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.

When is there interference if two quantum systems are entangled?

Our derivation of the electron spin depends on the fact that the (hidden) spin rotation angles of an electron and its brother positron are entangled, so that destructive interference cannot happen in the electron rotation.

There certainly happens interference in the double slit experiment, though the passing photon is entangled in many ways with other particles. What exactly governs the appearance of interference?

In the double slit experiment, if we "can know" through which slit the photon passed, there is no interference pattern on the screen. What does it mean that we "can know"?

The passing photon always interacts with the walls of the slits. It is diffracted by the walls. The interaction does not destroy the interference pattern.

But if the photon collides with another particle close to either of the slits, then the interference pattern does not appear.

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

The weak measurement concept by Yakir Aharonov and others is of relevance here. A weak measurement does not destroy interference, though a strong measurement does.

The double-slit experiment

The wave of a single photon passes through both slits and forms the familiar interference pattern on a screen.

Let us try to detect through which slit the photon passes. That will destroy the interference pattern. We put lots of free floating electrons at one of the slits.

Classical electrodynamics explains why there is no longer an interference pattern. The electrons disturb the wave which comes through the slit. The inertia of the electrons makes them a source of new waves of various frequencies. If we have enough electrons, the wavefront through that slit is badly scrambled, and there is no interference pattern on the screen.

Free electrons make the wave equation nonlinear. They act as a source in the wave equation. We say that a photon has been scattered from an electron if we observe a photon which has a new frequency or whose path has been disturbed.

A produced pair of particles in a ring

Suppose that we have some process which produces two particles which circle a ring at the same speed to opposite directions. If there were just one particle, an integer number of the wavelength should fit in the ring. It is like the Bohr atom model. But if we have two particles?

It is easiest to handle the problem in cartesian coordinates, where we identify points

         n C,

where n is an integer and C is the length of the circle. The x axis is a covering space for the circle.

For a single particle, suppose the solution is of the form

       exp(i p x) * exp(-i E t).

The value of the first factor must return to its x = 0 value at x = C, so that there is no destructive interference. Setting t = 0, for the lowest energy state, we get the equation

       i 2π = i p C

      <=>

       p = 2π / C.

What if we have two particles? How are two particles born together? Suppose that we have a particle whose energy is 2E, momentum 0, and for which we do not know the position at all. Its wave function is

      exp(-i (2E t - 0 * x)).

The particle decays very fast into two particles, each of which has the energy E.  How do we determine the wave function of the two particle combination?

Suppose that the particles were born at x = 0. We may redefine the laboratory coordinates if not. The wave function of the whole configuration might be

       exp(-i (E t + p x_1)) * exp(-i (E t - p x_2)) * δ(-x_1, x_2)

where δ(y, z) is 1 if y = z, zero otherwise. We may define x_2 = x and x_1 = -x, and we get

       exp(-i (2E t - 2p x)).

The periodicity condition then gives p = π / C. That is, we can treat the whole system as a single particle who has the energy E and the momentum 2p.

What were our assumptions and what did we really prove if anything? We assume that when the particles are "born", they already have a stationary wave function which occupies the whole ring. We assume that the wave functions of both particles are symmetric at 0. We assume that we can get the wave function of the whole system by multiplication, and we assume that we can project out the solution with the δ() function.

From those assumptions we proved the Principle of combination 1 of our previous blog post. It is not at all certain that this "proof" is any more convincing than the Principle of combination 1 taken as an axiom.

What happens if we let the 2 particles to move to separate rings?

Is it essential that they are circling the same ring? Or is the periodicity condition satisfied even if one of them moves to another ring?

Since the electron and the positron can exist independently, the answer empirically is yes.

If we move the particles to separate rings, and let ring B move fast, then special relativity says that the clocks at the rings no longer agree. The phase of the particle in ring B has shifted, as measured by an observer moving with B. Does that mean that the Principle of combination no longer works and there is destructive interference in both rings?

Let A describe the particle in B with a wave function of type

      exp(-i (E' t' - m' v' x)).

A and B agree on x, but not on the quantities marked with the prime symbol '.

A sees the particle in B move slower, v', but that is compensated by the bigger mass m'.

A sees the time t' go slower in B, but that is compensated by a higher energy E'.

It might be that according to A, the phase of the rotation in B goes at the same rate, regardless of the translational speed of B. That would mean that A "knows" the phase of B, and the Principle of combination works even when the rings are separated from each other.

The reasoning depends on the fact that nothing can affect the absolute "rotation speed" of the wave function of B, as seen by A. The rotation axis can be turned, but the absolute value of the angular momentum stays the same.

