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.

A heuristic derivation of the Pauli exclusion principle from the spin magnetic interaction

As we showed in the previous blog post, it is doubtful if the current proofs of the Pauli exclusion principle are correct.

The magnetic potential of the spin up-down electrons is

       -10^-53 / r^3   meter^3 J,

when the electrons are in the x,y plane, and r is their separation, and r is large enough so that we can consider the electron as a pointlike magnetic dipole.

The Coulomb potential is

       3 * 10^-28 / r   meter J.

For a "typical" separation in a helium atom of 2 * 10^-11 m, the magnetic potential is 10^-21 J and the Coulomb potential is 1.5 * 10^-17 J. Thus, the Coulomb potential has a 7,000 X larger absolute value.

When the electrons are not close, we can ignore the magnetic potential. When the separation of the electrons is one Compton wavelength 2 * 10^-12 m, the ratio of potentials is much lower, only 70 X.

We calculated in a previous blog post that the electron spin can be modeled as a classical particle moving in a circle of a radius 2 * 10^-13 m at the speed of light. In that model, when the distance is that, the magnetic attractive potential dominates greatly. Its potential is of the order -511 keV, while the Coulomb potential is 10 keV.

If spins are parallel, then an electron cannot tunnel through the potential of another electron

Let us assume that the electron is a pointlike magnetic dipole, an assumption which may not be true.

Let us assume that electron B stays at a fixed position. Electron B forms a repulsive potential barrier for electron A. If the spins are parallel, the barrier height grows as 1 / r^3 when r decreases. But if the spins are antiparallel, the barrier levels off at about 3 * 10^-13 m and becomes attractive at distances less than 2 * 10^-13.

Let us calculate the potential when r = 10^-15 m and we have electrons B and A with parallel spins. The potential of the magnetic repulsion is 10^-8 J, that is, some 70 GeV. How far can an electron tunnel through such a potential barrier? Our new energy-momentum relation

       E^2 = (p c)^2 + (m c^2 + 10^-8 J)^2

has E roughly zero. Then

       p c = i 10^-8  J
      =>
       p = i * 3 * 10^-17   kg m/s,

where i is the imaginary unit. For a relativistic particle,

       p c = h c / λ
      =>
       λ = h / p = i * 2 * 10^-17 m.

The imaginary wavelength means that the wave function decays by a factor e = 2.718 during that distance. Since the barrier width in our case is 2 * 10^-15 m, we see that tunneling through the barrier is negligible.

What about the case where the spins are antiparallel? The potential levels off at about r = 3 * 10^-13 m, where the potential is of the order 10^-15 J. The momentum p is i * 3 * 10^-24 kg m/s. We have then

       λ = i * 2 * 10^-10 m.

Electron A can easily tunnel through the potential of B if the spins are antiparallel.

Modeling two electrons with a single particle in 6 dimensions

We can model electrons A and B as a single particle which moves in 6 spatial dimensions. If the spins of A and B are parallel, that means that the wave function

      Ψ(t, x, y, z, x', y', z') = 0,

      if x = x', y = y', and z = z',

because A cannot tunnel into the potential barrier of B. The probability of finding A and B very close to each other is zero.

The equation above restricts the set of possible wave functions. The restriction may force the lowest energy state much higher than in the antiparallel spins case.

The setup has 6 spatial dimensions and a 3-dimensional plane where Ψ is restricted to be zero. It is hard to visualize what that means for the wave. But if we consider a wave in a 3-dimensional spherical cavity, then restricting the wave function to be zero at a 0-dimensional object, that is, at a point, will obviously raise the lowest resonant frequency greatly.

An analogue is a drum skin.
 
             drum skin
           ____________
          |             /\        |
                     fixed
                   support

If we attach one point of the skin to a fixed support, so that the displacement of the skin is always zero at that point, then the lowest resonance frequency of the skin will be much higher. The fixed point forces a node to be in the oscillation at that point.

Suppose that we are adding the 2n +1'th electron as spin up. There are already n spin up electrons. We will add 3 new spatial dimensions and introduce n new constraints for the new electron.
For 2n electrons, there are

        2 + 4 +... + 2(n - 1)
        = (n - 1) * n

constraints. For 2n + 1 electrons, there are n^2 constraints

The number of constraints increases rapidly with n. Nature seems to cope with them by placing electrons in different shells.

If the two lowest energy electrons are close to the nucleus, then all other electrons must have the wave function close to zero there. The 2s orbital of lithium would not be like the 2s orbital of hydrogen. In hydrogen, all s orbitals are non-zero at the nucleus.

Solving the helium and lithium atoms hierarchically

Let us try to solve the helium atom hierarchically, so that we let electron B to be in the lowest energy state. B will fly around on the 1s orbital. B will mostly stay rather close to the nucleus. If the spins of A and B are parallel, then the wave function of A will have a zero amplitude at B, and a small amplitude at the nucleus. The energy level of A is then a lot higher than in the orbital 1s.

If the spins of A and B are antiparallel, then A can easily tunnel through the potential of B, and A can choose its wave function relatively freely. This is the electron structure of helium.

In lithium, let us add a third electron C with spin up, for instance. If B has spin up, then the wave function of C must have a zero amplitude at B, and a small amplitude at the nucleus. The energy level of C will be substantially higher than the energy level of A and B.

The Pauli exclusion principle

Electrons tend to fill orbitals in pairs of antiparallel spins. Each electron has to avoid other electrons which have a parallel spin. That gives rise to the shell structure of electron orbitals.

Such organization of electrons will appear to fulfill the Pauli exclusion principle. Our reasoning did not assume anything about the spins being 1/2 or integer values. What is needed is a very strong magnetic repulsion between parallel spins. We did not assume anything about the wave function being antisymmetric under "interchanges" either.

Are there examples of bosons having strong magnetic repulsion, but still occupying the same quantum states? An internet search does not reveal anything.

The spin-statistics theorem: the proof of the Pauli exclusion principle is flawed

"Why is it that particles with half-integral spin are Fermi particles whose amplitudes add with the minus sign, whereas particles with integral spin are Bose particles whose amplitudes add with the positive sign? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved. For the moment, you will just have to take it as one of the rules of the world."

                    - Richard P. Feynman, in the Feynman lectures on physics.

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

We could, and maybe should, adopt as an axiom the Fermi-Dirac statistics of a spin 1/2 particle. There are various proofs of the statistics on the Internet. We try to analyze in the following what the proofs actually show.

The Pauli exclusion principle is the very concrete consequence of the statistics. Though, we need to analyze in what sense Fermi-Dirac statistics really bans two fermions in the identical quantum state.

In our blog post Nov 17, 2018 we developed a model for the electron spin.

The energy 511 keV of the electron travels at the speed of light in a circle of radius 2 * 10^-13 m. The electron only does 1/2 of the Compton wavelength in a rotation. There would be a destructive interference which would prevent such state in a single loop, but there apparently is some hidden degree of freedom which makes the rotation actually to happen in a "coil" with two loops, or, equivalently, along the edge of a Möbius strip.

We can make a coil with two loops by taking a loop, like "O", bending it to an "8", and then bending the two loops on top of each other, resulting in a double loop "o".

