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The spin current $\mathbf{J^\gamma}$ due to the electric field $\mathbf{E}$ is mathematically given by $$ \mathbf{J^\gamma = \sigma^\gamma E} $$ here $\gamma$ shows spin-component, and $\mathbf{\sigma^\gamma}$ represent spin conductivity tensor. I read in an article that SHE $\mathbf{\Omega^\gamma}$ is mathematically represented by the anti-symmetric part of the conductivity tensor $\sigma^\gamma$ i.e. $$ \mathbf{\Omega^\gamma} = \frac{\sigma^\gamma-(\sigma^{\gamma})^T}{2} $$ The publication is:

Origin of the Magnetic Spin Hall Effect: Spin Current Vorticity in the Fermi Sea by Alexander Mook, Robin R. Neumann, Annika Johansson, Jürgen Henk, Ingrid Mertig

The exact words (on the first page, 2nd paragraph) are

While the anomalous Hall effect (AHE) in a magnet produces a transverse charge current density upon applying an electric field $E$, the SHE in a nonmagnet produces a transverse spin current density $\langle J^\gamma \rangle = \sigma^\gamma E$, $\gamma$ indicates the transported spin component). Mathematically, the SHE is quantified by the antisymmetric part of the spin conductivity tensor $\sigma^\gamma$ . For example, the $\sigma_{xy}^z$ element comprises z-polarized spin currents in the x-direction as a response to an electric field in the y-direction

Why SHE has to be the anti-symmetric part and not the full $\sigma^\gamma$ tensor?

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The Hall effect, whether Spin, anomalous, or regular variety, is a transverse current due to a longitudinal electromotive force. That is, a current in $y$ due to a voltage across $x$. This component is represented by the off-diagonal elements of the conductivity tensor. Moreover, knowing how, for example, the SHE transforms with inversion, we can specify that it will be antisymmetric (only the off-diagonal elements can possibly be antisymmetric). The diagonal elements of the tensor refer to a current in $x$ due to a voltage in $x$, and similarly $y$ current for $y$ voltage. This is an important part of the tensor, it’s just not a Hall effect. Rather, the diagonal represents regular conductivity.

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  • $\begingroup$ thank you for your answer. So, Hall effects are connected with $\sigma_{ij}$ with $i\ne j$ components of conductivity tensor. According to the article that I've mentioned, for a 2D system, the SHE conductivity tensor is $$ \Omega^\gamma = \frac{1}{2} \begin{bmatrix} 0&(\sigma_{xy}^\gamma - \sigma_{yx}^\gamma)\\ -(\sigma_{xy}^\gamma - \sigma_{yx}^\gamma)&0 \end{bmatrix} $$ my confusion is, why SHE is $\sigma_{xy}^\gamma - \sigma_{yx}^\gamma$ and not only $\sigma_{xy}^\gamma$? $\endgroup$ Commented Jul 18, 2021 at 17:00

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