Two-layer System
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Now we can consider the two-layer system. The corresponding ground state wavefunction is the Halperin state, which reads $$\tilde{\Psi}_{H}\left({z_{i}},{w_{j}}\right)=\prod_{i<j}^{N}(z_{i}-z_{j})^{p}\prod_{m<n}^{M}(w_{m}-w_{n})^{q}\prod_{i,m}^{N,M}(z_{i}-w_{m})^{r}\,.\label{eq:Halperin}$$ The complexity of this wave-function lies in that it involves an interaction between two layers. Let us first define the bosonic fields $$\begin{aligned}
\phi^{1}(z) & = & q_{0}^{1}+a_{0}^{1}\ln z+\sum_{n\neq0}\frac{a_{-n}^{1}}{n}z^{n}\,,\\
\phi^{2}(w) & = & q_{0}^{2}+a_{0}^{2}\ln w+\sum_{n\neq0}\frac{a_{-n}^{2}}{n}w^{n}\,,\end{aligned}$$ with commutation relation $$[a_{n}^{I}\,,\, a_{m}^{J}]=n\delta^{IJ}\delta_{n+m,0}\,,\, n,m\in\mathbb{Z},\, I,J\in{1,2}\,.$$ The related coordinate system is defined as $$\begin{aligned}
Z_{I} & = & z_{I},\,\mbox{if}\,\ I\leq N\,,\\
Z_{I} & = & w_{I-N}\,,\, if\ \, N<I\leq N+M\,,\\
\partial_{I} & = & Z_{I}\partial_{Z_{I}}\,,\end{aligned}$$ we can now write down the Hamiltonian of this system as $$\begin{aligned}
H^{Hal} & = & \sum_{I}(\partial_{I}-2\partial_{I}(\ln\tilde{\Psi}_{H}(Z_{I}))\partial_{I}\\
 & = & \sum_{i=1}^{N}\left(z_{i}\partial_{z_{i}}-2z_{i}\partial_{z_{i}}\left(\ln\prod_{i<j}(z_{i}-z_{j})^{p}\right)\right.\\
 &  & +\left.2z_{i}\partial_{z_{i}}\left(\ln\prod_{i,m}(z_{i}-w_{m})^{r}\right)\right)z_{i}\partial_{z_{i}}\\
 & + & \sum_{m=1}^{M}\left(w_{m}\partial_{w_{m}}-2w_{m}\partial_{w_{m}}\left(\ln\prod_{m<n}(w_{m}-w_{n})^{q}\right)\right.\\
 &  & +\left.2w_{m}\partial_{w_{m}}\left(\ln\prod_{i,m}(z_{i}-w_{m})^{r}\right)\right)w_{m}\partial_{w_{m}}\\
 & = & H_{L}(p)+H_{L}(q)+2r\sum_{m,i}\frac{z_{i}^{2}\partial_{z_{i}}}{z_{i}-w_{m}}-2r\sum_{m,i}\frac{w_{m}^{2}\partial_{w_{m}}}{z_{i}-w_{m}}\,,\\
 & = & H_{L}(p)+H_{L}(q)+2r\sum_{n\geq0,m,i}\left(\frac{w_{m}}{z_{i}}\right)^{n}z_{i}\partial_{z_{i}}-2r\sum_{n>0,m,i}\left(\frac{w_{m}}{z_{i}}\right)^{n}w_{m}\partial_{w_{m}}\,\,,\end{aligned}$$ where $H_{L}(p)$ is the Laughlin Hamiltonian defined before in which $b^{2}\rightarrow p\,.$ The vertex operators in this system are $$V_{\sqrt{p}}^{1}(z)=e^{\sqrt{p}\phi^{1}(z)}\,,\: V_{\sqrt{q}}^{2}(w)=e^{\sqrt{q}\phi^{2}(w)}\,.$$ The differential operators relate to operators and power-sum polynomials are defined as follows $$\begin{aligned}
z_{i}\partial_{z_{i}}\prod_{i=1}^{N}V_{\sqrt{p}}^{1,-}(z_{i}) & = & \sum_{n>0}\sqrt{p}a_{-n}^{1}z_{i}^{n}\prod_{i=1}^{N}V_{\sqrt{p}}^{1,-}(z_{i})\,,\\
w_{m}\partial_{w_{m}}\prod_{j=1}^{M}V_{\sqrt{q}}^{2,-}(w_{j}) & = & \sum_{n>0}\sqrt{q}a_{-n}^{2}w_{j}^{n}\prod_{n=1}^{M}V_{\sqrt{q}}^{2,-}(w_{j})\,,\\
a_{n}^{1}\prod_{i=1}^{N}V_{\sqrt{p}}^{1,-}(z_{i})|0\rangle & = & \sqrt{p}\sum_{i}z_{i}^{n}\prod_{i=1}^{N}V_{\sqrt{p}}^{1,-}(z_{i})|0\rangle\\
a_{n}^{2}\prod_{j=1}^{N}V_{\sqrt{p}}^{2,-}(w_{j})|0\rangle & = & \sqrt{q}\sum_{i}w_{j}^{n}\prod_{i=1}^{N}V_{\sqrt{p}}^{1,-}(w_{j})|0\rangle\end{aligned}$$ Therefore we have $$\begin{aligned}
2r\sum_{n\geq0,m,i}\left(\frac{w_{m}}{z_{i}}\right)^{n}z_{i}\partial_{z_{i}} & \rightarrow & 2r\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}\,,\\
2r\sum_{n>0,m,i}\left(\frac{w_{m}}{z_{i}}\right)^{n}w_{m}\partial_{w_{m}} 