To recap the previous post we began by combining the SU(2) spin-1/2 and U(1) operators into 2×2 matrices. We then showed that the leptons were solutions of the idempotency equation UU = U for 2×2 matrices subject to the additional requirement that the solutions be eigenstates of electric charge Q. For pure density matrix formalism, individual particle states are represented by primitive idempotents (with trace = 1), so we then converted these idempotents into primtive form by embedding them into 4×4 matrices. In doing this, we found that the idempotents given by the 2×2 matrices were composite, each being composed of two sub particles.

**1-Circulant and 2-circulant matrices**

In this post, we add the quarks to the picture. To do this, we need to use the 1-circulant and 2-circulant 3×3 matrices Kea talks about. We will write the general 1-circulant and 2-circulant matrices as follows:

Where I, J, K, R, G, and B are complex numbers. Note that there are only 6 complex degrees of freedom in the 1-circulant and 2-circulant matrices, one cannot create an arbitraray 3×3 matrix, with 9 complex degrees of freedom, from 1-circulant and 2-circulant matrices. In addition, setting R=G=B=1 gives a matrix of 1s, the same as setting I=J=K=1. Consequently, the 1-circulant and 2-circulant matrices together, have only 5 complex degrees of freedom, about half that of the 3×3 matrices in general. Writing a 3×3 matrix as a sum of a 1-circulant and a 2-circulant matrix is very restrictive; to write it as just a 1-circulant is even more so.

One obtains the basis for these matrices by setting one of the elements to 1 and the rest to zero. For example, putting I=1, J=0, K=0 gives the unit matrix. These 3×3 basis matrices correspond to permutations on three elements. We will think of the three elements being permuted as red, green, and blue, hence the labels R, G, and B for the 2-circulant matrices (i.e. R labels the permutation that leaves red unchanged and swaps green and blue, etc.). Similarly, “I” labels the permutation that leaves nothing changed, while J and K are the non trivial even permutations.

The 1-circulant and 2-circulant matrices are closed under multiplication. The square of either type is a 1-circulant, and the cross product is a 2-circulant. For this reason, to generalize a 2×2 matrix into 1,2-circulant form, one naturally places the 1-circulants on the diagonal and the 2-circulants on the off diagonals. The result is a 6×6 matrix:

The above defines a subgroup of the 6×6 matrices. That is, it is closed under addition and multiplication. Since we are working in pure density matrix theory, we are interested in the idempotents of these matrices; these will correspond to the quarks and leptons. If desired, we can convert our idempotents into primitive idempotents in a method similar to that of the previous post, the result will be 12×12 matrices where the quarks and leptons are the primitive idempotents.

Squaring the above matrix and requiring that it be unchanged (i.e. idempotent) gives us a set of 36 coupled quadratic equations, however 30 of these equations are duplicates and can be ignored. The remaining 6 equations are:

The general solution for the above 6 coupled quadratic equations is difficult to find, but they are also subject to the requirement that the solutions be eigenstates of electric charge. Note that these quadratic equations encode the permutation group on three elements; for instance, the three odd permutations R, G, and B each square to the identity and the quadratic equation for the identity includes terms for . In general, if XY = Z, then there will be an XY term in the quadratic equation for Z.

These equations define the rules that a set of Feynman diagrams giving color permutations must satisfy in order to be self consistent; that is, one requires that the Feynman diagrams be defined in terms of themselves. One is reminded of the old bootstrap version of particle theory.

We already know four solutions from the 2×2 matrix lepton case:

We have some choice in how to convert the above matrices into 6×6 form. We’d like to keep weak hypercharge as the diagonal “I” element, so we’ll convert the diagonal values on the old matrices into the new unchanged. We can’t do this with the off-diagonal elements, so we will divide those elements by 3 and spread them equally into the R, G, and B components. The resulting 6×6 lepton idempotent matrices are:

which the user can verify are idempotent. Note that the 1/6s appear when one splits the weak isospin 1/2 into three equal parts. It is also possible to split these up in various other ways, and this is natural for the quarks (where it can be used to represent the three different colors of quarks), however, when one does this, the sum, R+G+B is preserved.