The electron spin 1/2 derived from basic principles of quantum mechanics

Spin of the photon derived from classical physics plus the Planck-Einstein relation

Let us study the classical limit of polarized photons. The limit of many coherent photons has to make sense classically.

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

For a photon, the spin is equivalent to the circular polarization of the electromagnetic wave.

Suppose that we have a rotating electric dipole. It will emit electromagnetic radiation which carries away angular momentum. Actually, all the energy that the dipole radiates is originally mechanical rotation energy. We may have a motor which makes the dipole to rotate, and the mechanical energy is emitted as electromagnetic waves.

       electromagnetic
               waves
      \            |             /
        \          |           /

        rotating electric      ______
               dipole
          e- --------- e+
                    |
                    |           \
                __|__         \
              |          |
               motor

The electromagnetic waves thus contain angular momentum. When a photon is absorbed by an antenna, the angular momentum moves to the antenna and tries to make the antenna to rotate to the same direction as the dipole which sent the wave.
Above we have a classical interpretation for the photon spin. Linearly polarized waves do not carry angular momentum, but quantum mechanics states that each photon still has spin 1 or -1.

If the length of the rotating dipole is 2r, and the motor has to exert a force F to one end of the dipole, then the power of the motor is

       P = E / t = F ω r,

where ω is the angular velocity. The motor transfers angular momentum to the field at the rate

       L / t = F r.

The ratio E / L is ω.

According to the Planck-Einstein relation, a photon has the energy

       E = h f.

We know empirically that its angular momentum is

       L = h / (2π).

The ratio E / L is 2π f, which is equal to ω. We were able to derive the spin = 1 of the photon from classical physics plus the Planck-Einstein relation. Our result generally implies that if we can pump energy to a field with a rotating object, then the quantum of the field has the spin 1 or -1.

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

A question is, if a circularly polarized wave in some sense is a vortex, too.

Why is the photon spin +1 or -1, why not 0? That is, why is a single absorbed energy quantum of the field always circularly polarized and not linearly polarized?

In rowing, an oar which is moved linearly in water will produce two vortices spinning to opposite directions. Could the circular polarization of photons have an analogous origin? The vortices made by an oar can be interpreted as a "particle" and an "antiparticle". For a photon of spin +1, the antiparticle is a photon of spin -1.

A linear dipole radio transmitter

       <--->                 <--->

          e-  -------------  e+

Suppose that we have a horizontal electric dipole which we alternately make longer and shorter. It will emit a linearly polarized electromagnetic wave. The dipole is a radio transmitter.

One can certainly extract angular momentum from a linearly polarized wave, too. Just put another dipole vertically in the field. The dipole is an antenna:

                     e-

                      |

                      |

                     e+

The alternating horizontal electric field of our horizontal radio transmitter will alternately turn the vertical dipole clockwise and counter-clockwise.

If we assume that the the vertical dipole can absorb a quantum of energy, E = h f,  from the field, then that quantum also carried angular momentum.

What about a quantum which carries no angular momentum? If our radio antenna is horizontal, then it might absorb a quantum E = h f which had no angular momentum. Do we have any argument against the existence of such spin = 0 quanta?

The radio transmitter cannot produce all its energy output in spin = 0 quanta, because then the vertical antenna would not turn. But can it emit at least some of its output in spin = 0 quanta? Classically, the vertical antenna can receive energy and angular momentum from any linearly polarized radio wave, even if the wave is extremely weak. Suppose that our radio transmitter produces some spin = 0 quanta. If we can filter all spin = 1 and spin = -1 quanta out, what is the remaining wave? It cannot be a classical wave because a classical wave can always be absorbed by our vertical antenna. This is a heuristic argument against the existence of spin = 0 photons. They would contradict the classical theory of electromagnetism.

Longitudinal electromagnetic waves

If we put a test charge in line with our horizontal antenna:

       test charge     e-         e- ----------- e+

then some of the energy that we feed in our radio transmitter will go to moving the test charge. The static electric fields of the charges in the antenna will move it. We could say that the test charge is receiving "longitudinal" electromagnetic waves. The intensity of such waves falls down very rapidly when the distance from the transmitter grows. That is why they normally are not called waves at all, but just an electric force. Transverse electromagnetic waves can carry energy over vast distances.

What is the spin of a hypothetical quantum of longitudinal waves? We can extract angular momentum from longitudinal waves, too. If there exists a quantum of longitudinal waves. it cannot have the spin = 0.

