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?
Answer
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∑δ(zi−zj) where zi is a complex number describing the position of the ith boson. The zero energy state Ψ(z1,...,zN) satisfies Ψ(z1,...,zN)=P(z1,...,zN)exp(−∑i|zi|2/4) where P is a symmetric polynomial that satisfy ∫∏id2zi Ψ(z1,...,zN)†VΨ(z1,...,zN)=0.
Now it is clear that all the zero energy state are given by symmetric polynomial that satisfy P(z1,...,zN)=0 if any pair of bosons coincide zi=zj. For symmetric polynomial this implies that P(z1,...,zN)∼(zi−zj)2 when zi is near zj. The Laughline wave function $P_0=\prod_{i
However, a physically more relevant math question is: Every energy eigenstate below a certain finite energy gap Δ is of the form of a symmetric polynomial times the Laughlin wave function for any number N of particles. (Here Δ does not depend on N.)
We only have numerical evidences that the above statement is true, but no proof.
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