check if we understand the logic in the reasoning and calculations which are based on the physical model.

We are staunch supporters of the "consensus" in physics. We are not contrarians. More precisely, we support generally accepted principles of physics in item A. We do not support models which break these principles.

Some generally accepted principles:

1. Conservation of momentum. Unruh radiation seems to break this.

2. Conservation of energy. Hawking radiation may break this. Certain cosmological models break this.

3. Conservation of the speed of the center of mass. Certain self-energy diagrams in quantum electrodynamics may break this.

4. Non-existence of singularities and infinite physical values. Traditional black hole models claim that the horizon is "infinitely strong" and there is a singularity at the center. They break this principle.

5. The behavior of a system should be determined by initial values. Roger Penrose proved that gravitational waves in general relativity have an "unusual causal structure". They break this principle.

6. Unitarity. Hawking radiation breaks this. Many-worlds interpretations of quantum mechanics may respect this, but interpretations based on some kind of a "collapse" do not.

7. A physical process is always reversible. This is associated with unitarity. The traditional model of a one-way black hole horizon breaks this.

8. Information must not leak into an invisible "shadow world". This is associated with 5 and 6. The Dirac hole model of vacuum polarization with invisible negative energy electrons breaks this.

9. There should be no closed causal loops. Certain cosmological models break this.

10. We should not extend the physical model past the infinite value of the global time coordinate of the universe. Gullstrand-Painleve coordinates for the event horizon break this. In physics we do not use "nonstandard models" of arithmetic.

Item C says that we must be able to understand the reasoning. We do not understand the reasoning of the "proof" of the spin-statistics theorem in Wikipedia. The proof uses rotations of the spins of spin-z 1/2 particles. Our recent blog post about spin rotations may clarify various errors in the proof.

So far we have not found a good model for the electron spin and the magnetic moment. Our toolbox is not magical. It cannot solve all problems.

Another open problem is how Nature implements conservation laws. Are "events" some kind of transactions?

The collapse of the electron wave must always result in pair production?

How do we create the new wave function after the collapse of the old wave function?

An example is a photon which is emitted by a rotating dipole, and after that absorbed by another rotating dipole, and then re-emitted. The old wave function maybe was vastly diluted, but it becomes concentrated again after the absorption and re-emission.

A photon has the spin-z 1, and can be created as a standalone particle.

How to absorb and re-emit an electron: the semiconductor case

Let us study this process first in a semiconductor where there are bound electrons in valence bands and "free" electrons in conduction bands.

If we excite an electron and it jumps to a conduction band, we produce a pair: an electron and a hole, which behaves somewhat like a positron.

Now we have an interpretation for absorbing and re-emitting an electron. Let us have a free electron wave function which has spread wide in a block of a semiconductor

                ● -----------------> O -----------------> ●

     free electron            hole      new free electron

The free electron wave function collapse might be this process:

1. The free electron drops to a hole: the free electron is "annihilated with a positron".

2. The hole emits a new electron. A new "pair is produced".

The end result of this process is a new free electron, but no new hole.

Alternatively, we may interpret that in the Huygens principle, the two-phase process above creates a new wave.

      e-  --------

                   /

                 / e+

                 ----------- e-

        

The Feynman diagram of the process is like the letter Z.

Why we cannot simply make the wave function of the free electron to collapse? Why should we include a complicated two-phase process?

The reason might be the following: we can create a new, concentrated wave function for an electron if we produce a new pair. But it is not clear if we can do the same trick without pair production - using just the electron wave.

Analogously, we can create a new, concentrated wave function for a photon using absorption and emission. But it is not clear if we can do the same trick without absorption and emission.

In a Feynman diagram, a Green's function is used to create a new electron wave. It is somewhat suspicious if such a wave can be created without the associated positron.

Also, Erwin Schrödinger and others observed that a standard Dirac electron wave packet contains positron components. As if one would not be able to treat the electron as a standalone particle - the positron is always present.

Let us assume that we have an "atom" in a semiconductor. We may have a separate positive charge, a free electron orbits it. It might be that zones of positive charge appear in the semiconductor as the free electron moves. These zones can be interpreted as "positron" wave functions.

The electron in otherwise empty space

The semiconductor might be analogous to empty space. Pair production can be treated as the excitation of the electron from the -511 keV energy state to the +511 keV energy state.

The gyromagnetic ratio 2 of the electron

In our February 23, 2022 blog post we speculated that the gyromagnetic ratio 2 of the electron comes from the fact that we "see" the magnetic field of the associated positron, too.

The spin-z 1/2 of the electron defies imagination: how can an object possess a wave function which seems to cancel itself out after a 360 degree rotation? The hypothesis that a positron is tightly associated with the electron may help us to accept this strange state of affairs.