If the spin is rotation of the electric field, what is the hidden degree of freedom? A possibility is that the electron jumps back in time before completing a full 360 degree rotation.

Suppose that besides the electron, some unknown particle is rotating in the same loop at 1/2 the speed of light. Then that unknown particle would be the extra degree of freedom which would cause the the wave function return to the original value when the electron rotates 720 degrees.

https://en.wikipedia.org/wiki/Spin–statistics_theorem

In the spin-statistics theorem, in the Julian Schwinger proof, it is unclear what exactly the 360 degree rotation in it means. It cannot be a spatial coordinate rotation at a fixed time t_0, because then the wave function necessarily returns to the original value after a 360 degree rotation.

If the time coordinate is the same, then the state has to be the same when the spatial coordinate axes point to the original direction - unless it is a multiverse where a 360 degree rotation takes us to another universe. That is actually an interesting hypothesis. Suppose that we live in a double universe where in most cases the two sides are identical. An exception would be spin 1/2 particles, whose wave function can be defined antisymmetric relative to an arbitrary "center line" of the double universe.

The rotation topology of the double universe is like the double coil, or the edge of a Möbius strip.

For bosons, the two copies of the Möbius universe are identical, but for fermions, the wave function on the other side has the sign flipped.

Maybe it is too complicated to introduce such a double universe. Can we accomplish the same with the double particle model?

A triple particle model might work better: in the same circle with the main electron, we have an extra electron going at 1/2 of the speed of light to the same direction, and an extra positron going in the opposite direction at 1/2 the speed of light. That would explain why the magnetic moment is double of what we would expect classically.

The triple model explains the angular momentum and the magnetic moment, but how do we explain that the whole triple system has just the energy of a single electron? Can we assume that the extra electron-positron pair is tunneling with zero energy?

The electron in this model is really two electrons plus a positron. The extra particles are observable only in the spin and the magnetic moment of the electron.

Something called the Compton "scattering radius" of the electron is roughly double the radius of the circle in our model. Is that a coincidence?

What do the 180 and 360 degree rotations mean for the spin?

https://en.m.wikipedia.org/wiki/Spin_(physics)

In Wikipedia, the authors consider the probability amplitudes in a given state, for spin_z being up or down. If we do an infinitesimal rotation, we can calculate the amplitudes for spin_z up or down in the new coordinates.

If we let the amplitudes change only infinitesimally, that is, the mapping to new amplitudes does not contain big jumps, then apparently, a 360 degree rotation does flip the sign of the probability amplitudes.

How to interpret that? The physical system itself cannot change when we just turn our coordinate system.

The Pauli exclusion principle is a very real law for a real physical system. Why would it follow from something we prove for a coordinate system?

Maybe we should assume that we measure the spin of the electron repeatedly with a physical device, turning the device slightly after each measurement. We calculate the path integral for various outcomes after our device has made a 360 degree rotation. The measurements do a real change to the physical system. Let us try to repeat Schwinger's proof after this clarification.

We calculate the vacuum expectation value

       < 0 | R ϕ(x) ϕ(-x) | 0 >.

…

What does it mean for two particles to be "in the same quantum state"?

The Pauli exclusion principle states that two half-integer spin particles cannot exist in the "same quantum state". If we have an atom with several electrons, what does it mean that electrons A and B are in the "same quantum state"? The wave equation has to be solved for all particles simultaneously. The wave function exists in a space with lots of spatial dimensions and maybe also many time dimensions. Since electrons are indistinguishable, it is not even clear what we mean by "an electron" or "two electrons" being in a certain quantum state.

Consider the classical particle in a box. The ground state is well-defined and clear. If we put two particles in the box, and the particles have repulsion, then most probably they cannot occupy the wave function of the ground state of a single particle. What do we mean by the statement that "the two electrons now have different quantum states"? We probably cannot measure one electron without disturbing the other.

We should describe a realistic physical measurement where we measure both electrons, and what we claim about the measurement outcome. For example, if we measure the spin of one of the electrons of the helium atom and then remove it by some method, what does a spin measurement give for the remaining electron?

Is there a classical analogue for the Pauli exclusion principle? A setup where the wave gets disturbed if its amplitude exceeds some limit? Then the wave energy has to be divided among various low-amplitude waves.

John Baez's ribbon proof

http://math.ucr.edu/home/baez/spin_stat.html

The ribbon proof contains a concrete statement: if we have an electron, and it is subsequently annihilated by a positron in a pair, then the pair electron is like the original electron rotated by 360 degrees. We assume that there is no extra disturbance in the annihilation or pair creation, which could turn the spins.

       511 keV
       photons

             \   /          ^

               \/          /

               /\        /

              /  \      /

            /     \   / e-

           /  e+ \/
         /         /\

 t     ^        /   \

 ^    e-      511 keV photons
 |
  ----------> x

The Feynman way of interpreting the above path is that there is just one electron which scatters first back in time, and then forward in time. It is scattered from the 511 keV photons in the diagram.

Is the above path equivalent to a 360 degree rotation in some sense? If time and space are fully symmetric at very short distances, then scattering back in time might be equivalent to bouncing back from a spatial wall. What happens to the electron spin when it is reflected from a potential? Probably not much because the electron in the hydrogen atom gets reflected all the time and keeps its spin.

In pair production, the electron and the positron often have opposite spins. Positronium annihilates much faster if the spins are opposite.

In the above diagram, we may assume that in the laboratory frame, the handedness of the electron's linear movement versus its rotation (chirality) stays the same.

What does the CPT symmetry mean in this setup? The charge changes the sign at the sharp points and time changes the direction. The parity transformation means flipping the sign of one spatial coordinate. If we change the sign of both t and x, it means a 180 degree rotation of the t, x -plane.

If t and x are completely symmetric at short distances, we may as well assume that x is the time axis and t is the space axis. If we use t as the parameter for the rotation of the electron, the the positron line segment has spin_z flipped.

But if we use x as the (time) parameter of the rotation, then spin_z stays the same sign throughout the journey. If x is the time axis, then the path describes two electron bounces from two photons. If an electron scatters from a photon, both have to keep their spin_z, since for the electron its absolute value is 1/2 and for the photon 1.

The journey seems uneventful for spin_z. Is there a reason why the probability amplitude of the electron wave function would flip sign? Does it flip the sign when an electron scatters from a photon when t is the time coordinate?

If the electron scatters from a low-energy photon, then it would be strange if it would change the phase of the electron wave function significantly. What about a high energy photon in the 511 keV range?

Generally, in wave equations, reflection from an optically dense material causes a 180 degree phase shift. Reflection from thinner material does not cause a phase shift. What kind of a reflection is it when an electron bounces from an energetic photon? If we interpret that the electron is reflected from a steep electric potential barrier, then there is a 180 degree phase shift in the electron wave function. But we can decide if the phase shift is +180 or -180 degrees. This does not prove anything.

Scattering does change the momentum and the wavelength of the scattered particle. This is the reason why measuring the particle path spoils the interference pattern in the double slit experiment.

