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?

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Bizarrely, we can ping various sites to ask other people for help, but not users of this site? Since the author recently joined Physics.SE, it seems appropriate for him to answer. – genneth May 29 '12 at 10:37

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.

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Thanks for clearing that part up, Prof. Wen. I am glad that I can directly ask you question this way. I do have one more question on that fascinating paper. The wave function $\prod_i (1-\frac{z_i}{\xi})^n$ you proposed for the interaction between a outside 'charge' n at $\xi$ and the FQH droplet, can we just thought that as a 'charge' n/m quasihole, since the wave function will be almost the same. – huyichen May 29 '12 at 14:11
The operator $\prod(\xi-z_i)^n$ only generate a quasiparticle when $|\xi|$is less than the radius of the FQH droplet. When $|\xi|$is larger than the radius of the FQH droplet, the operator only create an deformation of the FQH droplet. – Xiao-Gang Wen May 29 '12 at 14:50