as $$\begin{aligned}
\tilde{H}_{CS} & = & -\frac{1}{2}\sum_{i=1}^{N}\partial_{i}^{2}+b^{2}(b^{2}-1)\sum_{i<j}\frac{1}{\sin^{2}(x_{i}-x_{j})}-\frac{1}{6}b^4(N-1)N(N+1)\,,\nonumber \\
 & = & H_{CS}-\frac{1}{6}b^4N(N-1)(N+1)\end{aligned}$$ so Theorem $\ref{theo:isospectrum}$ is proved. Q.E.D.

It is now nature to consider the $\tilde{H}_{CS}$ rather than $H_{CS}$ since the later one, when acting on the ground state, will have a large energy (proportional to $N^{3}$) contribution to the spectrum. The ground state of $\tilde{H}_{CS}$ is simply $$\Psi_{CS}=\prod_{i<j}\sin^{\beta}(x_{i}-x_{j}),\quad \tilde{H}_{CS}\Psi_{CS}=0\text{\,.}\label{eq:GroundCS}$$ To extract the spectrums as well as corresponding excitation states, we need to eliminate the contribution of ground state. It implies the Jacobi transformation $$2H'_{CS}=\Psi_{CS}^{-1}\tilde{H}_{CS}\Psi_{CS}\,.$$ In this way, we have $$\begin{aligned}
2H'_{CS} & = & -\frac{1}{2}\sum_{i}(\partial_{i}+2\partial_{i}\ln\Psi_{CS})\partial_{i}\nonumber \\
 & = & -\frac{1}{2}\sum_{i}(2iz_{i}\partial_{z_{i}}+i2b^{2}\sum_{i\neq j}\frac{z_{i}+z_{j}}{z_{i}-z_{j}})(2iz_{i}\partial_{z_{i}})\nonumber \\
 & = & 2\sum_{i}(z_{i}\partial_{z_{i}})^{2}+2b^{2}\sum_{i<j}\frac{z_{i}+z_{j}}{z_{i}-z_{j}}(z_{i}\partial_{z_{i}}-z_{j}\partial_{z_{j}})\,.\label{eq:complexHcs}\end{aligned}$$

Bosonic oscillator formalism of $H'_{CS}$
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The ground state as in ($\ref{eq:GroundCS}$) can be understood as a CFT correlation function, that is $$\Psi_{CS}({z_{i}})\simeq\langle k_{f}|\prod_{i=1}^{N}V_{b}(z_{i})|k_{in}\rangle\,,$$ with the vertex operator defined by $$V_{b}(z)\equiv :e^{b\phi(z)}:\,,$$ and the bosonic field has the standard mode expansion $$\begin{aligned}
\phi(z) & = & q_{0}+p_{0}\ln z+\sum_{n\neq0}\frac{a_{-n}}{n}z^{n}\,,\\
{}[a_{n},a_{m}] & = & n\delta_{n+m,0},\quad [p_{0},q_{0}]=1\,.\end{aligned}$$ We can show that briefly. The OPE of vertex operators reads $$V_{a}(z)V_{b}(w)=(z-w)^{ab}:V_{a}V_{b}(\frac{z+w}{2}):\,.$$ If we choose the initial (final) momentum of right (left) vacuum $k_{in}=\frac{b}{2}(1-N)$ ($k_{f}=k_{in}+Nb$) , the correlation function is charge neutral and gives the result $$\begin{aligned}
\langle k_{f}|\prod_{i=1}^{N}V_{b}(z_{i})|k_{in}\rangle 