Consequently, we define the weak hypercharge to be the value “I” down the diagonal as before, and we define the weak isospin to be R+G+B. As with the leptons, we will require that the weak hypercharge quantum numbers be positive, and define the negative weak hypercharges to be the quantum numbers of anti-particles. This shuffles the definition of particle and antiparticle a bit, but they still form pairs as is required. Then the desired quarks / antiquarks are:

The weak isospin values +- 1/2, become +- 1/6 in the top right and bottom right quadrants as before. The weak hypercharge values, 1/6, 1/3, and 2/3, go on the diagonal. The values for J and K are then determined. We find:

**Necessary and Sufficient Quantum Numbers**

Thus all the quark and lepton weak hypercharge and weak isospin quantum numbers (as pairs of numbers) can be put into 6×6 matrix form. In addition, these assignements exhaust the possiblities for the 1-circulant / 2-circulant matrices. That is, if a 6×6 matrix has the I, J, K, R, G, B form, and is idempotent, then the pair (I, R+G+B) is a pair of (weak hypercharge/2, weak isospin) that is observed in the standard model. That is, the 1-circulant / 2-circulant matrix form defines precisely the weak hypercharge, weak isospin quantum numbers.

This is reminiscent of how Schroedinger’s equation, combined with Pauli’s spin-1/2 for the electron, precisely defines the structure of the excitation states of the hydrogen atom (and comes close to giving the periodic table of the elements). One supposes that there must be something fundamental to the 1-circulant / 2-circulant form, probably having to do with the permutation group on 3 elements.

Perhaps there is a theorem in the QCD literature that requires this form for weak hypercharge and weak isospin on the basis of renormalization or gauge theory; I’ve not seen it but the literature is large, I’m small.

One thing that is missing from the “necessary and sufficient” defintion of the weak hypercharge and weak isospin quantum numbers are the solutions for the antiparticles; that is, the solutions which have negative weak hypercharge. We can obtain these by generalizing our idempotency equation from requiring UU = U to be the more general . The solutions with the minus sign can be obtained by negating the solutions for the positive sign and this negates the quantum numbers as well. This doubles the number of particles defined by the idempotency equation and what’s more, the quantum numbers are still represented by all the possible solutions.

If a particle did satisfy the equation UU = – U, and we interpret U as the Feynman diagrams corresponding to the permutations of the colors of the preons (snuarks) making up the particle, then the existence of only positive weak hypercharge solutions to the usual idempotency equation, UU = +U, can be interpreted as the sign change being eliminated by repeated squaring. Since the signs (and phases) of quantum states are arbitrary, more generally we may be interested in solutions of the form where k is some real number. As it turns out, this generalization will lead us to the method for extending this model to include

the fermion generations (and therefore 3x as many particles) in the same structure as the quarks and leptons.

**Generations and Koide Formulas**

We now have 6×6 matrices representing the 4 leptons and 4 anti-leptons, the 4 quarks and 4 anti-quarks, and we can represent the various quark colors by breaking the symmetry of R, G, and B (which, it should be noted, can cause the J and K values to change) and so have 12 quarks and 12 anti-quarks. This is a grand total of 32 particles which is quite a lot. What’s more, our definitions are necessary and sufficient from the idempotency rule. What remains is to allow the modeling of the particle generations in the same structure.

There are three particle generations, in this sort of thing Kea and I model generations as a form of triality. Triality can be thought of (in my simple mind) as the three cubed roots of unity. We will use the three roots of unity to modify the 6×6 matrices in ways that leave them idepotent, but give three different solutions for each of the 32 matrices defined above.

Let w be a cubed root of unity. Then the following transformation of a 1-circulant / 2-circulant matrix preserves idempotency:

Other transformations on 6×6 matrices could be considered but the above is somewhat special in that it exists on the 3×3 submatrices as an identical transformation. In addition, it is the transformation that is implied by the Koide formula for the lepton masses. An earlier post, where this method of modeling the generations was first introduced Quarks, Leptons, and Generations, discusses the method at much greater length, in relation to the Koide formulas for the leptons.

**Circulancy, Koide, and the Quarks**

The quarks presumably also follow Koide mass formulas but quarks are not observed as bare particles and so their masses are difficult to measure. The mesons, which are composed of quark / anti-quark pairs, provide the simplest composite particles containing quarks. The standard approach to the problem is to fit curves based on QCD theory to the lowest meson resonances however this method universally fails to obtain the higher resonaces.

The cleanest results on are in “heavy quarkonium”, in particular the Upsilon and J/Psi resonances. These each have 6 observed resonances and standard QCD is compatible only with the lowest two or three. See, for example, chapter 3 of Heavy Quarkonium Physics, the 2004 paper by the quarkonium working group. On the other hand, all 6 of the resonances quite exactly fit two pairs of Koide triplets as shown on this blog. Compare Koide Splittings and Heavy Quarkonium with the column of figure 3.1 (page 91/73) of the above linked “Heavy Quarkonium Physics” paper.

