The minimal coupling p - qA

UPDATE Jan 11, 2019: The Landau quantization

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

solves the problem of an electron in a uniform magnetic field. The solution hangs on the fact that the hamiltonian only depends on coordinate x in the Wikipedia article. The solution is then symmetric on translations along the y axis, which means that p_y must be constant, an eigenvalue of the p_y operator.

The operator

      p_y = -i d/dy

gives the canonical momentum which is not the measured kinetic momentum. Our earlier discussion in the text below mixed the canonical momentum with the kinetic momentum. This led to confusion in our text.

https://physics.stackexchange.com/questions/281687/why-is-p-y-conserved-in-the-landau-gauge-when-we-know-the-electron-moves-in-ci

At the Physics Stack Exchange there are several questions that rise from the confusion.

If we break the symmetry along the y axis by adding an electric potential which confines the electron into some y interval (y_0, y_1), what happens then?

---

The Pauli equation contains a hamiltonian where the kinetic energy term is something like

       (p - qA)^2 / (2m).

There p is the momentum vector, q is the charge, and A is the magnetic vector potential.

If we have a wire with a current, then the vector potential points to the direction of the current and is less when we go farther from the wire.

We assume that the wire has no electric field. Its magnetic field lines are circles around the wire.

If we let a charge fly freely closer to the wire, the magnetic force is at a right angle relative to the kinetic momentum of the electron.

---

NOTE: p is the canonical momentum. It does not change. The kinetic momentum p - qA does change, as it should.

The curl of the magnetic vector potential

https://en.wikipedia.org/wiki/Curl_(mathematics)

The magnetic vector potential A is defined as the vector field whose curl is the magnetic field B:

       ∇ × A = B.

The component of the curl pointing to the direction of the thumb at a point x is defined as the path integral of A on the circular path pointing to the direction of the fingers of the right hand, divided by the area enclosed by the path.

Let us consider an electric wire of a finite length:

           •        •        •  

        •   •   •   •   •   •   •

       -----------------------------

            I -->

The dots • mark magnetic field B lines pointing out of the page.

A vector potential A which reproduces the field B is like:

           -->    -->    -->

          ---->    ---->   ---->  A

        -------->   -------->

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

           I -->

|A| is larger close to the wire. A points to the direction of the electric current.

Now, if we have an electron approaching the wire in the diagram from up, the electron will draw a path like:

   e-              ^

      \            /

        \ ____/

       -----------------------------

         I  -->

  ^ y
  |
  |
   ------> x

The electron will turn counterclockwise in the magnetic field. Since the magnetic force

       F = q v × B

is orthogonal to the velocity vector v, it will not change the absolute value of v but its direction. The electron will leave the magnetic field with the same absolute velocity at which it arrived.

In the diagram above, the quantity

      H = (p - eA)^2 / (2m)

is not larger when the electron is close to the wire because p is not the kinetic momentum but the canonical momentum. Note that the charge e of the electron is negative.

Hamilton's equations

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

The equations are

       dp/dt = -dH/dq

       dq/dt =  dH/dp,

where p is the canonical momentum and q is the (canonical?) position.

The lagrangian

The hamiltonian is derived from the lagrangian. Let us check if the lagrangian gives the correct circular orbit for the electron in our example case.

https://lampx.tugraz.at/~hadley/ss1/IQHE/cpimf.php

The action is defined as the path integral of the lagrangian over a time interval [t1, t2] over a path q(t). If the path is a correct time evolution of the system, then the action should remain constant under "very small changes" of the path.

The path integral, of course, suffers from the fact that the set of allowed paths does not have an exact mathematical definition. Do we allow fractals? Let us consider just paths q(t) which are built from intervals of analytical functions.

Does the action change under a "very small" deformation of the circle orbit q(t)? We assume that t_1, t_2, q(t_1), and dq(t_1)/dt remain fixed.

https://en.wikipedia.org/wiki/Lagrangian_mechanics#Electromagnetism

The lagrangian of a massive charged particle with charge e in magnetic field is

       L = m/2 * v^2 + e v • A,

where v is the velocity vector and A is the magnetic vector potential. The dot marks the inner vector product.

Let us consider a circular path in the diagrams above. Let us assume that the magnetic field B is constant and points out of the page.
                 
