the fermionic picture, which we will explore in next section.

Fermionization
==============

Fermionization of free term
---------------------------

We now rewrite the CS Hamiltonian as $H\equiv H_0 + H_{int}$, here $$\label{eq:freefermion}
H_0 = \sum_{n,m>0}(\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}+\tilde{a}_{-n-m}\tilde{a}_{n}\tilde{a}_{m})$$ is the free part, while $$\begin{aligned}
\label{eq:intpart}
H_{int} = (1-b^{2})\left(\sum_{n>0}n\tilde{a}_{-n}\tilde{a}_{n}-\sum_{n,m>0}\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}\right)\end{aligned}$$ is the interaction part which could be separated into two components $$\begin{aligned}
H_int &=&H_1+H_2, H_1 = (1-b^{2})(-\sum_{n,m>0}\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m})\\
H_2&=&(1-b^{2})\sum_{n}n\tilde{a}_{-n}\tilde{a}_{n} \end{aligned}$$ for further convenience. Now we want to fermionize the deformed bosonic Hamiltonian $H$ by introducing $$\tilde{a}_{n}=\sum_{r\in\mathbb{Z}+\frac{1}{2}}:\psi_{n-r}\psi_{r}^{*}:\,,$$ and also the free Virasoro generator $$\begin{aligned}
\tilde{T}(z) & = & \frac{1}{2}(\partial_{z}\tilde{\phi}(z))^{2}=-\frac{1}{2}[\psi\partial\psi^{*}+\psi^{*}\partial\psi](z)\\
\tilde{L}_{n} & = & \frac{1}{2}\sum_{m}:\tilde{a}_{n-m}\tilde{a}_{m}:=\sum_{r>0,r\in\mathbb{Z}+\frac{1}{2}}(r+\frac{n}{2}):\psi_{-r}\psi_{n+r}^{*}:\end{aligned}$$ Firstly, let us consider the free part $H_0$. Notice that, $$\begin{aligned}
\sum_{n>0}\tilde{L}_{-n}\tilde{a}_{n} & = & \sum_{n,m>0}\tilde{a}_{-n-m}\tilde{a}_{m}\tilde{a}_{n}+\frac{1}{2}\sum_{n>m>0}\tilde{a}_{-n+m}\tilde{a}_{-m}\tilde{a}_{n}\\
\sum_{n>0}\tilde{a}_{-n}\tilde{L}_{n} & = & \frac{1}{2}\sum_{n>m>0}\tilde{a}_{-n}\tilde{a}_{n-m}\tilde{a}_{m}+\sum_{n,m>0}\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}\,.\end{aligned}$$ It gives rise to $$\begin{aligned}
H_{0}=\sum_{n,m>0}(\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}+\tilde{a}_{-n-m}\tilde{a}_{n}\tilde{a}_{m})=\frac{2}{3}\left(\sum_{n>0}\tilde{L}_{-n}\tilde{a}_{n}+\tilde{a}_{-n}\tilde{L}_{n}\right)\\
\sum_{n,m>0}\tilde{a}_{-n}\tilde{a}_{-m}\tilde{a}_{n+m}=\frac{2}{3}\left(\sum_{n>0}[2\tilde{a}_{-n}\tilde{L}_{n}-\tilde{L}_{-n}\tilde{a}_{n}]\right)\,.\end{aligned}$$ Notice that the $H_{0}$ is just the zero mode of the OPE of $$\frac{1}{2\pi i}\oint\frac{dz}{z-w}\tilde{T}(z)\partial_{w}\tilde{\phi}(w)\,.$$ In fermionic representation, we have $$\begin{aligned}
H_{0}(w) & = & \frac{2}{3}\oint\frac{dz}{2\pi i(z-w)}\left(-\frac{1}{2}[\psi\partial_{z}\psi^{*}+\psi^{*}\partial_{z}\psi])(z)[\psi\psi^{*}](w)\right)\\
 & = & \frac{2}{3}\oint\frac{dz}{2\pi i(z-w)}\left\{ \frac{1}{2}\left(\frac{1}{(z-w)^{3}}+\frac{\psi(z)\psi^{*}(w)}{(z-w)^{2}}-\frac{\partial_{z}\psi^{*}(z)\psi(w)}{z-w}\right)\right.\nonumber \\
 &  & +\left.\frac{1}{2}\left(-\frac{1}{(z-w)^{3}}-\frac{\psi^{*}(z)\psi(w)}{(z-w)^{2}}+\frac{\partial_{z}\psi(z)\psi^{*}(w)}{z-w}\right)\right\} \nonumber \\
 & = & \frac{1}{3}\oint\frac{dz}{2\pi i(z-w)}\left\{ \frac{\psi(z)\psi^{*}(w)-\psi^{*}(z)\psi(w)}{(z-w)^{2}}\right.\nonumber \\
 &  & +\left.\left(\frac{\partial_{z}\psi(z)\psi^{*}(w)-\partial_{z}\psi^{*}(z)\psi(w)}{z-w}\right)\right\} \nonumber \\
 & = & \frac{1}{2}\left([(\partial_{w})^{2}\psi(w)]\psi^{*}(w)-[(\partial_{w})^{2}\psi^{*}(w)]\psi(w)\right)\,.\nonumber\end{aligned}$$ The operator formalism $H_{0}$ is $$\begin{aligned}
H_{0} & = & \frac{1}{2\pi i}\oint w^{2}dwH_{0}(w)\\
 & = & \frac{1}{2}\left([(\partial_{w})^{2}\psi(w)]\psi^{*}(w)-[(\partial_{w})^{2}\psi^{*}(w)]\psi(w)\right)\nonumber \\
 & = & \frac{1}{2}\sum_{r\in\mathbb{Z}+\frac{1}{2}}:\psi_{-r}\psi_{r}^{*}:\left((-r-\frac{1}{2})(-r-\frac{3}{2})+(r-\frac{1}{2})(r-\frac{3}{2})\right)\nonumber 