How do we prove the Pauli exclusion principle from the "interchange" property of the wave function?

In quantum field theory, we decide in which order we apply creation operators to the vacuum. We can decide to create the "first" a particle at a spacetime point x and "after" that at y. If x and y are spacelike separated, then there is no temporal order for x and y. The order of creation is just a property of the formal model which we are using, it is not a physical thing.

There is no isolated creation operator in nature. An electron-positron pair is always created together.

How do we prove the Pauli exclusion principle in our formal model? We should show that the two electrons in the helium atom ground state cannot both have the spin up.

Suppose that we have two electrons x and y far away from an alpha particle. How do we prove that the electrons cannot end up in the same quantum state around the alpha particle?

The usual reasoning seems to go like this:

1. We can describe the initial setup either as the x electron being created the "first" and the y electron created the "second". But we can as well describe the setup with y created the first.

2. Since the wave functions in these two descriptions have the sign flipped, then the sum of the two wave functions in the hypothetical final state around the alpha particle is zero.

3. Alternatively, we can "interchange" the electrons in the final state. Because the sign of the wave function is flipped, but the physical setup stays exactly the same, we conclude that the wave function has to stay the same. The wave function has to be zero.


The reasoning above is flawed. If we have just one electron, we can describe it with a wave function Ψ or -Ψ, or any

       exp(i φ) Ψ,

where φ is a real number. If the electron ends in a quantum state, the sum of the two alternative wave functions is zero. Did we show that the electron cannot end up in any quantum state? No. The reasoning is wrong. We cannot use two alternative descriptions for the same physical setup and later sum those two descriptions to prove something.

The whole concept of the wave function flipping sign in an "interchange" is nonsensical. Quantum mechanics is agnostic about changing the phase of the wave function. We are always allowed to flip the sign of any wave function by multiplying it by -1, and we get an equivalent wave function. There is no need to flip the sign when we perform an "interchange".  The interchange itself is not a physical operation but just a change of the description of the system. If we change the description in one way, what prevents us from changing the description again by multiplying by -1? Then there is no flip of the sign.

Is there a flip of the sign of the wave function when electrons are interchanged physically?

What about interchanging electrons x and y physically? Suppose that we have the electrons in a closed box and we describe the initial state with a wave function. After some time has passed, we have no way of knowing which electron is which, because they are indistinguishable particles.

Then we let these two electrons approach the alpha particle and form a helium atom. Can we prove that the electrons cannot have the same quantum state in the end? When the electrons approach the alpha particle, they will emit photons. We will have a system of at least 4 particles, probably many more.

Is there a destructive interference which prevents the two electrons ending up in the same state?

The exchange interaction

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

The Wikipedia article above explains the Pauli exclusion principle in a simple setup through the energies of the states. Spins pointing in the same direction make the energy of the state higher. The magnetic repulsion of the two tiny magnetic needles (electrons) is negligible, but there is another effect.

Looking at energies of the states is a concrete way of deriving the Pauli principle and should not suffer from the conceptual error which we pointed out in the previous section. Does the argument in Wikipedia derive the result without making the conceptual error in the ordinary proofs of the Pauli principle?

Other proofs

https://press.princeton.edu/titles/6938.html

Streater and Wightman have an axiomatic proof in their 1964 book. We will look at it.

https://journals.aps.org/pr/abstract/10.1103/PhysRev.58.716

Wolfgang Pauli has a heuristic proof in a 1940 paper in Phys. Rev.

Pauli shows that for fields of an integer spin, it is impossible to define a particle density in a satisfactory way, and for fields of a half-integer spin, it is not possible to define an energy density which is positive definite.

Pauli concludes that field theory cannot be interpreted as a one-body problem. Pauli says that relativistic electromagnetic theories have charged particles which are created and absorbed in pairs, and uncharged particles which are emitted and absorbed singly.

Pauli argues that half-integer spin fields change sign if we "interchange" the particles. In that, he seems to follow the typical path in the proof.

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

Richard P. Feynman has presented an argument based on scattering.

Optical gravity and cosmology

Optical gravity claims that spacetime is the flat Minkowski space. Gravitational fields cause weakening of forces, and moving any object in a gravitational field causes energy flow, which in turn makes the inertial mass larger. Furthermore, the inertial mass in the vertical direction is bigger than in the horizontal direction.

In cosmology, the standard cosmological model is the inflating balloon model.

Can we reproduce the balloon geometry within the flat Minkowski space? If one cannot escape from the balloon, then we must be in some kind of a black hole geometry.

Optical gravity predicts that a forming black hole will eventually reflect back all material. Maybe we are close to the event horizon of a huge black hole, and seeing the back reflection phase?

Why is the matter density of the visible universe close to the critical density of the standard cosmological model? Alternatively, we could make a black hole from all the available matter, and its radius would roughly be the radius of the visible universe.

What is the matter density in the reflection phase of a black hole?

Can optical gravity explain dark energy? If the universe is becoming more and more Minkowski, would that create the illusion of an accelerating expansion?

Electron spin as a classical rotating object; electrons on a neutron star have larger spin?

The intrinsic angular momentum of the electron is

       1/2 h / (2π) = 5 * 10^-35 kg m/s.

Suppose that the electron is a classical mass of 0.9 * 10^-30 kg spinning around 1/2 its Compton wavelength at a speed b * c. Then its angular momentum is

       L = m_e * 10^-12 m * c
          = b * 0.9 *10^-30 * 1.2 * 10^-12
             * 3 * 10^8 kg m/s
          = b * 3.2 * 10^-34 kg m/s,

from which we get the speed b c = 1/6 c.

A classical model as the electron as a rotating ball of size the Compton wavelength is quite reasonable.

The spin kinetic energy in our model is

       E = 1/2 m_e v^2
       = 1/2 m_e c^2 / 36
       = m_e c^2 / 72.

That is, 1.5 % of the electron total energy would be spin energy.

The electron charge circulating in a loop of radius 1.2 * 10^-12 m makes an electric current of

       I = 1.6 * 10^-19 / (2π * 1.2 * 10^-12 m)
             * (0.5 * 10^8) A
         = 1 A.

The magnetic field strength inside the current loop I is huge:
       B = μ_0 I / (2r)
          = 1.3 * 10^-6 / (2.4 * 10^-12) T
          = 5 * 10^5 T.

The magnetic moment of our model electron is

       m = I S
           = 1 * π * (1.2 * 10^-12)^2 A m^2
           = 5 * 10^-24 J/T.

The measured magnetic moment of the electron is

       m = 9 * 10^-24 J/T,

that is, it is roughly double to our classical model.

If we put the electron in a strong magnetic field B, flipping it will give an energy

      E = 2 * |m| B.

For a field of 1 T, the energy is only 2 * 10^-23 J, while the total energy of the electron is 10^-13 J and the rotation energy of the electron is 1.5 * 10^-15 J. If we put the electron to a magnetic field of 10^8 T, then the electron might speed up its rotation to acquire a larger magnetic moment, and consequently, a larger spin. Neutron stars have magnetic fields up to 10^11 tesla.

The electron orbit radius is only 1/(4π) Compton wavelengths?

