The wave functions of the electron and the positron have a different handedness with respect to increasing time.
t
^
|
| e- e+
| \ /
| \ / string
| \________/
| \\O// spider sits on the string
| and rotates it
-----------------------------------> x
This suggests the following model: when we excite the electron-positron field to produce a quantum, we let an agent (a spider in our earlier blog posts) to rotate the string while sitting on the string in the middle. The spider will create a right-handed wave to one direction and a left-handed wave to the other. The waves are strictly correlated, "entangled".
A quantum of the field contains both waves.
A quantum of the field has the energy h f. The "total absolute sum" of angular momenta in the quantum corresponds to spin 1. That total amount is divided between the electron and the positron who each have the absolute value of the spin 1/2.
What if the spider wants to create photons? The photon can be absorbed and emitted alone. The spider sits at the end of that string and rotates.
t
^ photon
| /
| / string
| /
| \\O// spider rotates the end of the
| string
| string
------------------------------->
We proved in an earlier blog post that a rotating electric dipole will make quanta whose spin is 1. That is why the photon has the spin 1.
Thus, the reason why the electron has the spin 1/2 is:
1. The produced positron and the electron are "the same particle", which means that they live on the same string.
2. A quantum contains both the positron and the electron.
3. The total absolute sum of the angular momenta which a quantum puts to the string corresponds to spin 1.
4. The total absolute angular momentum is divided evenly between the electron and the positron.
The above model is more intuitive than our earlier tube model of the produced pair. Having the electron and the positron as waves on the same tense string stresses the wave nature of particles.
Fermions are particles which can bounce back and forth in time. Their string has no end. Either it is a loop, or the string extends to infinity. Boson strings may have ends and their strings probably cannot zigzag in time.
Actually, a string is not the right analogue for a particle which has a non-zero rest mass. Its waves have dispersion, where long waves move faster. Water waves are a better analogue. Thus, a fermion is like a water wave which zigzags in time and is always produced to two directions at once in the pool.
How does a positron go back in time and transform to an electron?
The spider on a string model allows us to investigate what happens in a pair production. Let us follow the positron backward in time first.
1. The positron wave packet moves closer to the spider and touches the spider. The positron wave packet is right-handed in the time-backward perspective.
2. The positron wave packet is completely absorbed by the spider.
3. We change the direction of time to the normal.
4. The spider emits the electron wave packet. The electron wave packet is right-handed.
5. The spider can stay static in its place. It will use two legs to rotate the left and the right side of the string in unison.
The wave packet of the positron was first completely absorbed by the spider, and then re-emitted as the electron wave packet. The wave packet did not continue through that part of the string undisturbed, but the spider was the middleman.
<-- wave packet
wall |--------------------- string
An analogous process happens when an electron wave is reflected from a potential wall. The simplest such case is when we have a tense string and a wave meets the endpoint which is a solid wall. There is a 180 degree phase shift in the reflected wave. The wall at no point absorbs the wave completely. That is a difference from the spider model. There is something similar, though. The wall makes a 180 degree phase shift to the reflected wave. The wave does not continue undisturbed.
The spacetime diagram of the reflection is something like this:
wall |
|\ \/ /
| \ \/ waves
| /\ \
The wave is never completely absorbed by the wall.
/ \
/ \
______/___\_____ "time wall"
\\O// spider
The spider "time reflection" case has no waves at the start of the process. The diagrams look quite different. Maybe pair production is not analogous with ordinary scattering, after all?
Does a source in the Dirac equation produce both electron and positron waves?
The crucial question in the string model is that if we put a disturbance to the Dirac equation, that is, add a source to the homogeneous equation, does it automatically produce an equal amount of electron and positron waves?
An equal amount is required by charge conservation, too.
We do not know if the Dirac equation correctly describes pair production, but let us try to analyze it.
