We solved the Klein paradox

UPDATE Nov 12, 2018: see our potential step calculations from the Nov 12, 2018 post.

...

Theorem 1. The correct energy-momentum relation under a potential is

         E^2 = p^2 + (m + V)^2.

That is, the potential energy is added to the rest mass.

Proof. See our example at the end of our previous blog post, of a man living on a plane of negative charge. If the electron in his hand would not behave like an object with a rest mass m + V, then the conservation of the center of mass would be broken. If we assume that very basic conservation law of newtonian mechanics, the potential energy has to be added to the rest mass.

Also, the rest mass of the electron is actually potential energy that it acquired in pair creation. QED.


The Dirac equation which we introduced in the previous blog post should give reasonable results when a flux of electrons hits a step potential. For now, we omit the term V Ψ. We calculate the momentum after a potential step with the energy-momentum relation and do not use the wave equation in that.

Let us use the 1+1 dimensional equation. Let there be a step potential which is 0 if x < 0 and V if x > 0. Let a flux of electrons hit the step from the left. The rest mass of the electron is m.

A solution to the Dirac equation is

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

Let us then solve the step potential exactly. We assume that the wave function at x < 0 is a sum

        A Ψ(p, m) + B Ψ(-p, m),

where A and B are complex numbers. The incoming flux is A and the reflected flux is B.

The transmitted flux is

       C Ψ(q, m + V),

where C is complex and q may be real or imaginary.

The energy-momentum relation for the incoming flux is

        E^2 = p^2 + m^2,

and for the transmitted flux

        E^2 = q^2 + (m + V)^2.

We may assume A = 1. The continuity of component 1 of the wave function at 0 requires:

       1 + B = C,

and component 2:

      p / (E + m) - B p / (E + m)
      = C q / (E + m + V)
      =>
      1 - B = C q (E + m) / (p (E + m + V)).

Let us denote C = 1 + C'. We have:

       1 + B = 1 + C'

       1 - B = (1 + C') (q / p) (E + m) / (E + m + V).

Let us denote
 
       1 - 2 R = (q / p) (E + m) / (E + m + V).

R comes from "Reflection". If V is small and positive, then R is small and positive.

We have  B = C' and

        1 - C' = (1 + C') (1 - 2 R)
        =>
        2 = (1 + C') (2 - 2 R)
        =>
        1 + C' = 1 / (1 - R)
        =>
        C' = R / (1 - R).

To summarize, the reflected flux is

        B = R / (1 - R),

and the transmitted flux is

        C =1 + R / (1 - R).

What if we require dΨ/dx to be continuous?

If the potential V(x) is continuous, then it might make sense to require our wave function to have a continuous derivative in space. Let us see what equations we get for A, B, C, where A = 1.

Component 1 gives:

          p (1 - B) = C q.

Component 2 gives:

          p^2 (1 + B) / (E + m)
                              = C q^2 / (E + m + dV),

where we assume that dV is close to zero. Let us again denote C = 1 + C'.

          p (1 - B) = q (1 + C'),

          (1 + B) p^2 / (E + m)
       = (1 + C') q^2 / (E + m + dV).

Continuity of the wave function earlier produced equations:

         1 + B = 1 + C',

         1 - B = (1 + C')
                    * (q / p) (E + m) / (E + m + dV).

Thus, B = C'. Let us see if the equations yield anything sensible.

         q = p (1 - B) / (1 + B),

         q^2 = p^2  (1 - dV / (E + m)),

         q = p (1 - B) / (1 + B).

In the two last equations we used the fact that dV is close to zero. The energy-momentum relation gives:

         q^2 = p^2 + m^2 - (m + dV)^2
                = p^2 (1 - 2m dV)

The energy-momentum relation, of course, gives a sensible result. On the other hand, the requirement that the derivative of the second component of the wave function be continuous produces a nonsensical result. If we used the old Dirac equation, then this continuity requirement would imply q = p, which does not make sense, either.

Destructive interference in wave equations

Suppose that the potential step is rectangular:
             ____
   _____|       |____

If the width of the step is suitable, then the reflections from each end of the rectangle will have a destructive interference. Then the reflected flux is zero, and transmission is 100%. This phenomenon is used in antireflective coatings of eyeglasses.

But how do we describe this in a linear wave equation? What mechanism causes the transmitted flux to increase if there is a destructive interference at the other end of the barrier? The next blog post concerns this fundamental problem.

The Klein paradox

We also need to study what happens when V goes to infinity. The energy-momentum relation gives then

         q = i V.

A wide rectangular barrier will let through an infinitesimal flux because q is imaginary. That is a consequence of our new energy-momentum relation where we add the potential energy to the rest mass.

The physical setup of a discontinuous potential is not realistic. We should calculate the result with a steep continuous potential.

https://www.hep.phy.cam.ac.uk/theory/webber/GFT/gft_handout2_06.pdf

The Klein paradox for the Klein-Gordon equation matches the wave function value and the spatial derivative at the barrier borders.

Our new energy-momentum relation does remove the Klein paradox there, because the momentum inside barrier is imaginary, while it is real for the traditional energy-momentum relation.

Old "solutions" to the Klein paradox claimed that a pair is created at the other end of the wall and that a positron mysteriously finds the incoming electron and annihilates it. The positron is bound inside the strong field of the barrier. Why and how would it find an incoming electron?

Our solution says that the transmission is, indeed, very small. The Dirac equation and the Klein-Gordon equation do not describe pair creation. There should be no tunneling by pair creation if we solve the problem using those equations.

A new Dirac equation with a potential

We uncovered a new problem in our study of zitterbewegung. See our blog post on Oct 14, 2018.

The problem is a fundamental one and deserves a blog post of its own: how to add a potential term to the Dirac equation?

http://rspa.royalsocietypublishing.org/content/117/778/610

Dirac himself, in his famous 1928 article, added the potential energy of the electron as a simple term V Ψ to his equation. His choice is, at least in some cases, equivalent to subtracting the potential energy from the total energy of the electron. But that is a strange choice because the potential energy "moves" along with the electron, and should be added to the rest mass of the electron.

Intuitively, an electron under a constant potential should act like a free electron.

How is the matter in the Schrödinger equation? A particle which moves to a lower potential zone gains kinetic energy and acts like a free particle with more kinetic energy. Since the Schrödinger equation is nonrelativistic, we do not need to add the (small) potential energy to the rest mass of the particle.

In the Dirac equation, we may have something like

        dE / dx = -dV(x) / dx,

        d/dx sqrt(p(x)^2 + m^2) = -dV(x) / dx,

        2 p(x) dp(x) / dx
        * 1/2 * 1/sqrt(...) = -dV(x) / dx.

We get a complicated formula for d^2 Ψ / dx^2, that is, dp(x) / dx.

        dp(x) / dx = -dV(x) / dx
                            * sqrt(p(x)^2 + m^2)
                            / p(x).

If m = 0, we get simply dp(x) / dx = -dV(x) / dx, which implies p(x) = -V(x) + C.

The massless case m = 0

The Dirac equation with m = 0 is

        α * -i dΨ/dx = i dΨ/dt,

or in the component form:

       -i dΨ_2/dx = i dΨ_1/dt
       -i dΨ_1/dx = i dΨ_2/dt.

Let us try a solution where Ψ_1 = exp(-i (E t - p x))

                    dΨ_2/dx = i E exp(-i (E t - p x))
  -i p exp(-i (E t - p x)) = dΨ_2/dt.

We get

         Ψ_2 = i E / (i p) exp(-i (E t - p x))
         Ψ_2 = i p / (i E) exp(-i (E t - p x)).

This implies E^2 = p^2 => E = +-p.

Thus, the solution is:

          Ψ_1 = Ψ_2 = exp(-i E (t - x))
or
                 
          Ψ_1 = exp(-i E (t + x))
          Ψ_2 = -exp(-i E (t + x)).

For a light-speed particle, E = |p|, assuming c = 1.

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

The Wikipedia article uses these two solutions to calculate the reflection and the transmission at a potential step. In the massless case, the equation

             α p Ψ + VΨ = i dΨ/dt

does yield a somewhat reasonable solution, but it produces the Klein paradox.

Should we rotate the phase of the wave function?

Suppose that a single electron approaches a step potential at x = x_0. If we represent the electron as a wave of a type

              exp(-i (E t - p x)),

there is no way we can make a continuous wave function by gluing together that wave with

             exp(-i (E' t - p' x))

at x_0. Should we allow a discontinuous wave function? Or is there a total reflection of the wave?

Since the phase of the wave function is not an observable, it does little harm to have a discontinuity of the phase at x_0. Suppose that we rotate the phase of the wave function in the way that the phase is always 0 at x = 0, regardless of t. Then we have a function

           exp(-i p x),

which we can easily glue together with a function

           exp(-i p' x)

at x_0, and we obtain a continuous wave function. But does this make sense? Let us consider a more realistic setup.

A path integral approach

     e- ----->
                    \     |
                      \   |
proton O           x
                      /   |
      e- ------>   
                           screen

The 1+1-dimensional case is too restricted. Let us look at a 1+2-dimensional case. Suppose that there is a static 1 / r potential of a proton. A flux of electrons hits it and forms an interference pattern on a screen. We should calculate the interference pattern.