What is the difference between a linearly polarized wave and a vortex?

We have an instinctive grasp of what is a vortex in water and what is a wave. What exactly is the difference? If we have a short block of floating wood, we can make it rotate with a vortex, but with a wave it will just turn a little one way and then move back.

We might imagine the rotating dipole transmitter forming a huge vortex which fills the whole space and makes all dipole antennas rotate the same way. Or we may imagine that the transmitter send "mini-vortices", photons, which are absorbed by the antennas.

The linear dipole transmitter is classically quite close to two rotating transmitters which rotate to opposite directions. We might say that its linear wave is formed by two superimposed vortices.

Do the quanta of all force fields carry angular momentum?

An oar makes vortexes to water. The photon has a non-zero spin. Is there any way to transfer energy in any force field with quanta which have spin = 0?

What about classical physics? Is there a way to transfer energy over a distance in a way where no device can extract angular momentum from the energy transfer? In water, waves as well as vortexes and turbulence involve changes in angular momentum, at least locally.

If we have a perfect linear flow of water, then no floating object can extract angular momentum from it, but neither can it extract any energy. The same holds for a homogeneous gravitational field.

What about longitudinal sound waves? We can extract angular momentum by attaching a light plate to a heavy object. The plate will move with the sound waves while the heavy object stays still.

In classical physics, does transferring energy through a periodic motion allow extraction of angular momentum in all possible cases?

Why do the quanta of the electron field carry angular momentum?

Above we found a heuristic reason why photons must carry angular momentum. Is there any reason why in a produced pair, the electron and the positron must carry angular momentum, that is, have the spin non-zero?

In classical physics, particles do not need to carry angular momentum. We can throw grains of sand, and there is no need for the grains to rotate.

What exactly happens in pair production? We support the Feynman hypothesis that a pair production is like an electron scattering from a high-energy photon, where a space coordinate has been flipped with the time coordinate.

Is there a reason why a scattering electron must carry the spin 1/2?

In scattering, in a Feynman diagram, the electron first absorbs the photon and later emits it. Suppose that the spin of the photon is 1. If the electron has the spin -1/2 before the absorption, then its spin after the absorption is 1/2, until it emits the photon, and then its spin is again -1/2.

How does an electron actually scatter from a photon or a rotating dipole field? Is the magnetic interaction much stronger than the electric interaction?

We had the classical model where the electron flies at the speed of light in a circle of radius 2 * 10^-13 m. That explains its angular momentum. The magnetic moment is double of this classical model.

A 511 keV photon is born from an electric dipole where an electron and a positron circle their center of mass at the speed of light, and the length of the orbit is one Compton wavelength 2 * 10^-12 m. The radius is then 3 * 10^-13 m.

The magnetic fields of the dipole and the electron which comes close to the dipole are of the same order of magnitude. The magnetic potential reaches 511 keV at a distance of the order 10^-13 m, while the electric potential would require a distance of the order 10^-15 m.

The rotating tube model of the electron-positron pair explains spin 1/2

Classically, the energy density of the electric field is E^2, where E is the electric field strength. Can we interpret that the electric field is a material object which can move and rotate? Sure. We can measure the magnetic component at different places and determine if the electric field is moving or rotating.

The uncertainty principle in quantum mechanics states that if an object can rotate, then its angular momentum in the z direction is some n * h-bar, where n can also be 0. We cannot determine the angular momentum components to the x, y, z directions simultaneously, which means that the object always has rotational uncertainty. One cannot put an extended object in a non-rotating state.

Thus, the electric field of the electron has to rotate. But why is its spin 1/2 or -1/2?

What is the wavelength of a complex object in quantum mechanics? We can give the wavelength of an atom or even a molecule, though it contains also very light parts like electrons which individually would have a wavelength much longer than the atom as a whole? Quantum mechanics seems to include the principle:

Principle of combination 1. If we have a complex object of mass m whose parts are bound to each other, "entangled" in some sense, then we can treat it as a single wave of a pointlike object of mass m. In periodic motion, there has to be an integer number of wavelengths of the wave of the combined object. Destructive interference prohibits other periodic orbits.

Each individual part in the complex object contains the information of the location and the momentum of the whole complex object. But what about parts which are far away and contain information of the object?