                   <-- v  e-
                 ______
                /            \         -->    -->   -->
               |              |
                \______/        ----->     ----->    A
  ^ y
   |          
   |
   ------> x
 
The electron e- does a perfect circle. It leaves position x at a velocity v at time t_1, and returns back at time t_2.

We assume that there is no electromagnetic radiation out of the system. The radius of the circle is determined by the absolute speed |v| of the electron and the strength of the magnetic field B = ∇ x A.

What happens to the path integral of the lagrangian L over the time interval [t_1, t_2]? The action is

               t_2
       S = ∮      m/2 * v^2 + e v • A,
             t_1

where the path integral is done counter-clockwise.

We may assume that |A| is zero at the upmost point of the circle. Let us increase the speed of the electron by 1% for the whole circle, so that its radius is 1% larger. The starting point of the electron is kept constant. How does the action S change?

The kinetic part grows 2 %. The circle extends further down. In the lower half of the circle, |A| is 1% larger and |v| is too. The contribution is negative and grows by about 2%. There is no contradiction in these numbers. The lagrangian seems to work ok.

If the energy would vary, could a hamiltonian work for the Pauli equation?

If the energy of the electron would be larger in the lower part of the circle in the diagram, then the phase of its wave function would rotate faster in the lower part. The phase difference between different points in the circle would tend to infinity, which would mean an infinite momentum.

The Schrödinger equation under a scalar potential works beautifully because the total energy is constant, and the phase of the wave function rotates at the same rate everywhere.

The Biot-Savart law and magnetic interaction

http://teacher.nsrl.rochester.edu/phy122/Lecture_Notes/Chapter30/chapter30.html

For slowly moving charges, the magnetic force is

      F = μ_0 / (4π) * q_1 q_2 / r^2 * v_1 x (v_2 x r),

where r is the vector from charge q_1 to q_2. The formula is a consequence of the Biot-Savart law.

To which direction does the force point? In our example above, we have electrons moving to left inside the wire, and our test electron does a circular path. That is, the force is always orthogonal to the velocity vector v_2 of the test electron.

Can we model the force simply as a force between two point objects? The forces on the charges q_1 and q_2 will generally exert a torque on the system q_1 & q_2. Conservation of angular momentum requires that there is an opposite torque on something. Apparently, the magnetic field can store angular momentum.

If we stop both charges with some device, the magnetic field disappears. If there is no radiation out, then the angular momentum stored in the magnetic field has to be returned back to the system q_1 & q_2 & the device.

The minimal coupling is approximately right for the classical electron?

In our example case of an electron circling in a loop, some angular momentum is stored in the combined electromagnetic field of the wire and the electron. Can we neglect this effect in the classical treatment of the electron?

Probably yes, because a hypothetical self-interaction of a single electron is much weaker than the the force which is caused by the external magnetic field B.

Wave function collapse is like recreation of the particle

Suppose that we have a particle which is described by the Schrödinger equation. When we measure its position at some accuracy Δs to be the point x, we collapse its wave function.

To describe the further development of the particle, we form a wave packet from plane waves, such that its initial configuration has a width of roughly Δs and it is centered at x. The wave packet is built by summing plane waves with various momenta p. The wave packet is built such that it has no angular momentum, it has a mirror symmetry around any plane which contains x.

We assume that a particle is born from applying a source to a wave equation. When we build the new wave packet, we kind of recreate the particle. If we build a packet which contains no angular momentum, then the spin of the particle is zero.

But the electron contains h-bar/2 of angular momentum. We have two options:

1. Add the angular momentum to the particle as a spin quantum number and use a wave packet which contains no angular momentum.

2. Or, build a new wave packet which contains angular momentum. For example, we could build "mini-packets" which are launched from a ring around x in a fireworks pinwheel fashion. This would imitate the original creation of the particle wave by a rotating source.


In an earlier blog post, we had the idea that a rotating electric dipole is sending rotating mini-dipoles around, and such a mini-dipole is the photon.

The Huygens principle states that "every point in space becomes a producer of new waves". This sounds somewhat like option 2 above.

There is a complication in the case of the electron: any source which produces an electron will also produce a positron. How to get rid of the created extra positron? Maybe the positron is created "virtual" in some sense, and it proceeds to annihilate the original electron.

The gyromagnetic ratio of the electron is 2 because the electron sees a magnet double?

