Ternary Ladder Operators

Abstract

We develop a triplet operator system which encompasses the structure of quark combinations. Ladder operators are created. The constants \(\beta\) are currently being found.

Ternary Ladders

Based on quarks

Define: \[a|\alpha>=\beta^{\alpha}_a|\alpha-\frac{1}{3}> \\ a^+|\alpha>=\beta^{\alpha}_{a^+}|\alpha+\frac{2}{3}> \\ \hat{a}|\alpha>=\beta^{\alpha}_{\hat{a}}|\alpha+\frac{1}{3}>\\ \hat{a}^+|\alpha>=\beta^{\alpha}_{\hat{a}^+}|\alpha-\frac{2}{3}>\]

So in Triplet operations exists operations which have an eigenstate of \(|\alpha>\) and counting triplets and also binary operations in the same way quarks combine. \[aaa^+|\alpha>=\beta^{\alpha}_{a^+}\beta^{\alpha+\frac{2}{3}}_a\beta^{\alpha+\frac{1}{3}}_a|\alpha> \\ a^+a^+a|\alpha>=\beta^{\alpha}_{a}\beta^{\alpha-\frac{1}{3}}_{a^+}\beta^{\alpha+\frac{1}{3}}_{a^+}|\alpha+1> \\ aaa|\alpha>=\beta^{\alpha}_{a}\beta^{\alpha-\frac{1}{3}}_{a}\beta^{\alpha-\frac{2}{3}}_{a}|\alpha-1> \\ a^+a^+a^+|\alpha>=\beta^{\alpha}_{a^+}\beta^{\alpha+\frac{2}{3}}_{a^+}\beta^{\alpha+\frac{4}{3}}_{a^+}|\alpha+2>\\\]

So considering the \(\beta\) terms. The only time \(\beta^{\alpha-\frac{2}{3}}_a\) might arise is if an a has operated on a state that is already filled by \(\alpha-\frac{2}{3}\). So, it should seem appropriate that if \(\alpha=0\) this coeficcient drops to zero to prevent negative integer fillings. This proposes a trial function of \[\beta^{\alpha}_a=\sqrt[3]{\frac{2}{3}+\alpha}\]

Then the operation \[aaa|\alpha>=\sqrt[3]{\frac{2}{3}+\alpha}\sqrt[3]{\frac{2}{3}+\alpha-\frac{1}{3}}\sqrt[3]{\frac{2}{3}+\alpha-\frac{2}{3}}|\alpha-1>\]

and for \(\alpha=0\) the coefficient drops to zero.

One could forsee that if the is a base number operator \(N=aaa^+\) then combinations of these transitions could be such that \[(aaa)(a^+a^+a)|\alpha> = a(aaa^+)a^+a|\alpha>\]

For doublet (meson) operations take an operator and it’s anti-operator\[\hat{a}^+a^+|\alpha>=\beta^{\alpha}_{a^+}\beta^{\alpha +\frac{2}{3}}_{\hat{a}^+}|\alpha> \\ a^+\hat{a}^+|\alpha>=\beta^{\alpha}_{\hat{a}^+}\beta^{\alpha -\frac{2}{3}}_{a^+}|\alpha>\]

Question? Does \[\beta^{\alpha}_{a^+}\beta^{\alpha +\frac{2}{3}}_{\hat{a}^+}=\beta^{\alpha}_{\hat{a}^+}\beta^{\alpha -\frac{2}{3}}_{a^+}?\]

Another Method

Another method to try is 3 (or 6 really) operators, for example \[a^r_{+}|\alpha>=\sqrt[3]{\alpha}|\alpha+\frac{2}{3}> \\ a^r_{-}|\alpha>=\sqrt[3]{\alpha}|\alpha-\frac{1}{3}> \\ a^g_{+}|\alpha>=\sqrt[3]{\alpha-\frac{1}{3}}|\alpha+\frac{2}{3}> \\ a^g_{-}|\alpha>=\sqrt[3]{\alpha-\frac{1}{3}}|\alpha-\frac{1}{3}> \\ a^b_{+}|\alpha>=\sqrt[3]{\alpha+\frac{1}{3}}|\alpha+\frac{2}{3}> \\ a^b_{-}|\alpha>=\sqrt[3]{\alpha+\frac{1}{3}}|\alpha-\frac{1}{3}> \\\]

Then is was found that the combination \[a^r_+a^g_-a^b_+|\alpha>=(\alpha+\frac{1}{3})|\alpha+1>\]