If an electron and a positron are always born as a pair and the positron always preserves the information of the electron rotational state, then in rotation, we have to treat the combined object the electron & the positron as one. A full rotation of the combined object consists of a full rotation of both the electron and the positron. If we just look at the electron wave function, it looks like there was just a 1/2 wavelength in the rotation. This is the origin of the strange spin value 1/2.

                rotating tube
       e- ============= e+

But does the positron preserve information of the rotational state of its brother electron? What if we use a magnetic field to turn the positron? One may imagine that the rotational angle is a hidden variable in both the electron and the positron. Maybe turning the positron does not affect the angle but rather the orientation of the rotational axis? Maybe the electron and positron are the ends of a flexible rotating tube. One can turn the ends to any direction, but the low-level rotation of the tube stays the same.
                                 
     e- ========= e+ e- ========= e+
                                    B
                         annihilation

What if we annihilate the positron with an electron B? If we accept the Feynman interpretation that annihilation means scattering backward in time, then the annihilation really means gluing together the two tubes. Our electron will still have a brother positron. In the diagram, the two electrons and two positrons are just one particle bouncing back and forth in time.

What if we speed up the electron, so that its time slows down relative to the laboratory frame, consequently its rotation slows down, and the positron loses the knowledge of the electron rotation angle?

Problem 2. When does the Principle of combination 1 lose its validity if time in part of the system is slowed down by increasing the velocity of that part? How can we cancel the entanglement of two quantum mechanical systems?


Why is the electron spin 1/2 or -1/2, and not 1 or 0? In the hydrogen atom, the electron may have the orbital angular momentum 0. That is probably because then the electric field outside the atom is zero. One cannot know if the atom is rotating by measuring the magnetic component of the electric field.

The electron total spin cannot be 0, because we can measure the rotational state of its electric field. But can the z component of the spin be 0? Then we would know that the spin axis is in the x,y plane. That is probably prohibited by an uncertainty principle.

An electron spin 1 or -1 is possible in our rotating tube model. Maybe it has not been observed because it would be stable only under a magnetic field of 10^8 tesla or more.

We did not need to use special relativity at all in deriving the electron spin 1/2. It is enough to assume that the electric field can be a rotating object, and that the rotation of the electron and its brother positron forms a "bound state" or entangled state where the (hidden) rotation angle of both particles is perfectly correlated.

Why does the spin 1/2 follow from the Dirac equation? Dirac does not need to assume that the electron is entangled with a positron.

Why is the g factor in the electron magnetic moment 2? We calculated in an earlier post that if the electron goes in a circle at the speed c * sqrt(3) / 2, then the gyromagnetic ratio is 2, according to special relativity. Is there a reason why the electric field of the electron should rotate just at that speed?

A spinor is a "square root" of a vector?

https://en.wikipedia.org/wiki/Four-current

The spin 1 or -1 of a photon is explained in literature from the fact that the 4-dimensional current density, the four-current is a vector.

When a vector is rotated 360 degrees, it is again identical with the original vector. In contrast, the Dirac spinor has to be "rotated", or more precisely, has to move on a Möbius strip a path of 720 degrees to return to the original value. In a sense, a spinor is a square root of a vector.

The rotation of an ordinary vector can be understood as moving it along a straight strip which is obtained when a ring is cut from a circular disk by removing the center.

The Möbius strip is obtained by cutting the ring open, flipping one end 180 degrees, and gluing together.

Rotation of a single particle can be understood as moving a vector along a straight ring for 360 degrees.

Our rotating tube model describes two particles rotating in unison. The rotation of the whole system contains one full cycle of the wave function. If we just look at one end of the tube, maybe we can describe its behavior with a spinor? The combination of two spinors should in some sense be equivalent to a vector.

If we have two particles rotating strictly in unison, at the same location or separated by some distance, can we describe one particle as a spinor, and the combination of the two spinors is an ordinary vector which describes the rotation of the combined system?

Since we can change the direction of the electron spin and its brother positron spin independently, the model is more complicated than what we get by just gluing two particles together. That gives rise to the spinor.

A spinor is a vector in a 2X sphere where we identify opposite points?

If the electron wave only does 1/2 wavelengths in one round, we have the problem of destructive interference. But we get rid of the interference by making the orbit 2X in spatial size. The electron can orbit on the surface of such a sphere.

Then we identify each point in the sphere with the opposite point, to get a new topology. Can we project the 2X sphere to a normal 1X sphere in the space, such that identified points are mapped to the same point in the 1X sphere? The mapping is certainly possible if we want to map a 2X circle to a 1X circle. What about the 2-dimensional case?

If we have a point on the 2X sphere, with a latitude c degrees and a longitude d degrees, where would we map it on the 1X sphere? 

If we map (c, d) to (2c, 2d), does that work? Probably yes.

