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Why in quantum hall effect longitudinal resistance and conductivity can be simultaneously zero? I am puzzled about it. What's the physics meaning of it?

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The resistivity and conductivity are tensors \begin{align} \rho &= \begin{pmatrix}\rho_{xx} & \rho_{xy}\\\rho_{yx} & \rho_{yy}\end{pmatrix}\\ \sigma &= \begin{pmatrix}\sigma_{xx} & \sigma_{xy}\\\sigma_{yx} & \sigma_{yy}\end{pmatrix} \end{align} linearly relating the electric field and current density: \begin{align} E_a &= \rho_{ab}J_b,\\ J_a &= \sigma_{ab}E_b. \end{align} As matrices, $\rho$ and $\sigma$ are inverses, \begin{align} \sigma = \rho^{-1} \end{align} so that \begin{align} \sigma = \frac{1}{\det \rho}\begin{pmatrix}\rho_{yy} & -\rho_{xy}\\-\rho_{yx} & \rho_{xx}\end{pmatrix}. \end{align} If the longitudinal components $\rho_{xx}$, $\rho_{yy}$ of the resistivity are zero, and if the transverse $\rho_{xy}$ and $\rho_{yx}$ are non-zero so that $\det \rho \neq 0$, then the longitudinal components of $\sigma$ are also zero.

Since the longitudinal conductivity is zero, an electric field applied along the $x$ (or $y$) direction does not lead to a current in this direction. Instead, a transverse current is produced. Since the longitudinal resistivity is zero, if we pass a current along the $x$ (or $y$) direction, we don't observe a potential difference along this direction. Instead, we would measure a transverse (Hall) voltage.

Quoting Steve Girvin:

Notice that, paradoxically, the system looks insulating since $\sigma_{xx} = 0$ and yet it looks like a perfect conductor since $\rho_{xx} = 0$.

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