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Longitudinal resistivity in the Quantum Hall Effect

  1. Feb 21, 2010 #1
    My only problem with a basic conceptual understanding of the Quantum Hall Effect is the relation between longitudinal conductivity and resistivity when the magnetic field is such that the filling factor is an integer, and the Hall resistance is quantized. I fully understand the splitting of the 2D DOS into Landau levels, and can see why the longitudinal conductivity will go to zero since the Fermi level is now at an energy where the DOS is approximately zero.

    My confusion is that the longitudinal resistivity also goes to zero. In reading the literature, one can often see in the same paper, plots of ro_xx and sigma_xx as functions of magnetic field, and both show the zeros at the same fields! Perhaps I do not understand the physics behind:

    sigma_xx = ro_xx / (ro_xx^2 + ro_xy^2)

    Which clearly explains ro_xx = 0 implying sigma_xx. But fundamentally, how can resistivity AND conductivity be zero? Thanks for any light someone can shed.
  2. jcsd
  3. Feb 21, 2010 #2
    Well, that is sort of beauty of the quantum Hall effect. The resistivity [itex]\rho[/itex] and conductivity [itex]\sigma[/itex] are really connected through the matrix relation

    [tex]\rho^{-1} = \sigma[/tex]
    which comes from the relations
    [itex]\mathbf{E} = \mathbf{\rho}\cdot\mathbf{J}[/itex]
    [itex]\mathbf{J} = \mathbf{\sigma}\cdot\mathbf{E}[/itex]

    In the end, the conductivity is simply defined as the inverse of the resistivity (or vice versa).

    So in principle, the resistivity and conductivity are tensors, namely
    [tex]\rho = \begin{pmatrix} \rho_{xx} & \rho_{xy}\\ \rho_{yx} & \rho_{yy}\end{pmatrix}[/itex]
    [tex]\sigma= \begin{pmatrix} \sigma_{xx} & \sigma_{xy}\\ \sigma_{yx} & \sigma_{yy}\end{pmatrix}[/itex]

    On top of that, the matrices are each others inverse. But the inverse of any 2x2 matrix is always
    [tex]\sigma^{-1} = \frac{1}{\det(\mathbf{A})} \begin{pmatrix} \sigma_{yy} & -\sigma_{xy}\\ -\sigma{yx} & \sigma_{xx}\end{pmatrix}[/tex]

    with [itex]\det(\mathbf{A}) = \sigma_{xx}\sigma_{yy} - \sigma_{xy}\sigma_{yx} = \sigma_{xx}^2-\sigma_{xy}^2[/itex]. That last equation arises due to isotropy of the sample.

    So this explains the very general relation you have written down. Note that when you have an off-diagonal structure for the conductivity tensor (i.e. all diagonal elements are zero, but the off-diagonals are not), then the resistivity is automatically also off-diagonal. It still has non-zero off-diagonal components though, so saying that the resistivity is zero is simply wrong.

    In "normal" metals you would expect that the response of the system, which is the current, is in the same direction as the applied electric field -- this is simply Ohm's law. But here, the Hall effect comes into play: the magnetic field acts as a Lorentz force, and causes the response current to be aligned perpindicular to the electric field. The presence of the magnetic field always causes the effect that there are some off-diagonal components. What makes the Quantum Hall effect so special is that a mechanism causes the diagonal components to remain zero -- even though you expect them not to be.

    Hope this doesn't create more questions for you... ;)
    Last edited: Feb 21, 2010
  4. Feb 21, 2010 #3
    Thanks for the response, xepma. That clears up the mathematical side of things a little bit, but what would be the response to an electric field applied longitudinally? Since ro_xx and sigma_xx are both zero if we have an appropriate filling factor, do we have current? Thanks.
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