If the rotational state of the electron lives on the 2X sphere, and it is projected to the 1X sphere in our universe, then the properties of spinors might find an explanation.

The combined system the electron & the positron would have a rotational state on a 1X sphere, but when the system is split into two particles, each of those particles has to be described on the 2X sphere with the modified topology.

The photon can decay into an elecron-positron pair?

Our rotating dipole model of the first section in this blog post suggests that the photon is a small pointlike rotating electric dipole of spin 1. It is like a miniature copy of the rotating dipole which produced the electromagnetic field.

The rotating dipole of the photon can split, decaying into the negative charge part and the positive charge part. Since these new particles start from the same spacetime point, they have to carry away the angular momentum in their own spin, which is 1/2. This would be the origin of a produced pair.

The photon wave is (at least roughly) described by the Klein-Gordon equation. We need to find out why a "square root" of the Klein-Gordon equation describes the ends of the dipole once they manage to break free. The square root is the Dirac equation.

Water waves are molecules in a vertical circular motion, vortices are horizontal circular motion

There is an intriguing analogue between water and the electromagnetic field.

A wave in water is a temporary "hill" or "valley". But a vortex is a semi-permanent valley. The electron may be analogous with a vortex because it is a permanent valley in the electric field.

If the surface of water is sloping, then a vortex may tend to move upward. That would be analogous to electric attraction.

The positron might be an "anti-vortex" which is a permanent hill. There are no anti-vortices in water, though.

When powerful waves in water meet head on, they probably do not produce vortices. Waves hitting an obstacle may produce vortices.

The vortex analogy might explain why there are permanent charges in the electromagnetic field.

The Poynting vector, 1/μ_0 * E × B, does make circles around the electron. If there is energy circling the electron, there has to be a centripetal force which keeps it in the orbit. We need to check what is the Poynting vector. The energy of the electric field outside the classical radius, 3 * 10^-15 m, is the electron mass 511 keV. The energy of the magnetic field is probably much more.

If we model the electron as the current loop whose radius is 3 * 10^-13 m, then close to the loop, the magnetic field is of the order 10^8 tesla and the electric field is 10^17 V/m.

The Poynting vector is

      10^6 * 10^8 * 10^17 = 10^31 W/m^2.

Does this make sense? The electron moves at almost the speed of light. The energy transfer in the loop has the power

       m c^2 c / C = m c^3 / C
       = 10^-30 * 27 * 10^24 * 10^12 / 3 W
       = 10^7 W,

where C is the loop circumference.

If the area through which the Poynting vector moves is in the ballpark (3 * 10^-13 m)^2, the Poynting power is 10^6 W. The figures look reasonable.

The centripetal force to keep the electron in the loop is

       m v^2 / r = 10^-30 * 10^17 * 10^13 / 3 N
       = 0.3 N.

The electric repulsion between two electrons at the distance 3 * 10^-13 m is

       k e^2 / r^2
       = 10^10 * 3 * 10^-38 * 10^25 N
       = 3 * 10^-3 N.

The centripetal force which we need is roughly 1,000X the electric force.

If we assume that a water whirlpool has the velocity of the water constant on the surface, then the depth of the vortex will follow the 1/r law which is like the Coulomb potential of electric force.

A simple calculation shows that the kinetic energy in a water whirlpool is roughly the same as the potential energy in the valley which the vortex makes in water.

A vortex in water is always a valley. In a hypothetical water where the centrifugal acceleration has the sign flipped to the normal case, a vortex would be a hill in water.

http://www.worksofvalerychalidze.com/uploads/1/0/2/8/102863812/1126751print1a.pdf

The Soviet dissident Valery Chalidze (1938 - 2018) developed some kind of a vortex hypothesis for electromagnetism. His model is quite complicated and probably does not help us much. Fluid models and vortices in them were researched in the 19th century, before elementary particles were discovered. Apparently, no one was able to create a satisfactory fluid model of electromagnetism.

How to formulate a mathematically precise Pauli exclusion principle?

The standard way of stating the Pauli exclusion principle is that two fermions cannot "share the same quantum state". As we noted in the Nov 20, 2018 blog post, the concept "the quantum state of a single electron" is unclear if many electrons are present and interact.

The Pauli exclusion principle was formulated by Wolfgang Pauli in 1924 to explain the placement of electrons in orbitals in many-electron atoms. Empirically, the electrons seem to fill the pigeonholes that we find when we solve a one-electron atom, the hydrogen atom.