If the flux consists of electrons of a fixed momentum p, then there are just 2 paths through which an electron can end up at a point x on the screen. We can calculate the phase of each of these 2 paths and their interference.

From this case we see that we are not allowed to rotate the phase of the incoming flux arbitrarily at different spacetime points. It would spoil the interference pattern.

Does the above diagram give us a clue how to add a potential to the Dirac equation?

The two paths to the point x are classical in the sense that we know the initial momentum p of the electron exactly and we can measure its position at x with an arbitrary precision. It is like a classical particle with a precisely defined rotating phase hitting the screen. Is there any sense in trying to derive a differential equation which describes the phase at x? The phase is a function of two classical paths. It does not matter what happens outside those paths, as long as no new paths to x appear.

Wave equations describe a smooth process where what happens at x is determined in a smooth way from what happens in its surroundings.

We may imagine that the flux which hits the proton is just a single electron, whose momentum we know exactly, but whose position has a huge uncertainty. Then it is obvious that the different paths of the electron are independent. The paths do not interact in any way. This suggests that there is really no sense in describing the process as a wave equation. Why does the Dirac equation work then?

A potential step

Let us consider in 1+1 dimensions the case where a flux of electrons of momentum p hits a low potential step at x =0, of height V.

 ^ t           X

 |          |  /  

 |          | /

            |/

           /|

         /  |

       /    |

     /      |

   e-

            0             ----> x       

            step         

                        

The flux of free electrons, or a single free electron whose position we do not know, is described by the 2-vector

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

where E^2 = p^2 + m^2. How does the electron behave once it has crossed the potential step? Its kinetic energy decreases by V. A crucial question is what happens to E?

Only one path leads to X. If we translate the path upward by Δt, the phase of the start of the path advances by E Δt. The end of the path is a spacetime point X'. How much does the phase advance in X' relative to X?

If we assume that the phase of the electron rotates at a fixed speed per distance after crossing 0, then the phase in X' relative to X has advanced the same E Δt as at the start of the path.

Even though p^2 + m^2 is reduced after the potential step, the potential energy V seems to make for it. Thus, we may write:

           E = sqrt(p^2 + m^2) + V.

What is the phase of the electron with respect to x, if x > 0? In the Schrödinger equation, it is determined by p x, where p is the momentum. We may guess that the same holds for the Dirac case.

The phase in the area x > 0 would then be:

           exp(-i (E t - p x)),

where

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

Let us try to add the potential term to the Dirac equation in the same way as in the Schrödinger equation:

          α * -i dΨ/dx + β m Ψ + VΨ = i dΨ/dt.

Let us try a solution exp(-i (E t - p x)). From our zitterbewegung blog post, we get the formula

         p^2 / (E + m) + m + V = E,

         p^2 + m^2 + VE + Vm = E^2,

         p^2 = E^2 - m^2 - VE + Vm.
 
The formula does not look right. The relation between E, m, p, V is complicated.

Has anyone found a natural way to add V to the Dirac equation? Maybe we should look at the paper by Oskar Klein from 1929, which introduces the Klein paradox.

Add the potential energy to the rest mass of the electron?

                e-

            O/
           /|
            /\
  =======================
   -  -  -  -   -  -   -   -  -   -  -  -  -  -
           x                        y

Suppose that we have a fixed planar negative charge. An inhabitant living on the plane pulls an electron close to the plane. The electron acquires a potential energy V. Does it then behave like a particle whose rest mass is m + V, where m is the rest mass of the electron in empty space?

In the pulling, the inhabitant can use an energy store of size V at location x. The inhabitant carries the electron to location y and fills another energy store there with energy V by letting the electron move away.

In the process, energy m + V moved from x to y.

The conservation of the center of mass requires that the electron has to behave just like a particle with a rest mass of m + V. Otherwise the center of mass of the system the plane & the electron would change.

This suggests that we should define

          m(x) = m + V(x).

The Dirac equation would then be

        α * -i dΨ/dx + β (m + V(x)) Ψ = i dΨ/dt.

That would make a lot of sense, because if we move the electron under a potential V > 0, we move an inertial mass which would imply a higher rest mass.

We dropped the term V(x) Ψ because the purpose of that term was to subtract the potential energy from the total energy. We add the potential energy to the rest mass, and the total energy remains the same.

The rest mass of the electron itself can be considered as potential energy which it gains when it in pair production is separated from the positron.

But what if the electron descends into a negative potential whose absolute value exceeds its rest mass? If the electron would be able to give up its kinetic energy in that case, there would be a breach of energy conservation.

In the plane example, the inhabitant can let a positron approach the plane and fill an energy store V at location x. We assume V > m. Then he carries the positron to y where he uses an energy store V to push the positron away. Energy V moved y -> x and energy m moved x -> y. The positron should have a negative rest mass in this process, which would be an utterly strange experience for the inhabitant.

The energy-momentum relation suggests that the mass becomes imaginary if the kinetic energy exceeds the rest mass of a particle. The particle is then a tachyon.

The proton decays because positive frequencies slowly turn to negative

Our blog post about charge nonconservation predicts proton decay. A quark will eventually turn into an antiquark, because its positive frequencies slowly turn to negative. There are 3 quarks in a proton plus a number of gluons, and the interaction is not strictly between two particles. There are random accelerations, which necessarily produce negative frequencies.

The radius of the proton is 9 * 10^-16 m.

Quark masses are only 2 or 5 MeV, while the mass of the proton is 938 MeV.

The Compton wavelength of a quark is around 2 * 10^-13 m. Most of the proton mass comes from gluons. If we assign the proton mass evenly to the three quarks, the Compton wavelength of a quark is 4 * 10^-15 m.

Since a quark is contained within 2 * 10^-15 m, its wavelength inside proton potential well has to be around 2 * 10^-15 m or less.

A quark will bump into the potential wall surrounding the proton at most some 10^24 times per second if it moves at the speed of light.

If the gluon cloud inside the proton is essentially independent of the position of an individual quark, then the quark will move in an essentially static potential and will not get much negative frequencies per bump.

Can we somehow estimate the production rate of negative frequencies?

Let us first calculate the ballpark figure what magnitude of acceleration "significantly" deforms a single wavelength of a reflected wave. Then there should be a significant negative frequency component in the reflection.

The acceleration must be such that it will move the mirror almost a wavelength during one cycle of the wave.

                    1/2 a t^2 = λ,

where t = λ / c. We get

                    a = 2 c^2 / λ.

Conversely, we get λ =  2 c^2 / a. Let us calculate the black body temperature associated with acceleration a. Wien's displacement law gives

                     b / T = 2 c^2 / a

                     T = 1/2 b a / c^2 = 1.5 * 10^-20 a

kelvins. The Unruh temperature is 0.4 *10^-21 a kelvins. The coefficient is in the same ballpark as our 1.5 * 10^-20.

What is the acceleration when a quark bounces back? There is asymptotic freedom inside the proton, and the bounce happens from a steep potential wall. The process will take around 10^-24 seconds and the speed change is 6 * 10^8 m/s. The acceleration is 6 * 10^32 m/s^2. The Unruh temperature is T = 2 * 10^11 K, which corresponds to a wavelength of 10^-14 m.

Unruh calculated that the (nonexistent) Unruh radiation has approximately the black body spectrum. Let us make an educated guess and conjecture that the portion of negative frequencies produced in a reflection goes like the black body spectrum.

If the proton would consist just of three quarks, then each reflection of a quark would produce a significant portion of negative frequencies, because there would be a significant amount of random acceleration a in each reflection. But since the mass of an individual quark is just 1 / 200 of the proton mass, the random acceleration is just 1 / 200 of a. The temperature is thus 1 / 200 of the temperature.

Planck's law says that black body radiation is proportional to

            f^3 / (exp(h f / (k T)) - 1),

where f is the frequency, h is the Planck constant and k is the Boltzmann constant. The maximum is when h f / k T is roughly 5. In our quark case, the frequency of the quark was 5X the frequency of the maximum Unruh effect. Thus, h f / (k T) might be 25. When we drop T by a factor 200, the ratio is 5,000. We conclude that the flux of negative frequencies is not significant by a factor 10^2,500. The proton lifetime is in the ballpark of 10^2,500 years.

How can the Compton wavelength of a quark be much bigger than the proton radius?

There is a problem in the proton model. If the Compton wavelength of a quark, or 1 / 3 of the proton mass, is much larger than the proton radius, how can it be localized inside the 9 *10^-14 m radius of the proton? Maybe the particle is in a potential well, and its wavelength inside that well is shorter than it would be as a free particle. Localization inside 10^-15 m suggests that the energy of each particle with respect to the floor of the potential well is actually in the GeV range.

If we try to pull a quark out of a proton, then according to the rubber band model of QCD, the rubber band which pulls the quark to the proton will eventually break, and two hadrons will be formed. This means that each quark inside the proton really is sitting in a potential well whose depth is in the GeV range, if we set the zero potential at the rubber band break position. But why should we set the zero level there?

An explanation for the matter-antimatter asymmetry in the visible universe

Our previous blog post raised the question of baryogenesis where matter for an unknown reason is more abundant than antimatter. How is this possible if the laws of nature are perfectly symmetric for matter and antimatter?

