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?


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 $\rho_{xx} = 0$.

| 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.