The angular momentum of a rotating electron is only h-bar/2. If we imagine it making a circular orbit, its wave function can only complete half a cycle during a full circle of the orbit.

Let us put a covering space around the full circle, such that in the covering space points are identified when their angular distance is 720 degrees. In a normal circle, points are identified after 360 degrees.

In the covering space, the electron can orbit in the normal way such that its wave function does a full cycle during a round of the orbit.

The covering space is like a coil which consists of two loops. The covering space is mapped non-bijectively to the 360 degree circle in the natural way, such that every point in the circle is an image of 2 points in the covering space.

The rotating electron lives in the covering space. An external magnet lives in the normal 1+3D space.

The electron will see the magnet double: the magnet appears at two points in the covering space. It will also see the magnetic field lines double. This might explain why the interaction of the rotating electron with the external magnetic field is 2X of what one would expect - its gyromagnetic ratio g is 2.

The question is what is physics like in the covering space. Do external interactions that depend on rotation have a double strength?

It is not at all clear that the physics in the covering space should produce results that differ from the ordinary 1+3D space. We can certainly build physics which agree 100% with the ordinary space.

Quantum mechanically, the strange angular momentum 1/2 h-bar is the outstanding feature of the electron. Therefore, it probably is responsible for the equally strange gyromagnetic ratio 2.

The Dirac equation, and probably the Pauli equation, too, are aware of the angular momentum as well as the gyromagnetic ratio. The equations do not give us an intuitive model of the rotation and the associated waves. We have been trying to sketch a model, but cannot yet connect it to the equations.

https://arxiv.org/abs/1207.5752

David Delphenich in his arxiv.org paper writes about the history of the Pauli and Dirac equations. There were efforts to find an intuitive model for the electron spin 1/2, but no one has succeeded in it.

We in this blog claim that the existence of a spin is a necessary consequence of a quantized field  if the source can feed angular momentum into the field. The challenges are linking this idea to the known equations, and showing the intuitive reason for g = 2. The idea of a covering space is not too intuitive, and even less intuitive is the claim that the magnetic field is seen as having a double strength by an electron living in the covering space. We did not find a plausible explanation.

The Pauli equation revisited

The fact that in the Pauli equation, the whole kinetic term, including the linear momentum p, is subject to the Pauli matrices, suggests that the "correct" momentum space is the space of 2 x 2 Hermitian matrices. Rather than being a vector

          (1, 0, 0) * p_x
       + (0, 1, 0) * p_y
       + (0, 0, 1) * p_z,

the momentum is represented as a 2 x 2 matrix

          σ_1 * p_x
       + σ_2 * p_y
       + σ_3 * p_z,

where the sum matrix operates on a two-component wave function

        (Ψ_+,
         Ψ_-).

The Pauli equation is a natural mathematical generalization of the kinetic term to a 2-component wave function. Mathematically, the Pauli equation is "intuitive" in this sense. But physically, it is hard to comprehend.

We explained in the previous blog post that if we put angular momentum in a field, then the wave function collapse cannot be modeled with a simple one-component wave function of a point particle.  The angular momentum is a global property of the field and cannot be formulated in a hamiltonian which only looks at a single point in spacetime of a one-component wave function.

It is then natural to ask how we should add more "degrees of freedom" to the wave function, so that it describes the angular momentum, too. The Pauli equation shows that adding another component to the wave function does the trick. But why does it work?

If angular momentum is stored in a field, the wave function collapse requires the hamiltonian to have global information

Suppose that an electron-positron pair is born from a rotating source, and the source stores angular momentum to the electron wave function.

           rotor
           -----------
               |
             _|_
           |      | motor

The wave function far away may look almost like a plane wave. But since it contains rotation, it should react to a magnetic field like a magnetic needle. How can the hamiltonian at a point far away know that the wave contains rotation?

Classically, there is no problem because the information of the rotation is stored in the global wave function. But in quantum mechanics, the wave function collapse will erase the global wave.

The hamiltonian in quantum mechanics should be able to calculate the energy at a spacetime point. It cannot use global information of the field.

The mysterious spin of particles may be due to the fact that angular momentum is not a local phenomenon, and it has to be added artificially to the local wave function of the particle as an extra quantum number.

The above reasoning does not yet explain the gyromagnetic ratio g = 2 of the electron, though. We need some further insight.