Theorem 1. Assume that a cooling primordial soup after the Big Bang favors creation of new baryons, that is, baryons become more abundant when the soup cools.

Assume that there exists an unknown attractive force which pulls together matter baryons, and pulls together also antimatter baryons.

Then there will form areas where matter baryons dominate and areas where antimatter baryons dominate. This is spontaneous symmetry breaking, or in a sense, crystallization of the soup.

Proof. Because of the unknown attractive force, distributions where matter baryons and antimatter baryons are separated have a lower potential than mixed distributions. When the soup cools it favors a lower potential.

It is like cooling of water where two types of ice can form, but the types cannot coexist close to each other. When the water cools, ice must form. Since the ice cannot be mixed, the two types of ice have to appear as separate "crystals". QED.


We do not know how to calculate the crystallization process. We seem to live inside one huge crystal.

Crystallization takes a long time. Is this compatible with a fast Big Bang? If we assume extremely heavy particles which can decay into either matter or antimatter, the crystallization might happen very quickly. The visible universe is then the decay product of one or more such superheavy particles.

Actually, we do not need Theorem 1 at all if we assume a superheavy particle which can decay into an entire visible universe.

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

Andrei Sakharov in 1967 wrote three "necessary" conditions for baryogenesis. It turns out the first two conditions are not necessary. There is no need for baryon number violation or any symmetry violation. The crystallization process causes apparent symmetry breaking.

The third condition of Sakharov is that interactions are out of thermal equilibrium. The third condition is trivial: if something is "born", then the system must have been out of equilibrium.

Lumping together even when there is no attraction

Even if there is no unknown attraction between matter baryons, there will be some lumping together of matter. The following toy model clarifies the thing.

Suppose that we have n baskets and a random flux of baryons and antibaryons which fall in them. We assume that in each basket, antiparticles immediately annihilate. In each basket, the number

       N_baryons - N_antibaryons

will perform a random walk. After n particles, its distribution is

       2 * B(n, 1/2) - n,

where B is the binomial distribution.

The typical number of particles in a basket grows like sqrt(n) after n particles have fallen into a basket.

We had to assume that individual particles choose their basket at random. If a produced pair baryon-antibaryon would always fall in the same basket, then there would be no lumping together. Since antiparticles attract each other, the cross section for annihilation is bigger for slow particles. If the particles in a pair are born with a high kinetic energy, they will often fall in different baskets.

Lumping together is reduced by the motion of particles from a basket to an adjacent basket.

Electric charge is not conserved in reflection from an accelerating potential

UPDATE: Our post on Oct 23, 2018 about proton decay contains a numerical estimate of how quickly quarks will turn into antiparticles because of reflections inside the proton: in 10^2,500 years.

http://meta-phys-thoughts.blogspot.com/2018/10/the-proton-decays-because-positive.html

...

The spider model of an accelerating mirror raised an interesting question. If reflection converts right-hand waves to a mixture of right-hand and left-hand waves, is it possible that electric charge is not conserved?

Or is the flux of positrons balanced by a flux of new electrons from the acceleration motor?

A symmetric collision of two electrons does not produce a chirp wave. The reduced mass method which is used in two-particle collisions transforms the problem to a single charge moving in a static electric field. Reflection from a static field does not produce a chirp.

If there is an inertial coordinate system where the electron appears to move in a static field, then there is no chirp.

The accelerating mirror has to be propelled by something else than the collision of just 2 particles to uncover phenomena that is associated with chirps.

The mathematical question is: if a flux of positive frequency waves is reflected from an accelerating mirror, does that create negative frequency waves which are not balanced by an equal amount of new positive frequency waves? The wave equation in question is the Dirac equation.

                               p
           e- ------------------------>                | <---- accelerating potential

Let us assume that we have a flux of electrons of momentum p hitting an accelerating potential. It is like a mirror. An observer sitting on the mirror will see some of the electrons as positrons of various momenta. This is because the electron wave will appear as a chirp, and a chirp contains negative frequencies.

Let us draw a spacetime diagram.

t
^              ^
|                 \
|                    \    \
|                      ^     \
|                    /         |
|                  /            |
|                /              |  mirror
|              e-   p
|
----------------------------------> x

It would be a little miracle if a wave can reflect from an accelerating mirror without acquiring negative frequencies. But is it possible that an equal amount of positrons and electrons are produced in the complicated reflection?

If we again think of the spider, it will mostly produce right-hand waves and a little flux of left-hand waves. Since the spider is accelerating, also the right-hand waves contain negative frequencies in the laboratory frame. If the spider thinks that it produces a flux F of left-hand waves, the observer in the laboratory frame will think it produces a flux 2F. Is there any way to show that the left-hand waves are not balanced by an equal amount of new right-hand waves?

Let us think of a simpler example. First the mirror is static and then suddenly starts to move at some speed v.

The question is what is the Fourier decomposition if the wave

        exp(i (E t - p x))

is changed to exp(i (E' t - p' x)) at a certain point.

If we have an arbitrary point z in spacetime, it will receive the first flux in a time interval t_1 to t_2. It will receive the second flux from some time t_3 to t_4. What is the Fourier decomposition of the flux that z sees?

The integral of
   
       exp(i (E t - p x)) exp(i (-E'' t - q x))

over an interval will differ a little bit from zero for most combinations of E'', q. Thus, there is almost always some negative frequency flux mixed with positive frequencies. Almost always is an electron mixed with a little bit of positron.

It is hard to analyze a single reflection, but by iterating the process we found a proof of charge nonconservation. The process of converting electrons to positrons is extremely slow: the flux is comparable to the (nonexistent) Unruh radiation. But in principle, we can change the charge, if the wave equation of the electron is not very exotic.

Theorem 1. Electric charge is not conserved if we allow arbitrary accelerating motion for electric potentials. This holds for the Dirac equation and most wave equations.

Proof. Suppose that we have a vessel where a mirror performs oscillation.

                     motor
   -------------- /--------
  |               /            |  vessel
  |       <--  |  -->       |
  |                            |
   ------------------------

Let us put an electron into the vessel. It will bounce from the accelerating or decelerating mirror many times. Most wave equations will create some negative frequencies from a positive frequency flux which bounces from an accelerating mirror. That is, a fixed frequency wave will reflect as a chirp. A chirp contains negative frequencies.

The wave function of the electron will slowly grow into a mixture of 50% electron and 50% positron. Does that prove that charge is not conserved? The electron may by accident gain so much energy that a pair is created. The pair may get annihilated later.  Probably there is a steady state where the number of electrons or positrons does not grow any more. Let this number be N.

What if we initially put N electrons into the vessel? After some time we will have N / 2 electrons and N / 2 positrons. Charge is not conserved.

There is still the possibility that the driving motor of the oscillating mirror might produce electrons or positrons that balance the created charges. But how would the motor know if we are turning electrons into positrons or the other way around inside the vessel? Thus, the motor cannot save conservation of charges. QED.


Theorem 1 is in conflict with the hole theory of Dirac. If the positron were just a hole in the Dirac sea of electrons, charge would be conserved. Paul Dirac apparently did not notice that his wave equation is not compatible with the hole theory.

Corollary 2. It is possible to change the baryon number, the electric charge, or any quantum number of a system where particles are subject to arbitrary irregular accelerations. QED.

Fermions turn into a boson soup?

Corollary 2 opens the possibility that all falling matter is reflected back from a black hole, according to optical gravity. We had the problem that baryons cannot be reflected back. But if there is intense heat in the falling matter, particles will feel intense random acceleration. Corollary 2 implies that some baryons turn into antibaryons and annihilate. The output from such a hot soup would be bosons, like photons. Zero rest mass bosons can always reflect back from the forming horizon.

But are falling baryons close to the forming horizon subject to intense heat for a long time? They will bathe in the heat of the reflecting bosons.

Noether's theorem and conservation of charge

On the Internet there are vague claims that the U(1) gauge symmetry of the QED lagrangian implies the conservation of electric charge through Noether's theorem. The U(1) symmetry means that the phase of the wave function can be rotated by an arbitrary amount, locally, and the physics does not change.

https://en.m.wikipedia.org/wiki/Noether%27s_theorem

Noether's theorem looks exact, at least in classical physics where we can assume an infinite signal speed. In classical physics, there is no paradox if positive frequency waves turn into negative frequency waves. Noether's theorem cannot forbid it.

It seems that on the Internet no one claims having an exact proof for charge conservation in QED, through Noether's theorem or otherwise.

The free Dirac equation apparently conserves the number of particles, and consequently, also charge. But we are interested in interacting systems.

In Feynman's diagrams, charge conservation is enforced. But Feynman's diagrams are just a perturbative approximation of the physical process.

Can we write wave equations which would satisfy a charge conservation law? It looks difficult or impossible to block the leak from positive frequencies to negative frequencies.

Matter-antimatter imbalance in the universe

We found a way to turn particles to antiparticles. Can this in some way explain why there seems to be much more matter than antimatter?

The process is completely symmetric. If there is a Big Crunch, the process may equalize the amount of matter and antimatter.

