Suppose I consider an infinite, non-interacting (so no FQHE should happen) 2DEG in the magnetic field $\vec B=B\hat z$ with a non-integer filling factor, say 0.13 or whatever. Suppose I apply an electric field $\vec E=E\hat y$, apparently I should expect response current in both $x$ (Hall) and $y$ (longitudinal) directions.

However, I do not understand the very existence of the longitudinal response along $y$. Here I provide a simple argument against it. I can perform a Galilean transformation to a reference frame with $\vec v$ along $\hat x$ such that $\vec E=-\vec v\times \vec B$. In this reference frame electric field should disappear. In the non-relativistic limit, in this frame there is only magnetic field $\vec B$. So there should not be any net current in this frame. However we know there is longitudinal $y$ current! I cannot seem to find where this contradiction comes from. Any help? Am I missing something simple?

PS: I think I understand the "standard" explanation about gapless excitations etc. But I would appreciate it if someone can point out to me the inconsistency in my logic above, or mistakes in assumption.


1 Answer 1


Your argument is correct, in fact you can use it to show that the Hall conductivity is set by the density. However, there is an underlying assumption when you apply Galilean transformation, that is translation invariance. In reality there are impurities that can backscatter electrons and cause the current to dissipate. So your argument immediately fails for any real systems with disorder.

EDIT: In fact, one can use a Drude-type classical model to calculate the conductivity tensor. Assume the charge and mass of the carrier is $e$ and $m$, with carrier density $n$, the result is

$ \begin{pmatrix} J_x\\ J_y \end{pmatrix}=\frac{\sigma}{1+(\omega_c\tau)^2} \begin{pmatrix} 1 & \omega_c\tau\\ -\omega_c\tau & 1 \end{pmatrix} \begin{pmatrix} E_x\\ E_y \end{pmatrix}, \sigma=\frac{ne^2\tau}{m} $

Here $\tau$ is the relaxation time and $\omega_c$ is the cyclotron frequency. For (unrealistic) clean system, $\tau\rightarrow \infty$, the longitudinal conductivity vanishes.

  • $\begingroup$ Thanks for your answer, but I think I'm still confused. For the moment let's take this unrealistic model. You said at the end that the longitudinal conductivity is infinite. However, if my argument were correct, then it should be zero - i.e. there shouldn't be any longitudinal response. BTW, As you can see this question is nothing "quantum" :) $\endgroup$ Aug 30, 2015 at 1:16
  • $\begingroup$ OK, I should be more clear that the last sentence was for clean electron gas in the absence of magnetic field, just Drude conductivity. For Hall system, you'll get exactly what you said. And you are right nothing is quantum, this is purely classical. $\endgroup$
    – Meng Cheng
    Aug 30, 2015 at 1:18
  • $\begingroup$ So, in this case it is fair to say that one of the following, (a) finite size of sample and presence of lattice, (b) impurity scattering, that makes $\sigma_{xx}$ in a magnetic field from zero to nonzero? Hmm, this is counterintuitive... $\endgroup$ Aug 30, 2015 at 1:26
  • $\begingroup$ What's counterintuitive about it? Classically we know that for a steady current to flow in $x$ direction, there must be an electric field in $y$ to cancel the Lorentz force. That's essentially equivalent to the calculation using Lorentz/Galilean transformation. Now if you want to have current along $y$ direction, there must be some force to balance the Lorentz force along $x$, which can not be provided by the applied electric field along $y$, so it has to come from somewhere else - like impurities. $\endgroup$
    – Meng Cheng
    Aug 30, 2015 at 1:47
  • $\begingroup$ Yes I thought about the same. Maybe counter-intuitive was too strong a word :) It was just interesting to see that impurities can both bring $\sigma_{xx}$ from infinity to finite (without $B$), and zero to finite (with $B$). $\endgroup$ Aug 30, 2015 at 1:58

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