The bijective correspondence between a symmetric polynomial and edge excitation of the fractional quantum hall droplet I am recently reading Xiao-Gang Wen's paper (http://dao.mit.edu/~wen/pub/edgere.pdf) on edge excitation for fractional quantum hall effect. On page 25, he claimed that it is easy to show that there exist a bijective correspondence between a symmetric polynomial and edge excitation of the fractional quantum hall droplet. As we all known that Laughlin state is a zero-energy eigenstate for Haldane pseudopotential. And it is easy to see that if a symmetric polynomial times the Laughlin wave function, then that increases the relative angular momentum for particles, thus that wave function is still a zero-energy eigenstate for Haldane pseudopotential. However, Wen claimed that the reverse also holds, but I am not quite convinced by his argument in his paper. Does anybody know how to rigorously show that the reverse is also true, that is every zero-energy eigenstate is of the form of a symmetric polynomial times the Laughlin wave function?
 A: 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.
