formalism $$\begin{aligned}
H=\sum_{n,m>0}b(a_{-n}a_{-m}a_{n+m}+a_{-n-m}a_{n}a_{m})\\
+(1-b^{2})\sum_{n>0}na_{-n}a_{n}+b^{2}N\sum_{n>0}a_{-n}a_{n}\,.\nonumber\end{aligned}$$ The last term involves the level of corresponding excitations, when $N\rightarrow\infty$, it overwhelms the excitation spectrum since it is much larger than other contributions in $H$. In our analysis, we treat it as the background and we ignore this term. Besides, if we set $$\tilde{a}_{-n}=\frac{a_{-n}}{b},\,\,\tilde{a}_{n}=a_{n}b\,,{\text for}\,\,n>0$$ then we rewrite $H$ as $$\begin{aligned}
H & = & \sum_{n,m>0}b(b\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}+\frac{1}{b}\tilde{a}_{-n-m}\tilde{a}_{n}\tilde{a}_{m})\label{deformcs}\\
 & + & (1-b^{2})\sum_{n>0}n\tilde{a}_{-n}\tilde{a}_{n}\nonumber \\
 & = & \sum_{n,m>0}(\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}+\tilde{a}_{-n-m}\tilde{a}_{n}\tilde{a}_{m})\nonumber \\
 & + & (1-b^{2})\left(\sum_{n>0}n\tilde{a}_{-n}\tilde{a}_{n}-\sum_{n,m>0}\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}\right)\,.\nonumber\end{aligned}$$ It is easy to see that $\tilde{a}$ still holds the Heisenberg algebra so that the fermionization is exact. We split the Hamiltonian into free part (the first line of last equality of (\[deformcs\])), which is the same as free fermions, and the interacting part (the second line of last equality of (\[deformcs\])).

Eigenstate and spectrum
-----------------------

The CS model is exactly solvable. To see that, we first classify the Fock space expanded by bosons by its level $\mathcal{N}=\sum_{n>0}a_{-n}a_{n}$ such that an arbitrary state $$|n_{1},n_{2},\cdots,n_{l}\rangle=a_{-n_{1}}a_{-n_{2}}\cdots\, a_{-n_{l}}|0\rangle\,,\label{eq:Fockstate}$$ has a level $$\mathcal{N}|n_{1},n_{2},\cdots\, n_{l}\rangle=\sum_{i=1}^{l}n_{i}|n_{1},n_{2},\cdots\, n_{l}\rangle\,.$$ In this classification, there are $P(k)$, the partition number of $k$, states at a given level $k$. It is easy to check that $H$ commutes with $\mathcal{N}$. Hence they can have common eigenstates. Therefore, 