Neutrinos Again II
Recall that a renormalised circulant matrix is a kind of magic square, where we don't worry about summing along diagonals. In neutrino physics, the unitarity of mixing forces the (squared) mixing matrix to be a magic square with rows and columns summing to 1. The tribimaximal case was first discussed by Harrison et al, where the whole matrix follows from the entries $U_{13}$, $U_{23}$ and $U_{12}$. Labelling columns by $\nu_{1}$, $\nu_{2}$, $\nu_{3}$ and rows by $e$, $\mu$, $\tau$ the matrix $U^{2}$ is
$\frac{2}{3}$ $\frac{1}{3}$ $0$
$\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{2}$
$\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{2}$
In terms of the standard mixing angles this corresponds to $\theta_{13} = 0$, $\theta_{23} = \frac{\pi}{4}$ and $\textrm{sin} \theta_{12} = \frac{1}{\sqrt{3}}$ with no additional (Dirac) CP violating phase. Given the excellent experimental agreement with this case, the question is, what is the justification for choosing $U_{13} = 0$, $U_{23} = \frac{1}{\sqrt{2}}$ and $U_{12} = \frac{1}{\sqrt{3}}$? Most physicists expect some deviation from tribimaximal mixing, but perhaps there is a good reason for things being so simple. For instance, observe that we can reorder the columns arbitrarily so that $U^{2}$ is derived (assuming one democratic column) from a diagonal
$\frac{2}{3}$, $\frac{1}{2}$, $\frac{1}{3}$
which is the length 3 Farey sequence. That is, it has the modular group property that for consecutive fractions $\frac{a}{b}$ and $\frac{c}{d}$, one has $bc - ad = 1$.
On the other hand, what mixing do we get if we substitute Carl's neutrino Koide rule for the one assumed by Harrison et al? Note that Harrison et al use the $3 \times 3$ circulant mass matrix for the charged leptons. On using the same quantum Fourier diagonalisation operator for both the charged leptons and neutrinos (see page 7 in Harrison et al) one would find that $U^{\dagger} U = 1$, so the tribimaximal mixing matrix would be replaced by the identity! It is the interplay of $3 \times 3$ circulants and $2 \times 2$ circulants that gives rise to the observed tribimaximal mixing.
$\frac{2}{3}$ $\frac{1}{3}$ $0$
$\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{2}$
$\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{2}$
In terms of the standard mixing angles this corresponds to $\theta_{13} = 0$, $\theta_{23} = \frac{\pi}{4}$ and $\textrm{sin} \theta_{12} = \frac{1}{\sqrt{3}}$ with no additional (Dirac) CP violating phase. Given the excellent experimental agreement with this case, the question is, what is the justification for choosing $U_{13} = 0$, $U_{23} = \frac{1}{\sqrt{2}}$ and $U_{12} = \frac{1}{\sqrt{3}}$? Most physicists expect some deviation from tribimaximal mixing, but perhaps there is a good reason for things being so simple. For instance, observe that we can reorder the columns arbitrarily so that $U^{2}$ is derived (assuming one democratic column) from a diagonal
$\frac{2}{3}$, $\frac{1}{2}$, $\frac{1}{3}$
which is the length 3 Farey sequence. That is, it has the modular group property that for consecutive fractions $\frac{a}{b}$ and $\frac{c}{d}$, one has $bc - ad = 1$.
On the other hand, what mixing do we get if we substitute Carl's neutrino Koide rule for the one assumed by Harrison et al? Note that Harrison et al use the $3 \times 3$ circulant mass matrix for the charged leptons. On using the same quantum Fourier diagonalisation operator for both the charged leptons and neutrinos (see page 7 in Harrison et al) one would find that $U^{\dagger} U = 1$, so the tribimaximal mixing matrix would be replaced by the identity! It is the interplay of $3 \times 3$ circulants and $2 \times 2$ circulants that gives rise to the observed tribimaximal mixing.
posted by Kea | 6:19 PM
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