power sum symmetric function $$\begin{aligned}
 &  & \tilde{a}_{-n}\exp\left(\sum_{i,n}b\frac{a_{n}}{-n}z_{i}^{n}\right)|0\rangle=\exp\left(\sum_{i,n}b\frac{a_{n}}{-n}z_{i}^{n}\right)\sum_{i}z_{i}^{n}|0\rangle\nonumber \\
 &  & \Rightarrow\tilde{a}_{-n}\simeq\sum_{i}z_{i}^{n}=p_{n}(z_{i})\,.\end{aligned}$$ It in turn determines the eigenstate itself as a symmetric function. The eigenvalue of $H$ is $$\begin{aligned}
H|P_{\lambda}(\psi\psi^{*})\rangle=E_{\lambda}^{\beta}|P_{\lambda}(\psi\psi^{*})\rangle\,.\label{eigeneqn}\end{aligned}$$ Let us consider the leading Schur function $s_{\lambda}$. Because $H_{t}$ changes basis, the eigenvalue should come from the action of $H_{0}^{\beta}$ on $$|s_{\lambda}\rangle=s_{\lambda}(\psi\psi^{*})|vac\rangle\,.$$

The equation (\[eigeneqn\]), has an explicit solution (up to a constant similarity transformation) that $$|P_{\lambda}\rangle \propto  R(E)|s_{\lambda}\rangle\,,$$ where $$R(E)  =  \frac{1}{1-\frac{1-b^{2}}{E_{\lambda}^{\beta}-H_{0}^{\beta}}H_{t}}\,.\label{explicitsol}$$ This solution can be derived as follows. We first rewrite $H$ as $$H=E-(E-H_{0}^{\beta})\left(1-\frac{1-b^{2}}{E-H_{0}^{\beta}}H_{t}\right)\,\label{decompotrick}$$ then $$\begin{aligned}
H|P_{\lambda}\rangle & = & \left(E_{\lambda}^{\beta}-(E_{\lambda}^{\beta}-H_{0}^{\beta})\left(1-\frac{1-b^{2}}{E_{\lambda}^{\beta}-H_{0}^{\beta}}H_{t}\right)\right)R(E)|s_{\lambda}\rangle\\
 & = & E_{\lambda}^{\beta}R(E)|s_{\lambda}\rangle-(E_{\lambda}^{\beta}-H_{0}^{\beta})|s_{\lambda}\rangle=E_{\lambda}^{\beta}|P_{\lambda}\rangle\,.\nonumber\end{aligned}$$

Notice that till now we have used the deformed bosonic oscillators $\tilde{a}$. It is not convenient when we consider the standard symmetric function formulae. We need to introduce a similarity transformation that transforms $\tilde{a}$’s back to $a$’s.\
$$D=\exp\bigl(-\log(b)(\tilde{q}\tilde{a}_{0}+\sum_{n>0}\tilde{a}_{-n}\tilde{a}_{n}/n)\bigr)\,,$$ or equivalently, we have a fermionic formalism $$D_{\pm}=b^{\pm\frac{1}{2}\sum_{r>0}(\psi_{-r}\psi_{r}^{*}+\psi_{-r}^{*}\psi_{r})}\,,$$ where $\pm$ means acting on bra(left) or ket(right) state respectively.

Laughlin state and its Hamiltonian
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The Laughlin state is defined as $$\begin{aligned}
\Psi_{L}(\{z_{i}\})=\prod_{i<j}(z_{i}-z_{j})^{b^{2}}\exp\left(-\sum_{i}\frac{|z_{i}^{2}|}{4\ell}\right)\,,\end{aligned}$$ where $b^{-2}=\nu\in\mathbb{Z}$ is the filling fraction. In the formula $$\ell=\sqrt{\frac{\hbar}{eB}}$$ is the basic magnetic length scale. Usually it is normalized to be 1. $e$ stands for the electron charge and $B$ is 