**W and Higgs Bosons**

The above definition of the quarks and leptons into I, J, K, R, G, B form suggests a similar attack on the Higgs and W bosons. First the Higgs.

The Higgs is the boson associated with the transformation of a left-handed particle to its right handed form and vice versa. It would be ideal if we could have the Higgs be defined to be the difference between these pairs. In this endeavor, the obvious pair to begin with is the because our model of the $\nu_R$ is entirely zero. Accordingly, the Higgs is defined to have the same I,J,K,R,G,B quantum numbers as the . This gives the quantum numbers for the Higgs as:

The value of the J and K quantum numbers are not fully determined in that there are three generations of these neutrinos; each of the has the same J and K quantum number, i.e. 0, but the carry different quantum numbers for J and K. (And the R, G, and B are a bit of a mess.) But the J and K quantum numbers will carry the same magnitude, with phases differing by factors of cubed roots of unity.

The remaining pairs of leptons and quarks give the same Higgs quantum numbers, again with choices as to the phase of the J and K portions (each of which can be arranged to have the same magnitude, as with the neutrinos).

The important thing to note here is that, along with the fundamental fermions, the Higgs falls under the model of the elementary particles as 6×6 1-circulant / 2-circulant idempotents. This is consistent with all these objects as being composed of subparticles which can be modeled as permutations of the colors R, G, and B. The nature of the Higgs is sufficiently mysterious that we will leave it here.

The action of the W boson is to swap the left handed leptons or the left handed quarks. Looking at the (I,J,K,R,G,B) quantum numbers for these, it is apparent that a solution for the W is (I,J,K,R,G,B) = (0,0,0,1/3,1/3,1/3). This is not an idempotent solution, but it can be put into this form if it is taken to be the sum of left and right handed W bosons, for example:

**More Stuff to Do**

Since the above gives a unified description of the fundamental elementary particles in terms of very simple 6×6 matrices with 1-circulant and 2-circulant portions, including the generation structure, it’s natural to hope that it will allow the calculation of all the various properties of the elementary particles.

As mentioned above, the formulas are naturally compatible with the Koide formulas and these formulas provide very powerful simplifications of the arbitrary components of the elementary particles.

The other part of the elementary particles are the mixing angles expressed in the CKM (MNS) matrices, which apply to the quarks (leptons). These arise when a left-handed quark (lepton) is convert to the other left-handed quark (lepton) by the emission or absorption of a W boson. These are all included in the above model so it is natural that one expects that the mixing angles should be determined from the model as well.

Work on fitting the mixing matrices to this form has proceeded apace recently. Beginning with Kea’s brilliant observation that the MNS matrix is approximately in the form of the sum of a 1-circulant and 2-circulant matrix, we rewrote the CKM matrix into that form in this blog: Doubly magic matrices and the MNS The interesting result is that the MNS matrix can be written as a “doubly magic” 3×3 matrix; that is, it can be written as a complex matrix where the sum of the rows and columns are all 1, and the sums of the squared magnitudes of the rows and columns are also all 1.

When written as the sum of 1-circulant and 2-circulant matrices, the MNS matrix is peculiarly simple:

The above simple form has rows and columns whose sums of squared magnitudes give 1. The rows and columns also all sum to the same value, but “this value” differs from 1 by a complex phase. To get the form of the matrix which is doubly magic, divide by “this value”.

So this is the current state of applying 1-circulants, 2-circulants, 3 permutations, and Koide formulas to the elementary fermions. My version is a little plebian; Kea will sharpen the picture with her usual insight and hopefully we will write up a paper. Among the things I need to put together for this is a good statistics package for the fitting of heavy quarkonium to Koide splittings. I’m thinking of an automated fitting program and a Monte-Carlo based on some distribution of mass values.

In addition, I’ve got some clues on how to pull more degrees of freedom out of these heavy quarkonium Koide fits. The basic idea is that the neutrino-like triplets have to appear in two forms, which have opposite angles (but the cosine hides this detail). This means that the neutrino-like triplets are actually a degenerate doublet. Then the electron-like triplets might be taken as sums of pairs of neutrino-like triplets. It seems that this gives the approximate values for the “s” values of the electron-like triplets. And the “s” and “v” values for the neutrino-like triplets are particularly simple in form.

Excellent stuff, Carl! I’m afraid I’m not aware of any papers that contain these observations, but perhaps some knowledgeable old physicists could point us in the right direction. Much to do.

Hi Carl,

“I’m thinking of an automated fitting program and a Monte-Carlo based on some distribution of mass values.”

How many variables does the fitting program need to simultaneously optimize?