Our model of the electron circling at a radius 1/2 of its Compton wavelength would mean that the electron actually would complete 3.14 waves in its orbit. That does not sound reasonable.

In our model, the electron is only moving at 1/6 the speed of light. What if we let the electron move at the speed of light? The radius is then only 1/6, and the electron does roughly 1/2 wavelengths during its orbit, which corresponds to the spin quantum number 1/2.

The Compton wavelength is

       λ = h / (m_e c).

Let us calculate again the angular momentum if the electron circles at a radius λ / (4π):

       L = m_e c h / (m_e c 4π)

          = 1/2 h / (2π).

Thus, the electron moving at the speed of light at a radius of 1/(4π) Compton wavelengths is a natural model for the angular momentum of the electron. The spin kinetic energy in this case is 100 % of the total energy of the electron.

The associated electric current would be roughly 40 amperes and the magnetic field inside the current loop some 10^8 T.

The magnetic moment stays the same, at 5 * 10^-24 J/T.

We conclude that a better classical-relativistic model for the electron is an object circling with a radius of

       r = λ / (4π)

         = h / (4π m_e c)

         = 2 * 10^-13 m.

The spin 1/2 corresponds to the electron doing 1/2 Compton wavelengths in its orbit.

Note that zitterbewegung has the speed of light and the orbit radius of 1/2 the Compton wavelength.

In the hydrogen atom, in the ground state, the electron does one "spatial wavelength" in its orbit around the proton. The spatial wavelength is determined by the momentum p of the electron in the non-relativistic Schrödinger equation. The electron does one rotation around the proton in 1.5 * 10^-16 seconds. The Compton frequency of the electron is

       f = 3 * 10^8 / (2.4 * 10^-12) Hz

         = 1.2 * 10^20 Hz.

The wave function of the electron will rotate 1.8 * 10^4 (= α^2?) times in the time it completes one rotation around the proton.

The electron in the hydrogen atom can be considered a particle which makes a small circle at the speed of light, so that its wave function at a fixed point rotates 1/2 rounds per circle. It also makes a large slow circle around the proton, such that its wave function at a fixed point rotates 18,000 times during the circle, but its "spatial wave function" only rotates 1 round in the circle.

The spin in this model is very much a relativistic effect, while the orbit in the hydrogen atom is non-relativistic. We model the electron in this small circle as a wave

       exp(-i (E t - E x)),

where we have set c = 1, and E is the rest mass of the electron. For a particle moving at the speed of light, E = p.

The electron must make two complete rounds around the small circle, for the wave function phase to return to its original value at a fixed point. This is probably the origin of the strange 720 degree rotation symmetry of the electron wave function.

Magnetic fields in particle collisions

The electric field in a head-on collision of an electron and a positron reaches at least

       511 kV / (2 * 10^-12 m) = 2 * 10^17 V/m,

that is, it is close to the Schwinger limit.

A Lorentz transformation of the electric field to a speed close to the speed of light produces a magnetic field strength of the same order of magnitude, 10^17 tesla. The magnetic field in a collision might temporarily excite an electron to a higher spin state.

In our model of the previous section, all electron energy is spin energy. An electron in the spin state 3/2 would have an energy 1.5 MeV.

Optical gravity and the inertial mass in a gravitational potential well

Our optical gravity concept is based on the hypothesis that all forces grow weaker in a gravitational potential well, when the forces are measured by an observer who is himself not in the potential well.

Why are forces weaker in a gravitational potential well?

Why do the forces grow weaker for an outside observer? Because if the outside observer sends in a light pulse 1 joule of energy to another observer inside the well, the receiver thinks he gets 1 + E joule. Gravitational acceleration adds to the energy. We then use the equivalence principle here: 1 + E joule should do the same work for the observer inside the potential well as 1 + E joule does for the observer outside.

From the point of view of the outside observer, the 1 joule he sent was able to do work for 1 + E joule. He thinks forces are weaker in the well.

Another way to show the weakening of forces is this: suppose that we have two electrons in a low gravitational potential on a neutron star. The repulsive force is a consequence of the increase in the electric field energy if we push the electrons closer to each other. But in a low gravitational potential, the field energy increases less when we move the electrons a distance ds. In this we used the assumptions that:

1) forces are a result of field energy;

2) field energy is worth less for an outside observer, if the field resides in a gravitational potential well.

Inertial mass for horizontal movement

Mechanic clocks tick slower inside the well because a weaker force accelerates the weight of the clock less. In this it is crucial that also the inertial mass of the weight is suitable inside the well.

Our discussion of the inertial mass of the electron revealed that it is by no means simple to determine the inertial mass of an object under a potential.

                      m -------------------->
                           \O
                             |
                            /\
          =======================
                      x                         y
                       <------------------- |V|

We employ once again our model of a man standing on a surface. In this case, it is a surface of a planet and the force is gravitation.

The man lets the test mass m approach from far away, and fills an energy store at x with the energy |V| that the test mass gets in the approach. Then the man moves the test mass to y and uses an energy store |V| there to push the test mass far away again. The inertial mass of the test mass might be

       m + |V|,

but we need to think more carefully how the field energy of the gravitational field moves when we move the test mass.

The field is weaker in a wide area between the test mass and the center of the planet. Everywhere else, the field is stronger.

The inertia in an electric field versus gravitational

The electric field has a positive energy density of E^2, where E is the field strength. The energy density of the gravitational field is negative. Suppose that we add a small field dE to a strong electric field E. The energy density change is

       (E + dE)^2 - E^2
       = 2 dE E + dE^2.

For the gravity field G, the change is

       G^2 - (G + dG)^2
       = -2 dG G - dG^2.

Let us then compare the fields of an electron close to a large positively charged sphere, and a test mass m close to a planet, such that their field strengths have equal numerical absolute values. The direction of the field of the electron is opposite to the field of the test mass. The change in the energy density then has the same sign in both cases far away from the electron or test mass, and is approximately equal

Close to the electron, the field has positive energy, while it has a negative energy close to the test mass.

Speed of a horizontal clock in gravitational potential

Suppose that the gravitational potential on the surface of a neutron star is -0.01 m for an object of mass m. We have set c = 1.

Let observer A be far away and observer B stand on the surface of the star.

A thinks that all forces between objects which are close to B have weakened by 1 %.

         A

         B   1 kg    1 kg

  ==================

          neutron star

A and B want to compare the speed of their time. B puts two 1 kg weights (which were weighed far away and exported to the neutron star) horizontally at the distance of 1 meter from each other. Both weights will carry 1 coulomb of negative electric charge. B measures the end velocity of the weights as they push each other far. A uses a telescope and measures the velocity, too.

The electric force is 1 % weaker on the surface as measured by A. The kinetic energy will be 1 % less. Furthermore, the inertial mass is 1 % larger as measured by A, if our hypothesis from a previous section is valid.

       E_kin = 1/2 m_inertial v^2

      <=>

       v^2 = 2 E_kin / m_inertial.

The velocity that A sees from his telescope is 1 % less than A would see if he himself would perform the same experiment in space. A concludes that the time of B has slowed down by 1 %. That 1 % is the redshift of light from B to A, according to general relativity.

