= Quantum Hall conductivity tensor inversion
{title2=$\boldsymbol\rho_s=\boldsymbol\sigma_s^{-1}$}
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 $s\geq0$, $k>0$:
$$
\mathbf K=\frac{e^2}{h}\begin{pmatrix}s&-k\\k&s\end{pmatrix}\mathbf E,
\qquad
\mathbf E=R_K\frac1{s^2+k^2}\begin{pmatrix}s&k\\-k&s\end{pmatrix}\mathbf K.
$$
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 $L$, $I=WK_x$ gives
$$
\boxed{\frac{|V_H|}{I}=R_K\frac{k}{s^2+k^2},\qquad
\frac{V_{xx}}{I}=R_K\frac{L}{W}\frac{s}{s^2+k^2}.}
$$
At $s=0$, $k=N$, 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 $V_{xx}/I\simeq129.061\,\Omega$, while the Hall magnitude is below $R_K/2$ by the relative amount $s^2/(k^2+s^2)\simeq25.0$ 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.
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