Looks like I have to answer this question :-)

Let me first answer the math question: Every zero-energy eigenstate is of the form of a symmetric polynomial times the Laughlin wave function.

To be concrete, let us consider an $N$ boson system, with delta-potential
interaction $V=g\sum \delta(z_i-z_j)$ where $z_i$ is a complex number
describing the position of the $i^{th}$ boson.
The zero energy state $\Psi(z_1,...,z_N)$ satisfies
$\Psi(z_1,...,z_N)=P(z_1,...,z_N)exp(-\sum_i |z_i|^2/4)$ where $P$ is a symmetric polynomial that satisfy
$\int \prod_i d^2 z_i \ \Psi(z_1,...,z_N)^\dagger V \Psi(z_1,...,z_N) =0$.

Now it is clear that all the zero energy state are given by symmetric polynomial
that satisfy $P(z_1,...,z_N)=0$ if any pair of bosons coincide $z_i=z_j$.
For symmetric polynomial this implies that
$P(z_1,...,z_N) \sim (z_i-z_j)^2$ when $z_i$ is near $z_j$.
The Laughline wave function $P_0=\prod_{i<j}(z_i-z_j)^2$ is one of the symmetric
polynomials that satisfies the above condition and is a zero energy state.
Since any other zero-energy symmetric
polynomial must satisfy $P(z_1,...,z_N) \sim (z_i-z_j)^2$, $P/P_0=P_{sym}$ has no poles and is a well defined symmetric
polynomial. So every zero-energy eigenstate $P$ is of the form of a symmetric polynomial $P_{sym}$ times the Laughlin wave function $P_0$.
More discussions can be found in the first part of arXiv:1203.3268.

However, a physically more relevant math question is: Every energy eigenstate
below a certain finite energy gap $\Delta$ is of the form of a symmetric polynomial times the Laughlin wave function for any number $N$ of particles.
(Here $\Delta$ does not depend on $N$.)

We only have numerical evidences that the above statement is true, but no proof.

This post imported from StackExchange Physics at 2014-04-05 17:31 (UCT), posted by SE-user Xiao-Gang Wen