What did we assume?

1. We used the fact from general relativity that horizontal distances look the same for A and B.

2. We used the equivalence principle to deduce that forces are weaker in the potential well, or alternatively that field energy is worth less in a gravitational potential well.

3. We used a hypothesis of gravitational field energy distribution to determine the inertial mass in the potential well.

We did not need to assume that the geometry of spacetime is anything else than the flat Minkowski.

The Schwarzschild geometry and vertical distances

In the Schwartzschild geometry, if the time measured by an observer A on a neutron star has diminished by a factor b, then vertical distances measured by him have increased by a factor 1 / b. In this, we compare his measurements to the global Schwarzschild coordinates.

A spacetime element in the global coordinates is squeezed when A measures it, but the spacetime volume measured by A agrees with global coordinates.

Can we explain with the gravitational potential why the ruler of A is shortened when it is vertical, but not shortened when it is horizontal?

When we move the test mass horizontally, there is energy flow within the field but not to/from the test mass.

If we move the test mass vertically, there is energy flow also between the test mass and the field. The inertial mass for the test mass might be larger in the vertical direction.

Let the observer A throw the test mass horizontally so that it hits a wall at a 45 degree angle, and bounces vertically up. If the inertial mass of the test mass is bigger in the vertical direction, it will fly slower.

Any object, an electron in an atom or anything, will move slower to the vertical direction. That might be the reason why the ruler of A is shorter in the vertical direction.

We need to estimate the energy flow in a vertical movement.

Let us have a neutron star of radius 1. The potential V of the test mass m is

       V = -k / r,

for some k.

                     V
             <---- m ---->
   
             ==========
            neutron star

In horizontal movement, if we move the test mass a distance 0.01, the negative potential energy V, in a sense, seems to cause positive energy |V| to flow a distance 0.01. The total energy displacement is

        m * 0.01 + |V| * 0.01.

The inertial mass is m + |V| for horizontal movement.

                   ^
                   |
                  m    V
                   |
                   v
            ==========
            neutron star

If we move the test mass vertically +0.01, we move|V| the same distance. Our test mass has stolen its extra inertial mass |V| from the field of the neutron star. When the test mass moves up 0.01, it returns 0.01 |V| of the inertial mass to the field of the neutron star. Maybe the energy displacement for that 0.01 |V| is the radius (= 1) of the neutron star? The total energy displacement may be

       m * 0.01 + |V| * 0.01 + 0.01 |V| * 1.

We see that the inertial mass in vertical movement is m + 2 |V|.

The movement of electrons in an atom is 1 % slower in the vertical direction. That squeezes the ruler 1 % vertically.

If the gravitational potential |V| is much less than m, our model explains the Schwarzschild geometry from the potential of a newtonian gravity force.

The behavior close to the horizon of a black hole requires a separate analysis. Since time has slowed very much there, forces are very weak, and the inertial mass is very large.

Optical gravity and Tuomo Suntola's Dynamic universe

In the spring of 2018 we introduced the concept of optical gravity. The idea is that the geometry of spacetime is always a flat Minkowski space. Low gravitational potential reduces the speed of light, that is, the space is more "optically dense" at such areas. Also clocks tick slower there.

The idea is originally due to Isaac Newton, who tried to unify gravitation with optics.

Optical gravity is equivalent to general relativity. It is just another way of looking at a physical system.

https://www.physicsfoundations.org/1_5_dynamic-universe.html

Doctor Tuomo Suntola has developed a different model where the "tilt" of space affects the speed of light. Looking at his diagrams, the tilt seems to be the strength of the gravitational field, not the potential. According to Suntola, clocks tick the slower and the speed of light is the slower, the greater the tilt.

A brief browsing of the Internet did not reveal any published paper about the speed of clocks in a deep mine. It should be measured. According to general relativity, the speed is dependent on the potential and the clock ticks slower in the mine. But if the speed is dependent on the gravitational acceleration, then the clock ticks faster.

The new, corrected energy-momentum relation

In the past few days we have studied the effect of different configurations for the electron inertial mass.

The simplest case was the man standing on a negative plane manipulating an electron.

The man inside a positively charged box was a very different scheme.

We can now present the new version of the energy-momentum relation for a static external electric field and an electron in it:

       E^2 = p^2 + (m + b |V|)^2,

where b can be calculated by considering the distribution of the electric field energy when we move the electron.

The field energy density for a static electric field is E_V^2, where E_V is the electric field strength.

The Poynting vector is one method of calculating the energy flow in the electric field when we move the electron a little bit. The momentum of the energy flow tells how much extra inertia the electron gets from the field.

If metals are present, they affect the electric field change greatly and may dominate the calculation of b. The method of the "mirror electron" shows that m + b |V| is actually 2m inside a metal shell if the electron moves slowly.

Since the hydrogen atom does not present a different spectrum inside a metal shell, the orbital frequency of 10^18 Hz is too fast for mirror electrons to take part in action. Or, the quantum mechanical fact that the atom is in its ground state bans all energy-consuming interaction with the environment.

We should calculate the inertial mass of the electron in the hydrogen atom. Since the electric fields of the proton and the electron partially cancel each other out, our first guess is that the inertial mass is slightly less than 511 keV. The inertial mass depends on the distance from the proton. Close to the proton, we need to decide how close to each particle we let the field to extend. As is well known, the electric field energy is infinite for a point particle, and the integral may diverge.

The system electron-proton is a dipole. Most of the field energy is concentrated to a cigar-shaped area which surrounds the particles. That area determines the inertial mass of the electron.
           ____________
          /                       \
         |   e-          p       |
          \____________/

If the inertial mass of the electron does not change, that breaks newtonian mechanics?

Let us think more about the van de Graaff box and the electron inertial mass inside it.

V. F. Mikhailov. Influence of an electrostatic potential on the inertial
electron mass. Annales de la Fondation Louis de Broglie, 26:33–38, 2001.

V. F. Mikhailov observed a change in the oscillation frequency of electrons in a Barkhausen-Kurz oscillator which is placed inside a charged spherical shell. But subsequent experiments have not confirmed his result.

https://www.researchgate.net/publication/316716539_Experimental_Investigation_of_the_Influence_of_Spatially_Distributed_Charges_on_the_Inertial_Mass_of_Moving_Electrons_as_Predicted_by_Weber's_Electrodynamics

Our thought experiments, on the other hand, suggest that the inertial mass has to change. If not, the center of mass of the system would not be conserved.

Mikhailov and others have thought they are testing Wilhelm Weber's electrodynamic hypothesis from 1848. We did not know of Weber when we designed our thought experiments. The thought experiments use just the basic electrodynamics, Newton's law, and Einstein's mass-energy equivalence.

1. Very basic electrodynamics claims that since the electron in the charged sphere sees a zero electric and zero magnetic field, there are no forces on it. This does not yet determine what is the inertia of the electron, but people seem to think the inertia is the same as for a free electron.

2. Poynting's vector, on the other hand, claims that the movement of the electron causes energy flow in the electric field surrounding the sphere. That energy flow should exert inertia on the movement of the electron.