The "complete set of orthogonal solutions" to the Dirac equation apparently has to include both the positive energy E solutions and the negative energy -E solutions. Only then we can construct an arbitrary wave packet. Since the Dirac equation is completely symmetric with respect to E and -E, then a random source to the homogeneous Dirac equation will produce both E and -E waves. But under what condition is there a perfect balance of those waves, so that charge is conserved?
A collision of two electrons can produce a pair. The setup is not physically symmetric for an electron versus a positron, but empirically we know that an equal number of electrons and positrons are produced. Maybe in nature, the sources to the Dirac equation are always symmetric for some unknown reason, or the Dirac equation needs to be modified to ensure an equal number of produced electrons and positrons.
For Maxwell's equations, we probably can form a complete set of orthogonal solutions from the positive energy solutions.
Let us define that positive energy solutions are emitted through one string which starts from the emitter, and negative energy solutions through another string. We have found the mathematical meaning for our diagrams above.
Our new Dirac equation has m + V, where V is a scalar potential, in the place of m.
The electron has m positive and E positive while the positron can be defined to have m and E negative. If V is the potential for the electron, then -V is the potential for the positron. There is a perfect symmetry. If a pair is produced in some way through an abrupt spike V, then there might be a perfect symmetry for the electron and the positron production.
We do not know if the Dirac equation correctly describes pair production, but let us try to analyze it.
The "complete set of orthogonal solutions" to the Dirac equation apparently has to include both the positive energy E solutions and the negative energy -E solutions. Only then we can construct an arbitrary wave packet. Since the Dirac equation is completely symmetric with respect to E and -E, then a random source to the homogeneous Dirac equation will produce both E and -E waves. But under what condition is there a perfect balance of those waves, so that charge is conserved?
A collision of two electrons can produce a pair. The setup is not physically symmetric for an electron versus a positron, but empirically we know that an equal number of electrons and positrons are produced. Maybe in nature, the sources to the Dirac equation are always symmetric for some unknown reason, or the Dirac equation needs to be modified to ensure an equal number of produced electrons and positrons.
For Maxwell's equations, we probably can form a complete set of orthogonal solutions from the positive energy solutions.
Let us define that positive energy solutions are emitted through one string which starts from the emitter, and negative energy solutions through another string. We have found the mathematical meaning for our diagrams above.
Our new Dirac equation has m + V, where V is a scalar potential, in the place of m.
The electron has m positive and E positive while the positron can be defined to have m and E negative. If V is the potential for the electron, then -V is the potential for the positron. There is a perfect symmetry. If a pair is produced in some way through an abrupt spike V, then there might be a perfect symmetry for the electron and the positron production.
The Dirac equation has 4 components because it codes the spin of both the electron and the positron?
Wolfgang Pauli originally added the spin to the Schrödinger equation by making the wave function two-component and defining the spin operators based on the Pauli 2 x 2 matrices.
Our thesis is that the quantum of the Dirac equation is actually the combination of the electron & the positron. It is logical that the wave function would then have 4 components, because the electron and the positron are allowed to reorient their spins independently of each other.
If the rotation state of the electron is independent from its spatial location, then the electron can be described by a triplet
(Ψ, s_d, s_u),
where s_d and s_u are complex numbers and Ψ is a one-component complex function. Let x be a spacetime point. If we rotate x infinitesimally around some axis A so that it is mapped to x', we have to update the triplet slightly to get the value of the triplet at x'.
The rotation state of the electron lives in the same spacetime as the wave function Ψ. In that way it is not completely independent.
If the rotation state of the electron is independent from its spatial location, then the electron can be described by a triplet
(Ψ, s_d, s_u),
where s_d and s_u are complex numbers and Ψ is a one-component complex function. Let x be a spacetime point. If we rotate x infinitesimally around some axis A so that it is mapped to x', we have to update the triplet slightly to get the value of the triplet at x'.
The rotation state of the electron lives in the same spacetime as the wave function Ψ. In that way it is not completely independent.
