https://en.wikipedia.org/wiki/Time_crystal
It has a connection to the time crystal concept of Frank Wilczek. Vacuum atoms might even form crystals under some conditions?
Definition 1. A vacuum atom is either a real or virtual pair of an electron and a positron which is first produced and then annihilates.
e+ ____
/ \
/ \
\ /
\ e- ____/
We demand that the wave function of the vacuum atom must be continuous in timespace. An obvious consequence is that a full number of wavelengths must fit around the loop which the electron and the positron form in timespace.
Does the Dirac equation work as a differential equation which governs vacuum atoms? The equation obviously has to fit with special relativity.
To form, a vacuum atom has to be in interaction with other particles. A collision of particles is a typical case.
If the vacuum atom is formed by a virtual pair, then the wave function amplitude decays to the direction of increasing time. The vacuum atom is a tunneling phenomenon, and entropy of the system will grow with time. That is the reason why the atom is time asymmetric.
The electron shell of a hydrogen atom in timespace can be visualized as a pipe. A vacuum atom may look like an elongated sphere.
Does the vacuum atom have a discontinuity at the point of the light cone?
When a pair is born or annihilated, the particles move at most at the speed of light.
t e- e+
^ \ /
| \/
|
|-------------------> x
The trajectories have a sharp 90 degree turn at the point where the pair is born. They move in the light cone. Does this make the wave function of the vacuum atom discontinuous?
Or should we allow "superluminal" movement? In practice, that would mean that the particles are not born at the same point but have a small spatial distance when they appear. The classical radius of the electron, 3 * 10^-15 m might be the spatial distance.
This is the mystery of a pair annihilation/creation. If they start from the same point, they need to fight an infinite Coulomb potential, which does not seem right. When the pair annihilates, it only releases 1.022 MeV of energy, which corresponds to a distance of 3 * 10^-15 m of the electron and the positron. If they would move closer, the released energy should be larger.
What do solutions of the Dirac equation look like? Are they smooth at distances 3 * 10^-15 m or less?
The Compton wavelength of an electron is 2 * 10^-12 m. The vacuum atom should look fuzzy at distances less than that.
We also need to consider the process which creates a vacuum atom. We have been looking at a collision of an electron and a nucleus. The collision happens in an area 10^15 m or less. A virtual pair screens some of the electric charge of the colliding particles.
The colliding pair, in a sense, induces a pulse of an electric current in space. The virtual pair is the carrier of that current.
^
e- | -------->
|
|
Z |
The vertical arrow depicts the induced pulse of the current.
The circuit of the current has to close. The virtual pair will annihilate later.
The crucial question is: can we build a smooth wave function for the vacuum atom by using path integrals where the path of the virtual pair stays in the light cone? The path will have a sharp 90 degree turn, but maybe the combined effect of all paths is smooth?
Classically, the electron and the positron had an infinite kinetic energy when they were born. How many cycles of phase does the path integral do then?
We may assume that an electron moves in a potential -1 / x towards the origin x = 0.
The path integral is on
i (E dt + p dx)
E is the constant total energy of the electron.
If the electron is moving at almost the speed of light, then p is approximately c times the relativistic mass of the electron. Most of that mass is kinetic energy, which is 1 / x. We get
p = c * 1 / (x c^2) = 1 / (x c).
The integral on 1 / x diverges. There would be an infinite number of cycles when the electron approaches the origin.
The sharp point in the light cone is the origin of the divergence. The closer we go to the point, the faster the phase in the path integral rotates. This reminds us of the "breaking into more degrees of freedom" discussion.
An obvious way to remove the divergence is to allow the path to take a superluminal shortcut past the point in the light cone:
e- e+
\____ /
\ /
\/
The horizontal line is the shortcut. In annihilation, the photons are emitted during the shortcut. That explains why they only have 1.022 MeV of energy. The pair never descends into the infinitely deep potential well at the sharp end of the light cone.
In Feynman diagrams, particles start at a spacetime point, and consequently, have an infinite kinetic energy and their wave function would typically perform an infinite number of cycles. Feynman gets rid of the infinite cycles by requiring the momentum p to be conserved at each vertex. If the diagram is treelike, p is constrained to low values at each vertex. That is the way how Feynman introduces "fuzziness" to point particles and avoids the problem of infinite cycles in treelike diagrams. The procedure does not work for loops.
The shortcut which we introduced above, corresponds to a momentum cutoff for the waves of the pair. One cannot form a sharp point in a wave function in spacetime unless one uses Fourier modes of an arbitrarily high momentum p.
The annihilation energy 1.022 MeV is a natural cutoff for the momenta. We conjecture that the vacuum atom must be fuzzy in the sense that its Fourier decomposition cannot contain much weight on modes whose energy is bigger than that.
The paths in the path integral formulation must be smooth in the sense that if they are Fourier-decomposed, the weight of components with energy > 1.022 MeV must be small.
Note 1. We are clearly building nonperturbative QED here. This is the new approach which Paul Dirac called for to remove renormalization from QED.