3. The Aharonov-Bohm effect says that also a constant potential affects electron behavior. There is no need for the field strength vector to be non-zero at the electron.

Sources of error in the experiments

Why were several experimenters unable to measure the change in the inertial mass? The obvious suspect is the influence of the electron on the charge distribution of the metal sphere surrounding it.

The metal sphere tries to keep its electric field uniform and normal to the surface. If we move an electron slowly inside the shell, then the electric field outside the shell does not change at all. There is no flow of energy in the electric field outside the shell, and thus no extra inertia from the electric potential of the shell.

The charge distribution in the shell polarizes to cancel any change in the outside electric field. We may model the polarization with a "mirror electron" which moves to the opposite direction from the test electron. The effective inertial mass of the test electron should thus be constant, twice the inertial mass of a free electron.

There is an electric current in the shell. That will cause resistive energy loss. In the oscillator, there is energy loss from electromagnetic radiation. Can we discern these losses? Did the experimenters calculate these?

At high frequencies of 1 GHz or more, the field of the oscillator will "mostly" be electromagnetic radiation. How does that affect the model?

The electron in the hydrogen atom has a frequency of some 10^18 Hz. Hydrogen does not show a distorted spectrum inside a metal shell. Maybe the electric neutrality of the atom cancels the effects on the inertial mass of the electron in the atom. This is probably a quantum effect. The mirror electron model would make the inertial mass double.

The shell should be made of an insulator. Even in that case, can we be sure that there is no current or significant polarization in the insulator?

There are several metal parts around the oscillating electrons inside the shell. The electron will polarize charges in these, and the influence tends to reduce the temporal change of the electron's electric field. Did the experimenters calculate these?

Polarization of air and non-metallic parts will shield some of the changes of the electric field.

We need to check the articles of the experiments.

A negative potential adds to the rest mass, does not reduce it?

Both the old and the new energy-momentum relation claim that a static electric potential affects the kinetic energy of an electron. The effect is big even at moderate potentials of a few kV.

The Stark effect is measured under an electric field, not in a constant potential.

What is the effect on the hydrogen spectrum?

According to our new energy-momentum relation, the electron has an imaginary mass if the potential is > +511 kV, and should behave in a weird way.

A van de Graaff generator can create voltages up to 25 MV.

What about observers inside the static electric field? How do they measure the inertial mass of the electron?

An electron inside a positively charged van de Graaff generator

If positive voltages over +511 kV can be generated, how can we explain that no weird physics appears? Could it be that the geometry of the situation assigns the negative inertial mass to something else than the electron? Or is the inertial mass measured by people inside the shell of the van de Graaff generator different than what people outside measure?

Let us put a man standing inside a van de Graaff generator which has a positive voltage V_b. The voltage can be low, too.


           +   +    +    +
          ____________
   +    |                       |   +
         |                       |
   e- --->  \O        ----------->
         |       |              |
   +    |___/\________|   +
           +    +     +    +
         x                    y
          <--------------- E

The man lets the electron come in and fills an energy store at x with energy E. Then he moves the electron to y and uses an energy E to push the electron out.

We assume that the man is weightless.

The net result is that a 511 keV electron moved from x to y and an energy E moved the other way.

The inertial mass of the electron from the point of view of the person moving it can be positive, though.

                  rod
                 --------------------------W   weight
                   --> lever
                  |
                 /|\      supporting
               /  |  \    structure
             /    E   \
          -/----------\------ box floor

If there is a lever which moves a mass E to the left when the man pushes it to the right, the lever certainly has an inertial reaction but still moves mass to the opposite direction. The lever is attached to the box. What if the man uses a rod which is attached to a weight outside the box, to push on the lever? He does not touch the box at all but uses the rod to win the inertia of the lever. The supporting structure of the lever pushes the box to the right.

The end result is that the weight W has moved to the left, as well as the mass E, and the box has moved to the right. The head of the lever "borrowed" some inertia from the box.

If the electron can borrow inertia from the the electric field of the box, then the electron can behave quite normally also in a potential which is higher than +511 kV.

Let the box have a voltage V_b > 0. We conjecture that the man inside the box will feel that the electron has an inertia of

       511 keV + |e V_b|,

that is, we need to add the absolute value of the (negative) potential to the inertial mass of the electron. Also an outside observer will think that the electron has that same inertia.

The conjecture assumes that the forces between the electron, the box, and the fields form a kind of a lever system which moves E.

If there are unknown forces between the electron and the box, then the inertial mass of the electron can be arbitrarily high. The electric field of the box might be very "viscous", such that moving the energy hole caused by the electron would require great force.

We definitely need empirical experiments to determine how the electric field behaves.

Classical solution

Classical electrodynamics claims that there are no forces between the box and the electron because the electric field is zero inside the box. Let us calculate what would happen if the man inside the box would feel that the inertia of the electron is M.

The electron performs a displacement of s m, where s = y - x. The man pushes the box left with his feet to win the inertia M. If there are no forces at all between the electron and the box, then the displacement of the box is -s M.

The displacement of E is -s E. We get an equation

       s m - s M - s E = 0
      <=>
       s M = s m - s E.

The formula is not sensible if E > m.

Another way to calculate things classically is to use the Poynting vector. Let us assume that the positively charged box is surrounded by a negatively charged box, such that its charge exactly cancels the field far from the box. We can then restrict ourselves to calculating the electric field energy inside the box and in its immediate vicinity.

The electric field energy of the electron inside the box is essentially 511 keV, because it is the only field there. When the electron moves from x to y, the Poynting vector describes the field energy flow from the right side of the outside of the box to the left side. Thus, the electron also acts as a kind of lever which moves the energy E from right to left.

We see that the Poynting approach of calculating the field energy gives roughly the same result as our conjecture in the previous section: the inertia is

       511 keV + |V|.

The inertia depends very much on the geometry of the setup. In the hydrogen atom, the Poynting approach says that the inertia is almost zero when the electron is in a potential -1.022 MeV close to the proton, because the energy of the combined electric field is close to zero. Here we assume that the whole rest mass of the electron is in its field and the field does not gain more energy when the electron comes close to the proton.

Experimental tests of the inertial mass of the electron under a potential

Let us check what literature says about the inertial mass of an electron under a potential.

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

Wikipedia says that Maxwell electrodynamics does not conserve particle momentum if there is radiation out. That is reasonable. Wilhelm Weber's 1848 theory is claimed to conserve momentum.

http://www.nrcresearchpress.com/doi/10.1139/p04-046#.W-28t_ZuLb0

Mikhailov (1999) measured the dependency of the electron inertial mass on the potential and confirmed that it changes. But further experimenters have disputed his result.

People are trying to break newtonian mechanics with Weber's theory. Since the classical theory, the Poynting approach, and our thought experiments give conflicting results, more experiments are needed.

Electron spin is a consequence of an uncertainty principle?

Dirac derived the existence of the electron spin from his relativistic equation. But, actually, does nonrelativistic quantum mechanics already imply the existence of the spin?

The electron has an electric field. If we would know that the field does not rotate, that would probably break an uncertainty principle of the rotation of an object.

