& = & \prod_{i<j}(z_{i}-z_{j})^{b^{2}}\prod_{i}(z_{i})^{\frac{b^{2}}{2}(1-N)}\\
 & = & \prod_{i<j}\left(\frac{z_{i}-z_{j}}{\sqrt{z_{i}z_{j}}}\right)^{b^{2}}\\
 & = & \prod_{i<j}(2i\sin(x_{i}-x_{j}))^{b^{2}}\,.\end{aligned}$$ Therefore up to a constant factor, it is the ground state of CS model. The excitation state, in principle, will be a state in the Fock space of the conformal field theory, which in general is a polynomial of bosonic oscillators. It implies there are one-to-one correspondence from the excitation wavefunction to an oscillator polynomial. The basic relation is the coherent relation such that $$a_{n}\prod_{i}V_{b}^{-}(z_{i})|k_{in}\rangle=b\sum_{i}z_{i}^{n}\prod_{i}V_{b}^{-}(z_{i})|k_{in}\rangle\,.\label{eq:coherent-1}$$ It relates the bosonic oscillator mode $a_{m}$to a symmetric polynomial (or symmetric function if $N\rightarrow\infty$). If we define the excitation state as follows $$\begin{aligned}
\frac{1}{2}\tilde{H}_{CS}\Psi_{\lambda}^{\beta}({z_{i}}) & = & \Psi_{CS}H'_{CS}P_{\lambda}^{\beta}({z_{i}})=E_{\lambda}^{\beta}\Psi_{\lambda}^{\beta}({z_{i}})\\
\Psi_{\lambda}^{\beta}({z_{i}}) & = & \Psi_{CS}({z_{i}})P_{\lambda}^{\beta}({z_{i}})\\
 & \simeq & \langle k_{f}|\prod_{i=1}^{N}V_{b}(z_{i})P_{\lambda}^{\beta}(a^{-})|k_{in}\rangle\,,\end{aligned}$$ then $$\langle k_{f}|P_{\lambda}^{\beta}(a^{+})H'_{CS}(a)\prod_{i=1}^{N}V_{b}^{-}(z_{i})|k_{in}\rangle=E_{\lambda}^{\beta}P_{\lambda}^{\beta}({z_{i}})\,.\label{eq:OpereqCS}$$ We have defined here the normal-ordered operator formalism of $H'_{CS}(a)\equiv H$ , such that

$$H|P_{\lambda}^{\beta}\rangle=E_{\lambda}|P_{\lambda}^{\beta}\rangle.$$ The next step is to translate the differential formalism Hamiltonian $H'_{CS}$ into operator formalism with the help of coherent relation (\[eq:coherent-1\]). We have the following relations $$\begin{aligned}
z_{i}\partial_{z_{i}}\bullet  =  b\sum_{n>0}a_{-n}z_{i}^{n}\bullet,& &
(z_{i}\partial_{z_{i}})^{2}\bullet  =  b^{2}\sum_{n,m>0}a_{-n}a_{-m}z_{i}^{n+m}\bullet
  +b\sum_{n>0}na_{-n}z_{i}^{n}\bullet\\
\sum_{i<j}\frac{z_{i}+z_{j}}{z_{i}-z_{j}}(z_{i}\partial_{z_{i}}-z_{j}\partial_{z_{j}})\bullet & = & b\sum_{i<j,n>0}a_{-n}\frac{z_{i}+z_{j}}{z_{i}-z_{j}}(z_{i}^{n}-z_{j}^{n})\bullet\\
=b\sum_{i<j,n>0}a_{-n}(z_{i}^{n}+2z_{i}^{n-1}z_{j} & + & \cdots2z_{i}z_{j}^{n-1}+z_{j}^{n})\bullet\\
=\left(b\sum_{n,m>0,i,j}a_{-n}z_{i}^{n-m}z_{j}^{m} \right.& + & \left. Nb\sum_{n>0,i=1}^N a_{-n}z_{i}^{n}-b\sum_{n>0,i=1}^N na_{-n}z_{i}^{n}\right)\bullet\end{aligned}$$ where $\bullet$ denotes $\prod_{j=1}^{N}V_{b}^{-}(z_{j})|k_{i}\rangle$. So we have the CS Hamiltonian in bosonic operator 