We are currently studying the detailed structure of the Pauli equation. It contains a trick of factoring the kinetic energy operator, which reminds us of the factoring of the Klein-Gordon operator in the Dirac equation.

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

The kinetic energy operator

     p^2 = (d^2/dx^2 + d^2/dy^2 + d^2/dz^2)
                / (2m)

can be factored into a product of sums by using the Pauli matrices:

       (σ_1 d/dx + σ_2 d/dy + σ_3 d/dz)^2
       = I d^2/dx^2 + I d^2/dy^2 + I d^2/dz^2.

The factorization can be seen as introducing a generalized momentum operator P, such that P^2 will react to a magnetic field B through cross terms.

The wave - particle duality "explains" the momentum and the spin of a particle

A particle is like a concentrated packet representation of a wave.

We can store a linear momentum p into a wave, and can associate p with the momentum of a particle. A particle in free space can have any momentum. The momentum is not quantized.

We can store angular momentum in a wave. If the wave just has one particle, then that angular momentum must be its spin. Since rotation happens in a restricted 360 degree space, the values of the spin are quantized.

The spin is not related to special relativity; neither is the existence of the positron

Our analysis of the Dirac equation is nearing completion.

We had the example where a rotating electric dipole stores angular momentum to the electromagnetic field. Did we need to use special relativity there? No. The only thing which matters is that we can use a rotating object as the source of waves. It could be a finger doing a circle on a drum skin as well.

The spin is the result of:

1. One can store angular momentum to a wave through a cyclic process.

2. That angular momentum is divided evenly to the energy quanta of the wave.

What happens in the derivation of the Dirac equation

We start from the Klein-Gordon equation and try to write it to an equivalent form where there are no second derivatives.

Since the Klein-Gordon equation is Lorentz covariant, there is a good chance that the equivalent form is, too.

We separate solutions which correspond to positive m and negative m. For that purpose, we use the matrix

        1   0   0   0

        0   1   0   0

        0   0  -1   0

        0   0   0  -1

as the coefficient of m Ψ. The "main components" of the negative energy solutions will appear in the lower components.

In the original Klein-Gordon equation, negative frequency solutions correspond to negative E. The Dirac spinor is just a new way to represent a basis of positive and negative frequency solutions of the Klein-Gordon equation.

The factorization of the Klein-Gordon operator, for some reason, also separates the spin_z up and spin_z down cases of solutions. We still need to analyze in more detail why that happens. The factorization has lots of symmetry. Maybe it is not that surprising that the separation happens.

A linearly oscillating source and spin 1/2

Since the Klein-Gordon equation is a general wave equation which allows a rotating object as the source, its quanta can have a non-zero spin.

Let us assume that the pair-producing source of the Klein-Gordon equation does linear back-and-forth movement. It is like a dipole radio transmitter. The energy which is pumped into the source contains no angular momentum.

In the case of photons, we needed to split the linear movement into two rotating dipoles to get the right model for quantum production. One dipole rotates clockwise and the other counter-clockwise when looked down from the positive z axis. Let us do that split in this case, too.

Let the frequency of the rotation of the source be f.

Because of the symmetry of negative and positive frequencies, the clockwise rotating source will produce both negative and positive frequency waves, and an equal amount.

Just as in the case of the photon, we can calculate that for work h f, the angular momentum stored in the field corresponds to spin 1. The energy and the angular momentum is evenly divided between the positive and negative frequency waves.

A quantum consists of the electron and the positron. The positive frequency wave corresponds to the electron. The energy of the electron is 1/2 h f and the spin_z is 1/2. The same for the positron.

The counter-clockwise rotating source produces electrons and positrons of spin_z -1/2.

Thus, the linearly oscillating source produces both electron and positron waves with spins up and down. What is the quantum of this whole process? It must be an electron-positron pair whose spins point at opposite directions, to ensure that the total angular momentum is zero.

Note that the rest mass of the quantum is 2 m_e, where m_e is the rest mass of the electron. If the momenta of the particles is zero, the quantum energy

       h f = 2m_e c^2.

The source has 2X the frequency of the wave of the electron alone.

In deriving the spin 1/2, we did not use special relativity at all. Our assumptions were:

1. The source in the Klein-Gordon equation does a linear oscillation.

2. The correct way to model linear oscillation of a source is two rotating sources.

3. A quantum of the process is the electron-positron pair and its energy is h f, where f is the frequency of oscillation.

