Recall the basic
MUB operator
We often say that $R_{3}^{3}$ is the identity, but strictly speaking it is $-i \cdot \mathbf{1}$. In quark boson
notation, where phases are written as powers of the angle $\pi / 6$, we have
and so on. That is, the true relation is $R_{3}^{12} = \mathbf{1}$. As always, the
Fourier matrix satisfies $F^{4} = \mathbf{1}$. This is analogous to the rule $R_{2}^{8} = \mathbf{1}$ for the $2 \times 2$ operator, which also relies on the complex number $i$. The Fourier matrix encourages the circulant $R_d$ matrices to take on that extra factor of four. So in dimension $6$, which is the best dimension for studying the particle spectrum, we would have an $R_6$ operator obeying $R_{6}^{24} = \mathbf{1}$. If we rephased the matrix $R_3$ by the factor $-i$, then the cubic rule is obeyed. This matrix is in fact precisely a
quark boson matrix for $\overline{u}_{L}$.
We often say that $R_{3}^{3}$ is the identity, but strictly speaking it is $-i \cdot \mathbf{1}$. In quark boson notation, where phases are written as powers of the angle $\pi / 6$, we have 
and so on. That is, the true relation is $R_{3}^{12} = \mathbf{1}$. As always, the Fourier matrix satisfies $F^{4} = \mathbf{1}$. This is analogous to the rule $R_{2}^{8} = \mathbf{1}$ for the $2 \times 2$ operator, which also relies on the complex number $i$. The Fourier matrix encourages the circulant $R_d$ matrices to take on that extra factor of four. So in dimension $6$, which is the best dimension for studying the particle spectrum, we would have an $R_6$ operator obeying $R_{6}^{24} = \mathbf{1}$. If we rephased the matrix $R_3$ by the factor $-i$, then the cubic rule is obeyed. This matrix is in fact precisely a quark boson matrix for $\overline{u}_{L}$.
In deleted comment I asked "what's a quark boson matrix?" and guessed wrongly.
ReplyDeleteFrom the vixra paper, it seems a "quark boson matrix" results from acting on a quark-associated matrix with the "twisted Fourier transform" T, and it gets called a boson matrix because it has the properties of a boson-associated matrix in this scheme.
Yes, that's right Mitchell. Probably not a good name choice, but while I'm the only one talking about this ...
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