Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What are the wave functions of the ground state of Integer Quantum Hall (IQH) and Fractional Quantum Hall (FQH) electrons?

share|cite|improve this question
The Laughlin ones. – wsc Oct 26 '12 at 1:47

Consider the lowest Landau level, and the simplest FQH state (namely the Laughlin state of 1/3 filling). The IQH ground state wave function is $$\Psi_1=\prod_{i<j}(z_i-z_j)e^{-\frac{1}{4}\sum_i|z_i|^2},$$ and the FQH state wave function is $$\Psi_3=\prod_{i<j}(z_i-z_j)^3e^{-\frac{1}{4}\sum_i|z_i|^2}.$$

share|cite|improve this answer
Why is your wavefunction for the IQH state a many-body wavefunction when it is a single-particle phenomena? – DaniH Oct 30 '12 at 22:23
@DaniH A single-particle phenomenon does not mean that the phenomenon only involves one particle. IQH state is a many-body state, which does have a many-body wave function. – Everett You Nov 10 '12 at 10:59

A few extra words complementing Everett You's answer.

The ground state depends on the "filling fraction", which is the number of electrons per flux quanta that thread the 2 dimensional electron gas, $\Psi_1$ is the ground state expected when the filling fraction is exactly 1, that is one electron per flux quanta, $\Psi_3$ is the ground state expected when the filling fraction is 1/3, that is three flux quanta per electron.

The ground state in principle depends on the details of the inter-particle potential too, but it is believed that for a vast class of repulsive interactions the ground states should be very similar to those Everett wrote down.

It is worth saying that nobody really knows the exact form of the fractional ground states for very relevant interactions like the Coulomb potential, even after projection into a single Landau level, but there has been enormous amount of evidence, including detailed numerical studies, showing that the Laughlin state is a very good "trial state", perhaps even the exact ground state in the thermodynamic limit, but no rigorous proof of that yet.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.