Does delta-function always enter the real part of conductivity?

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?

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: $$Im(\sigma_{\mu\nu} - \sigma_{\nu\mu}) = \frac{\hbar}{iV} \sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\delta(\hbar\omega + \varepsilon_n - \varepsilon_m) \left(\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle - \langle m \lvert j_{\mu}\rvert n\rangle \langle n \lvert j_{\nu}\rvert m\rangle \right)\;.$$ $$= \frac{2\hbar}{V} \sum_{mn} \frac{f_n-f_m}{\varepsilon_m-\varepsilon_n}\delta(\hbar\omega + \varepsilon_n - \varepsilon_m) {Im}\left(\langle n \lvert j_{\mu}\rvert m\rangle \langle m \lvert j_{\nu}\rvert n\rangle \right)\;.$$

• 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. Dec 10, 2022 at 7:59
• You're welcome.
– hft
Dec 10, 2022 at 18:42