# Long Lived Snuark Bound States

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 original left to right and right to left interactions, “J” signified that +x went to +y (red went to green), with the other letters in cyclic permutation. “K” signified that +x went to +z, etc. In the above, we see that the combination JJ gives the “K” color change, while KK gives the “J” color change. And JK and KJ both give no color change at all. We will label the “no color change” as “I”. These labels are given on the right side of the equal signs in the above diagrams.

The JK and KJ interactions both give the “I” color change. While their internal details are different, the two interactions are identical on the outside: they have the same inputs and outputs, and they have the same scalar amplitude. We may as well combine them into a single diagram. As with the usual QED Feynman diagrams, the scalar amplitudes for identical diagrams add and we can replace those two diagrams with a single diagram with a scalar amplitude of 0.125 + 0.125 = 0.25.

The Even Permutation Group on Three Elements

The color changes we have labeled as “I”, “J”, and “K” form a group. The product of two group elements is how colors are changed when two consecutive changes are made of the given types. In our Feynman diagrams above, we found that JK = KJ = I, JJ = K, and KK = J. The missing operations are IJ = JI = J, and IK = KI = K. This is the even permutation group on three elements. Unlike more general permutation groups, it is Abelian. (Later, when we work on quarks and baryons, we will use the full permutation group on three elements which is non Abelian.) The group operation is: According to snuark theory, the electron, muon, and tau are bound states, our objective is to understand these bound states. While the electron is stable, the muon and tau are not, and to this extent our calculations are only first order approximations because we will assume that they are precisely stable. A good project for a graduate student would be to extend these first order calculations to the next order by including decay probabilities in the bound states.

To the extent that we can characterize these bound states as stable, we can imagine extremely complicated Feynman diagrams that modify the colors of the snuarks. According to the assumptions we made about the mass interaction in the previous post, even these extremely complicated Feynman diagrams will be classified into one of the three color permutation group elements, “I”, “J”, or “K”.

In order to allow us to add two Feynman diagrams together, we need to make sure that they have the same input and output states. For qubits, this means that we need to assume that the initial and final state times are the same. So let’s assume that the Feynman diagrams all have a time T between their initial and final states. We want T to be short compared to the decay times of the states we are studying, but long enough for the Feynman diagrams to reach “stability”.

Each of the complicated “I” type Feynman diagram has the same three incoming propagators and the same three outgoing propagators. So just like the JK and KJ (LRL) Feynman diagrams, we can combine all these various “I” Feynman diagrams into a single equivalent diagram. This diagram will look just like the JK or KJ diagrams in the illustration at the top of this post, but the complex scalar amplitude could be different.

The same thing can be done with the “J” and “K” type Feynman diagrams. Thus we end up with three infinite collections of Feynman diagrams. We sum over each group. This turns each infinite set of Feynman diagrams into a single effective Feynman diagram of the “I”, “J”, or “K” type, each with a single complex amplitude. We will call the scalar amplitude I, J, or K. For example, the I amplitude is given by an infinite series of diagrams that looks like this: In choosing the time between initial and final states T, we wanted the time to be long enough that the Feynman diagrams reached “stability”. Now we see what we mean by this. T needs to be long enough that the sum over the Feynman diagrams, when reduced to the three complex numbers I, J, and K, does not depend much on T. In other words, we want there to be no detailed information on the exact time delay T in the number I, J, and K.

If the sums for the various bound states we are modeling are periodic, this might not be possible. Then we require that T be a multiple (or arbitrarily close to a multiple) of all these periods, and we wish to have T be large enough that when it is increased by another such period, the Feynman sums do not change much.

Thus we assume that the Feynman sums for a time delay of T are the same as the Feynman sums for a time delay of 2T. And we can (approximately) use the complex numbers for the time delay of time T to compute the complex numbers for the time delay of time 2T. To do this, we break the 2T time interval into two time intervals of length T. There are three possible cases (I,J,K) for the transformation in the first interval and three possible similar cases for the transformation in the second interval.

