the Hamiltonian
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As we have done in the case of Calogero-Sutherland model, we now propose a Hamiltonian exactly has this Laughlin state as its ground state, that is, $$\begin{aligned}
H_{L} & = & \sum_{i}(\partial_{i}+\partial_{i}(\ln\tilde{\Psi}\{z_{i}\}))(\partial_{i}-\partial_{i} (\ln\tilde{\Psi}\{z_{i}\})).\end{aligned}$$ Here $\partial_{i}=\partial/\partial x_{i}\equiv z_{i}\partial_{z_{i}}$, or equivalently, $z_{i}=e^{x_{i}}$. By separating out the ground state contribution as follows, we have

$$\begin{aligned}
\tilde{{H}} & = & \tilde{(\Psi}\{z_{i}\})^{-1}H_{L}\tilde{\Psi}\{z_{i}\}\nonumber \\
 & = & \sum_{i}\left(z_{i}\partial_{z_{i}}+2z_{i}\partial_{z_{i}}(\ln\tilde{\Psi}\{z_{i}\})z_{i}\partial_{z_{i}}\right)\label{eq:Htilde}\\
 & = & \sum_{i}(z_{i}\partial_{z_{i}})^{2}-2b^{2}\sum_{i<j}\frac{z_{i}^{2}\partial_{z_{i}}}{z_{i}-z_{j}}\nonumber \\
 & = & \sum_{i}(z_{i}\partial_{z_{i}})^{2}-b^{2}\sum_{i<j}\frac{z_{i}^{2}\partial_{z_{i}}-z_{j}^{2}\partial_{z_{j}}}{z_{i}-z_{j}}\,.\nonumber\end{aligned}$$

When acting on the normal ordered vertex operators, we have $$\begin{aligned}
z_{i}\partial_{z_{i}}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle & = & b\sum_{n}a_{-n}z_{i}^{n}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\,,\nonumber \\
(z_{i}\partial_{z_{i}})^{2}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle & = & b^{2}\sum_{n,m}a_{-n}a_{-m}z_{i}^{n+m}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\nonumber \\
 & + & b\sum_{n}na_{-n}z_{i}^{n}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\,,\nonumber \\
a_{k}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle & = & \sum_{i}bz_{i}^{n}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\,,\label{eq:coherent}\\
\frac{z_{i}^{2}\partial_{z_{i}}-z_{j}^{2}\partial_{z_{j}}}{z_{i}-z_{j}}\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\nonumber \\
=\frac{b\sum_{n}a_{-n}\left(z_{i}^{n+1}-z_{j}^{n+1}\right)}{z_{i}-z_{j}} &  & \prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\nonumber \\
=b\sum_{n}a_{-n}(z_{i}^{n}+z_{i}^{n-1}z_{j}+ & \cdots & +z_{i}z_{j}^{n-1}+z_{j}^{n})\prod_{j}^{N}\exp\left(b\sum_{n,j}\frac{a_{-n}}{n}z_{j}^{n}\right)|0\rangle\,,\label{eq:interactionterm}\end{aligned}$$ and substituting ($\ref{eq:coherent}$) into ($\ref{eq:Htilde}$), we have the free part of ($\ref{eq:Htilde}$) as $$\tilde{{H}_{0}}=\sum_{n>0}na_{-n}a_{n}+b\sum_{n,m>0}a_{-n}a_{-m}a_{n+m}$$ We need to be careful with the interaction term ($\ref{eq:interactionterm}$). The summation on $i,\, j$ is not arbitrary. To turn this term into bosonic operator formalism, we have to do a trick as follow. Firstly, we change the summation from $i<j$ to $i\neq j$. It gives rise to a simple $\frac{1}{2}$ factor. Secondly, we add $i=j$ terms into the summation and then finally subtract these terms. Follow 