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  translation invariance of the Laughlin wavefunction?

+ 4 like - 0 dislike

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 in Theoretical Physics by André Henriques (210 points) [ no revision ]
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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