total charge from contour integration 2

TRANSIESTA uses a different approach. eq.(75) can be written as

$$\rho_{ij} = \int_{-\infty}^{\infty} d\omega A_{ij}(\omega ) f(\omega -\mu)$$ (80)

$$= -\frac{1}{\pi} {\rm Im} \int_{-\infty}^{\infty} d\omega G_{ij}^{R}(w) f(\omega -\mu)$$ (81)

where $f(\omega -\mu)$ is the Fermi distribution function. using the thermal Green function and the pole of $f(\omega )$ is at $i\omega _n$,

$$(\int_{L_1} + \int_{L_2} + \int_{L3}) d\omega G^R(\omega ) f... ...-2 \pi i \frac{1}{\beta} \sum_{i\omega _n}^{iQ} G(i\omega _n)$$ (82)

where the path $L_1$ is ($E_{min}$,0) $\rightarrow$ ($\infty$,0), $L_2$ is ($\infty$,$iQ$) $\rightarrow$ ($\mu$,$iQ$) and $L_3$ is ($\mu$,$iQ$) $\rightarrow$ ($E_{min}$,0). $E_{min}$ is smaller than the minimum of the eigenvalues. $Q$ is a small value enough to stabilize the integration. Then

$$\rho_{ij} = -\frac{1}{\pi} {\rm Im} \left[ \int_{L_2+L_3} d... ...\frac{1}{\beta} \sum_{i\omega _n}^{iQ} G(i\omega _n) \right]$$ (83)

practical upper limit to calculate $\int_{L_2} d\omega$ is $\sim \mu+10T$ due to the presence of $f(\omega -\mu)$

Note that $G^R$ has no pole in the upper half plane of $\omega$