Conclusions

It might be that the correct way to manipulate the electron wave function is through pair production, and we cannot treat the electron as a standalone particle at all.

In a semiconductor, the system of electrons and nuclei is very complex, and it is not surprising if a free electron and a hole cannot be treated as standalone (quasi)particles.

Richard Feynman considered the hypothesis that positrons are electrons traveling back in time. The semiconductor analogue might be more fruitful: positrons are holes in a semiconductor.

A new question: are electromagnetic waves polarization in the "semiconductor" of the vacuum?

Where is the angular momentum "stored" in an electromagnetic wave?

Let us consider a rotating electric dipole.

                       ^

                       |      motor

                         + === O === -

                                                 |

                                                 v

A motor makes the dipole to rotate. The electromagnetic wave carries away angular momentum from the motor.

If we put an absorbing dipole close to the emitting one, the absorbing dipole starts to rotate and we can harvest energy from it. The absorbing dipole harvests angular momentum, too.

What kind of a model might describe the  angular momentum and its absorption? In a wave, parts of the wave are coupled together through the wave equation. We probably cannot describe the system with particles flying independently of each other.

                        __________

                  _ /                      \_

A jumping rope of children does carry angular momentum from one end to the other. Could this describe the coupling of the two rotating dipoles? The rope carries energy, angular momentum, and also linear momentum which tries to push the receiving end away. Qualitatively, the rope looks like a good model.

We believe that the Poynting vector of the electromagnetic wave describes the angular momentum. However, does it tell us in an intuitive way how the angular momentum is transmitted?

The angular momentum of a photon

The absorbing dipole receives angular momentum in units of the reduced Planck constant ħ. Do we have an intuitive model for this?

Classically, angular momentum is a continuous quantity. The receiving dipole would accelerate gradually. There would not be a discrete unit ħ. At the same time, the emitting dipole would lose angular momentum gradually.

What if we do not have receiving dipole? Then we interpret that the angular momentum lost by the emitting dipole went to the electromagnetic wave.

In quantum mechanics, a measurement makes the wave function of the receiving dipole to collapse. The probability of measuring an additional unit ħ in its angular momentum grows with time.

We interpret that the receiving dipole "absorbed a photon" if we see that its angular momentum grew.

The classical limit states that the if very many photons are emitted and absorbed, and if the dipoles are macroscopic, then the classical model must correctly describe the process. Consequently, we can calculate in classical physics, and only at the end map the result to the language of quantum mechanics.

Conclusions

Where is the angular momentum of a classical electromagnetic wave stored? We believe that the Poynting vector describes the phenomenon correctly. The angular momentum is distributed smoothly in the wave.

The jumping rope of children might be a qualitative classical model.

Where is the angular momentum of a single photon stored? It is distributed in the wave. In a wave function collapse, that angular momentum is found in a receiving dipole. There is no need to describe the storage of angular momentum in a more detailed way. We do not need to talk about the photon as a particle, or claim that the photon somehow "rotates" or "orbits in a circular path". The "photon" is essentially the wave function collapse.

How many photons does an orbiting electron send per a cycle?

Let us have an electron moving in a circular orbit at a constant velocity. How many photons does it radiate per cycle?

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

For a relativistic electron, the radiated power grows as γ⁴, where γ is the Lorentz factor.

Let us calculate with the non-relativistic Larmor formula. The acceleration is

       a = v² / r

          = ω² r.

According to the Larmor formula, the radiated power is

       P = e² ω⁴ r² / (6 π ε₀ c³).

The cycle time is t = 2 π / ω.

The energy per a single cycle is

       E = e² ω³ r² / (3 ε₀ c³). 

The energy of a single photon is

       E' = h f = h ω / (2 π).

The number of photons per cycle is

       n = 2/3 π e² ω² r² / (h ε₀ c³)

          = 2/3 π e² v² / (h ε₀ c³)

          = 0.031 * v² / c².

That is, if the electron is mildly relativistic (v ~ c), it takes ~ 33 cycles to emit one photon. The number 33 is not very far from 1. Is it a coincidence that a relatively small number of cycles is able to produce a single photon?

If the velocity v = 0.01 c, like for the electron in the hydrogen atom, it takes 330,000 cycles to produce a photon.

The energy of a single photon for a mildly relativistic electron is

       E'' = h ω / (2 π)

            ~ h c / (2 π r)

            = h / t,

where t is the cycle time. This is the energy - time uncertainty relation:

       ΔE Δt ≥ h.

We can derive the uncertainty relation by building a wave packet where we assume that the photon energy of a wave of a frequency f is h f.