& \rightarrow & 2r\frac{1}{\sqrt{p}}\sum_{\begin{array}[t]{c}
n>0\\
m>0
\end{array}}a_{-m}^{1}a_{-n}^{2}a_{n+m}^{2}\,.\end{aligned}$$ It is now easy to write down the bosonic operator formalism, we have $$\begin{aligned}
H^{Hal} & = & H_{L}(p,a^{1})+H_{L}(q,a^{2})+H_{int}\,,\\
H_{int} & = & 2r\left(\frac{1}{\sqrt{q}}\sum_{\begin{array}[t]{c}
n\geq0\\
m>0
\end{array}}a_{-n}^{1}a_{n-m}^{1}a_{m}^{2}-\frac{1}{\sqrt{p}}\sum_{\begin{array}[t]{c}
n>0\\
m>0
\end{array}}a_{-m}^{1}a_{-n}^{2}a_{n+m}^{2}\right)\\
 & = & r\left(\frac{1}{\sqrt{q}}\sum_{m>0}L_{-m}^{1}a_{m}^{2}+\sum_{n,m>0}(\frac{1}{\sqrt{q}}a_{-n}^{1}-\frac{2}{\sqrt{p}}a_{-n}^{2})a_{-m}^{1}a_{n+m}^{2}\right)\,.\end{aligned}$$ Notice that the $H_{int}$ is a fascinating interaction. It is always a triangular term such that it subtracts boxes in Young diagram $\mu$ on layer-2 and adds the same number of boxes into Young diagram $\lambda$ on layer-1. The Hilbert space of this Hamiltonian is expanded by coupled bi-Jack functions, that is, $$|\Omega_{\lambda,\mu}^{0}\rangle=|P_{\lambda}^{1}(p/2)\rangle\otimes|P_{\mu}^{2}(q/2)\rangle\,.$$ It is not an eigenstate of the Halperin Hamiltonian, but only a highest weight state of the eigenstate, which can be obtained as following formula $$|\Omega_{\lambda,\mu}^{r}\rangle=\frac{1}{1-\frac{1}{E_{\lambda,\mu}-H_{L}(p)-H_{L}(q)}H_{int}}|\Omega_{\lambda,\mu}^{0}\rangle\,,$$ where the eigen-energy is written as $$E_{\lambda,\mu}=\sum_{i}^{\lambda_{1}^{t}}(\lambda_{i})^{2}+\sum_{j}^{\mu_{1}^{t}}(\mu_{j})^{2}-\frac{p}{2}\sum_{k}^{\lambda_{1}}(\lambda_{k}^{t})^{2}-\frac{q}{2}\sum_{l}^{\mu_{1}}(\mu_{k}^{t})^{2}\,.$$ The fermionization of the Hamiltonian as in ($\ref{eq:Halperin}$) is not hard. Let us first consider the first term in $H_{int}$ . It is $$\begin{aligned}
H_{int}^{1} & = & \frac{2r}{\sqrt{q}}\sum_{\begin{array}[t]{c}
n\geq0\\
m>0
\end{array}}a_{-n}^{1}a_{n-m}^{1}a_{m}^{2}\\
 & = & \frac{2r}{\sqrt{q}}\left(\sum_{\begin{array}[t]{c}
r,s,k,l,u\in\mathbb{Z}+\frac{1}{2}\\
r+s+k+l=-m<0
\end{array}}:\psi_{r}^{1}\psi_{s}^{1*}\psi_{k}^{1}\psi_{l}^{1*}\psi_{-u}^{2}\psi_{m+u}^{2*}:\right.\\
 &  & \hspace{1em}\hspace{1em}+\left.\sum_{u\in\mathbb{Z}+\frac{1}{2},m>0}:\psi_{-u}^{2}\psi_{m+u}^{2*}:(\text{Contractions})\right)\,,\end{aligned}$$ where the contractions is calculated as $$\begin{aligned}
\text{Contractions} & = & \sum_{\begin{array}[t]{c}
\begin{array}[t]{c}
n,r>0\end{array}\end{array}}:(\psi_{-r}^{1}\psi_{-n+r}^{1*}-\psi_{-r-n}^{1*}\psi_{r}^{1})(\psi_{-s}^{1}\psi_{n-m+s}^{1*}-\psi_{n-m-s}^{1*}\psi_{s}^{1}):\,,\\
 & = & \sum_{r>0}(\psi_{-r}^{1}\psi_{r-m}^{1*}+\psi_{-m-s}^{1*}\psi_{s}^{1})\,.\end{aligned}$$ Therefore $$\begin{aligned}
H_{int}^{1} & = & \frac{2r}{\sqrt{q}}\left(\sum_{\begin{array}[t]{c}
r,s,k,l,u\in\mathbb{Z}+\frac{1}{2}\\
r+s+k+l=-m<0
\end{array}}:\psi_{r}^{1}\psi_{s}^{1*}\psi_{k}^{1}\psi_{l}^{1*}\psi_{-u}^{2}\psi_{m+u}^{2*}:\right.\\
 &  & \hspace{1em}\hspace{1em}+\left.\sum_{\begin{array}[t]{c}
