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Looking for a formula which gives good values for the charged lepton masses and which satisfy the Koide relationship, I derived this formula (note the square):

$$\{m_e, m_\mu, m_\tau\} \propto eigenvalues \left( \begin{bmatrix} 1 & 1 & \frac{1}{2} \\ \frac{2}{3} & 1 & 1 \\ \frac{1}{3} & \frac{2}{3} & 1 \end{bmatrix} ^2 \right) $$

This gives mass ratios of approximately $1:208:3498$ which is quite close to the real ratios of $1:206.768:3477.15$ (Within 0.6%).

This could come from a Lagrangian as a term like: $$\overline{\psi_L} M^2\psi_R$$ where $M$ is the above matrix and $\psi$ is a vector containing the three families.

There are other similar matrices which give the same eigenvalues. I'm disappointed I couldn't find a symmetric matrix with similar properties. My question is, is this an acceptable formula for charged leptons masses? Or would something rule this out such as it not being symmetric? Is there a way to make it symmetric and keep the same eigenvalues?

I think this is one of the simplest squared rational matrix that is this close and satisfies the Koide formula unless you can prove otherwise.

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  • $\begingroup$ I am divided about this question. I think that "should such a mass matrix be symmetric" may make for an interesting physics question. But "I contend ... etc." (v2) suggests that this might fall under our non-mainstream close reason. $\endgroup$ – rob Apr 11 '18 at 2:26
  • $\begingroup$ You could say it is numerology, but you could say there is good reason to try and find a simple rational matrix which gives close eigenvalues to the masses. In this case also one where the eigenvalues satisfy the Koide formula. Probably the question could be posed better. $\endgroup$ – zooby Apr 11 '18 at 2:44
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    $\begingroup$ Note that the ratios given in the formula, while, close, are way, way, way off considering how well we know the masses of these particles. The electron and muon masses (pdglive.lbl.gov/Viewer.action) are known to about 1 part per billion, and even the tau mass is known to one part in 10,000, so your formula is off by hundreds of thousands of sigma, at least. $\endgroup$ – probably_someone Apr 11 '18 at 2:54
  • $\begingroup$ Well I'm only looking at mass ratios not their absolute value in whatever units. And so just dividing the real ratio by the calculated one it is within 0.6%. That's pretty good and surprising considering the simple fractions. Obviously different from the real values though. $\endgroup$ – zooby Apr 11 '18 at 3:05
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    $\begingroup$ @zooby No, it's still not good, because the uncertainty on a ratio of two quantities is dependent on the uncertainty on those quantities in a well-defined way. In particular, if the numerator has an uncertainty of 1 part per billion and the denominator has an uncertainty of 1 part per billion, then the ratio will have an uncertainty of $\sqrt{2}$ parts per billion. So you're still off by hundreds of thousands of sigma for the ratio of muon mass to electron mass. en.wikipedia.org/wiki/Propagation_of_uncertainty $\endgroup$ – probably_someone Apr 11 '18 at 3:23
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This is certainly a "better" formula than most physics numerology out there, if we judge it by the prospects of embedding it into a fully functioning quantum field theory.

There are hundreds of papers out there which do something similar, by postulating a "texture" for the yukawas. The structure of the texture is then usually derived from the new symmetries, interactions, fields, and charges of an underlying Lagrangian.

The main peculiarity here is that the yukawas come from the square of your matrix. Koide has written perhaps a dozen papers of this form, in which the yukawas come from the square of the vevs of fields he calls yukawaons.

For expert commentary, I would suggest writing to Koide, and to a professional model-builder like Stephen F. King.

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  • $\begingroup$ Yes, I had a look at some of those papers. I never really understood why particular "textures" were chosen. A "texture" just seems to say that some of the matrix elements are equal and so there are less free parameters just because there are too many free parameters. $\endgroup$ – zooby Apr 11 '18 at 2:40

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