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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?

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

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