While studying Hall conductivity with The Quantum Hall effect written by S.M.Girvin, I read a sentence

"We have shown that the random impurity potential (and by implication Anderson localization) is a necessary condition for Hall plateaux to occur"(page 7)

As you know, in QHE experiment, we can see regions where two conductivity $\sigma_{xx}$ and $\sigma_{xy}$ is constant, especially $\sigma_{xx}=0$ and $\sigma_{xy}=\nu$ $e^2\over\hslash$ where $\nu$ is the filling factor of Landau levels.

Many literature say that the plateus appear since there are localized states (which comes from impurities in real samples) between the extended states ( in DOE vs E graphs ).

I understand this statement in this way : When the Fermi energy is set to be in energy range of a localized state, there is no states carrying current about the fermi energy in the 'bulk' , but there are current carrying states on edge as Landau level's energy level vent upward due to edge confining field. Therefore the number of crossing between fermi level and energy level on the edge appears in the result $\sigma_{xy}=\nu$$e^2\over\hslash$.

In this way of understanding, I concluded that there should be a region in which $\sigma_{xy}$ is constant whether the sample is perfectly pure or not, because the number of current channels on the edge is the only important one. (Is it right?)

Then why does the QHE require the existence of impurity in the sample?

[1] : http://arxiv.org/abs/cond-mat/9907002

  • 2
    $\begingroup$ Your quote says "for the Hall plateau" to occur - not for the Hall effect to occur. You might want to edit your title accordingly - or clarify what it is you are asking (you are mixing the two). $\endgroup$ – Floris Mar 10 '15 at 15:52
  • 1
    $\begingroup$ I've changed. However, is there difference between quantum Hall effect and the appearence of Hall plateus? $\endgroup$ – Changhee Lee Mar 10 '15 at 16:29

First of all, the Hamiltonian doesn't have to be the Landau Hamiltonian for the IQHE. So assume you have a generic Hamiltonian.

Impurities are (that I know) one way to achieve a region of localization in the spectrum of the Hamiltonian, this region is called a "mobility gap". Perhaps there are other ways?

At any rate, what you need is the "mobility gap". Without it, there are no plateaus.

The mechanism is as follows: Let's say you start with the density of electrons such that the Fermi energy is in a spectral gap of the Hamiltonian, and you start raising the Fermi energy. Within the spectral gap, the Hall conductivity indeed doesn't change, but also density of electrons doesn't change, since there is a gap. However, as you go higher you get into the mobility gap, a region in which the density of electrons does increase, but the conductivity remains constant because, by definition, the mobility gap states do not contribute to conductivity, as they are localized. It is this region that generates the plateaus. As you go yet higher with the Fermi energy, you get out of the mobility gap and into a conduction band, in which case both the density and the conductivity change.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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