# Koide Splittings and heavy quarkonium

Work on my paper on the Koide formula continues. I’m in the meson section. Things are going okay, but…., so many mesons, so little time. Here’s a short description of some coincidences in the heavy mesons. I already put up a formula for the angular excitations of the pions on physics forums.

The bound states of hydrogen are, to lowest order, described by spherical harmonics, $\psi_{nlm}$. The energy levels of these wave functions depend only on $n$. The energies of these wave functions depend only on $n$, they are approximately $-13.6/n^2$ electron volts. Since the energies do not depend on l or m, the energy states are degenerate. However some of these degeneracies are split at higher orders.

At the lowest order, one ignores the spins of the proton and electron. Taking these into account we find a spin-spin effect. When the electron and proton spins are parallel, the energy is slightly different from when they are anti-parallel.

In the preon model I’m working on, the electron/muon/tau, and the neutrinos are composed of three preons each. The preons come in two types, charge $Q = \pm 1/3$ and neutral. Naturally, the electron has three charged preons while the neutrino has three neutral preons. The three generations of charged and neutral leptons follow two slightly different forms:

The quarks, having charges between 0 and $\pm 1v$, are supposed to be composed of a mixture of charged and neutral preons. The masses of the quarks cannot be determined experimentally because colored states never appear alone. The simplest particles made from quarks are the mesons, which are composed of a quark and an antiquark. And the simplest mesons are the q-qbar mesons, where the quark and the anti-quark are of the same type. In such a q-qbar meson, there is a mixture of neutral and charged preons and we can expect the same sort of interactions between these preons as breaks the degeneracy in the hydrogen atom.

Of the q-qbar mesons, the ones that are cleanest in the Particle Data group’s figures are the heaviest. Unfortunately, due to difficulty producing it, we have no data on the heaviest q-qbar meson, that from the top quark, toponium. The next heaviest are the b-bbar and c-cbar known as the Upsilon and J/psi, or, sometimes, bottomomium and charmomium.

The Upsilon and J/psi are analogies to the hydrogen atom (and positronium, where the proton is replaced by a positron) in that they are composed of a pair of oppositely charged particles. But the primary force holding them together is the color force instead of the electric force that holds together the hydrogen atom.

Both the Upsilon and J/psi come in 6 energies. These are labeled as if they were radial excitations like the $n$ excitations of the hydrogen atom, but they do not look like radial excitations. In hydrogen, the radial excitations are infinite in number, rather than 6, and they become closer and closer together as the energy increases. With the Upsilon and J/psi, there are only six energies and they seem to appear with random gaps between them. They smell like splittings of a single wave function rather than radial excitations:

Different splittings present different patterns in the energy spectrum. A spin-1/2 splitting will separate the energies into pairs of energies. Spin-1 gives triplets of energy values that are equally spaced. When one is faced with six energies:

In the literature, the higher excitations of the J/psi are labeled as psi alone for historical reasons. In the context of Koide mass formulas, it’s natural to see if the above six masses can be written as a Koide split.

For the leptons, the Koide formula is simple in that it includes a sqrt(0.5). For a splitting, these parts cannot apply and we can look only at the two angles, $2/9$ for the charged leptons and $2/9 + \pi/12$ for the neutral ones. This means that the general formula for six particles following a Koide splitting is:

That is, for n=1,2,3, the above gives six completely arbitrary particles that satisfy Koide’s relations. If we put the right numbers into the above, the formulas will spit out the masses of the leptons, and we will find the additional restriction that the v’s differ from the s’s by a factor of the square root of 2. Perhaps we will notice some coincidences on fitting the Upsilon and J/psi mesons.

In fitting six particles to a Koide splitting, we have choice in how we assign the six particles to the two groups of three. Then, since each group of three is ordered in mass by n=1,2,3, the assignment is defined. Therefore, there are 6 choose 3 = 20 ways we can assign the particles. So it is not much of a surprise to find some sort of a fit this way. What matters here is how significant the fit is.

I’ve written some java code to find Koide splittings in groups of six numbers. One way of determining the goodness of fit is to look at the 10 ways of splitting the six states into two groups of three, and then fitting a more general Koide formula, that allows any multiple of the angle $\pi/12$. That is, one can fit $\sqrt{m_n} = v + s\cos(2/9 + k\pi/12 + 2n\pi/3)$, and then see if a fit with k=0 for one group of 3, and k=1, for the remaining 3 energies, stands out as a particularly good fit.

Because of how the equations work, when one allows s negative as well as positive, the values of k that are of interest are k=0,1,2, and 3. If one were to choose two of these cases at random, one would find the case k=0 and k=1 to happen 2 ways in 4×4 = 2/16 = 1 eighth of the time. Doing this twice, once with the Upsilon mesons and again with the J/psi, reduces the chance of a Koide formula winning to one 64th.

In addition, the resulting equations for the Upsilon and J/psi masses is a remarkably good fit to the masses. Not only are the Koide fits better than the best fits of alternative selections, in terms of maximum error, they are considerably better than the average. I’m guessing that running a simulation of random masses will show the Koide fit is remarkably rare. The splitting fit for these mesons is:

where an overall factor of 25.05 sqrt(MeV), as used in the lepton formulas, has been left off.

The fits to the masses are as follows, for the Upsilon neutrino type masses, (measured, Koide fit):
9460, 9451.8
10023, 10041.0
10579, 10569.1
Upsilon electron type:
10355, 10355.1
10865, 10864.3
11019, 11019.5
J/psi neutrino types:
3096.9, 3096.9
3771.1, 3773.8
4421.0, 4418.4
J/psi electron types:
3686.1, 3686.1
4039.0, 4040.3
4153.0, 4149.8

In both cases, the electron fits are better than the neutrino fits. And the neutrino fit particles are generally lower in energy than the electron fits.