That is, there has to exist a spin for all objects which have some kind of spatial extension, small and large. The spin is not constrained to elementary particles.

The electron is a point particle. One might conjecture that it cannot rotate, because it is a point. However, the electric field of the electron is not a point, and has an orientation in space. When we measure the electron spin, we measure the rotation state of its electric field.

https://aapt.scitation.org/doi/10.1119/1.11806

David Hestenes has a similar idea in his 1979 paper.

The rotation axis is confined to the surface of a unit sphere. Does the spin 1/2 correspond to half a wave circling the sphere? That might explain the strange 720 degree rotation symmetry of the spin.

Is there a way to make the electron electric field rotate more rapidly? The spin 1/2 may correspond just to the lowest rotation energy state.

In an electric potential, the electric field of the electron may have much more energy than 511 keV, and it would be natural if new rotation states appear.

What about the spin of the muon? The muon is 200 times heavier than the electron. Could we make the muon to spin more rapidly?

What is the spectrum of hydrogen like under an electric potential?

https://en.m.wikipedia.org/wiki/Aharonov–Bohm_effect

The Aharonov-Bohm effect for an electric potential has not been measured yet.

But, according to our new energy-momentum relation as well as the old one, a constant electric potential changes the inertial mass of the electron. It should be easy to measure under a potential of, say, a few kilovolts.

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

In the Stark effect, the spectrum of an atom is affected by a static non-zero electric field. What about a constant potential? Our Nov 15, 2018 post studies this question.

Space and time are interchangeable at short distances; antiparticles come from this

If we have a relativistic electron, then the uncertainty in its energy E is of the same order of magnitude as the uncertainty in the momentum p.

We can combine the Heisenberg uncertainty principles into a spacetime uncertainty principle:

       Δ sqrt( t^2 + x^2 ) ΔE ≥ h / (4π).

Time and space appear completely symmetric in the relation and we can use the euclidean metric.

This is probably the reason why a "natural" wave packet of a Dirac electron always contains also negative frequencies, that is, the positron. The positron is an electron traveling back in time. If time and space are interchangeable, we cannot ban such paths, just like we cannot ban paths that go to the negative x direction.

We cannot ban superluminal paths either. The path integral for the electron should contain all kinds of paths, zigzagging in any direction in spacetime.

Richard P. Feynman in his 1949 papers stressed the symmetry of his diagrams when t and x are switched.

Large objects are, for some reason, confined inside the light cone (why?), but particles are not when the distance is less than their Compton wavelength. This allows the wave functions of particles be smooth in spacetime, which in turn may remove the need for regularizarion in the vacuum polarization loop of QED.

If a particle has a zero rest mass, or m + V is zero, then p = E, and the Heisenberg spacetime uncertainty principle is perfectly symmetric in time and space. Photons, too, can zigzag in time. Is there zitterbewegung of photons?

The symmetry of space and time at short distances explains why particles have to have antiparticles: there is no mechanism which prevents a particle from colliding in a way which sends it back in time, that is, it becomes an antiparticle.

Does the new energy-momentum relation break relativistic gauge theories?

Our new energy-momentum relation is equivalent to an old version when |p| and |V| are small.

       (E - V)^2 = (p + A)^2 + m^2
      <=> (|p|, |V| small)

       E - V = m
             sqrt((p + A)^2 / m^2 + 1)
                = (p + A)^2 / (2m) + m
      <=> (|p|, |V| small)

        E = (p + A)^2 / (2 (m + V)) + m + V
           = (m v + V v)^2 / (2 (m + V))
              + m + V.

The choice A = v V gives the same result as treating V as rest mass.

However, in relativistic settings, the two energy-momentum relations are inequivalent. A question is if relativistic gauge field theories are flawed as they use the old relation?

A brief study of the material on the Internet reveals that gauge field theories are usually nonrelativistic, or are expressed through a lagrangian density and Feynman diagrams. Feynman diagrams have interactions only at vertexes, and the particles are assumed to be free in the lines between them. The question whether potential energy and rest mass should be identified never comes up.

Rest mass in the Standard model is potential energy in the Higgs field. Thus, the Standard model does identify at least the Higgs potential energy with rest mass. Before the Higgs field assumes its vacuum expectation value, fermions have a zero rest mass. The physics would be weird in such a setting because the value m - V would often be negative and the fermion would be a tachyon. Maybe the Higgs field always has had the same expectation value? Is there evidence that fermions can be rest-massless at high energies? If the rest mass of the electron would be zero, would its acceleration under a potential be infinite?

The new energy-momentum relation and Lorentz transformation of the potential

We were perplexed in our Nov 4, 2018 post about how to Lorentz transform the potential and the rest mass of a particle.

The electric scalar potential ϕ and the magnetic vector potential A are prime examples.

We work in 1+1 dimensions.

The Lorentz transformation (ignoring the gamma parameter, we assume v is small) is

       ϕ = ϕ - v A

       A = A - v ϕ.

Our new energy-momentum relation is

       E^2 = p^2 + (m + V)^2,

where V is the scalar potential of the particle. When we Lorentz transform the relation, should we Lorentz transform m and V, too?

Traditionally, m is considered a scalar which stays the same in a Lorentz transformation. If we identify V with m, then V should not be transformed either. The energy-momentum relation transforms then like:

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

If we try to Lorentz transform also m and V, we end up with a relation like:

       (E - v p)^2 = (p - v E - v m - v V)^2

                              + (m + V)^2.

If we have p = 0, that relation would indicate that the inertial mass of the particle is double of what we are used to. It may be more logical to think that the transformation of E in the momentum term already includes the transformation of m and V.

We are left with the question in which frame should we measure m or V. The simplest case is when we measure in the rest frame of the object which produces the electric potential V. We assume there is no magnetic field.

Conjecture 1. The correct energy-momentum relation is

       E^2 = p^2 + (m + V)^2,

and we must not Lorentz transform m or V when we move to a new frame. V is the work we need to do to bring the particle from infinity to its current position, in the rest frame of the object that produces the field. We assume V does not change with time and that we move the particle under a constant potential V.


Suppose that we have an electron at rest. Then

       E = m + V.

If we Lorentz transform, we get

       E^2 = (-mv - v V)^2 + (m + V)^2.

We can interpret -v V either as the vector potential of V or as the momentum of the rest mass V.

How does Conjecture 1 differ from a traditional way to handle the potential in the energy-momentum relation?

Some people write

       (E - V)^2 = (p + A)^2 + m^2,

where A is the vector potential associated with the scalar potential V. The value E' = E - V is considered the energy of the particle.

If A = 0 and p = 0, and we Lorentz transform the above, we get

       (E - V)^2 = (-v (E - V) - v V)^2 + m^2
                       = (-v E)^2 + m^2,

where we Lorentz transformed V to get a vector potential A = -v V.

The Lorentz transformation of our new relation is

       E^2 = (-v E)^2 + (m + V)^2.

The kinetic term -v E is the same as in the traditional case. Our new energy-momentum relation gives roughly the same results as a traditional one when |V| and |p| are small.

All rest mass is potential energy?