Rather than proving a general exclusion principle, whose statement is unclear, we can start from a much more concrete problem.

Problem 1. Prove that electrons in many-electron atoms, in some sense, fill the orbitals of the hydrogen atom in a way where just one electron can occupy one orbital of hydrogen. An orbital of hydrogen is determined by its four quantum numbers n, l, m, s.


Problem 1 contains the phrase "in some sense fill the orbitals", which is unclear mathematically. When someone solves Problem 1, he should also clarify what that phrase means.

Problem 1 is a many-body problem, and such problems are notoriously hard to solve. A brief Internet search does not reveal any picture of the orbitals of helium. Even the simplest case of two electrons remains unsolved.

Our own heuristic derivation of the Pauli exclusion principle in the Nov 24, 2018 post is by no means a mathematical proof. Furthermore, it does not explain why electrons should fall in the pigeonholes determined by the single electron in the hydrogen atom. We can present a more modest problem:

Problem 2. Give a heuristic argument which explains why the orbitals of hydrogen determine the pigeonholes where electrons fall in a many-electron atom.


Solving the many-body problem of mutually repulsive particles might be easier for particles in a box. A step forward in solving Problem 2 is to find any setup where a heuristic analysis shows that electrons fall into the pigeonholes determined by the ground state and the excited states of a single electron system.

Our Nov 24, 2018 post does give some clues that repulsive electrons in a box might fill states that resemble single-electron states.

The spin-statistics theorem

The spin-statistics theorem is claimed to imply the Pauli exclusion principle. The spin-statistics theorem states that the "sign of the wave function flips at an interchange of two fermions".

Since in quantum mechanics, one is allowed to multiply an arbitrary wave function by any exp(i φ), where φ is real, what prevents us from flipping the sign of the wave function back?

Also, the "interchange of two fermions" is not clear. What does that mean physically and how can we be sure the two fermions were interchanged when fermions are indistinguishable?

In most sources, "interchanging" seems to mean that we change the order of two creation operators in our mathematical description of the system. The creation operators are applied at spacetime points x and y where the separation of points is spacelike. If the creation operators would be real physical events, then there is no physical ordering of them in special relativity, and consequently, it makes no sense to "interchange" the order of these events.

But it may be sensible to interchange the order of operators in our description of the system. Can we calculate something by manipulating our description of the system in such a way?

Is there an empirical experiment where particles are "interchanged" in some sense, and the interference pattern shows that their wave function flipped the sign?

http://www.feynmanlectures.caltech.edu/III_04.html

Feynman's lectures contain a diagram of scattering experiment, where the interchange depends on how close the particles come to each other. It is not just an interchange, but the paths of the particles in their mutual potential differ in a significant way.

The electron and the positron are the "square roots" of a Klein-Gordon particle?

The spin 1/2 is perplexing. As if the electron would rotate around its axis in some extra dimension.

The Dirac equation has been called the "square root" of the Klein-Gordon equation. If we take this seriously, then the electron and the positron are "square roots" of a Klein-Gordon particle.

     e+ <-------> e-

The electron and the positron are always born together. Maybe we should always treat them together, as a combination of two particles. The wave function of the combined particle might be the product of individual wave functions.

We had the classical model where the electron at the speed of light makes a circle whose radius is 2 * 10^-13 m. That would give the right spin angular momentum.

The strange thing was that the plane wave representation of the electron only completes half a cycle in one round. Why there is no destructive interference which would spoil the wave? The reason may be that the complete system includes also the positron. The product of the electron and the positron waves does complete a full cycle when the electron makes a single circle. There is a constructive interference when we look at the whole system electron & positron.

In quantum mechanics, there is no interference of the wave of a single particle if enough information of the wave is copied somewhere else. The cardinal example is the double slit experiment.

When the electron rotates around its axis, we may imagine that information of the rotation is present in the positron, too.

When an electron and a positron are born, we may think that their spin rotation is entangled in some sense. We cannot treat them as separate particles when it comes to the rotation.

The positron may annihilate with another electron B, but then the information about the rotation might be preserved in the created photons, or alternatively, in some way in the positron which is the pair of electron B.

The photon has some similarity to the Klein-Gordon equation. Maybe the electron and the positron are square roots of the photon in some sense?

Does this new idea cast light on the question in which way the positron is an electron traveling back in time?

Taking the square root of the Klein-Gordon equation

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

Let us try to figure out in which sense the solutions of the Dirac equation are square roots of solutions for the Klein-Gordon equation. We work in 1+1 dimensions. Paul Dirac in his book Principles of quantum mechanics (1930) writes a "relativistic hamiltonian":

       (p_0^2 - p_1^2 - m^2) Ψ = 0,

where p_n = i d / dx_n is the momentum operator. It is the Klein-Gordon equation.