The next mesons down are the s sbar, which are the phi particles. There are only two of these, however, one of what is now classified as an omega, which has the same quantum numbers as the phis, completes a (neutrino-type) fit for three particles. The extra particle is what used to be the X(2290), now the omega(2290). It is still a not very well defined meson and is not “included in the summary table”. Under this, the Koide fit is v0,s0 = 1.604143, -0.372535.

Without the omega(2290), there are seven remaining omega mesons. About half of them are not included in the summary tables, but by assuming one of them is in error, we can fit the remaining six well with a Koide type splitting with the parameters: v0,s0 = 1.491067, -0.423105 and v1,s1 = 1.808595, -0.185822. The particles involved range from 782 to 2330 MeV. By and large, the lighter mesons have masses that are not as accurately determined and the formulas are not so convincing.

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### 8 responses to “Koide Splittings and heavy quarkonium”

1. Formulas for vaguely known masses may be seen to give theoretical predictions for later mass determinations, which is a good thing.

2. carlbrannen

Kea, what I need are formulas for the v and s. Some of the numbers are kind of suggestive, and I do have various preliminary guesses, but to predict masses that’s what you need.

The Koide splitting takes two degrees of freedom out of six masses. If you had a spin-1/2 splitting on a group of six masses, they would presumably split into three pairs of masses. If you also had that each pair was separated by the same gap, then a spin-1/2 splitting would also take two degrees of freedom out of six particles.

Thus a Koide splitting is equivalent, in degrees of freedom, to the spin-1/2 splitting that would be so obvious to us all, but a Koide splitting only shows up when you write the masses in Koide form.

So here’s the question. Is there a transformation you can taken on six objects that will make it as obvious that it is a Koide splitting as the spin-1/2 case? I’m wondering if there’s a discrete Fourier transform like what you’ve talked about.

3. nige cook

This is very interesting. I get lost in the fourth paragraph where the first two equations are introduced. Are these obtained anywhere from the model described in the earlier paragraphs, i.e. the energy level model?

The numerical agreements for predicted and observed masses of many particles which you give later in the article are very impressive. I’m trying to grasp the details of the mechanisms involved.

Koide’s formula always gets me thinking about averaging in statistics, because in its initial form it is applying a special kind of averaging to the lepton masses. To find a standard deviation, you take the square root of the sum of the squares of the error of each data point from the mean. One reason to square the errors here is that the data points can be either above or below the mean, and if you just subtract the data point from the mean you will get a summation that on average contains as many positive as negative terms, giving zero. That’s a mistaken error analysis. Squaring the amounts of error for each data point included in the standard deviation calculation automatically gets rid of the negative signs for components where the data point is less than the mean which is subtracted from it. However, you could just take the modulus (square root of the square) of each instance of a data point minus the mean, which would get rid of minus signs in the sum of terms. The really special function of adding up the sum of the squares of the errors from the mean, and then taking the square root of the result, is that the standard deviation you obtain emphasises or exaggerates those contributions to the error which come from the largest individual errors in data points. The Koide formula does the exact opposite, by summing the square roots of data points (masses) and then squaring the result, you are doing an averaging which exaggerates the contributions from the smallest data points (the smallest masses). In a qualitative way, this makes sense because the lower the mass of the lepton, the more stable it is and the more predominant it is the the virtual particle soup: virtual electrons occur in the vacuum out to the greatest distance from real particles, but virtual muons and tauons require more field energy and only occur much closer, so populate a smaller space and so have less contribution to the total mass of a real particle.

I’m wondering about the mathematical connection between the model analogy of the hydrogen atom with energy levels of E = -13.6n^2 eV, and the Koide formula. Taking your first paragraphs literally, if we put E=mc^2 into the hydrogen atom energy level formula E = -13.6n^2 eV, then the integer number of the energy level, n, is directly proportional to the square root of the mass. This will explain the use of the square root of mass terms in the Koide formula, but you have three preons rather than just the two particles in hydrogen, so the first couple of formulae you give have the square root of mass proportional to a constant plus a cosine expression containing n.

4. carlbrannen

Nige, the formulas for the leptons (in the 4th paragraph) are already in the literature in slightly different form, with a square root of 2 instead of a square root of 1/2. See my paper, The Lepton Masses.

Gerald Rosen uses the same formula in Mod. Phys. Lett. A 2007, 22, 4 283-288. In his abstract, he’s squared the formula, and the number 313.85773 is half of my 25.054, squared.

About the RMS formula, the fits I give are calculated by computing the RMS error for the square roots of the mass. This means that the formulas were chosen to minimize not the average squared error of mass, but instead the average of the squared square root of mass which amounts to what you’re talking about. But this is done just because it is the simplest way to get the computer to do the work.

The finished paper will have dozens of fits. What I need is a decent measure for goodness of fit. I’m not sure what this should be. Since I’m doing RMS fitting to the Koide formulas, I guess one way is to compute some sort of chi-squared error. Another way is to vary the 2/9 and see if there is a dip at 2/9. And another way is to make up random data and see how many times it gives a fit better than that observed.

5. Carl, some sort of chi squared is probably expected here. I don’t think you need to worry about anything else.

Is there a transformation you can taken on six objects that will make it as obvious that it is a Koide splitting as the spin-1/2 case? I’m wondering if there’s a discrete Fourier transform like what you’ve talked about.

Sure, there’s a Fourier transform on any number of objects. Of course, I haven’t looked at your six-plets yet, but I guess we’ll have to do this.