Can we consider also the rest mass as potential energy? We can bring an electron from nonexistence by creating a pair. What is the role of the antiparticle in this?

                e+
             
                e-
                   \O
                     |
              E    /\  ----------> E
     =========================
     -    -    -    -    -    -    -     -    -     -    -
               x                        y

Let us introduce once again the man standing on a finite charged plane, who we had in our Nov 3, 2018 blog post. This time, the man uses an energy store E to create the electron far away from the plane by "detaching" it from the corresponding positron. That is, the man uses energy to create a pair.

We claim that both particles have an inertial mass 511 keV before the man brings the electron close to the plane, where the inertial mass of the electron is 511 keV plus the potential V.

The man moves the positron and the electron horizontally from x to y. After that, he fills an energy store at y by letting the electron move away and annihilate.

The inertial mass of the positron was 511 keV, that we know. The inertial mass of the electron had to be 511 keV plus V, otherwise the center of mass would have moved.

In this example, it makes sense to say that the rest mass of the electron is potential energy.

If the universe contains electrons that do not have an antiparticle, is it still reasonable to say that their rest mass is potential energy?

The new Dirac equation and a potential step; Klein paradox solved again

We learned about the conserved current in our previous blog post, as well as about Lorentz covariance. But it was not clear if the potential of the electron should be Lorentz transformed.

Let us analyze once again what happens at a potential step.

Potential step of the Schrödinger equation

The standard plane wave solution for the Schrödinger equation is

       exp(-i (E t - p x)).

Let us match the solutions for a potential step of height V. The incoming wave is 1, the reflected wave B, and the transmitted wave is 1 + C.

         p                q
     1 -->      1 + C -->
    -p <-- B __________
   ________| V
              x = 0

From the continuity we get:

       1 + B = 1 + C => B = C.

From the continuous derivative d/dx:

       p - B p = q + B q
      =>
       B = (p - q) / (p + q).

The "current" is proportional to p times the square of the amplitude. We should have

       (1 - B^2) p = (1 + B)^2 q
      <=>
       (1 - B) p = (1 + B) q

which is true.

If V is high, then q is imaginary and

       B  = exp(-i * 2 φ),

where φ is the phase of the complex number (p, q). The plane wave solutions of the Schrödinger equation work very well over a potential step.

Potential step of the new Dirac equation

The standard plane wave solution for our new 1+1-dimensional Dirac equations is

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

and its conserved current (divided by 2) is

      J = p / (E + m + V).

Let us solve the potential step in the diagram of the previous section. From the continuity of component 1 we get

       1 + B = 1 + C.

Continuity of component 2 gives:

       (1 - B) p / (E + m) = (1 + B) q / (E + m + V).
      <=> (def.)
       (1 - B) J_p = (1 + B) J_q
      <=>
       B = (J_p - J_q) / (J_p + J_q).

The conservation of currents requires

       (1 - B^2) J_p = (1 + B)^2 J_q
      <=>
       (1 - B) J_p = (1 + B) J_q,

which is true. Thus, our new Dirac equation behaves very well.

Let us then study what happens when V is large. The energy-momentum relation is

       E^2 = q^2 + (m + V)^2.

If (m + V)^2 > E^2, then q is imaginary and also J_q is imaginary.

       B = (J_p - J_q) / (J_p + J_q)
          = exp(-i * 2 φ),

where φ is the phase of the complex number (J_p, J_q).

We have shown that our new Dirac equation is very well-behaved. There is no Klein paradox in our equation.

Could it be possible to solve the hydrogen atom with the new equation?

Is the electron very close to the proton a tachyon, a positron, or a neutral light-speed particle?

UPDATE Nov 15, 2018: In our Nov 15, 2018 post we try to analyze the impact of a negative potential energy more carefully. The inertial mass may stay positive, after all.

…

We have already discussed the case m + V < 0, in which we think an imaginary  i |m + V| is the correct interpretation for the inertial mass, and the electron is a tachyon very close to the proton. Another possibility is that the electron moves at exactly the speed of light but fails to see the force of the proton.

Yet another possibility is that the electron changes into a positron close to the proton and sees a repulsive force from the proton.

If the electron is superluminal, some observers see it right-handed, some observers left-handed (= positron). We may think that it is in an intermediate stage of turning into a positron. In pair annihilation, the electron would turn back in time and be a positron.

Does the new Dirac equation cast light on this? What happens if m + V < 0 or when m + V is imaginary?

 t
 ^
 | 511 keV    511 keV photons
 |      ^              ^
 |        \          /
 |          \ ___/
 |          /       \
 |        /           \
 |     e-              e+
 --------------------------------> x

When an electron approaches a positron to annihilate, then at some point, m + V = 0 and p = E. The plane wave in that case is

       (1, 1) exp(-i (E t - p x)).

That would be a good point to glue it to the positron plane wave, since the vector is (1, 1) for both particles?

But the electron has to emit photons. If the electron emits a 511 keV photon at that point, the plane wave after that might be something like

       (1, 2p / (i 2p)) exp(-i (0 t - 2p x)).

The mass of the electron at that stage is i 2p and it moves at an infinite speed because E = 0 and the wave fronts in a spacetime diagram "move" horizontally.

The electron emits a 511 keV photon again and the wave is

       (1, 1) exp(-i (-E t - p x)).

That is, the electron has become a positron. In the annihilation, the electron became a tachyon in the intermediate stage. If you need to turn back in time, it is not that strange to be a tachyon during one line segment.

If we flip the roles of t and x, the diagram above shows an electron in an elastic collision with a 511 keV photon.

Annihilation allows superluminal communication. That may be ok if we cannot measure the speed accurately enough to prove that superluminal communication took place. If the electron and the positron collide with a high energy, then the superluminal phase has to be shorter.

We may need a superluminal phase to make the wave function smooth in the spacetime diagram.

A basic principle of special relativity is that time and space are similar entities. Allowing superluminal travel makes them even more similar, because the path of a particle is not confined inside the light cone. Allowing traveling back in time still improves the similarity.

The diagram above suggests that at short distances, time and space become interchangeable, they have an "euclidian metric". But why at macroscopic scales we are bound inside the light cone and time and space most definitely cannot change places?

Maybe at short distances, the uncertainty of quantum mechanics allows particles to "tunnel" back in time. This might be the explanation for zitterbewegung. If we paint a fuzzy line of width one Compton wavelength to the the spacetime diagram above, it looks quite natural to allow the electron path zigzag not just in space but also in time. During the trips back in time it is a positron.

Thus, the electron is allowed to make superluminal trips as well as trips back in time as long as it stays inside the fuzzy path whose width is one Compton wavelength 2 * 10^-12 m or the corresponding time, 7 * 10^-21 s.

The annihilation case does not tell us what the electron does in the vicinity of a proton. The process might be an uneventful superluminal trip.

Note that in a head-on collision with a superluminal particle, the proton cannot exchange momentum with the particle. This is because the proton does not have time to react before the collision is already over. The potential of the proton will appear static to the colliding particle, that is, it stays in the same place as if the proton were infinitely heavy. If the superluminal particle passes from the side, then the proton will exchange momentum with it.