The absence of "sharp points" and arbitrarily high momenta is characteristic of Schrödinger type quantum mechanics. Feynman and others introduced the perturbative QED to explain creation of new pairs. But perturbative QED has the problem of divergences, which comes from arbitrarily high momenta, which in turn come from sharp points that are in the vertexes.
In the Schrödinger treatment of the hydrogen atom, we cut off the infinitely deep potential wells that are formed by the pointlike (?) quarks in the nucleus. The cutoff is considered the natural way to solve the Schrödinger equation. Similarly, QED we must cut off singularities that would exist in pair production and annihilation.
Regularization procedures in QED work because they cut off the sharp point in the light cone. Superluminal movement is required in the paths to remove the sharp point. We have uncovered new physics: superluminal movement happens in pair production and annihilation.
Why a superluminal path does not descend into the infinitely deep potential well in a pair annihilation? The electron "sees" the movement of the positron from the light speed signals that the positron sends. But in the superluminal part of the electron path, there is nothing in the light cone that the electron sees. The Coulomb force has disappeared. The electron moves with a constant kinetic energy on the x-axis.
e- e+
\ /
\ /
t \______/
^
|
----------------->
x
In the diagram, there is nothing in the downward light cone when the electron is in the horizontal path. No Coulomb force.
In the spring we introduced the concept of optical gravity, which can tame the singularity at the horizon of a black hole. A similar argument might tame the Landau pole?
In practice, a black hole forms at much lower energies than a QED Landau pole. Thus, black holes save us from the collapse of the vacuum which could result from Landau poles of QED.
Freeman Dyson said that if the Coulomb force would be a purely attractive force, then the QED vacuum would clearly be unstable.
In a black hole, the horizon is the point where a mass which is lowered into the black hole has given all its mass-energy in the work of the gravitational potential. If we would be able to lower the mass even lower, we could recover more energy than the mass-energy that we lowered. Then the vacuum could be unstable, that is, any physical system could potentially release an infinite amount of "free energy" by forming black holes.
The Landau pole probably is a similar device. If we could shoot an electron to within 10^-292 meters from a positron, then the Coulomb potential would release more energy than we used to shoot the electron. The energy which we need to focus an electron grows inversely to the precision. That is,
E = constant / r,
where r is the precision in meters.
If the potential between the electron and the positron falls faster than -1 / r, then there is a point where we can gain more energy than we use to shoot the electron. That point is probably the Landau pole.
We used the uncertainty principle to derive E = constant / r. The uncertainty principle dictates that if we have a potential which falls faster than -1 / r, then vacuum would be unstable.
In the case of a black hole, we are saved from a vacuum collapse by the fact that time freezes at the horizon. The freeze is a consequence of optical gravity. Also the interior of the horizon is frozen in time.
For QED, there is no freezing of time. Fortunately, the black hole horizon will save us from a naked Landau singularity.
We are now approaching quantum gravity. The singularity in the light cone graphs above is protected by the black hole horizon that would form above the sharp point of the singularity. A path that would wander within 10^-292 m of the light cone point would be frozen in time.
Classically, the electron and the positron had an infinite kinetic energy when they were born. How many cycles of phase does the path integral do then?
We may assume that an electron moves in a potential -1 / x towards the origin x = 0.
The path integral is on
i (E dt + p dx)
E is the constant total energy of the electron.
If the electron is moving at almost the speed of light, then p is approximately c times the relativistic mass of the electron. Most of that mass is kinetic energy, which is 1 / x. We get
p = c * 1 / (x c^2) = 1 / (x c).
The integral on 1 / x diverges. There would be an infinite number of cycles when the electron approaches the origin.
The sharp point in the light cone is the origin of the divergence. The closer we go to the point, the faster the phase in the path integral rotates. This reminds us of the "breaking into more degrees of freedom" discussion.
An obvious way to remove the divergence is to allow the path to take a superluminal shortcut past the point in the light cone:
e- e+
\____ /
\ /
\/
The horizontal line is the shortcut. In annihilation, the photons are emitted during the shortcut. That explains why they only have 1.022 MeV of energy. The pair never descends into the infinitely deep potential well at the sharp end of the light cone.
In Feynman diagrams, particles start at a spacetime point, and consequently, have an infinite kinetic energy and their wave function would typically perform an infinite number of cycles. Feynman gets rid of the infinite cycles by requiring the momentum p to be conserved at each vertex. If the diagram is treelike, p is constrained to low values at each vertex. That is the way how Feynman introduces "fuzziness" to point particles and avoids the problem of infinite cycles in treelike diagrams. The procedure does not work for loops.
The shortcut which we introduced above, corresponds to a momentum cutoff for the waves of the pair. One cannot form a sharp point in a wave function in spacetime unless one uses Fourier modes of an arbitrarily high momentum p.
The annihilation energy 1.022 MeV is a natural cutoff for the momenta. We conjecture that the vacuum atom must be fuzzy in the sense that its Fourier decomposition cannot contain much weight on modes whose energy is bigger than that.
The paths in the path integral formulation must be smooth in the sense that if they are Fourier-decomposed, the weight of components with energy > 1.022 MeV must be small.
