Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  translation invariance of the Laughlin wavefunction?

+ 4 like - 0 dislike
747 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 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

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):
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:
p$\hbar$ysicsOve$\varnothing$flow
Then 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).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...