How does the Dirac equation know that the electron spin is 1/2?

In the preceding section we showed that a linearly oscillating source produces waves with both positive and negative frequencies, and both the spin up and down.

How does the Dirac equation "know" that the spin of the electron is 1/2?

It may be that the Dirac equation describes correctly a half of the Klein-Gordon wave which we produced with the linearly oscillating source. We need to investigate this in more detail.

Conjecture 1. A spatially circularly polarized wave solution of the Klein-Gordon equation stores h-bar/2 of angular momentum per m of "rest mass energy", if m is non-zero. A linearly polarized wave must be broken down to circularly polarized components to reveal the "hidden" angular momentum.

A rotating source which radiates electron-positron pairs

A basic principle of quantum mechamics is: If we have a rotating source, its angular momentum is quantized in amounts of h-bar. It can radiate a quantum whose energy is h f, where f is the frequency of rotation.

Let us work with the Klein-Gordon wave equation. We assume that it can handle the type of source we have.

We conjecture that if h f is greater or equal to twice the electron rest mass, the source can radiate pairs. Each particle in the pair carries h-bar/2 of angular momentum.

In practice, the produced electron and positron have opposite spins. We conjecture that the source in such a case is doing linear oscillation.

The annihilation of a pair usually produces two photons of opposite spin. If we invert time, we produce a pair. It looks sensible that two photons with opposite spins make a linearly oscillating source to the electron-positron field. The source may through some mechanism be the strong electric field.

The quantum of the Klein-Gordon electron-positron field is the pair electron-positron. If we have an arbitrary wave, we can separate it to positive and negative frequency components to know the electron wave component and the positron wave component. Here the positive and negative frequencies correspond to the sign of E in a basis wave

       exp(-i (E t - p x)).

The "rotation" which differentiates the electron and the positron happens in an abstract complex space.

The spin is rotation which happens in the familiar 3-dimensional space. The rotating source stores angular momentum to the Klein-Gordon field.

We conjecture that the Dirac equation separates the Klein-Gordon waves which have positive versus negative angular momentum when looked down from the positive z axis.

That is, if the waves were born from a source which rotates clockwise or counter-clockwise, the Dirac equation separates them to different components. If we look at the wave very far away, it looks almost like a plane wave. But the big difference is that the wave contains stored angular momentum. The wave can be approximated by a plane wave plus information about the angular momentum. That is probably the origin of the Dirac spinor. The basis waves of the Dirac equation are like plane waves which contain a little more information in the spinor.

The stored angular momentum, that is, the spin, is present already in the familiar Klein-Gordon waves. The Dirac equation just pinpoints it.

The gyromagnetic ratio g = 2

Our model should be able to explain why the magnetic moment of the electron is 2X of the most straightforward orbiting particle model.

If the electron is a point particle, and it can be described with plane waves of type

       exp(-i (E t - p x)),

then it is hard to build a model where the gyromagnetic ratio is anything else than 1. If the "mass charge" and the electric charge are bound in the same way into the point particle, then the angular momentum and the magnetic moment should have the same ratio in all kinds of movement, whether it is translation or rotation.

When a pair is produced, there first is an electron present and a positron with an opposite spin present. If a particle B interacts with the pair, it will measure that the magnetic field is 2X of what one would expect from a lone electron.

But according to the Feynman interpretation, the electron and the positron are just one and the same particle. The particle B interacts with two copies of the same electron.

Suppose then that the positron has flown away. If B comes to interact with the electron, does it now interact just with one copy of the electron? If we are allowed to switch the time and space dimensions, then B should interact both with the "arriving" electron and the "departing" electron. This may be the reason why the magnetic moment of the electron is twice of what one would expect.

In Feynman diagrams, the pair production and photon scattering are drawn in a symmetric fashion:

  ^
  | time

  photon      e-
          \        /
            \    /
              \/     emission
              |
              |     absorption
              /\
            /    \
           /        e-
photon

      e+                e-
        \                 /
          \              /
            \______/      pair production
           /             \
         /                 \
  photon         photon

Why is the angular momentum just the spin 1/2 then? If we measure the angular momentum, should we measure both the arriving electron and the departing electron?

When we measure the electric charge of the electron, why is there no doubling if the model above is right?

