# Laughlin state unique ground state?

In the FQHE, one typically encounters the statement that the $\nu = 1/3$ Laughlin state is a unique exact ground state of a model Hamiltonian where the Haldane pseudopotentials $V_1 \neq 0$ and $V_m = 0$ for $m = 3, 5, \cdots$.(See link to a book with such a statement)

Qn 1) Is this the same as a hard-core interaction in real space $\sum_{i<j} V(r_i - r_j)$ where $v(r) = \partial_r^2 \delta(r)$ as considered by people like say in this article by XG Wen, page 25? He only says that the Laughlin state is an exact ground state of this Hamiltonian, and doesn't say it is unique.

Qn 2) If they are equivalent, then don't we have a contradiction? In XG Wen's article he identifies any state that is of the form a symmetric polynomial times the Laughlin state to have energy $0$ too of this model Hamiltonian. That contradicts the uniqueness statement made above.

I suspect that the two potentials are not the same representations, is that obvious to see?

Thanks.

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On the torus Laughlins state (like every other eigen state) is non unique : This comes from the center of mass motion. On the sphere Laughlins state is unique. – jjcale Jun 30 '13 at 8:27