# Rotational invariance of the conductivity tensor (Classical Hall Effect)

In classical Hall effect, the conductivity tensor is given as

$$\sigma = \frac{\sigma_{DC}}{1+\omega_B^2 \tau^2} \begin{pmatrix} 1 & -\omega_B \tau \\ \omega_B \tau & 1 \end{pmatrix}$$

where the author suggests that since it is rotationally invariant, it must be in the form of

$$\sigma = \frac{\sigma_{DC}}{1+\omega_B^2 \tau^2} \begin{pmatrix} 1 & -\omega_B \tau \\ \omega_B \tau & 1 \end{pmatrix} = \begin{pmatrix} \sigma_{xx} & \sigma_{xy} \\ -\sigma_{xy} & \sigma_{xx} \end{pmatrix}$$.

My understanding of rotational invariance is that of

$$\forall A \in SO(2), \quad A^T \sigma A = \sigma$$ , i.e. $$[\sigma, A]=0$$, which would result in that form as outlined in http://bit.ly/2Lq4FsR.

However, I want to know what it really means physically; $$\sigma$$ is a conductivity, hence a tensor, which used to be a scalar when magnetic field is absent. So what does it mean that the conductivity is rotationally invariant?

• "the author suggests..." - which author? – SuperCiocia Sep 9 '19 at 0:20
• This is from Lecture notes on Quantum Hall Effect by David Tong. It is from the first chapter, 1.2.2 The Drude model. – Nuri Sep 10 '19 at 6:09