derivation 2. time-independent

from eq.(129)

$$\sum_{ \alpha i} t_{ \alpha i} G_{ i a}^<(t, t) = \sum_{ \alpha i} t_{ \alpha i} \sum_j \int d t_1 \left[ \int \fra... ...\frac{d \omega _2}{2 \pi} g_\alpha ^<(\omega _2) e^{-i\omega _2(t-t_1)} \right.$$

$$+ \left. \int \frac{d \omega _1}{2 \pi} G_{ij}^<(\omega _1) e^{-i... ...\frac{d \omega _2}{2 \pi} g_\alpha ^A(\omega _2) e^{-i\omega _2(t-t_1)} \right]$$ (135)

$$= \sum_{\alpha i j} t_{\alpha i}t_{j\alpha } \int \frac{d \omega _1... ..._1) g_\alpha ^<(\omega _1) + G_{ij}^<(\omega _1) g_\alpha ^A(\omega _1) \right]$$ (136)

time independent $\rightarrow$ (eigenvalue of the isolated L region)

Green functions become

$$g_\alpha ^<(\omega ) = i \int dt e^{i\omega t} f(\epsilon _\alpha ) e^{-i \epsilon _\alpha t} = 2 \pi i \delta (\omega -\epsilon _a)$$ (137)

$$g_\alpha ^R(\omega ) = -i \int dt e^{i\omega t-\delta t} \theta(t) e^{-i \epsilon _\alpha t} = \frac{1}{\omega -\epsilon _\alpha +i \delta }$$ (138)

$$g_\alpha ^A(\omega ) = i \int dt e^{i\omega t-\delta t} \theta(-t) e^{-i \epsilon _\alpha t} = \frac{1}{\omega -\epsilon _\alpha -i \delta }$$ (139)

$J_L$ becomes

$$J_L = 2e {\rm Re} \sum_{\alpha ij} t_{\alpha n} t_{m\alpha } \int \frac... ... _1) + G_{ij}^<(\omega _1) \frac{1}{\omega -\epsilon _\alpha -i\delta } \right]$$ (140)

$$= 2e {\rm Im} \int d \omega _1 \left(\sum_\alpha t_{\alpha n} t_{m\... ...t\{ G_{ij}^R(\omega _1) f(\omega _1) + \frac{1}{2} G_{ij}^<(\omega _1) \right\}$$

$$+ 2e {\rm Re} \int \frac{d \omega _1}{2\pi} \left(\sum_\alpha t_{... ...alpha } {\rm P} \frac{1}{\omega -\epsilon _\alpha } \right) G_{ij}^<(\omega _1)$$ (141)

Here define

$$\Gamma_{ij}(\omega ) = 2 \pi \sum_\alpha t_{\alpha i} t_{j\alpha } \delta (\omega -\epsilon _\alpha )$$ (142)

$$= i \sum_\alpha t_{\alpha i} t_{j\alpha } \left( \frac{1}{\omega -\epsilon _\alpha +i\delta } - \frac{1}{\omega -\epsilon _\alpha -i\delta } \right)$$ (143)

$$= i \sum_\alpha t_{\alpha i} t_{j\alpha } \left( g_\alpha ^R(\omega )-g_\alpha ^A(\omega ) \right)$$ (144)

$$= i (\Sigma^R(\omega )-\Sigma^A(\omega ))$$ (145)

Use ,

$$J_L = -e i \int \frac{d \omega }{2\pi} \sum_{ij} \Gamma_{ij}(\omega ) \left\{ (G^R_{ij} -G^A_{ij}) f(\omega ) + G_{ij}^< \right\}$$

$$+ 2e {\rm Re} \int \frac{d \omega }{2\pi} \sum_{ij} \left( \sum_\... ...\alpha i} t_{j\alpha } \frac{\rm P}{\omega -\epsilon _\alpha } \right) G_{ij}^<$$ (146)

If region described by $\{{d_i}\}$,$\{d_i^\dag\}$ is non-interacting, then $G_{ij}^<$ is pure imaginary.

$$J_L = -e i \int \frac{d \omega }{2\pi} {\rm Tr}\left[ \Gamma(\omega ) \left\{ (G^R -G^A) f(\omega ) + G^< \right\} \right]$$ (147)

Note that $\Gamma(\omega )$ and $f(\omega )$ are defined for the equilibrium L region.

Let's connect L-C-R. Use equations,

$$G^< = G^R \Sigma^< G^A$$ (148)

$$\Sigma^< = i f_L(\omega ) \Gamma_L + i f_R(\omega ) \Gamma_R$$ (149)

$$\Gamma_L + \Gamma_R = i ({G^A}^{-1} - {G^R}^{-1} )$$ (150)

$$G^R - G^A = -i G^R(\Gamma_L + \Gamma_R) G^A$$ (151)

$f_L(\omega )$ means $f(\omega -\mu_L)$ where $\mu_L$ is the chemical potential in the L region. $f_R(\omega )$ means $f(\omega -\mu_R)$. Note again that eq.(150) is evaluated assuming the system is in the equilibrium.

$$J = J_L = -J_R$$ (152)

$$= (J_L-J_R)/2$$ (153)

$$= \frac{e}{2i} \int \frac{d\omega }{2\pi} {\rm Tr} \left[ \Gamma_L (G^R-G^A) f_L + \Gamma_L G^< - \Gamma_R (G^R-G^A) f_R - \Gamma_R G^< \right]$$ (154)

$$= \frac{e}{2i} \int \frac{d\omega }{2\pi} {\rm Tr} \left[ (\Gamma _... ... G^A + i(\Gamma _R -\Gamma _L) G^R (f_L \Gamma _L + f_R \Gamma _R ) G^A \right]$$ (155)

$$= \frac{e}{2} \int \frac{d\omega }{2\pi} {\rm Tr} \left[ -f_L \Gamm... ... G^A +f_R \Gamma _R G^R \Gamma _L G^A + f_R \Gamma _R G^R \Gamma _R G^A \right.$$

$$\left. +f_L \Gamma _L G^R \Gamma _L G^A - f_L \Gamma _R G^R \Gamm... ... G^A +f_R \Gamma _L G^R \Gamma _R G^A - f_R \Gamma _R G^R \Gamma _R G^A \right]$$ (156)

$$= -\frac{e}{2} \int \frac{d\omega }{2\pi} {\rm Tr} \left[ (f_L-f_R) (\Gamma _L G^R \Gamma _R G^A ) + (f_L-f_R)(\Gamma _R G^R \Gamma _L G^A) \right]$$ (157)

$$= - \frac{e}{2\pi} \int d\omega {\rm Tr} \left[ (f_L-f_R) (\Gamma _L G^R \Gamma _R G^A ) \right]$$ (158)

Now $\hbar$ was 1. correct prefactor,

The prefactor in the definition of the current in the tight binding scheme (eq.(113)) is imaginary, but the evaluated current is real.