The Dirac equation "separates" the positive and negative frequency solutions of the Klein-Gordon equation
Let us look at the 1+1-dimensional Dirac equation. The spin does not come up there, but the positron does appear.
The basis of the Klein-Gordon solutions is probably all plane waves
exp(-i (E t - p x)),
where E is allowed to be negative or positive. A negative energy corresponds to a "negative frequency" with respect to time.
The basis for the Dirac solutions consists of
(1, p / (E + m)) * exp(-i (E t - p x)),
where E is positive, and
(p / (E - m), 1) * exp(-i (E t - p x)),
where E is negative. We see that the Dirac equation separates the positive and negative frequency solutions, based on the coefficient spinor (1, ...) or (..., 1).
Next we should analyze what happens in 1+3 dimensions. Why does the spin come up?
Our photon generator, the rotating electric dipole, was able to pump angular momentum into the electromagnetic field. That is the ultimate reason why the photon has a spin: if the field is absorbed in quanta, then both the energy and the angular momentum have to be absorbed.
Is a source able to pump angular momentum into the electron field? If yes, then the electron must have a spin.
The symmetry of the Dirac equation suggests the following principle: any source has to be symmetric in the way that it creates perfectly symmetric waves to the electron and the positron fields. Then charge is conserved. The spins of the produced pairs have to be perfectly antisymmetric.
Suppose that we have the following setup:
e- e+
^
| rotating rod
----------o---------
|
v
e- e+
A rod rotates clockwise and shoots electron-positron pairs to the directions marked by the arrows. In that way we do store angular momentum into the electron-positron field. Any field which can store linear momentum can also store angular momentum. But do we also store some "intrinsic angular momentum", that is, spin? Could it be that the electron and the positron both have the spin 0 and only store angular momentum in their linear momentum p?
In the case of photons, can the angular momentum be stored in the linear momentum p of the photons if we shoot them from the ends of the electric dipole? When the dipole loses its rotational energy, its both ends lose a lot of momentum p. Photons have p / E the minimum possible value for any material object. If the dipole rotates so that its ends move at the speed of light, then two photons shot from the ends of the dipole would be able to carry away the energy and the momentum.
In practice, nature seems to store the angular momentum into the spin of a single photon. Maybe one photon is a better quantum than two strictly correlated photons?
Next we should analyze what happens in 1+3 dimensions. Why does the spin come up?
What fields can store angular momentum and spin?
Our photon generator, the rotating electric dipole, was able to pump angular momentum into the electromagnetic field. That is the ultimate reason why the photon has a spin: if the field is absorbed in quanta, then both the energy and the angular momentum have to be absorbed.
Is a source able to pump angular momentum into the electron field? If yes, then the electron must have a spin.
The symmetry of the Dirac equation suggests the following principle: any source has to be symmetric in the way that it creates perfectly symmetric waves to the electron and the positron fields. Then charge is conserved. The spins of the produced pairs have to be perfectly antisymmetric.
Suppose that we have the following setup:
e- e+
^
| rotating rod
----------o---------
|
v
e- e+
A rod rotates clockwise and shoots electron-positron pairs to the directions marked by the arrows. In that way we do store angular momentum into the electron-positron field. Any field which can store linear momentum can also store angular momentum. But do we also store some "intrinsic angular momentum", that is, spin? Could it be that the electron and the positron both have the spin 0 and only store angular momentum in their linear momentum p?
In the case of photons, can the angular momentum be stored in the linear momentum p of the photons if we shoot them from the ends of the electric dipole? When the dipole loses its rotational energy, its both ends lose a lot of momentum p. Photons have p / E the minimum possible value for any material object. If the dipole rotates so that its ends move at the speed of light, then two photons shot from the ends of the dipole would be able to carry away the energy and the momentum.
In practice, nature seems to store the angular momentum into the spin of a single photon. Maybe one photon is a better quantum than two strictly correlated photons?
No comments:
Post a Comment