So the set of nine unscaled Koide eigenvalues for the lepton masses
may now be expressed as the matrix product
where the left hand factor is the tribimaximal mixing matrix. The right hand factor does not look like a probability matrix, but we could take a fixed constant off all the $\lambda_{ij}$ so that this factor, and also the tribimaximal matrix, take the form
The two factors use the values $(\delta, \theta) = $ $(0, \omega)$ and $(2/9, \pi/12)$ respectively. Similarly, the democratic matrix offset uses $(\delta, \theta) = $ $(0, \pi/4)$. Note that we need the $\sqrt{2}$ coefficients to make magic matrices, when the angles are right.
where the left hand factor is the tribimaximal mixing matrix. The right hand factor does not look like a probability matrix, but we could take a fixed constant off all the $\lambda_{ij}$ so that this factor, and also the tribimaximal matrix, take the form 
The two factors use the values $(\delta, \theta) = $ $(0, \omega)$ and $(2/9, \pi/12)$ respectively. Similarly, the democratic matrix offset uses $(\delta, \theta) = $ $(0, \pi/4)$. Note that we need the $\sqrt{2}$ coefficients to make magic matrices, when the angles are right.
Note that we don't always get magic matrices for the two rows of $\sqrt{2}$ coefficients. It will be magic if the sum of cosine (sine) squares is $3/2$. For example, the angles $(\pi/4, \pi/12)$ give a solution.
ReplyDeleteSimilarly, for any $\delta$ the angle $\theta = \pi/3$ (sixth root) gives a magic matrix.
ReplyDeleteAh, so for the $\sqrt{2}$ coefficients and for ANY $\theta$ and any $\delta$ we have magic matrices for $(\pi/4, \theta)$ and $(\delta, \pi/3)$. Sixth roots and eighth roots.
ReplyDelete