Neutralizing Background and Fractional Quantum Hall ground state

Question

The idealized many-body Hamiltonian describing FQH is given by $$ H = \sum_i \left\{\frac{[\vec{p}_i -e/c \vec{A}(\vec{r}_i)]^2}{2m}+V(\vec{r}_i)\right\} + \frac{1}{2}\sum_{i\neq j} \frac{e^2}{|\vec{r}_i-\vec{r}_j|} $$ where $V(\vec{r}_i)$ is the neutralizing background potential. In a small number of charges case, we discard the background potential. In this case (zero background) with the three main assumptions one can find the form of (F)QH ground state. These assumptions are

1. Weak repulsive core: If the magnetic field is strong enough and $e$ (the charge) small enough one can treat the charge-charge Coloumb interaction as a perturbation.

2. Temperature low enough so the Lowest Landau level is the host of the ground state.

3. The ground state is an eigenstate of total angular momentum.

Then we can prove that the ground state will be

$$\Psi(z_1, \cdots, z_N) = P(z_1, \cdots, z_N) e^{-\frac{1}{4\ell_B^2}\sum|z_i|^2}$$ such that $P$ is a polynomial which only depends on relative coordinates (no center of mass dependence), it is homogeneous and its degree is equal to the total angular momentum and depending on whether the charge carriers are bosonic or fermionic it is sysmmetric or anti-symmetric. (There is also a minimal angular momentum condition which is not important for my question)

As far as I know, the above wavefunction is treated as the definition of (F)QH ground state. But then we found it when there is no neutralizing background potential present. Why is neutralizing background not important in finding the ground state?

Answer 1

If the background charge density is uniform, then the interaction of the electrons with the background charge just adds a constant to the hamiltonian, which we can then discard. So why even introduce the background potential at all? Well, if we want the system to have a well-defined thermodynamic limit (system size $\rightarrow \infty$ and number of particles $\rightarrow \infty$ at fixed density) we need to make it overall charge neutral.