The Condensate Scale II

Yesterday we didn't write out the $(W^{+},W^{-},Z)$ bosons as a Koide triplet, so let us do that now. This triplet is clearly of the form

$(1,1,\frac{\alpha }{2})$

where $\alpha {\cos \theta }_W=2$, for the Weinberg angle ${\theta }_W$. The eigenvalue set at the Koide phase $\phi =0$ is

$\lambda \in \{1+\alpha ,1-\frac{\alpha }{2}\}$

with multiplicity $2$ for the second eigenvalue. At ${\theta }_W=\pi /6$ we obtain

$\lambda \in \{1+\frac{4}{\sqrt{3}},1-\frac{2}{\sqrt{3}}\}$.

Squaring and summing the rest mass triplet for the Higgs, we obtain $m_H=11/2$ exactly, in the units chosen, which just happen to be the Koide scale for the top quark triplet.