Kris

Kris,

Six masses have 6 real degrees of freedom (dof). A Koide neutrino-like fit takes 2 real dof, and the electron-like fit takes 2 more, for a total of 4. Each of these gives three masses, so fitting the Koide removes 2 dof.

By the way, as far as counting dof, this is the same as fitting the masses by three particles with identical spin-1/2 splittings. That is, given 4 real parameters A, B, C and d, the 6 masses from a spin-1/2 kind of splitting would be A-d, A+d, B-d, B+d, C-d, C+d.

There are 6 choose 3 ways of splitting the 6 masses into three electron-like and three neutrino-like. I already have the program to optimize the Koide fit, given 3 masses, but I haven’t programmed it to look at all the 6 choose 3 ways of splitting things up.

Funny thing, when you fit 6 particles as three pairs of spin-1/2 splittings, you also get 6 choose 3 ways of doing it. That is, you get to choose which 3 particles are the lower mass of the pairs; A-d, B-d, and C-d.

Anyway, I’m guessing that the statistics on either of the heavy quarkonium cases will work out to be around 1% or so, and the probability of getting both of them to fit would be the square.

Hi Carl,

Just wanted to mention a free version of the optimization software from Frontline Systems:

http://www.solver.com

A cripple-ware version comes with Microsoft Excel. That one is limited to 10 degrees of freedom. (In paid versions, you get lots of dof and some very sophisticated optimization routines, but I think they don’t come cheap.)

To use the free Excel version, you need the Excel source disc, since it isn’t installed by default. It’s referred to as an “add-in.” (I think you can also download it from Frontline.)

I’m not a big Microsoft fan, but for me this thing has been handy. You just put the squared residual of your fit into a spreadsheet cell, and tell it to optimize that by changing other cells. Probably you know about it already.

Cheers,

Kris

No, I wasn’t aware of it. I have a tendency to like to do things on my own. I guess I don’t mind reinventing the wheel a bit. It turns out that I get along well with java and can write code fairly quickly.

For the neutrino-like and electron-like fits, the calculation to minimize least squares is easy, but an effect is that instead of minimizing the least squared mass, I end up minimizing the least squared (square root of mass). That is, I minimize the sum of the absolute values of the mass errors.

The code I’ve got also allows you to put in different variances on the mass values. This is useful if you’ve got two masses that are accurately known and one that is not; you let the program figure out how to predict the not very well known value from the well known ones. (This is what Koide effectively did in 1982 with the tau mass as a function of the electron and muon, so it’s a nice extra feature for the code.)

Hi Carl,

If you put the sum of the absolute values of the mass errors into a spreadsheet cell, you can optimize that just as easily. A spreadsheet runs slowly, but mainly serves as the front end. The optimization part goes fast. They also have a Java library, though I haven’t used that. (Their other stuff is free too, but only for 15 days.)

To deal with mass variances, I guess you are weighting related items in your residual error formula accordingly? (More weight on agreement with the masses which are better known.) In computer lens design, we call that formula the “merit function.”

It’s easy to get stuck in a local minimum of the merit function if you’re trying to optimize a lot of nonlinear or discrete variables. Frontline also has fancy software that will explore and seek the best global solution, without getting stuck.

Kris, it’s too simple to get stuck. Calculations are done in square root of mass. The predicted square root masses depend linearly on two parameters v and s:

lambda_n = v + s cos(2/9 + 2n pi/3 + d),

where d = pi/12 for the neutrino and zero for the charged leptons. These (Koide) formulas fit the leptons exactly. More remarkably, for the leptons, in both the neutrino and charged lepton cases, s/v = sqrt(2). Also see the factors of sqrt(2) in the lepton mixing matrix.

The sum of the squared errors is just the sum of

(m_n lambda_n)^2. To minimize this, you set the derivative (with respect to s and v) to be zero. Uh, there’s a little detail that when s is negative, you reverse the order of the masses.

To allow different variances, just divide the squared errors by the standard deviations (there’s a square root going on here) and sum, then differentiate with respect to s and v and set equal to zero. Either way you get a pair of equations, linear in s and v, that you solve for s and v. All this takes about as much time to solve and program as it does to read my bad notes on it. The calculation is second nature to those who’ve been drilled in what (back in the day) used to be a college junior statistics class but now is probably taught at better high schools.

For the case of heavy quarkonium, the experimental measurements are quite tight, so I set their variances all equal. This is the same as not using them at all. The experimental error in these masses is quite small. So the Koide fit is not perfect, just much, much better than random chance would give. Ignoring the imperfect fit is similar to making the usual assumption that the pi+ and pi0 have the same mass (and the resulting error is about the same size).

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