# 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.

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 $a^r_+a^g_-a^b_+|\alpha>=(\alpha+\frac{1}{3})|\alpha+1>$