Note 1. We are clearly building nonperturbative QED here. This is the new approach which Paul Dirac called for to remove renormalization from QED.
The absence of "sharp points" and arbitrarily high momenta is characteristic of Schrödinger type quantum mechanics. Feynman and others introduced the perturbative QED to explain creation of new pairs. But perturbative QED has the problem of divergences, which comes from arbitrarily high momenta, which in turn come from sharp points that are in the vertexes.
In the Schrödinger treatment of the hydrogen atom, we cut off the infinitely deep potential wells that are formed by the pointlike (?) quarks in the nucleus. The cutoff is considered the natural way to solve the Schrödinger equation. Similarly, QED we must cut off singularities that would exist in pair production and annihilation.
Regularization procedures in QED work because they cut off the sharp point in the light cone. Superluminal movement is required in the paths to remove the sharp point. We have uncovered new physics: superluminal movement happens in pair production and annihilation.
Why a superluminal path does not descend into the infinitely deep potential well in a pair annihilation? The electron "sees" the movement of the positron from the light speed signals that the positron sends. But in the superluminal part of the electron path, there is nothing in the light cone that the electron sees. The Coulomb force has disappeared. The electron moves with a constant kinetic energy on the x-axis.
e- e+
\ /
\ /
t \______/
^
|
----------------->
x
In the diagram, there is nothing in the downward light cone when the electron is in the horizontal path. No Coulomb force.
The Landau pole
The Landau pole results from the fact that the effective Coulomb force grows stronger at short distances, or equivalently, at high energies of colliding particles. At energies of some 10^286 eV, the fine structure constant α is > 1 and the contributions of complex Feynman diagrams grow exponentially without bounds. When α is the usual 1/137, the weight of an nth order diagram is (1/137)^n. Obviously, having α > 1 will spoil the Feynman machinery.
Can we fix the Landau pole divergence with our new ideas, or explain why it forms?
The Coulomb potential 1 / r potentially is a source of infinite energy.
The Compton wavelength of the electron is 2 * 10^-12 m, which corresponds to energy 511 keV.
The Landau energy corresponds to 10^-292 m.
The Landau pole is the "black hole" of the Coulomb force. The Coulomb force grows faster than 1/r^2 when r becomes smaller. The same holds for gravitational force in general relativity.
We conjecture that a black hole is actually the Landau pole of quantum gravity.
In practice, a black hole forms at much lower energies than a QED Landau pole. Thus, black holes save us from the collapse of the vacuum which could result from Landau poles of QED.
Freeman Dyson said that if the Coulomb force would be a purely attractive force, then the QED vacuum would clearly be unstable.
In a black hole, the horizon is the point where a mass which is lowered into the black hole has given all its mass-energy in the work of the gravitational potential. If we would be able to lower the mass even lower, we could recover more energy than the mass-energy that we lowered. Then the vacuum could be unstable, that is, any physical system could potentially release an infinite amount of "free energy" by forming black holes.
The Landau pole probably is a similar device. If we could shoot an electron to within 10^-292 meters from a positron, then the Coulomb potential would release more energy than we used to shoot the electron. The energy which we need to focus an electron grows inversely to the precision. That is,
E = constant / r,
where r is the precision in meters.
If the potential between the electron and the positron falls faster than -1 / r, then there is a point where we can gain more energy than we use to shoot the electron. That point is probably the Landau pole.
We used the uncertainty principle to derive E = constant / r. The uncertainty principle dictates that if we have a potential which falls faster than -1 / r, then vacuum would be unstable.
In the case of a black hole, we are saved from a vacuum collapse by the fact that time freezes at the horizon. The freeze is a consequence of optical gravity. Also the interior of the horizon is frozen in time.
For QED, there is no freezing of time. Fortunately, the black hole horizon will save us from a naked Landau singularity.
We are now approaching quantum gravity. The singularity in the light cone graphs above is protected by the black hole horizon that would form above the sharp point of the singularity. A path that would wander within 10^-292 m of the light cone point would be frozen in time.
A general observation about path integrals
If the energy scale of a phenomenon is E, then the Fourier decomposition of a wave function contains few modes with energy >> E.
In the path integral formulation, it is enough to use paths which are relatively smooth, no sharp turns. The Fourier decomposition of the paths needs to contain only few modes where the frequency is >> F, where F is somehow related to E, but how?
If the wave is smooth, then obviously the contributions of sharply turning paths will cancel each other out.
Can we make a virtual vacuum atom time symmetric?
If there is a shortage of energy, a virtual pair is formed by tunneling. We need to study virtual pairs with an energy excess later.
____
__ __ | |
/ \__/ \_ ---> __| |____
Suppose that we have a flux of particles hitting a high potential barrier from the left. Some of them will tunnel through to the space with few of such particles. Entropy will grow. Can we make the process time symmetric? Yes. We have to assume that an equal flux is hitting the barrier from the right. Then the entropy of the system does not change and the wave function is time symmetric.
To make a vacuum atom time symmetric, we should have equal fluxes of particles coming from the future as well as from the past. Such a setup cannot exist in our universe.
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