There is one eminent asymmetry in the universe. Entropy is increasing with time. Could that somehow create a preferred handedness? Matter is somehow associated with low entropy?

If the universe were a giant vortex, there would be a natural handedness. But why would that handedness be associated with matter rather than antimatter?

The explanation might be spontaneous symmetry breaking. The universe forms a "crystal" and in that crystal, matter is more abundant than antimatter. In the adjacent crystal it may be the other way around.

For the symmetry to break, matter should in some way favor matter, for example, by having a stronger attracting force to it. Antimatter would favor antimatter. This might be a result of handedness. Right-handed particles prefer right-handed?

UPDATE: Look at our next blog post for the answer.

Charge non-conservation and Feynman diagrams

Feynman diagrams enforce charge conservation. An electron cannot change to a positron on-the-fly, and they are always created in pairs. How can we reconcile the charge non-conservation with Feynman diagrams?

An electron wave packet can turn into a chirp in a collision with two electrons.

       e-  --------------------------------->
                        |       |          |          |
       e-  ------------------------------------->
                           |        |          |
       e-  -------------------------------->

The vertical lines mark virtual photons. The middle electron receives various momentum impulses from the other two. Since the end result is the middle electron being a chirp wave, there is a very slight chance that a positron comes out from the process. How do we model that in a Feynman diagram?

In a Feynman diagram, an electron line contains a definite momentum p. The waveform cannot be a chirp. The electron in a Feynman diagram is a classical point particle which can have a sharp momentum.

A diagram of type:

        e-
     /      \
   /          \
e- _____  e-

can symbolize the simultaneous effect of three electric fields. There would be a very small probability that an electron continues from the diagram as a positron.

If we conjecture that a positron is a negative energy electron traveling backward in time, then the conversion of an electron to positron is very natural. The electron brings to the spacetime point positive energy and the positron negative energy. Since the sum of energies is zero, they do not need to emit photons. The conversion, in a sense, is the true annihilation. The normal annihilation produces photons because there is surplus of energy at the spacetime point.

Thus, the conversion from an electron to a positron is in the spirit of Feynman diagrams, after all.

Is the electron conversion the only true annihilation? Suppose that we have a normal annihilation where the positron has the exact opposite energy to the electron. But such a process is equivalent to the electron simply moving in space.

Is the electric charge of a positron the same as of an electron?

It would be somewhat strange if the electric charge of an electron would change sign in a negative frequency component of its wave function. But is the electric charge of a positron actually the same negative elementary charge as of the electron? 

Let us think of the positron as a negative energy electron traveling back in time. Then a repelling impulse from an electron will not move the positron further away, but will attract the positron. This is because the mass of the positron is negative. A momentum pointing to the right will actually move the positron to the left.

In this interpretation, all electric charges are negative, but some particles are negative energy particles traveling back in time. If the electric charge is carried by such a particle, then it behaves like a positive electric charge.

Electric charge is not conserved in this interpretation. Pair production adds two elementary charges.

What about gravity? Why would gravity attract a negative energy particle traveling back in time? It should rather repel it. What do we know of gravity and different elementary particles?

We know that there is a gravitational pull between protons and neutrons, as well as atoms. Different atoms have a different composition of protons, neutrons, and electrons. Do we know that the inertial and gravitational mass are equal for different types of atoms? Apparently, there was a precision measurement in 1964.

Maybe the Dirac wave equation does not describe the electron and the positron correctly from the point of view of gravity? There are no negative energy particles traveling backward in time, as far as gravity is concerned.

Path integrals, Schrödinger's cat, and backward causation

Conjecture 5 of our Huygens post deserves a blog post of its own.

We claim that it does not make sense to calculate the sums of path integrals at an intermediate spacetime point. Only the path integral to the measuring apparatus matters.

In our blog post of Bell's theorem, we observed that we must not assume a wave function collapse at an intermediate stage, before the measurements are analyzed in the head of the scientist. If the collapse happens, for example, in the measuring apparatuses, then wave function information is lost, and the result is not compatible with a local version of quantum mechanics.

It is the interference pattern of wave functions which finally collapses in the head of the scientist.

John Stewart Bell proved that if we assume a collapse at two measuring apparatuses far apart from each other, then quantum mechanics is not compatible with local realism. Superluminal communication would be needed.

Albert Einstein was not satisfied with this "spooky action at a distance".

Conjecture 1. The correct quantum mechanics is a local version where the wave function only collapses inside the head of exactly one scientist.


Assuming the collapse in the heads of two scientists would require the spooky action at a distance. The collapse in just one head is a rather local thing. This may solve the problem of spooky action. Einstein's second problem "God does not play dice" remains. We need to think about the David Bohm model again.

Let us repeat our earlier Conjecture 5 in a different form here:

Conjecture 2. Quantum mechanics is defined by path integrals. It only makes sense to sum the amplitudes in the head of the one scientist. Any sum at an intermediate stage of an experiment is nonsensical.


In most cases we can calculate sums at intermediate stages, and that does help us in the mathematical computation. But there is no physical meaning in these sums.

According to Conjecture 2, the different paths do not exist in the same spacetime. There is no sense in summing them at a certain spacetime point. If we have two different planets, there is no sense in summing the temperatures at 30 degrees east, 60 degrees north, at 12 noon on the two planets.

Intermediate sums may diverge. It is no problem, according to Conjecture 2. A diverging sum of amplitudes happens in the middle of a virtual loop. One cannot meaningfully sum all waves of arbitrarily high momenta.

Claiming that Schrödinger's cat is both alive and dead in the box is talking about an intermediate stage in an experiment. Conjecture 2 says such talk is nonsense.

According to Feynman, a potential can cause an electron to scatter "back in time". That would be backward causation.

Having a loop in the path imposes an extra condition on the path: in the spirit of the vacuum atom, there must be an integer number of wavelengths in the path. It is like the future constraining the past. But another way to interpret this is that a particle and an antiparticle are emitted, and they are only allowed to annihilate when the wavelengths match. Then there would be no backward causation.

http://meta-phys-thoughts.blogspot.com/2018/10/how-to-renormalize-gravity-hawking.html

What about our claim in an earlier post (Conjecture 1 there) that all fermions will eventually scatter back in time? Surely a real electron cannot scatter back in time if we have in the experiment made sure that no positrons can enter. But can we make that sure? We need to perform measurements on the test system. Those measurements might produce positrons.

What if we have a universe where there are no positrons? Then electrons cannot scatter back in time. The wave equation of the electron would be linear then? This reminds us of the free electron. If there are no other particles around, we believe that the electron moves under a totally linear equation and cannot scatter back.

Is there a chicken and egg problem here? If we have a scientist at a location in spacetime, he may measure whether electrons tend to scatter back in time. If they do, he concludes there are a lot of positrons flying around. But if they do not? Then he concludes there are few positrons. Our own universe seems to lack antimatter.

Our conjecture that all fermions will reflect back was probably wrong. It was based on the claim that QED is nonlinear for the electron. But QED seems to be linear for the electron if there are no other particles around. This is consistent with the path integral approach. Only real particles which enter the experiment have an influence. There are no virtual particles popping out from space.

If the wave equation of the electron is linear in the absence of other particles, how can we reconcile this with the fact that the wave of the electron is coupled to the electromagnetic field, and the electromagnetic field is coupled to the electron field, making the wave equation of the electromagnetic field nonlinear? The solution is that an electron is not coupled to the electromagnetic field in the absence of other particles. This is consistent with our claim that the particles "create their own environment". There are no fields or vacuum fluctuations in spacetime which is devoid of particles.

Conjecture 3. The linearity of the wave equation of a particle depends on the existence of other particles in the neighborhood. If there are no other particles, the equation is linear, which means that the particle is not coupled to any other field.


The famous problem of the infinite energy of vacuum fluctuations would be solved with the following conjecture:

Conjecture 4. All fluctuations of fields are causally created by the real particles in the neighborhood, and the total energy of those fluctuations is the energy of the particles. There is no "vacuum energy" without particles.


A first step in proving Conjecture 4 is to show that Feynman diagrams do not require one to assume any vacuum fluctuations. But there might be "non-perturbative" effects of vacuum fluctuations. We will analyze the Casimir effect later. Robert Jaffe in his 2005 paper claimed that the effect does not have anything to do with vacuum fluctuations.

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

Our discussion above suggests that Feynman was wrong in speaking about electrons scattered back in time. There is no backward causation unless positrons are around, and in that case it is really not scattering back in time but annihilation.

Classical reflection from an accelerating mirror

Our blog post about a time mirror a few days back was based on a backward causation model. It seems to be wrong. If we let a laser beam reflect from an accelerating mirror, how do we explain the negative frequencies which appear in the reflected beam? It is a classical process. We should not need quantum mechanics in explaining this.

     wave ---->
   -----------------------|------------------
                      fixed point
   
Suppose that we have a string attached to a fixed point. We rotate the string with our hand to generate a circular right-hand wave. Since the end of a string is attached to a fixed point, we need to imagine a right-hand wave approaching from the opposite direction (reflection wave), to cancel out the circular wave at the fixed point, and keep it fixed. There is also a phase shift of 180 degrees in the reflection wave.

