# Laplacian of a delta function as an interaction potential for Laughlin state

I am reading Xiao-Gang Wen's paper "Pattern-of-zeros approach to Fractional quantum Hall states and a classification of symmetric polynomial of infinite variables", on page 8, he gives three interaction potentials for various Laughlin state and claim that these Laughlin state are exactly the zero-energy ground state of the given interaction potential. I do have some questions about these interaction potentials.

1. I do believe that the second term in the interaction potential for $\nu=1/4$ is from Kivelson-Trugman's delta function expansion ("Exact results for the fractional quantum Hall effect with general interactions"), but I think in their paper, the Laplacian is with respect to the relative coordinate which is $z_{rel}=z_1-z_2$, but why in Xiao-Gang's paper that Laplacian is only with respect to $z_1$?

2. Why the delta function only involves $z$, but no conjugate $z^*$? Isn't a two dimensional delta function usually defined as $\delta(z,z^*)$?

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– Qmechanic♦ Sep 11 '12 at 7:45

## 1 Answer

I haven't read your paper but is Laplacian not invariant by translation ? Z=z1-z2 For the notation for δ(z,z∗), you are right to say that to recover x, y you need z and z*, but it might just be a shorthening of the writing.

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The situation I am concerned is that when you treat the delta function as a limiting case of a Gaussian function(say if we want to investigate the interaction range for this Laplacian of the delta function), then if z is the only variable wouldn't make sense. – huyichen Sep 11 '12 at 6:16
– Shaktyai Sep 11 '12 at 8:21
I am aware of that definition, but I don't think that is quite equivalent to a two dimensional delta function, maybe a delta function for analytic complex functions. – huyichen Sep 11 '12 at 18:36
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