# Electrical conductivity and resistivity tensor

By definition of the conductivity tensor $\hat{\sigma}$ and the resistivity tensor $\hat{\rho}$, we have \begin{equation*} \begin{split} & j_{\alpha}=\sigma_{\alpha \beta}E_{\beta} \\ & E_{\beta}=\rho_{\beta \alpha}j_{\alpha} \end{split} \end{equation*} If we put the second relation into the first one, we would have $$j_{\alpha}=\sigma_{\alpha \beta}\rho_{\beta \gamma}j_{\gamma}$$ which simply means $\sigma_{\alpha \beta}\rho_{\beta \gamma}=\delta_{\alpha \gamma}$, i.e. the two tensors are matrix inverse of each other in this 2-D case.

However, I saw in some online notes on quantum hall effect relations like $\sigma_{yy}=\frac{\rho_{yy}}{\rho_{yy}^2+\rho_{xy}^2}$ and $\sigma_{xy}=\frac{\rho_{xy}}{\rho_{yy}^2+\rho_{xy}^2}$, which are obviously contradictory to the above fact that $\hat{\sigma}=\hat{\rho}^{-1}$. I have also seen similar relations in some other places. Did I miss anything?

• I think you missed a minus sign on your $\sigma_{xy}$ formula – Azad May 10 '15 at 10:33
• ok.. so if that's the case, do you know how to derive the relations? – M. Zeng May 10 '15 at 10:41

Assuming that the expressions given for a 2D electron gas here are valid, there really is no contradiction. Remember that there is an implicit sum in a repeated index (Einstein summation convention). Equivalently, an inverse matrix need not be the component-by-component inverse of the original matrix. For example, we get: $$[\sigma\cdot\rho]_{xx} = \sigma_{xx}\rho_{xx} + \sigma_{xy}\rho_{yx}$$ $$= \frac{\rho_{xx}^2}{\rho_{xx}^2+\rho_{xy}^2} - \frac{\rho_{xy}\rho_{yx}}{\rho_{xx}^2+\rho_{xy}^2}$$ Together with $\rho_{xy} = -\rho_{yx}$, we get: $$[\sigma\cdot\rho]_{xx} = 1$$ Similarly, $$[\sigma\cdot\rho]_{xy} = \frac{\rho_{xx}\rho_{xy}}{\rho_{xx}^2+\rho_{xy}^2} - \frac{\rho_{xy}\rho_{yy}}{\rho_{xx}^2+\rho_{xy}^2}$$ And using $\rho_{xx} = \rho_{yy}$, we get: $$[\sigma\cdot\rho]_{xy} = 0$$

And so on for the remaining two components. Overall, we then get: $$[\sigma\cdot\rho]_{ij} = \delta_{ij}$$ which is precisely what we require.

We can also derive the required expression directly from the matrix: $$\rho = \begin{pmatrix} \rho_{xx} & \rho_{xy} \\ -\rho_{xy} & \rho_{xx}\end{pmatrix}$$ The determinant is: $$\det\rho = \rho_{xx}^2+\rho_{xy}^2$$ and its adjugate is: $$\operatorname{adj}\rho = \begin{pmatrix} \rho_{xx} & -\rho_{xy} \\ \rho_{xy} & \rho_{xx}\end{pmatrix}$$ It's inverse is then: $$\sigma = \rho^{-1} = \frac{1}{\det\rho}\operatorname{adj}\rho$$ leading to the expression for $\sigma$.

Update:

The reason why $\rho_{xy} = -\rho_{yx}$ is because the magnetic field, $\mathbf{B} = B_z\hat{\mathbf{e}}_z$, always has a certain sense of orientation, being related to the force through a cross product. Let's say that it deflects the current anticlockwise (right handed coordinates). Then, a current in the $+x$ direction would be deflected in the $+y$ direction; that in the $+y$ direction would be deflected in the $-x$ direction. Thus, the $x$ component in the latter case is negative.

In the general case, this is contained in the relation $\rho_{xy} = -\rho_{yx}$, which effectively says that an electric field in one direction always results in a deflected (perpendicular) current which is in a specific orientation (clockwise or anti-clockwise) from it, with the same magnitude for a given electric field magnitude.

• I see.. I actually used the relation $\rho_{xy}=\rho_{yx}$, which was the reason I wasn't able to derive the correct results. But why should the two be negative of each other? – M. Zeng May 10 '15 at 12:29
• @Timo I've updated the answer. – AV23 May 10 '15 at 14:11