Earlier we postulated the snuark mass interaction for a transition from a left to a right handed state as a simple iq vertex stuck between two propagators:
We postulated that the mass interaction that converts a left handed lepton to a right handed lepton involved three of these transitions happening simultaneously. We put the left handed snuarks as +x, +y, and +z, and the right handed snuarks as -x, -y, and -z. In doing this, because the snuark interaction forbids transitions between incompatible quantum states (for example from +z to -z), we found that there were only two ways this could happen. We labeled these two complex interactions as J and K. These two interactions amounted to the even permutations on three objects (not including the identity). They were (x,y,z) goes to (y,z,x) or (z,x,y). To distinguish the right and left handed states, we wrote these as (+x,+y,+z) goes to (-y,-z,-x) or (-z,-x,-y). All other permutations on the three objects (x,y,z) were forbidden because they included a forbidden interaction such as +x goes to -x.
Our analysis was correct in that we did find all the final states (two of them), but it was incorrect in that we did not include all the possible intermediate states. In this post and the next, we will sum over these more complicated interactions and compute an effective mass interaction. The calculation is quite similar to a similar calculation in standard QFT: the treatment of soft photons as a vertex correction to the interaction between an electron and a hard photon. See section 6.5 of Peskin and Schroeder.
In the previous post, we looked at long term bound snuark states, came up with three coupled quadratic equations in complex numbers I, J, and K:
noted that these equations are equivalent to a matrix idempotency equation:
wrote down the eight solutions:
and promised to interpret them in the next (this) post. Okay.
In the previous post we showed how we can take the left to right snuark mass interaction, and combine it with the right to left interaction, to make a left to right to left interaction, which we will somewhat abusively call a “LRL propagator”. The reason for calling it a propagator is because we are going to ignore the gauge bosons; they’re pre gravitons and carry very little energy so they don’t matter much anyway. What’s left is fermions going in and fermions going out, just like a propagator. This was a Feynman diagram of order 6. That is, the coupling constant “iq” shows up 6 times.
The LRL propagators came in 4 types, two of which were identical. We wrote them separately for each of the three fermion lines. Each of these lines ended up with a vertex and a complex coupling constant. These complex coupling constants are scalars, so we may as well combine the three of them into a single scalar coupling constant. Putting the LRL propagators back together, the four cases are:
In the standard model, the proton is made up of three quarks. The individual quarks are spin-1/2 particles. As elements of a QFT, the quarks are represented by propagators that satisfy the Dirac equation. What’s a bit odd is that the proton is also a spin-1/2 particle and is also represented by that same Dirac equation propagator.
Snuarks have a similar attribute. Three snuarks are (more or less) spin-1/2 particles and are represented by Dirac propagators. In the qubit representation, where we ignore spatial dependencies, their propagators are Pauli projection operators (i.e. density matrices). Somehow these three qubit objects combine to make a lepton or quark, which is also a qubit object whose (virtual) propagator is again a Pauli projection operator. In Feynman diagrams, what we are looking for is something like this:
What we would like is a way of combining the propagators of the three snuarks into a single object that can fill the requirement of representing the propagator of the quark or lepton they combine into. This is the “bound state propagator problem” which we will return to more completely when we analyze the quarks. For the moment, let us consider the problem of how a bound state of three snuarks can model the lepton mass interaction that converts left handed leptons to right handed leptons and vice versa.
The snuark algebra is simply a finite subset of the Pauli algebra, the subset generated by spin in the +x, +y, and +z directions. If we were to do our calculations by the usual methods of the Pauli algebra, we would choose an orientation, typically +z, and define the +x and +y states in terms of the +z (spin up) and -z (spin down) states.
Our objective in this post is to relate the snuark algebra to the more traditional ways of calculating with the Pauli algebra. We will do this with the objective of turning the snuark algebra into a QFT, just as qubits can be turned into a QFT. And it will be the QFT theory of snuarks that will allow us to derive the Koide mass formulas.
Everybody knows that the quantum states of spin-1/2 particles come in two and only two independent spin orientations, for example spin +z and spin -z (up and down). My snuark theory uses spin-1/2 calculations to compute the Koide mass formulas, so it is a little inconvenient that snuarks need to come in six independent quantum states, +x, -x, +y, -y, +z and -z.
Three snuarks (no two antiparallel) combine to make one chiral handed lepton, the other three combine to make another chiral handed lepton, but traveling in the opposite direction. One chiral handed lepton could be a left handed electron moving in the +u direction, the other would be a left handed electron moving in the -u direction. Thus the six snuark degrees of freedom get turned into two spin-1/2 degrees of freedom.