What if the fixed point is accelerating to the left? Then the reflection wave must have a complex waveform, to keep the fixed point fixed. It has to be a chirp, and must contain left-handed component waves. The left-handed waves are the "negative frequencies" that we have blogged about.

From where does the energy come to the negative frequencies? It has to come from the incoming wave or from the fixed point movement.

        ^ string tension
          \
            \
             | ------------> string tension
    fixed point
     
The diagram shows the situation, greatly exaggerated. The sum of forces on the fixed point pulls to the right. In the reflection, there is a small impulse on the fixed point towards the right.

If the fixed point is accelerating to the left, it will feed more and more energy into the reflected wave. It is a blueshift. What is the role of the left-hand waves that the fixed point feeds into the reflection?

One way of modeling the reflection is to assume that a pair of waves is born at the fixed point, like a pair of particles. A left-hand wave travels to the right and "annihilates" the incoming wave. The right-hand wave that is born travels to the left.

If the fixed point is accelerating, then there also are very weak left-hand waves traveling to the left, besides the righ-hand main reflection. If these left-hand waves were born with a right-hand partner, what happened to the partner? Was it reflected from the fixed point? But then it would annihilate the left-hand wave. Strange.

The spider model of pair production; is charge conserved?

        wave                            

       ----->                   <----  //\O/\\ spider

       ------------------------------------------
             string

Let us replace the fixed point with a spider on a string. It will rotate the string with its one leg.

Pair production by a spider. When the spider rotates the string, a right-handed wave is produced in one direction and a left-handed one to the other. This corresponds to pair production in quantum field theory.


The spider may simulate a fixed point in the string by producing a left-handed wave which exactly cancels (annihilates) the incoming right-handed wave. The right-handed wave which the spider produces is the reflected wave.

Reflection by a spider. The spider rotates the string in a way which exactly cancels the incoming wave. The partner wave which the spider produces is the reflection.


What about a reflection from an accelerating fixed point? The spider runs on the string with an accelerating speed. The spider sees the incoming wave as a chirp. A chirp contains both positive and negative frequencies, that is, both right-hand and left-hand waves. To annihilate the incoming wave, the spider must produce both right-hand and left-hand waves.

Reflection from an accelerating mirror by a spider. The spider runs and produces a chirp to both directions. One of the chirps exactly annihilates the incoming wave. The other chirp is the reflection of the incoming wave.


Question 5. Is electric charge conserved when an electron bounces from an accelerating charge?


In the accelerating spider model a purely right-hand wave is converted to a mixture of right-hand and left-hand waves.

A right-hand wave corresponds to a positive frequency, that is, an electron. The spider model suggests that we can create positrons without the accompanied electron. The flux of such positrons is extremely weak. It is like the (nonexistent) Unruh radiation.

The mystery of zitterbewegung

Now that we have some understanding of Feynman diagrams, we may try to solve the mystery of zitterbewegung. In connection with Unruh radiation and Hawking radiation, we have investigated negative frequencies, which a collision with an accelerating object produces.

https://en.wikipedia.org/wiki/Fine_structure#Darwin_term

The Darwin term in the spectrum of hydrogen was derived by Charles Galton Darwin (1887 - 1962), who was a grandson of the famous Charles Darwin. The Darwin term is explained by the "smearing" which zitterbewegung causes in the location of the electron. In an s-orbital, the electron sometimes overlaps the proton in the center. If the electron behaves like a charge cloud of the size of the Compton wavelength, it cannot fall as low in the potential of the proton, as it could if it were a point particle. The energy level of the s orbital is therefore higher than one would expect from a point particle model. The energy difference for state 2s is 9 * 10^-5 eV.

https://en.wikipedia.org/wiki/Foldy–Wouthuysen_transformation

The Darwin term can be derived more rigorously (?) with a technique of Foldy and Wouthuysen.

A free spin-up electron is described in the Dirac equation as a 4-vector

                         1
                         0
          c p_z / (E_p + mc^2)
    c(p_x + ip_y) / (E_p + mc^2)

times

     exp(-i (E t - p • x) / h-bar).

If we have a free electron, then we can build a wave packet from pure positive frequencies, and the packet will not exhibit zitterbewegung. But there is a strange thing: to build a packet of a "typical" waveform, we need to use also negative frequencies. As if the packet would contain some mixture of a positron.

The two upper components in the 4-vector are called the "major" components for an electron and the two lower "minor". But for a relativistic electron, the minor components are not minor at all! Maybe we should accept the fact that the thing which differentiates an electron from a positron is the sign of the frequency in time, not the components of the 4-vector.

Is there zitterbewegung in the hydrogen atom?

Zitterbewegung is a result of a mixture of positive and negative frequencies in a Dirac wave packet.

But, actually, are there any negative frequency components in an electron orbital? If we use a reduced mass for the proton, then the electron moves in a static potential. A static potential does not make a positive frequency wave a chirp after the electron has left the collision area. That is, no negative frequencies are produced if the electron escapes.

If there are negative frequencies, they represent a positron. Why the positron does not escape from the atom? The proton repels it.

The wave is spatially distorted. It looks like a chirp spatially. But that does not make it negative frequency. A chirp in time contains negative frequencies, but not a chirp in space. Though what about different coordinate systems? A spatial chirp does appear as a time chirp to an observer moving at a constant speed. But there is a preferred frame in the hydrogen atom, the center of mass.

Since there are no apparent negative frequencies in the hydrogen atom, zitterbewegung may be a wrong explanation for the Darwin term.

The Foldy-Wouthuysen transformation

The idea in the transformation is to write the Dirac hamiltonian in terms of a new wave function "variable", so that the hamiltonian is diagonalized, that is, the 4 different components in the 4-vector, which are electron spin up/down, positron spin up/down, no longer interact in the hamiltonian.

Hsin-Chia Cheng has a field theory course online, with notes about the Foldy-Wouthuysen transformation:

http://cheng.physics.ucdavis.edu

http://cheng.physics.ucdavis.edu/teaching/230A-s07/rqm5_rev.pdf

The new wave function variable is related to the old through a unitary transformation exp(i S):

        ψ' = exp(i S) ψ,

where S is hermitian. The transformation is not exact. Only a couple of leading terms are used.

Once there is no interaction in the hamiltonian between the components of the 4-vector, we can set the major component 1 and all other components zero. The transformed hamiltonian H' turns out to be the Schrödinger hamiltonian plus some smallish terms. If we have a solution for the Schrödinger equation, we can perturb its wave function so that it is an approximate eigenfunction of H'. In that way we get the perturbed energy levels.

Cheng's manuscript says that the Darwin term is actually of the form

           -e / (8 m^2) * ∇•E.

That is, it depends on the charge density ∇^2 Φ. It is not associated with the singular -1 / r potential of a pointlike charge.

It is not due to the electron charge being "smeared" over an area of size one Compton wavelength. But why does the wrong derivation from zitterbewegung yield the right answer?

What if we take more terms from the Foldy-Wouthuysen formula? Does the series still give the same Darwin term?

The operator p in the Dirac hamiltonian causes the time evolution of the wave function to mix the major and minor components in the 4-vector of the wave function. The transformation has to be chosen such that H' = exp(i S) H exp(-i S) does not cause such mixing in the time evolution of the modified wave function.

It is no wonder that the final H' contains a term ∇^2 Φ. Namely, the operator p applied twice to Φ ψ will produce such a term, among others. Cheng's manuscript contains the detailed calculation.

Let us calculate the electric potential at the surface of the proton.

           V = k e / r
               = 10^10 * 1.6 * 10^-19 / 10^-15
               = 1.6 * 10^6 V.

The term Φe is there 3 * 10^-13 J.

Let us then calculate the Darwin term inside the proton.

            h-bar^2 e / (8m^2 c^2 ε_0)
             * ∇• E
          = 10^-68 * 1.6 * 10^-19
              / (10^-59 * 10^17 * 10^-11)
              * 1.6 * 10^-19 / (4/3 * 3.14 * 10^-45)
          =   10^-8 J.

That is, the "perturbation" in the Darwin term is 30,000 times larger than the major term of the electric potential!

We cannot treat such a huge term as a perturbation. If we could "smear" the proton charge over a radius of 10^-13 m, then it would make sense as a perturbation. But there is no known mechanism to smear the charge of a proton.

https://www.amazon.com/Quantum-Field-Theory-Dover-Physics/dp/0486445682

The treatment of the Darwin term in the textbook of Itzykson and Zuber is essentially identical to the manuscript of Cheng. It contains no comment that the Darwin term cannot really be treated as a perturbation.

We know that treating the electron as a charge cloud of size one Compton wavelength reproduces the experimentally measured Darwin term. But neither the zitterbewegung derivation nor the Foldy-Wouthuysen derivation is mathematically sound. Maybe we need to develop a "regularization" scheme for the central potential well of the proton, to create a robust way of deriving the Darwin term?

A general observation about solutions of wave equations in a potential

We are phenomenally lucky to have an exact solution for the Schrödinger hydrogen atom. Generally, wave equations are too complex to have exact solutions at all. The Schrödinger helium atom does not have an exact solution. The waveform may be, in some sense, chaotic.

