the external magnetic field strength. The Gaussian factor $$\exp\left(-\sum_{i}\frac{|z_{i}^{2}|}{4\ell}\right)\,,$$ will be ignored later on since it will bring in excessive clutters.

The Laughlin wavefunction can be understood as follows. Consider there are $N$ quasi-particles containing in the interior of a disk of area $A_{N}=2N\pi b^{2}$. The ground state of this system is a correlation function of these $N$ free quasi-particles in a background magnetic field, which in general makes the total charge to be zero, that is, a neutral correlation function. We define the vertex operator for a quasi-particle $$\begin{aligned}
 &  & V_{b}(z)=e^{b\phi(z)}\,.\,\end{aligned}$$ The density $\rho_{0}$ of the $\phi$ field on the disk, for a ground state, should be uniform anywhere. Otherwise a density wave will be excited. It is simply $$\begin{aligned}
\rho_{0} & = & \frac{N}{2\pi Nb^{2}}=\frac{1}{2\pi b^{2}}\,.\end{aligned}$$ Thus the background charge is $$S=e^{-b\int d^{2}w\rho_{0}\phi(w)}\,.$$ Without this factor, the correlation function will vanish. Now the correlation function is written as $$\begin{aligned}
\langle0\prod_{i=1}^{N}V_{b}(z_{i})S|0\rangle & = & \Bigg\langle\prod_{i=1}^{N}e^{b\phi(z)}\exp\bigg(-b\int d^{2}w\rho_{0}\phi(w)\bigg)\Bigg\rangle\nonumber \\
 & = & \prod_{i<j}^{N}(z_{i}-z_{j})^{b^{2}}\equiv\tilde{\Psi}_{L}(\{z_{i}\})\end{aligned}$$ Here we have defined the so-called reduced wave function $\tilde{\Psi}_{L}(\{z_{i}\})$ without the Gaussian factor. We now can consider the polynomial excitations of this ground state and moreover the behind integrability of this system.

From ground state to 