8 years ago

## Wednesday, October 6, 2010

### Koide Review II

In other words, we can now write the three Koide eigenvalues at $r = \sqrt{2}$ as displaying clearly the nine entries of the tribimaximal matrix. Observe the appearance of Hadamard type phases in the lower right block of terms. These occurred when we expressed the mixing matrix in terms of Fourier transform matrices, namely as $F_3 F_2$. A child could calculate the lepton masses. In the natural mass units fixed by these expressions, the mass triplet sums to $2$.

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Absolutely brilliant, and needs to be published quickly. Can it generalize to the CKM? Wish I could think about it more, but I'm cramming for the physics GRE on Saturday.

ReplyDeleteGood luck with the exam. Of course I am thinking about the CKM matrix ... but trying to publish is really just a waste of my time.

ReplyDeleteFor example, I get a mass sum of $2$ when I use the CKM sign assignments for $(1, \cos \theta, \sin \theta)$ as follows:

ReplyDelete$0.2252 + 0.9735 + 0.0406$

$0.9743 - 0.2250 - 0.0098$

$0.0035 - 0.0416 + 0.9991$

which is pretty cool, but referees typically view this as numerology.

Ah ... that's a cosine rule for the probability sums.

ReplyDeleteA possible justification for the square root of mass matrices identified as consisting of circulants might be in terms of super-symmetric quantum mechanics.

ReplyDeletehttp://en.wikipedia.org/wiki/Supersymmetric_quantum_mechanics

In the simplest SUSY quantum mechanics (having nothing to with super-symmetric QFTs) Hamiltonian can be expressed as H= QQ, Q the hermitian super generator. Hamiltonian would be replaced with mass matrix now and Koide eigenvalues would be eigenvalues of super-generator Q.

Why super-generator would be magic matrix is an open question.

Koide mass formula would say that the Hermitian super generator satisfies

Tr(Q^2)/Tr(Q)^2 = 2/3

Maybe this condition and magic matrix property express some special symmetry.