# translation invariance of the Laughlin wavefunction?

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The celebrated Laughlin wave function

$$\psi_{\mathrm{Laughlin}}(z_1,\ldots,z_N) = \prod_{i<j} (z_i-z_j)^n\prod_i e^{-|z_i|^2/4}$$

is a (very good) approximation of the ground state of a two-dimensional electron gas in a uniform background magnetic field when the filling factor $\nu$ of the lowest Laundau level is equal to $1/n$ (for some odd integer $n$).

Now, I have a fundamental problem with this. The two-dimensional electron gas is obviously translation invariant (more precisely, invariant under the action of magnetic translations). So $\psi_{\mathrm{Laughlin}}$ should also be translation invariant. Clearly, as $N$ is a fixed finite integer, $\psi_{\mathrm{Laughlin}}$ cannot be translation invariant... but it should be *almost* translation invariant, and I just don't see why that should be the case.

Can someone please help me with that?

asked Nov 19, 2014
I'm by no means learned on the subject, but why do you expect the Laughlin wavefunction is translation invariant? After all the ground states are degenerate(or are they?), so I don't see a reason for the ground state to share the symmetry of the Hamiltonian.
Jia Yiyang: The Laughlin wave function is associated to disc geometry. In that geometry, the ground state is not degenerate.

A brief comment paraphrased from Yang Xu, a friend,

It's about choice of gauge. Haldane had written a translation invariant version of the wavefunction, by putting a monopole at the center of the sphere and solving for the wavefunction at the surface. See Phys. Rev. Lett. 51, 605

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