# 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?

• 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. May 29, 2012 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.

• 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. May 29, 2012 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. May 29, 2012 at 14:50