the eigenstate of $H$ can be obtained by a superposition of states like (\[eq:Fockstate\]). A closer observation shows that the Hamiltonian $H$ acting on a state (\[eq:Fockstate\]) at level $k$ by certain times will definitely generate the lowest state $|1^{k}\rangle\equiv(a_{-1})^{k}|0\rangle$. We can choose the coefficients of the Fock state $|1^{k}\rangle$ of all states at level $k$ to be the same and equal to $b^{-k}$.[^2] By this choice, we have removed the irrelevant c-number common factor of each eigenstate. For example, at level 4, we assume an eigenstate has the following formalism $$|P_{\lambda}^{\beta}\rangle=b^{-4}((a_{-1})^{4}+\alpha_{1}a_{-2}(a_{-1})^{2}+\alpha_{2}a_{-2}^{2}+\alpha_{3}a_{-3}a_{-1}+\alpha_{4}a_{-4})|0\rangle\,,$$ Thus there are $P(4)-1=5-1=4$ unknown coefficients and also the eigen-energy $E_{\lambda}^{\beta}$ is not known. However, compare all the coefficients of the eigen-equation $$H|P_{\lambda}^{\beta}\rangle=E_{\lambda}^{\beta}|P_{\lambda}^{\beta}\rangle\,,$$ we have in total 5 independent equations. They in turn determine the eigenstate completely. The generalization to level $k$ is then straightforward.

However, this method does not provide a clear relation between the eigenstate and the Young diagram underlining the theory. In general, one can define by hand a sequence of eigen-energies at a given level so that each state is uniquely related to a Young diagram. But the reason is weak and unnatural. However, it is quite natural to see the Young diagram from 