Some sources claim that Walter Gordon in 1928 found an exact solution for the Dirac hydrogen atom. But on the Internet there is no mention of Gordon deriving the Darwin term.

https://www.uni-due.de/~sfjooert/downloads/msc_thesis.pdf

The Master's thesis of Johannes Oertel (Freiberg 2014) lists only a few very simple cases where an exact solution for the Dirac equation has been found. Oertel finds several more exact solutions by solving the potential for a given wave function.

The charge singularities inside the proton

The proton contains two up quarks with the charge +2/3 and one down quark with the charge -1/3. There are three electric potential singularities in the proton. In the case of a singularity, it does not make sense to talk about charge density at all.

In the classical Schrödinger solution of the hydrogen atom, the electron moves faster than light if its distance is less than 6 * 10^-15 m from the proton. The classical solution obviously needs fixing.

The solution of the 1s orbital is

          2 * exp(r / a_0) * exp(i E t).

A very simple formula. The potential singularity at r = 0 affects the solution very little.

The classical solution does not bother much about the infinitely deep potential well. But why would the relativistic Dirac equation essentially erase the potential well at distances less than 10^-12 m from the proton?

Our discussion about the vacuum atom and singularities in other blog posts suggests a solution to the Darwin term problem. A path integral solution to the hydrogen atom might have a characteristic energy of 511 keV. The paths have to be relatively straight. They cannot contain abrupt curves. This means that the paths do not see the attractive force of the proton at all when the electron is close to the proton.

The wavelength of the electron Dirac wave in the time direction in spacetime corresponds roughly to 511 keV. Suppose that the electron gains kinetic energy in a deep potential well. Then the wavelength in the space direction might correspond to much more than 511 keV. Then the wave would propagate superluminally because its wavefronts would be close to vertical in a spacetime diagram where time is vertical and space is horizontal.

t
^
|
|
 ------------------> x

Maybe we should use the rest mass plus the kinetic energy to calculate the wavelength in the time direction? But this is against the Schrödinger solution where the wave function is a stationary product of a function in space and exp(i E t).

If an electron moves to a lower potential which is in a wide spatial area, and gains kinetic energy, then we probably should use the rest mass plus the kinetic energy to determine the wavelength in the time direction. The electron Dirac wave would propagate within the light cone then.

The kinetic energy gained close to the proton is not "free energy" because the electron cannot give up that energy.

Can we find a reasonable principle which would entirely remove the potential well that the proton has withing the radius 1/2 * the Compton wavelength of an electron? If the electron is in the lowest energy state, then it has no free kinetic energy at all. Why would the area within 1/2 Compton wavelength be special?

Definition 1. The free kinetic energy of an electron is energy which it can give up without breaking the uncertainty principle and other rules of quantum mechanics. The electron is in an "excited state" and can move to a lower energy state.


Conjecture 2. In a path integral, the energy scale is the rest mass of the particle plus its free kinetic energy. The energy scale determines a cutoff of how sharply paths can bend in spacetime. The particle totally ignores potential wells which are too small to "see" at the energy scale. In the case of the hydrogen atom, the electron ignores the potential well whose size is less than the Compton wavelength of the electron. In the hydrogen atom, the electron does not have free kinetic energy at all if it is in the ground state.


In a hydrogen atom, the electron does not see features which are smaller than 2 * 10^-12 m, that is, the Compton wavelength. In a particle accelerator, we can give an electron a free kinetic energy of, say, 500 GeV. Such an electron will see features of size 2 * 10^-18 m.

In the Schrödinger equation, the electron does see even small features of the potential. Conjecture 2 claims that the electron does not see such small features at all. It will move past the miniature features on an essentially straight path, which is equivalent to not being affected by the potential at all.

The annihilation of an electron by a positron may break Conjecture 2. The electron must get within 10^-15 m of the positron, so that they can free 1.022 MeV of electromagnetic energy. Though, since the electron and the positron are actually the same particle, this cannot be directly compared to the hydrogen atom.

Conjecture 2 implies that we can think of the electron as a charge cloud of size one Compton wavelength. Its origin is in the uncertainty principle. Conjecture 2 says that we cannot circumvent the uncertainty principle by using non-free kinetic energy which the electron gains from the attraction of the proton.

Why is Conjecture 2 a consequence of applying special relativity to quantum mechanics? Abrupt curves in the path of a particle may involve short wavelengths in the spatial direction. If the wavelength in the time direction is determined by the rest mass plus the free kinetic energy, then those short spatial waves would move at a superluminal speed. That may be the fundamental reason why a relativistic particle must totally ignore potential wells which are smaller than its wavelength.

  e- --->   small kinetic energy
  _____    _____
           |  |
           |  |
           |_| deep potential well

Let us think more about Conjecture 2. Suppose that we have a very deep step potential well which is also narrow. If the well is deep enough, the electron can drop into it and release energy. This requires that the wavelength of the electron inside the well is short enough.

A Dirac electron wave is

       exp(-i (E t - p x)).

If we assume that a crest of a wave can only move within the light cone, then p cannot be larger than E. The question is what is E in various potentials? How does a relativistic particle behave in a box potential?

If the electron moves permanently from the potential 0 to a lower potential -V, then the kinetic energy of the electron grows by that amount V.

If we want to exhibit small features with a Dirac wave, the momentum p has to be large. But what if the potential well is not deep enough to give enough energy for a large p?

The big difference between the Schrödinger equation and the Dirac equation is that the former contains a second derivative with respect to space, d^2/dx^2, while the latter has the first derivative d/dx. If we try to fit a sine waves to the potential in the diagram above, then in the first approximation the Dirac wave will contain a discontinuity in the first spatial derivative, while the Schrödinger wave will have the discontinuity in the second derivative.

If we do a Fourier decomposition, it will probably contain more disturbance in the Dirac case. Does that imply that in the final fit, the Dirac wave should largely ignore a narrow potential well?

Schrödinger and Dirac waves meet a potential well

Let us try to calculate how each wave behaves.

            i dΨ / dt = (-d^2/dx^2 + V) Ψ.

The Schrödinger case is well known. Part of the wave will reflect back at the discontinuities of the potential. The main wave has a shorter wavelength in the potential well. Its phase changes relative to a wave which did not fall into the potential well.

How do we adjust a Dirac wave for a potential well?

             i dΨ / dt = (α d/dx + β m + V) Ψ.

Suppose that we have a solution:

            E^2 = (pc)^2 + (m_0 c^2)^2,

            exp(-i (E t - p x)).

How do we adjust it if the wave meets a potential well? In the Schrödinger case, E is constant. If we find a solution for t = 0, then the time dependence is simply the product with exp(-i E t).

The Dirac case is harder. If we simply increase p inside the well, then the

          exp(-i E t)

term will make the wave discontinuous at the temporal border of the potential well. If we glue together wave functions so that the wave is continuous in x at time t = 0, the wave will not be continuous at t = 1.

Let us consider the wave equation with two spatial dimensions and a time dimension for a while.

Something similar happens when water waves hit a shallow area in an otherwise open sea. The waves in the area move slower. The end result is a complex pattern of bent waves moving to various directions and interfering. If |V| is small, then maybe we should consider the term

         VΨ

in the Dirac equation as a source term which generates new waves?

What if |V| is large, but only in a small area relative to the wavelength of Ψ? Then it is like scattering from a small object.

In scattering in classical mechanics, if |V| is relatively large in a small area, it will scatter particles coming to that area to an arbitrary angle. It does not matter much if we make |V| a lot larger still. Thus, the depth of the proton potential well does not matter much as long as it is deep.

The scattering argument is the same for the Schrödinger wave as well as for the Dirac wave. Why would the Dirac wave be blind to small detail in V?

In the Schrödinger case, we get a good approximate solution just by increasing p inside the well. The solution is then the original solution, except that the phase of the waves has advanced in those waves that went through the shallow area. Equivalently, we can convert part of the main wave to a wave which has advanced 90 degrees relative to the main wave. The source in the wave equation is VΨ inside the well, and the produced wave is the difference of the new advanced wave and the old wave.

The source term solution of a Schrödinger wave

Let us return back to 1 spatial dimension and look at the problem of a narrow well from the point of a source term. An incoming wave exp(-i (E t - p x)) hits the potential well.

In the Schrödinger case, we have the equation

 d^2 Ψ/dx^2 + i dΨ/dt = V δ(0) exp(-i E t),

where on the right, |V| is small and it multiplies a standard plane wave solution of the equation. V is negative close to the origin; we have the Dirac delta there. E is the kinetic energy p^2 / (2m).

The term on the right is the source term. What kind of a wave does it produce? Everyday experience teaches us that if we exert a periodic force on a point in a string, then waves will start to propagate to the left and the right.

The solution should be something like

             C exp(-i (E t - p x - π/2))

for x > 0, and with + p for x < 0. Let us denote the solution by Ψ_V.

The function is not differentiable at x = 0. We may imagine that at x = 0 we have a Dirac delta function as the second derivative.

Close to the origin, the second derivative dominates the term dΨ/dt.  The first derivative is

              C * i p * exp(-i (E t - p x)) * i,

where we have used the fact that exp(-i π/2) = i. The second derivative at x = 0 is

              -2C p δ(0) exp(-i (E t - p x)).


