Quantum Hall conductivity tensor inversion (${\mathbf{\rho }}_s={\mathbf{\sigma }}_s^{-1}$)

Choose conventional surface current density along $+x$, magnetic field along $+z$, and negative Electron charge. An isotropic in-plane response has the following sheet conductivity and sheet resistivity tensors for $s\ge 0$, $k>0$:These are matrix inverses, rather than componentwise reciprocals. The sheet conductivity has units siemens, the sheet resistivity ohms. With $K_y=0$, the sign is $E_y/K_x<0$, agreeing with the negative Hall coefficient of a single Electron species. Contact polarity can reverse the measured voltage sign, so quote the Hall magnitude. For a homogeneous strip of width $W$ and longitudinal probe separation $L$, $I=W{K}_x$ givesAt $s=0$, $k=N$, these reduce to $R_K/N$ and zero. A nonzero transverse sheet resistivity creates no local Joule heating: the antisymmetric contribution cancels in $\mathbf{E}\cdot \mathbf{K}=(e^2/h)s|\mathbf{E}{|}^2$.

For an original nonideal example, choose $k=2$, $s=0.01$ and $L/W=2$. Then $V_{xx}/I\simeq 129.061\Omega$, while the Hall magnitude is below $R_K/2$ by the relative amount $s^2/(k^2+s^2)\simeq 25.0$ parts per million. An unwanted admixture $0.02V_{xx}$ in the Hall readout would instead introduce approximately $2.58\Omega$. The chosen calculation shows why a small intrinsic tensor correction does not by itself guarantee an accurate contact measurement.