Landauer Formula

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Landauer Formula

Use of scattering theory of transport as a conceptual framework for clarifying the meaning of electrical conductance and stressed its fundamental connection to the transmission function.

$I = \frac{q}{h}\int_{-\infty}^{+\infty} dE\,T(E) [f_0(E-\mu_1)-f_0(E-\mu_2)]$

Fermi function:

$f_0(E-\mu)=\frac{1}{1+e^{\frac{E-\mu}{k_BT}}}$

For $\mu_1=\mu_2$, $I=0$. Applying a small bias, $\delta V$:

$\delta I = \frac{q}{h}\int_{-\infty}^{+\infty} dE\,\delta T(E) [f_0(E-\mu_1)-f_0(E-\mu_2)]$

$+\frac{q}{h}\int_{-\infty}^{+\infty} dE\, T(E) \delta[f_0(E-\mu_1)-f_0(E-\mu_2)]$

$[f_0(E-\mu_1)-f_0(E-\mu_2)]=0$

$\delta[f_0(E-\mu_1)-f_0(E-\mu_2)]=-\frac{\partial f_0}{\partial E}\delta E$

$\delta E = q\delta V = \mu_1-\mu_2$

$-\frac{\partial f_0}{\partial E}=(1+e^{\frac{E-\mu}{k_BT}})^{-2}e^{\frac{E-\mu}{k_BT}}\frac{1}{k_BT}=\frac{1}{k_BT}\frac{e^{\frac{E-\mu}{k_BT}}}{(1+e^{\frac{E-\mu}{k_BT}})^2}$

$-\frac{\partial f_0}{\partial E}=\frac{1}{2k_BT}\frac{1}{1+\cosh(\frac{E-\mu}{k_BT})}$

$\cosh(2x)=2\cosh^2(x)-1$

$-\frac{\partial f_0}{\partial E}=\frac{1}{4k_BT}$ sech $^2\left(\frac{E-\mu}{2k_BT}\right)$

$F_T(E-\mu)=\frac{1}{4k_BT}$ sech $^2\left(\frac{E-\mu}{2k_BT}\right)$
$\delta I=\frac{q^2 \delta V}{h}\int_{-\infty}^{+\infty} dE\,T(E) F_T(E)$
$G=\frac{I}{V}=\frac{q^2}{h}T_0$
$T_0=\int_{-\infty}^{+\infty} dE\,T(E)F_T(E-\mu)$