How can the Dirac equation "know" about the mechanism we sketched above? It correctly predicts the magnetic moment.

The gyromagnetic ratio 2 seems to require some internal degree of freedom for the electron. A point particle would have it 1. The Dirac equation, for an unknown reason, conjures up the right internal degree of freedom. The four components of the Dirac spinor are interdependent, but allow some freedom.

We have not yet found an intuitive reason for the value 2, but let us keep trying.

Electron and positron spin is 1/2 because we rotate a string "from the middle"

The Feynman interpretation is that the positron is an electron going backward in time, and in a produced pair, the electron and the positron are really the same particle.

The wave functions of the electron and the positron have a different handedness with respect to increasing time.

t

 ^

 |

 |      e-                        e+

 |        \                       /

 |          \                   /   string

 |            \________/

 |                 \\O// spider sits on the string

 |                          and rotates it

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

This suggests the following model: when we excite the electron-positron field to produce a quantum, we let an agent (a spider in our earlier blog posts) to rotate the string while sitting on the string in the middle. The spider will create a right-handed wave to one direction and a left-handed wave to the other. The waves are strictly correlated, "entangled".

A quantum of the field contains both waves.

A quantum of the field has the energy h f. The "total absolute sum" of angular momenta in the quantum corresponds to spin 1. That total amount is divided between the electron and the positron who each have the absolute value of the spin 1/2.

What if the spider wants to create photons? The photon can be absorbed and emitted alone. The spider sits at the end of that string and rotates.

t

 ^             photon

 |             /

 |           /   string

 |         /

 |         \\O// spider rotates the end of the
 |                string

  ------------------------------->

We proved in an earlier blog post that a rotating electric dipole will make quanta whose spin is 1. That is why the photon has the spin 1.

Thus, the reason why the electron has the spin 1/2 is:

1. The produced positron and the electron are "the same particle", which means that they live on the same string.

2. A quantum contains both the positron and the electron.

3. The total absolute sum of the angular momenta which a quantum puts to the string corresponds to spin 1.

4. The total absolute angular momentum is divided evenly between the electron and the positron.

The above model is more intuitive than our earlier tube model of the produced pair. Having the electron and the positron as waves on the same tense string stresses the wave nature of particles.

Fermions are particles which can bounce back and forth in time. Their string has no end. Either it is a loop, or the string extends to infinity. Boson strings may have ends and their strings probably cannot zigzag in time.

We need to check if the Dirac equation in some sense corresponds to a string which is excited from the middle. Maxwell's equations would correspond to a string which is excited from the end.

Actually, a string is not the right analogue for a particle which has a non-zero rest mass. Its waves have dispersion, where long waves move faster. Water waves are a better analogue. Thus, a fermion is like a water wave which zigzags in time and is always produced to two directions at once in the pool.

How does a positron go back in time and transform to an electron?

The spider on a string model allows us to investigate what happens in a pair production. Let us follow the positron backward in time first.

1. The positron wave packet moves closer to the spider and touches the spider. The positron wave packet is right-handed in the time-backward perspective.

2. The positron wave packet is completely absorbed by the spider.

3. We change the direction of time to the normal.

4. The spider emits the electron wave packet. The electron wave packet is right-handed.

5. The spider can stay static in its place. It will use two legs to rotate the left and the right side of the string in unison.

The wave packet of the positron was first completely absorbed by the spider, and then re-emitted as the electron wave packet. The wave packet did not continue through that part of the string undisturbed, but the spider was the middleman.

                     <-- wave packet

       wall  |--------------------- string

An analogous process happens when an electron wave is reflected from a potential wall. The simplest such case is when we have a tense string and a wave meets the endpoint which is a solid wall. There is a 180 degree phase shift in the reflected wave. The wall at no point absorbs the wave completely. That is a difference from the spider model. There is something similar, though. The wall makes a 180 degree phase shift to the reflected wave. The wave does not continue undisturbed.

The spacetime diagram of the reflection is something like this:

   wall  |

            |\ \/  /

            | \ \/   waves

            | /\ \

The wave is never completely absorbed by the wall.

         /                     \

             /             \

    ______/___\_____ "time wall"

                 \\O// spider

The spider "time reflection" case has no waves at the start of the process. The diagrams look quite different. Maybe pair production is not analogous with ordinary scattering, after all?