        ____   ____ V
                \/

                /\        Ψ

We see that

           C = -V / (2p)

gives us the solution. C has to be much smaller than 1 for the solution to make sense. Produced new waves cannot contain more energy than the incoming wave.

Let us analyze the sum Ψ + Ψ_V for x > 0. Looking at the behavior in time, se wee that Ψ_V rotates to the same phase as Ψ at a later time. But looking at x, Ψ_V attains the phase of Ψ at a smaller value of x.

If we look at the real component of the wave function, Ψ + Ψ_V attains the maximum value at a smaller x than Ψ. This is expected since the wavelength is smaller inside the potential well.

A Dirac wave in 1 + 1 dimensions

Suppose that we have a flux of electrons

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

We denote a 2-component vector wave function by (Ψ_1, Ψ_2). We denote W = exp(-i (E t - p x)).

          α * -i * dΨ/dx + β mΨ = i dΨ/dt

is the 1-dimensional Dirac wave equation. The matrices

                  0      1
          α  =
                  1      0


                  1      0
          β  =
                  0     -1

have α^2 and β^2 equal to the identity matrix, and αβ + βα is the zero matrix. Let us write the two equations explicitly:

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

Let us check that the above electron plane wave function satisfies the Dirac equation.

The first product in the Dirac equation with the given Ψ is

        (p^2 / (E + m),                p)                    * W.

The second is

        (m,                                 -m p / (E + m)) * W.

The right hand side of the equation is

        (E,                                  E p / (E + m))  * W.

From the first component we get the equation:

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

From the second we get:

        p - mp / (E + m) = E p / (E + m)
       <=>
        E p + mp - mp = E p <=> 0 = 0.

The solution is E = +- sqrt(p^2 + m^2).

How to add a potential term to the Dirac equation?

How should we add a potential to the Dirac equation? E is the energy of a relativistic particle when we consider the potential V(x) zero.

In the Foldy-Wouthuysen calculation, a term V(x) Ψ is simply added to the Dirac equation to describe the electric potential.

If an electron permanently moves into a lower potential V, then we should obviously describe it as a free electron. In that case, the potential V should be ignored in the Dirac equation.

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

In the Klein paradox, the Wikipedia authors make a time-independent version of the massless (m = 0) Dirac equation by assuming that Ψ(t, x) = exp(-i E_0 t) * Ψ(x), where E_0 is constant.

Wikipedia then adds a term V(x) Ψ(x) to describe a potential barrier. The Klein paradox follows.

The mass gap and confinement in QCD

https://en.m.wikipedia.org/wiki/Yang–Mills_existence_and_mass_gap

The existence of a lowest energy excitation state of the gluon field in quantum chromodynamics forms a part of a Clay Millennium problem. We should be able to show that glueballs have a lowest possible mass. According to Wikipedia, computer simulations support this hypothesis, and there are heuristic arguments, too.

The photon field does not have a lowest energy for excitation: a photon is allowed to have an infinitesimal energy.

If glueballs of a very low energy would exist, then the strong force would be able to escape its "confinement" in hadrons and affect low-energy phenomena.

In this blog we have been developing the elastic rod model of a virtual photon. Let us look if we can say something about the confinement of the strong force.

Classically, the mass gap problem is the following: suppose that you have a complex gadget with drum skins, springs, rubber bands, weights, whatever. Show that it does not have low resonance frequencies. If you hit the gadget, it will let off only high-pitch sounds. Its possible low-frequency "sounds" contain no complete cycles. Rather, they are pressure impulses.

The gadget in question is the chromodynamic force field. We should study what kind of a "drum skin" it is.

Huygens principle gives a cutoff for Feynman diagrams

UPDATE Oct 25, 2018: We are working on our zitterbewegung post of Oct 14, 2018. It involves the handling of the charge singularity of a point particle. We will return back to this Huygens post when we have sorted out the charge singularity.

...

In an earlier blog post we asked if one can smoothen out the infinite momenta, or infinitesimal wavelengths in Feynman diagrams.

We found now an intuitive procedure where the sums of the path integrals are calculated on-the-fly and summed, so that short wavelengths never show up.

Let us model the annihilation with one space dimension x and time t. The electron is described as a wave of momentum p which moves to the right.

        e- ------------->

                             <------------- e+

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

Similarly, the positron is a wave which moves to the left with momentum -p.

The Huygens principle states that we can calculate the phase of the wave simply by following a crest of the wave and only using the paths of the "rays" of the wave. The ray at time t is the pair of momentum vectors (p, -p). We have one spatial dimension for each of the particles, as well as time. In a path, the momenta stay opposite numbers.

Note 1. Feynman's principle of using conservation of momenta in treelike Feynman diagrams to prune the integrals is actually an application of the Huygens principle. The input to the collision consists of waves of momenta p. It is enough to calculate paths where the momentum stays constant until the next vertex. This is like the principle: just calculate the movement of the crests of the waves along rays, by Huygens.


Note 2. Has anyone proved the Huygens principle for the Dirac equation, under an electric potential?


The electron will meet the positron at a spacetime point A_0 and its wave function will do an infinite number of cycles before that. We have there a singularity.

Let us assume that the electric potential between the particles is nonzero only when their distance is less than R.

The following diagram shows the strategy which we will use in the calculation of the sum of the path phases.

^                            A_0
|                      |       /  \     |
|                      |    /       \   |
|                      | /            \ |
|                      |                 |
|                    /|                 |\
|                   e-    R-area      e+
|
------------------------------------------> x

Let us keep A_0 as fixed spacetime point for a while, and let the actual annihilation point A vary. If A is to the right of A_0, there is still a path that goes through the spacetime point A_0.

The total amplitude at A_0 is an integral over all such paths, weighed on the distance that the paths have from each other outside the R-area.

Suppose that the total amplitude at A_0 is ψ. What is the amplitude if we move A_0 some time Δt later? From the geometry in the diagram above, we see that it just means shifting the whole picture Δt upward. The amplitude is

                   exp(i E Δt) ψ,

where E is the total energy of the electron. That is, there is just a simple phase shift!

Similarly, if we move A_0 a distance Δx to the right, the phase will be exp(i p Δx) bigger, where p is the original momentum of the electron. That is all.

The summation made the wave function of the electron to remain the same as it would be if there were no positron flying around. There are no high momenta. Just the momentum p with which the electron enters the experiment.

This sounds too good to be true. We assumed that the "input" to the R-area, the electron wave function, is a simple plane wave. But since the electron will eventually get annihilated, its wave function does decay at some rate, when we move to later times.

The real wave function is something like

                ψ = exp(i (E t - p x)) * f(t),

where f is a decreasing function on t. The Fourier decomposition of ψ is probably quite well-behaved.

What did we just show? Even though the annihilation of the electron and the positron is a violent act, the violent act happens at a random location of spacetime. Summing over all the paths yields a very smooth wave function.

Proposition 3. Let the wave function at a spacetime point x be a (very complex) function of its "inputs", the fluxes that enter its calculation area (R-area in the above example). If the inputs vary just by a phase shift factor, then the wave function value will vary just by that phase shift factor. QED.

In Feynman diagrams, this means that we can restrict ourselves to the momenta p which are given in the input particles.

Does Proposition 3 help us to get rid of the singularities in vacuum polarization loops?

The complex function in this case is the amplitude that a high-momentum virtual pair is born in some smallish zone inside the collision area. What are the input fluxes? They are the fluxes to the collision area.

In quantum mechanics, there is a princple that we can rotate the phase of any input flux by a constant angle, and we will get the same physical results. Does this help us?

Suppose that we are colliding two fluxes of electrons, with momenta p and q.

  e- ------------>             <----- e-
         p                              q

The spacetime diagram looks like this:

                               *
                 |         /    \   |
                 |      /        \  |
                 |   /            \ |
                 |/                \|
                /|                  |\
              /  |                  |  \
          e-   p                        e-   q

The star marks a collision. What is the amplitude of the left electron wave function at the star? It is a complicated integral of pairs of left electron and right electron paths, where the left electron goes through the spacetime point marked with the star.

The potential between the left electron and the right electron is agnostic to the phase of the right electron.

What is the amplitude if we move the star to the right some distance x? The phases of the right electron do not matter - the phase of the left electron changes by exp(i p x).

Similarly, the phase of the right electron changes by exp(i q x).

What if the collision produces some other particles? Let us have a path where the collision produces a real pair e- and e+. What is the amplitude of the e+ at a certain spacetime point within the calculation area?

            A  e+  k
             /
           *
         /   \
       /       \
  e-  p      e-  q


Since e+ can have any momentum k, we have various paths where the slope of the line between * and A varies. The amplitude of a path at A is probably something like a product of the amplitudes of the electrons and something that depends on k. This is in the spirit of Feynman diagrams.

What if we move A some distance x right? The paths that lead to the new position are like the old paths, but shifted by x. The amplitude at the new position will be the old value multiplied by exp(i p x + i q x).

We showed that the amplitude of the virtual positron behaves very well. Its Fourier decomposition does not contain much weight on high momenta.

Theorem 4.  A natural ultraviolet cutoff scale for any momentum in a treelike Feynman diagram is the total energy of the collision, as long as the running fine structure constant remains < 1.

