the fermionic triangularization and integrability. In sec. 4 and 5, we provide parallel analysis for Laughlin state and Halperin state. In sec. 6 we make a conclusion and discuss some further works.

The Calogero-Sutherland Model
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We start our analysis from the famous Calogero-Sutherland(CS) model. It is an exact solvable model, describing $N$ interacting charged particles on a unit circle, with two-body interaction $$\begin{aligned}
H_{int}=\sum_{i<j} & \frac{\beta(\beta-1)}{\sin^{2}(x_{i}-x_{j})} & \,,\end{aligned}$$ in which $x_{i}$ defines the $i$-th particle’s position on the circle. For simplicity, we substitute $\beta=b^{2}$. Then CS Hamiltonian is written as $$H_{CS}=-\frac{1}{2}\sum_{i=1}^{N}\partial_{i}^{2}+\sum_{i<j}\frac{b^{2}(b^{2}-1)}{\sin^{2}(x_{i}-x_{j})}\,.$$

\[theo:isospectrum\]$H_{CS}$ is isospectral to another Hamiltonian $$\tilde{H}_{CS}=-\frac{1}{2}\sum_{i=1}^{N}(\partial_{i}+\partial_{i}\ln\prod_{j<k}\sin^{b^{2}}(x_{j}-x_{k}))(\partial_{i}-\partial_{i}\ln\prod_{r<s}\sin^{b^{2}}(x_{r}-x_{s})\,.$$ up to a universal shift of eigen-energy.

$\!\textbf{Proof of Theorem }$$\ref{theo:isospectrum}$: Defining the complex coordinate $z_{i}=e^{i2x_{i}}$, we have $\partial_{i}=2iz_{i}\partial_{z_{i}}$. Therefore $$\begin{aligned}
\partial_{i}\ln\prod_{j<k}\sin^{b^{2}}(x_{j}-x_{k}) & = & b^{2}\sum_{\substack { j\\i\neq j}}\cot(x_{i}-x_{j})\\
 & = & ib^{2}\sum_{\substack { j\\i\neq j}}\frac{z_{i}+z_{j}}{z_{i}-z_{j}}\,,\end{aligned}$$ and the commutator $$\left[\partial_{i}, b^{2}\sum_{\substack{ k \\i\neq k}}\cot(x_{i}-x_{k})\right]=-b^{2}\sum_{\substack{ k \\k\neq i}}\frac{1}{\sin^{2}(x_{k}-x_{i})}.$$ We can rewrite the $\tilde{H}_{CS}$ as the following formula $$\begin{aligned}
\tilde{H}_{CS} & = & -\frac{1}{2}\sum_{i=1}^{N}\partial_{i}^{2}-b^{2}\sum_{i<j}\frac{1}{\sin^{2}(x_{i}-x_{j})}\\
 &  & +\frac{1}{2}b^{4}\sum_{i\neq j,i\neq k}\cot\left(x_{i}-x_{j}\right)\cot\left(x_{i}-x_{k}\right)\,.\end{aligned}$$ Using the identity $$\begin{aligned}
\sum_{\hbox{distinct \ }i, j, k}\cot\left(x_{i}-x_{j}\right)\cot\left(x_{i}-x_{k}\right)+{i, j, k}\text{{\ cyclic}}\\
=\sum_{\hbox{distinct\ } i, j, k}(-1)=-N(N-1)(N-2) & \,,\end{aligned}$$ and the $j=k$ contribution $$\sum_{i\neq j}\cot^{2}\left(x_{i}-x_{j}\right)=-\sum_{i\neq j}1+\sum_{i\neq j}\frac{1}{\sin^{2}(x_{i}-x_{j})}\,,$$ we now have the form of $\tilde{H}_{CS}$ 