Dirac realized that he can obtain an equation with just first derivatives by taking a "square root" of the operator:

       (p_0 - α_1 p_1 - β m) Ψ = 0,

where we can choose

                   0       1
       α_1 =
                   1       0

                   1       0
       β     =
                   0      -1

By choosing α_1 and β appropriately, the square of the operator above is equal to the operator in the upper equation. Suppose that we find a solution Ψ for the lower equation. It is trivially a solution of the upper equation, too. What about Ψ^2? Under what condition it is a solution of the upper equation?

Let us denote t = x_0 and x = x_1. The standard plane wave solution of the Dirac electron is

       Ψ_1 = (1, p / (E + m)) * exp(-i (E t - p x)),

where E^2 = p^2 + m^2. Let a positron solution be

       Ψ_2 = (-p / (E + m), 1) * exp(-i (-E t - p x)).

The squares Ψ_1^2, Ψ_2^2, as well as

       (Ψ_1 + Ψ_2)^2 = Ψ_1^2 + Ψ_2^2

are solutions for the Klein-Gordon equation for mass 2m. The electron and the positron are orthogonal solutions of the Dirac equation.

The gyromagnetic ratio 2

The Dirac equation apparently implies that the gyromagnetic ratio g of the spinning electron is 2. That is, if we imagine the electron as a classical particle making a loop at a slow speed, its magnetic moment relative to its angular momentum is 2X of what we would expect from a classical particle.

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

But if the electron moves at a fast speed, then the Lorentz transformation of its electric field has the fields multiplied by γ = 1 / sqrt(1 - v^2 / c^2). If the electron would move at the speed sqrt(3) / 2 * c, then the magnetic field would be double of what the nonrelativistic formula gives.

The electron magnetic moment from the Dirac equation

http://youngsubyoon.com/electronmagneticqft.pdf

The link above contains a derivation which compares the Dirac equation under an electromagnetic field to a hypothetical Klein-Gordon equation under an electromagnetic field.

https://en.m.wikipedia.org/wiki/Gauge_covariant_derivative

The electromagnetic field potential is incorporated into a free field wave equation by converting the partial derivatives to a "gauge covariant form":

       D_μ  = ꝺ_μ - i e A_μ.

For the free field equation, the "square" of the operator of the Dirac equation

       (i D-slash - m) * (-i D-slash - m)

is the Klein-Gordon equation operator

       D^2 + m^2.

But if the electromagnetic field is non-zero, then off-diagonal elements appear in the square of the Dirac operator:

       e F_μν S^μν.

The Klein-Gordon equation is a generalization of a 1+1-dimensional wave equation. The discussion above raises the question if a better generalization of the wave equation to 1+3 dimension should contain all second order derivatives and not just the diagonal derivatives d^2/dx^2. Then the new Klein-Gordon equation might be equivalent to the Dirac equation also in the case where an electromagnetic field is present.

Pauli matrices and squeezing n + 1 dimensions into n

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

The Pauli equation describes the electron spin interaction in the nonrelativistic case. The wave function is a two-component one.  Though, the full freedom of having two independent wave functions apparently does not come into reality. All examples seem to contain a two-component complex-valued vector multiplying a single wave function.

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

The Pauli matrices operate on the two-component wave function. They are the spin operators in the x, y, z directions. The eigenvalues for each of them are -1 and 1. An eigenfunction seems to be any product of an eigenvector and any function.

The spin axis should, by default, require a 3-component vector to specify its direction. Our uncertainty of the direction of the spin axis seems to be so profound that we can squeeze the mathematical structure of the spin axis into a 2-component vector. This is probably the mathematical origin of the strange 720 degree rotation symmetry of spinors.

The wave functions in the Pauli machinery do not contain any spinning or rotating motion. The electron seems to be a pointlike magnetic dipole or a magnetic needle.

We suggested in an earlier post that the electron spin comes from our uncertainty about the rotation of the electron's electric field. It is not clear if the Pauli machinery somehow incorporates this idea.

If we transform the Klein-Gordon equation into a group of equations that only contain first derivatives, the standard way is to introduce a new component for each derivative on t, x, y, z. Then we would have a 5-component wave function. The Dirac equation manages the same feat with a 4-component wave function. Somehow are 5 dimensions squeezed into 4. This reminds us of the Pauli machinery which squeezes 3 dimensions into 2. We need to find out what is going on with Pauli and Dirac.