Does a source in the Dirac equation produce both electron and positron waves?

The crucial question in the string model is that if we put a disturbance to the Dirac equation, that is, add a source to the homogeneous equation, does it automatically produce an equal amount of electron and positron waves?

An equal amount is required by charge conservation, too.

We do not know if the Dirac equation correctly describes pair production, but let us try to analyze it.

The "complete set of orthogonal solutions" to the Dirac equation apparently has to include both the positive energy E solutions and the negative energy -E solutions. Only then we can construct an arbitrary wave packet. Since the Dirac equation is completely symmetric with respect to E and -E, then a random source to the homogeneous Dirac equation will produce both E and -E waves. But under what condition is there a perfect balance of those waves, so that charge is conserved?

A collision of two electrons can produce a pair. The setup is not physically symmetric for an electron versus a positron, but empirically we know that an equal number of electrons and positrons are produced. Maybe in nature, the sources to the Dirac equation are always symmetric for some unknown reason, or the Dirac equation needs to be modified to ensure an equal number of produced electrons and positrons.

For Maxwell's equations, we probably can form a complete set of orthogonal solutions from the positive energy solutions.

Let us define that positive energy solutions are emitted through one string which starts from the emitter, and negative energy solutions through another string. We have found the mathematical meaning for our diagrams above.

Our new Dirac equation has m + V, where V is a scalar potential, in the place of m.

The electron has m positive and E positive while the positron can be defined to have m and E negative. If V is the potential for the electron, then -V is the potential for the positron. There is a perfect symmetry. If a pair is produced in some way through an abrupt spike V, then there might be a perfect symmetry for the electron and the positron production.

The Dirac equation has 4 components because it codes the spin of both the electron and the positron?

Wolfgang Pauli originally added the spin to the Schrödinger equation by making the wave function two-component and defining the spin operators based on the Pauli 2 x 2 matrices.

Our thesis is that the quantum of the Dirac equation is actually the combination of the electron & the positron. It is logical that the wave function would then have 4 components, because the electron and the positron are allowed to reorient their spins independently of each other.

If the rotation state of the electron is independent from its spatial location, then the electron can be described by a triplet

        (Ψ, s_d, s_u),

where s_d and s_u are complex numbers and Ψ is a one-component complex function. Let x be a spacetime point. If we rotate x infinitesimally around some axis A so that it is mapped to x', we have to update the triplet slightly to get the value of the triplet at x'.

The rotation state of the electron lives in the same spacetime as the wave function Ψ. In that way it is not completely independent.

The Dirac equation "separates" the positive and negative frequency solutions of the Klein-Gordon equation

Let us look at the 1+1-dimensional Dirac equation. The spin does not come up there, but the positron does appear.

The basis of the Klein-Gordon solutions is probably all plane waves

       exp(-i (E t - p x)),

where E is allowed to be negative or positive. A negative energy corresponds to a "negative frequency" with respect to time.

The basis for the Dirac solutions consists of

       (1, p / (E + m)) * exp(-i (E t - p x)),

where E is positive, and

       (p / (E - m), 1) * exp(-i (E t - p x)),

where E is negative. We see that the Dirac equation separates the positive and negative frequency solutions, based on the coefficient spinor (1, ...) or (..., 1).

Next we should analyze what happens in 1+3 dimensions. Why does the spin come up?

What fields can store angular momentum and spin?

Our photon generator, the rotating electric dipole, was able to pump angular momentum into the electromagnetic field. That is the ultimate reason why the photon has a spin: if the field is absorbed in quanta, then both the energy and the angular momentum have to be absorbed.

Is a source able to pump angular momentum into the electron field? If yes, then the electron must have a spin.

The symmetry of the Dirac equation suggests the following principle: any source has to be symmetric in the way that it creates perfectly symmetric waves to the electron and the positron fields. Then charge is conserved. The spins of the produced pairs have to be perfectly antisymmetric.

Suppose that we have the following setup:

          e- e+
            ^
            |    rotating rod
            ----------o---------
                                    |
                                    v
                                  e- e+

A rod rotates clockwise and shoots electron-positron pairs to the directions marked by the arrows. In that way we do store angular momentum into the electron-positron field. Any field which can store linear momentum can also store angular momentum. But do we also store some "intrinsic angular momentum", that is, spin? Could it be that the electron and the positron both have the spin 0 and only store angular momentum in their linear momentum p?