Proof. Let us consider a single Feynman diagram of the process. That is, we consider a set of paths with the same topology.

Suppose that the amplitude of a path is calculated from a function of type

          input1 * input2 * f (input1, input2),

where f is a function whose value does not depend on the phase of the inputs, only on the absolute value. The inputs are plane waves.

If we move the path by a translation in spacetime, the phase of the inputs changes but not their absolute value.

It is a general principle in quantum mechanics that physical processes are agnostic of the phase of inputs. Therefore, all f satisfy the condition above.

The wave function at a spacetime point depends in a complex fashion on a sum of different paths. But the sum only varies in spacetime according to

      input1 * input2
             = exp(i (Et - kx)) * exp(i (E't - k'x))
             = exp(i ((E + E') t - (k + k') x)).

The sum only contains one Fourier mode!

Momentum is conserved at the vertexes of a Feynman diagram. In a treelike diagram, all inputs to all vertexes have their momentum constrained by the momentum which entered in the original real particles into the experiment. That is why we can claim that the "natural cutoff scale" is the total energy. We can forget about the potentially infinite energy that is transiently available in pair annihilation.

If the diagram contains loops, then an arbitrary momentum can circle the loop and we cannot constrain momentums this easily. We will look at loops in the next section.

Where does the complexity then creep into processes? When we are summing over diagrams of different topologies, they have different Fourier modes and their sum becomes more complicated.
     
Also, the inputs are not perfect plane waves. Their Fourier decomposition contains all kinds of modes.

The proof of this theorem leaves open one possibility of divergence. Suppose that f can hugely amplify infinitesimal high frequency components in the inputs, so that their contribution becomes relevant.

If the fine structure constant is < 1, functions f in Feynman diagrams cannot hugely inflate amplitudes. QED.


What happens if the inputs have the energy of a Landau pole? Then the contribution from complex diagrams may grow without bounds. Our cutoff scheme removes infinities from a single diagram. But if complex diagrams contribute more than simple ones, then there is an explosion.

A vacuum collapse may happen if complex diagrams contribute more than simple ones, that is, if the coupling constant is > 1.

The Feynman integral on a virtual loop

Above we argued that in treelike Feynman diagrams, the wave functions are smooth for all vertexes. Our argument was symmetry: the inputs to the collision are plane waves. If we move any path in timespace, the amplitude from it changes only by a phase factor of the inputs.

If there is a loop, we cannot constrain the momenta of inputs. Let us study a simple virtual pair loop.

                    z

                    / photon

                  /

                ||

           e-  || e+

                ||

                /

              /  photon plane wave

The diagram above shows a loop in a more realistic fashion than usual. The electron and the positron are born at the same point and annihilate at the same point. They fly the exact same line. We want to calculate the photon amplitude at points close to z. 

The vacuum atom model says that there must be an integer number of electron wavelengths in the loop, but we ignore that for now and allow any momentum and length in the loop.

If the wave function of the input photon is a plane wave, then we can move any individual path and the amplitude of the path only changes by the phase factor of the input.

Energy and momentum conservation dictate that the output has to be the same plane wave as the input, except that a phase shift is possible.

But what about the diverging Feynman integral on the loop? Where did it go? The Feynman integral is ill-defined, and therefore one cannot really calculate with it. We may use any regularized version of it. We showed that any reasonable path integral just amounts at most to a phase shift on the photon plane wave. It may also be that the Feynman integral converges, after all, if the integration on the loop momentum k is performed last.

Feynman tried to calculate the amplitude at the annihilation point of a virtual loop. He noticed that the integral diverges. A regularization industry was born and has flourished for 70 years. But one can argue that since the amplitude at the annihilation point is not physically observable, the sum of the amplitudes at the annihilation point is not definable at all.

Conjecture 5. One should not try to sum the intermediate point amplitudes in a path integral. It does not make sense to talk about the amplitude of a wave function at a spacetime point if the wave function is not of a real observable particle.


If Conjecture 5 is true, then the regularization industry has been chasing a wild goose.

Conjecture 5 is relevant for the famous Schrödinger cat problem. When we put a cat and the radioactive atom in the box, it is, in a sense, an abstract scattering experiment. The system may take different paths until the box is opened, which corresponds to measuring the real particles that exit the scattering experiment. The path integral method suggests the following: the paths cannot be meaningfully summed during the experiment. It does not make sense to say that the cat is dead and alive "at the same time". The only meaningful wave function is the sum of the paths when we measure the cat afterwards.

What if the input is not a plane wave, but changes smoothly in timespace?

                __________
________|                  |______

              0              z  1

Let us calculate a 1-dimensional toy example, where the input is a constant 1 from 0 to 1 on the x axis, and the end of the loop is at x = z. The momentum |p| > 1. The point z may be anywhere in the x axis.

We assume that in the interval there is a photon flux of a negligible momentum, starting from 0. The photon is transformed into a virtual pair at some point of the interval, and the pair flies to z. The electron in the pair has an arbitrary momentum |p| > 1 and the positron the exact opposite momentum.

One may ask how the particles can meet at all, if they have opposite momenta. But here we imagine that a momentum corresponds to a plane wave which fills the whole spacetime.

Let us calculate the amplitude of the virtual electron at z. We cannot use the translation argument of Theorem 4, because the input varies in space.

The input to the loop comes from any point in the interval (0, 1) and the loop ends at z. We want to sum the amplitudes

    exp(i p * (z - x))

for all 0 < x < 1 and |p| > 1.

The integral on x is

     ψ(z) = i / p * (exp(i p (z - 1)) - exp(i p z)) = i / p * exp(i p z) * (1 / exp(i p) - 1).

If z is 0 or 1, then the integral on p diverges logarithmically. Otherwise, the integral is essentially of the form

     exp(i p) / p,

where |p| > 1. It is a Dirichlet integral and converges. We have shown that a simple path integral converges, unless the end point is at the border of the input area.

   k                k + p ->
~~~~~~>========= z
                     <- p
                virtual loop

What if we the input is a plane wave of momentum k, in one dimension? In the diagram, the double line === depicts the virtual loop. We need to integrate

       exp(i k x) * exp(i (k + p) (z - x))

       = exp(i (k z + p (z - x)))

over all x and |p| > 1. Let us integrate over a limited range x_0 < x < x_1. We require x_0 < z < x_1.

----------------------------------------> x
   x_0                  z       x_1

The integral over the range x0, x1 is

         i / p * exp(i (k z + p (z - x_0)))
       - i / p * exp(i (k z + p (z - x_1))).

The result rotates with the end points x_0 and x_1. The result is not very sensible. But let us integrate on 1 < p then. Let us look at the first term.

Let q = p (z - x_0). We need to calculate

                                               ∞
         ψ_1 (z)   = exp(i k z)    ∫  i (z - x_0) * 1 / q * exp(i q) dq.
                                         z - x_0

Let g(y) be the integral of the Dirichlet function

                       ∞                    
         g(y) =   ∫   sin(x) / x dx.
                    y

The real part of the integral above is

                        ∞
                 -y  ∫ sin(q) / q dq,
                    y

where y = z - x_0.

What happens when y tends to infinity? If y = 2n π, then g(y) is roughly

                   2 / (2n π)   -   2 / (2(n + 1) π)  + …
                   = 1 / π * (1 / n   -  1 / (n + 1)   +    …)
                   = 1 / π * 1 / 2n.

That is, at such points, g(y) is roughly 1 / y. Thus, g(y) generally is cos(y) / y with large y. The real part of the integral is thus -cos(y). We may guess that the imaginary part is -i sin(y). The integral would then be -exp(i y).

               ψ_1 (z) = - exp(i k z) exp(i (z - x_0)).

Unfortunately, ψ_1 (z) does not converge. We could have guessed that. A virtual loop can start at a remote position, and affect the amplitude at z without any damping. There is no obvious reason why remote areas should contribute to the amplitude at z less than close areas.

It is not physically reasonable that a high-energy virtual particle flies over great distances. Can we restrict the contribution of long loops?

The idea of assuming a infinite plane wave for the photon is bad. Let us look at a more realistic collision example.

   e- ----------------->
                   ^
                   |
                   z
                  || virtual pair
                  ||
                   |
        <------------------- e+

We are not interested in the amplitude of just any virtual electron at z. The virtual electron has to be from the collision with the nearby positron. We need to develop a way of handling the conditional probabilities that are involved in collisions.

Infrared divergences

Electron self-energy involves an infrared divergence. Is there some principle which allows us to establish a cutoff to low-frequency modes in the Fourier decomposition of any particle in a Feynman diagram?

The inputs themselves cannot contain an infrared divergence in them, because then the inputs would contain infinite energy. Can we prove that no process can create an infrared divergence if the inputs do not have an infrared divergence?

Can we use the proof of Theorem 4 to remove infrared divergences?

An infrared divergence would happen if the function f would hugely amplify the infinitesimal infrared components in the inputs. But we saw that such f do not exist if the fine structure constant is < 1.

Theorem 6. A natural cutoff scale for infrared divergences in a treelike Feynman diagram is the total energy of the collision. We assume that the running fine structure constant is < 1. QED.