When we connect two Feynman diagrams together to make a single Feynman diagram, their complex amplitudes multiply. And when we do this, the color permutation group element of the combined Feynman diagram is given by the color group multiplication. For example, the “I” diagrams for the 2T time interval can be built from the T time interval diagrams three different ways; as I then I, or as J then K, or as K then J: Among the 9 group multiplication rules, we have three that give a result of I; namely, II = JK = KJ = I. As shown above, under out assumptions about T, this statement about the color permutation group becomes a statement about the complex amplitudes I, J, K namely $I = II + JK + KJ = I^2 + 2JK$

Similarly, there are three ways that the 2T Feynman diagrams of type “J” can be built from T time interval Feynman diagrams. These give the equation $J = IJ + JI + KK = K^2 + 2IJ$. And the “K” diagrams of time interval 2T give the equation $K = J^2 + 2IK$.

We have three equations in three complex unknowns: These are three coupled quadratic equations.

For the moment, let us note that there are exactly 8 solutions to these equations: where $a,b = \exp(\pm 2i\pi/3)$. In our next post we will cut down these 8 solutions to three that are of interest in the present case. For now, I sense that some of my audience wants me to get to the very practical point of the Koide mass formulas.

Cyclic Matrices and Snuark QFT

It is sometimes convenient to assemble I, J and K into a cyclic 3×3 matrix: Then the three quadratic equations in 3 unknowns amount to the matrix equation $C^2 = C$. Thus C is an idempotent matrix.

Of course density matrices and the virtual propagators we use in snuark theory are also idempotent. And it should not be too much of a surprise that the Koide mass formula is conveniently written in cyclic matrix form. That is, if you square the eigenvalues of the following cyclic 3×3 matrix: you end up with three numbers that are very close to exactly proportional to the masses of the charged leptons, that is, the electron, muon, and tau. In this form, the Koide formula arises from the square root of 2. To get an exact proportionality, to current measurement error, one replaces 2/9 with “that damned number”, 0.22222204717(29).

Regardless of the constant 2/9, the three eigenvalues will satisfy: which is the original Koide relation. This is not a derivation of the formula, just a preview to whet the appetite. Similarly, the masses of the neutrinos are proportional to the squares of the eigenvalues of the matrix: Uh, the right hand column of the above matrix didn’t quite fit, but it’s a cyclic matrix so you can fill it out at home.

Stuff to come

The above is not at all a derivation of the Koide formulas, though we will eventually get to that. Before we get anywhere we need to look at the 8 solutions to the cyclic matrix idempotency equation and make sense of them. That will probably be the next post.

To get a real derivation of the lepton masses, there are three or four more posts after that required. The first problem is that the charged and neutral leptons differ in that one has the $3i\pi/4$ factor that we expect (which will be turned into a factor of $i\pi/12$ when you add $-2\pi/3$, a phases are contributed by the eigenvectors, thereby matching somewhat obsolete previous papers written by this author). This will be covered by the post after the next, I think, which will be devoted to the quantum numbers of the snuarks and the interpretation of “i = xyz” as an operator in the Pauli algebra, or more precisely, an operator: i = (xt)(yt)(zt) = xyzt in the 4+1 Clifford algebra.

The second problem is that a rational fraction has appeared as a complex phase, both in the charged leptons and the neutrinos (and will later appear in the baryons and mesons as well). The calculation we will use to derive this correction will be identical to the one used to account for undetectable soft photons in QED. Soft photons and gravitons have very low energies, so the QFT calculations are somewhat similar, with a slight difference in how one adjusts for over counting duplicated configurations. (QED divides by n!; we will have to divide by (n/2)! because our pre gravitons will appear in pairs). This will be the infrared correction to the bare snuark mass interaction we discussed a few days ago. At that time we will prove that the same correction works for the baryons as well.

The third problem is why the rational fraction that appears happens to be 2/9. The 2 will arise from the fact that each snuark is composed of two primitive idempotents, the 9 comes from the magnitude of the natural coupling constant. The solution to this problem will give clues for how it comes to be that the gravitational coupling constant is so weak.

After that, we will discuss the equivalents of the Koide mass relation in the structure of the baryon resonances, and after that, if I get lucky, we will turn to the meson resonances. The baryon and meson resonances include several hundred masses, we will be discussing a common formula that predicts them all. And somewhere along the line we have to give a derivation for the gravitional force as due to snuark QFT.