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Benedict Irwin edited Ternary Ladders.tex
almost 10 years ago
Commit id: 7d9f986ef94e6c3bd7cc1c26d584978c7a5faa46
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\beta^{\alpha}_a=\sqrt[3]{\frac{2}{3}+\alpha}
\end{equation}
Then the operation \begin{equation}
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}}
\end{equation}
One could forsee that if the is a base number operator $N=aaa^+$ then combinations of these transitions could be such that \begin{equation}
(aaa)(a^+a^+a)|\alpha> = a(aaa^+)a^+a|\alpha>