The phase $\pi /12$, which appears in the neutrino and antineutrino Koide triplets, has a nice relationship to the lepton $x$ parameter in a general eigenvalue set of the form

$\lambda_{i} = 1 + x \textrm{cos} (\phi + \delta + \omega_{i})$

where $\phi$ is often $2/9$, $\delta = \pi /12$ and $\omega_{i}$ are the cubed roots of unity. That is, at $\omega_2 = 2 \pi /3$, the angle $\delta + \omega_2$ is an eighth root of unity. Thus the corresponding eigenvalue may be expressed as

$\lambda_{2} = 1 - \frac{x}{\sqrt{2}} (\textrm{cos} \phi + \textrm{sin} \phi)$.

The expression

$R = \textrm{cos} \phi + \textrm{sin} \phi = \sqrt{2} \textrm{sin} (\phi + \pi /4)$

allows a simplification to

$\lambda_{2} = 1 - x \textrm{sin} (\phi + \pi /4)$.

$R$ has a maximum at $\phi = \pi /4$, namely $\sqrt{2}$, which results in a zero $\lambda_{2}$ at $x = 1$. Similarly, a zero $\lambda_{2}$ results for the lepton value of $x = \sqrt{2}$, when $\phi = 0$. At this $x$ value, the eigenvalue is a particularly simple expression.

At $\phi = 2/9$ and $x = \sqrt{2}$ we have $R = 1.196$, giving a negative $\lambda_{2} = - 0.196$. This is the only negative term (at $-0.0195$ $\sqrt{eV}$) in the neutrino Koide triplet. For all $\phi$ between zero and $\pi/4$, this $\lambda_{2}$ will be negative. Thus the difference between these particle and antiparticle masses (which occur at the same scale) comes down to the existence of the zero eigenvalue at $\phi$ an eighth root of unity.

7 years ago

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