# translation invariance of the Laughlin wavefunction?

+ 4 like - 0 dislike
329 views

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

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:$\varnothing\hbar$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.