In the case of photons, can the angular momentum be stored in the linear momentum p of the photons if we shoot them from the ends of the electric dipole? When the dipole loses its rotational energy, its both ends lose a lot of momentum p. Photons have p / E the minimum possible value for any material object. If the dipole rotates so that its ends move at the speed of light, then two photons shot from the ends of the dipole would be able to carry away the energy and the momentum.

In practice, nature seems to store the angular momentum into the spin of a single photon. Maybe one photon is a better quantum than two strictly correlated photons?

Analogous black holes do not produce Hawking radiation

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

In this blog we have presented reasons why the various derivations of Hawking or Unruh radiation are flawed. The main obstacle in the proofs is that they break conservation of energy or momentum. At this point it looks like Hawking or Unruh radiation does not exist. All the derivations use flawed quantum field theory.

We are not alone in our skepticism. Vladimir Belinski, Detlev Buchholz, and some other researchers have come to similar conclusions in the past 25 years.

Let us look at an analogous black hole, a sonic or acoustic black hole, where the "horizon" forms at the point where the speed of a liquid flow, measured in the laboratory frame, exceeds the speed of sound in the liquid.

https://www.nature.com/news/artificial-black-hole-creates-its-own-version-of-hawking-radiation-1.20430

"On one side of his acoustical event horizon, where the atoms move at supersonic speeds, phonons became trapped. And when Steinhauer took pictures of the BEC, he found correlations between the densities of atoms that were an equal distance from the event horizon but on opposite sides. This demonstrates that pairs of phonons were entangled — a sign that they originated spontaneously from the same quantum fluctuation, he says, and that the BEC was producing Hawking radiation."

Jeff Steinhauer from Technion, Haifa, claims to have observed Hawking radiation in the form of phonons.

The classical limit of sonic Hawking radiation

A basic result, or in some cases an assumption, in derivations of Hawking radiation is that a freely falling observer close to the horizon sees a vacuum, that is, no vibrations or sound.

An often cited derivation claims that a static observer close to the horizon must see the space as "warm", that is, energy quanta in space, if the freely falling observer sees a perfect vacuum.

Suppose that we have several microphones close to the horizon of a sonic black hole, and they register very many coherent phonons of Hawking radiation in a short time interval. Since the radiation is random, there is a non-zero probability of seeing such an event.

But many coherent phonons mean that there is a classical sound wave. A freely falling observer must see such a sound wave. The freely falling observer can measure the density of the liquid around him and detect it is varying around him. This is a contradiction, because we assumed that the freely falling observer sees a perfect vacuum.

In the quote above, Jeff Steinhauer says that he has "taken pictures" of density fluctuations around the horizon. Taking pictures suggests that the density fluctuations are a classical sound wave. That contradicts the basic assumption that the freely falling observer sees a perfect vacuum.

How would sound waves know where the horizon is?

How would atoms in the liquid know that they are at the horizon, so that they know to produce phonons?

The location of the horizon is defined as the point where sound waves stand still relative to the laboratory frame. If the fluid flows undisturbed, how do the atoms know what is the rest frame of the laboratory?

If the fluid is in a vessel whose walls are at rest relative to the laboratory, then the atoms might detect the laboratory rest frame by receiving some signal from the walls. But this is in contradiction with the assumption that the freely falling observer sees a perfect vacuum.

Entropy

If microphones receive phonons, where does the energy of these phonons come from? Obviously, the liquid must lose some of its kinetic energy to produce the phonons. The kinetic energy of the liquid is translational, it has an extremely low entropy. Produced phonons have a much higher entropy.

What process is able to increase the entropy in the experiment? Usually, it is friction which produces high-entropy energy, but we were assuming that the liquid is perfect and flows without any kind of disturbance. 

Momentum conservation

If the liquid loses some of its kinetic energy, where does the extra momentum go? It apparently has to be absorbed by the walls of the vessel, if the liquid is in a vessel. But if the acceleration of the liquid is done without any vessel walls, what in that case absorbs the surplus momentum?

Conclusions

It looks like no analogous Hawking radiation, in the form of phonons, can exist in a sonic black hole.

We still need to look at the papers by Unruh and Steinhauer, if they have addressed the contradictions we uncovered above.