The Klein-Gordon equation in 1+1 dimensions is

    d^2 Ψ/dx^2 - m^2 Ψ - d^2 Ψ/dt^2 = 0.

It is Lorentz covariant. The standard way to remove the second derivatives with new functions is (if m is not zero):

         dΨ_2/dx - mΨ        + dΨ_1/dt = 0
                           mΨ_1     +   dΨ/dt  = 0
          -dΨ/dx + mΨ_2                       = 0.

The Dirac equations in 1+1 dimensions are

     -i dΨ_2/dx + mΨ_1 - i dΨ_1/dt = 0
     -i dΨ_1/dx -  mΨ_2 - i dΨ_2/dt = 0.

Any solution of the Dirac equations is also a solution of the Klein-Gordon equation, but the converse is probably not true. Dirac adds some extra constraints which remove solutions of the Klein-Gordon equation. As if the Klein-Gordon equation would allow too much freedom for solutions if m is not zero.

Deriving the Schrödinger equation from the Klein-Gordon equation

Using the operator symbols p and E, the Klein-Gordon equation is

        (p^2 + (m + V)^2 - E^2) Ψ = 0.

We have added a scalar potential V to the rest mass, like we did in our earlier blog posts. If |p^4 / m^2| is very small and |V| is small, we can factor it like this:

        (p^2 / (2m) + m + V + E)
     * (p^2 / (2m) + m + V  - E)  Ψ = 0.

The lower factor is the Schrödinger equation. The upper factor of the operator would have Ψ rotating in time to the opposite direction. Is it the Schrödinger equation for the antiparticle?

Suppose that V is a linear potential which decreases to the right. Let us initialize the system with a wave function

       exp(-i (Et - px))

where p = 0. The particles start to slide to the right, which means that after some time, there will be components like

      exp(-i (Et - px)),

where p > 0, if we solve the Schrödinger equation.

If we solve the upper factor equation (we remove the Schrödinger operator from the equation above),  it will have components like

       exp(-i (-Et - px)).

If we increase t by dt, then to restore the phase in the exp() above, we need to decrease x by some dx. We see that the component describes a particle sliding to the left. The potential V works to those particles in the opposite way than in the Schrödinger equation. We can call them antiparticles.

Deriving the Dirac and Schrödinger equations from the energy-momentum relation

Let us factorize the energy-momentum relation

      E^2 - p^2 - m^2 = 0

with the matrices α_1, β above:

   1     0           0     1            1     0
(            E     -           p       -           m)
   0     1           1     0            0    -1

*

   1     0           0     1            1     0               0    0
(            E    +           p       +          m)    =
   0     1           1     0            0    -1               0    0.

In the Schrödinger factorization, we had

       E^2 = p^2 + m^2
     <=> (p^2 small)
       E      = +- (p^2 / (2m) + m).

Using the identity matrix I, we can write the above as

       I^2 E^2 = I^2 p^2 + I^2 m^2
      <=> (p small)
       I E         = +- (I p^2 / (2m) + I m).

In the Dirac factorization, we have

       I^2 E^2 = α_1^2 p^2 + β^2 m^2
                     = (α_1 p + β m)^2
      <=
       I E         = +- (α_1 p + β m).

The factorizations are completely different but both yield sensible results. Why?

The Schrödinger factorization produces the (almost) right result because of cross terms in the product. The Dirac factorization has cross terms zero. It has only implication into one direction.

Does every solution of the Schrödinger equation also solve the Dirac equation?

Let us use a two-component wave function Ψ and use the identity matrix I in our Klein-Gordon equation:

             I (p^2 + (m + V)^2) Ψ = I E^2 Ψ.

Above p and E are operators and V is a scalar potential. Both components of Ψ are individually solutions of the same one-component equation.

The corresponding Schrödinger equations are

       +- I (p^2 / (2m) + m + V) Ψ = I E Ψ.

For very small |p^4 / m^2| and small |V|, any solution of our Schrödinger equation is a solution of our Klein-Gordon equation. The converse is probably not true.

Our Dirac equations are 

       +- (α_1 p + β (m + V)) Ψ = I E Ψ.

If we have a solution of our Dirac equation, then it is a solution of our Klein-Gordon equation.

But are solutions of our Schrödinger equation solutions of our Dirac equation, and vice versa? Yes if the solutions only depend on time by a factor exp(-i E t), where E is constant, that is, they are energy eigenfunctions.