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A well-known general form of electrical conductivity for a free system is $$\sigma_{\mu\nu}(\omega)=\frac{i\hbar}{V}\sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n} \frac{\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle}{\hbar(\omega +i\eta)+\varepsilon_n-\varepsilon_m},$$ in which the infinitesimally small positive $\eta$ will lead to $\delta(\omega+\varepsilon_n-\varepsilon_m)$ in the real part $\mathrm{Re}\,\sigma$ (note the $\mathrm{i\hbar}/V$ factor).

In this paper, Eqs. (7,8) are just the $\mathrm{Re}\,\sigma_H$ and $\mathrm{Im}\,\sigma_H$ of a BdG system calculated from equivalent Kubo formula of the Hall conductivity $\sigma_H=(\sigma_{xy}-\sigma_{yx})/2$. But the $\eta$-induced $\delta$-functions obviously only enter the imaginary part $\mathrm{Im}\,\sigma_H$ instead. In the more detailed version of that paper, Eq. (10) of $\sigma_H(\omega)$ shows that it follows from terms like (other factors are real) $$\frac{1-n_F(\varepsilon_1)-n_F(\varepsilon_2)}{(\varepsilon_1+\varepsilon_2)[(\varepsilon_1+\varepsilon_2)^2-(\omega+i\eta)^2]}.$$

But it is nothing but the Kubo formula of conductivity applied to a specific case. It looks hard to imagine the subtraction $\sigma_{xy}-\sigma_{yx}$ can make such a huge difference. How to understand this apparent discrepancy?

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A well-known general form of electrical conductivity for a free system is $$\sigma_{\mu\nu}(\omega)=\frac{i\hbar}{V}\sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n} \frac{\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle}{\hbar(\omega +i\eta)+\varepsilon_n-\varepsilon_m},$$

I will use some abbreviations and conventions to re-write this expression as: $$ \sigma_{\mu\nu}(\omega)= \frac{i}{V} \sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n} \frac{ j_{\mu}^{nm}j_{\nu}^{mn}}{\omega +\varepsilon_n-\varepsilon_m+i\eta}\;, $$ where I use superscript $nm$ to avoid writing out bras and kets, and I set $\hbar=1$.

To make the notation even easier I will also define: $$ A_{nm} = \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\;, $$ which is symmetric and which I assume to be real (and I'll drop the indices).

And define: $$ \alpha_{nm} = \omega + \varepsilon_n - \varepsilon_m $$ (and I'll drop the indices).

So, with this further abbreviation: $$ \sigma_{\mu\nu}(\omega)= \frac{i}{V} \sum_{mn} A \frac{ j_{\mu}^{nm}j_{\nu}^{mn}}{\alpha+i\eta}\;. $$

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Or, re-writing a little more: $$ \sigma_{\mu\nu}= \frac{i}{V} \sum_{mn} A \frac{ j_{\mu}^{nm}j_{\nu}^{mn}(\alpha - i\eta)}{\alpha^2+\eta^2} $$

We also have: $$ \sigma_{\nu\mu}= \frac{i}{V} \sum_{mn} A \frac{ j_{\nu}^{nm}j_{\mu}^{mn}(\alpha - i\eta)}{\alpha^2+\eta^2} $$ and $$ \sigma_{\mu\nu}^*= \frac{-i}{V} \sum_{mn} A \frac{ j_{\mu}^{mn}j_{\nu}^{nm}(\alpha + i\eta)}{\alpha^2+\eta^2} $$ and $$ \sigma_{\nu\mu}^*= \frac{-i}{V} \sum_{mn} A \frac{ j_{\nu}^{mn}j_{\mu}^{nm}(\alpha + i\eta)}{\alpha^2+\eta^2} $$ and $$ \sigma_{\mu\nu} - \sigma_{\nu\mu} = \frac{i}{V} \sum_{mn} A \frac{(\alpha - i\eta)}{\alpha^2+\eta^2} \left(j_{\mu}^{nm}j_{\nu}^{mn} - j_{\nu}^{nm}j_{\mu}^{mn}\right) $$ and $$ (\sigma_{\mu\nu} - \sigma_{\nu\mu})^* = \frac{-i}{V} \sum_{mn} A \frac{(\alpha + i\eta)}{\alpha^2+\eta^2} \left(j_{\mu}^{mn}j_{\nu}^{nm} - j_{\nu}^{mn}j_{\mu}^{nm}\right) $$ and finally: $$ iIm(\sigma_{\mu\nu} - \sigma_{\nu\mu}) = \frac{1}{V} \sum_{mn} \frac{A\eta}{\alpha^2+\eta^2} \left(j_{\mu}^{nm}j_{\nu}^{mn} - j_{\nu}^{nm}j_{\mu}^{mn}\right)\;. $$

Restoring all the abbreviations and letting $\eta \to 0$ gives: $$ Im(\sigma_{\mu\nu} - \sigma_{\nu\mu}) = \frac{\hbar}{iV} \sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\delta(\hbar\omega + \varepsilon_n - \varepsilon_m) \left(\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle - \langle m \lvert j_{\mu}\rvert n\rangle \langle n \lvert j_{\nu}\rvert m\rangle \right)\;. $$ $$ = \frac{2\hbar}{V} \sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\delta(\hbar\omega + \varepsilon_n - \varepsilon_m) {Im}\left(\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle \right)\;. $$

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  • $\begingroup$ Thank you! The whole derivation looks correct. Now I see where I was wrong in the first place: $\delta$-function only enters the real part for the longitudinal conductivity, but not for transverse ones. $\endgroup$
    – xiaohuamao
    Dec 10, 2022 at 7:59
  • $\begingroup$ You're welcome. $\endgroup$
    – hft
    Dec 10, 2022 at 18:42

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