Respuesta Lineal
En esta sección estudiamos la respuesta lineal del sistema:
Con tales condiciones, la
expresión para
$I_{\eta}$ se reduce a:
\begin{align*}
I_{\eta} & =
{2\pi e \over h}\,\int_{-\infty}^{\infty}\dd\omega\,\rho\pars{\omega}
\sum_{\eta'}\bracks{%
{\Gamma_{\eta} \over \Gamma}\
\ob{\Im\Sigma_{\eta'}^{<}\pars{\omega}}
{\ds{2\Gamma_{\eta'}\fermi\pars{\omega - \mu + eV_{\eta'}}}}
-
{\Gamma_{\eta'} \over \Gamma}\
\ob{\Im\Sigma_{\eta}^{<}\pars{\omega}}
{\ds{2\Gamma_{\eta}\fermi\pars{\omega - \mu + eV_{\eta}}}}}
\\[3mm]&=
{4\pi e \over h}\,{\Gamma_{\eta} \over \Gamma}
\int_{-\infty}^{\infty}\dd\omega\,\rho\pars{\omega}
\sum_{\eta'}\Gamma_{\eta'}\bracks{%
\fermi\pars{\omega - \mu + eV_{\eta'}}
-\fermi\pars{\omega - \mu + eV_{\eta}}}
\\[3mm]&\sim
{4\pi e^{2} \over h}\,{\Gamma_{\eta} \over \Gamma}
\int_{-\infty}^{\infty}\dd\omega\,\rho\pars{\omega}\,
\partiald{\fermi\pars{\omega - \mu}}{\omega}
\sum_{\eta'}\Gamma_{\eta'}\pars{V_{\eta'} - V_{\eta}}
\\[3mm]&=
-\,{4\pi e^{2} \over h}\,{\Gamma_{\eta} \over \Gamma}\bracks{%
\partiald{}{\mu} \int_{-\infty}^{\infty}\dd\omega\,\rho\pars{\omega}
\fermi\pars{\omega - \mu}}\sum_{\eta'}\Gamma_{\eta'}V_{\eta'}
+{4\pi e^{2} \over h}\bracks{%
\partiald{}{\mu} \int_{-\infty}^{\infty}\dd\omega\,\rho\pars{\omega}
\fermi\pars{\omega - \mu}}\Gamma_{\eta}V_{\eta}
\\[5mm]
\end{align*}
tal que
$\ds{I_{\eta} = \sum_{\eta'}{\cal G}_{\eta\eta'}V_{\eta'}}$
donde
\begin{align*}
{\cal G}_{\eta\eta'} &\equiv
\pars{\Gamma_{\eta}\delta_{\eta\eta'} -
{\Gamma_{\eta}\Gamma_{\eta'} \over \Gamma}}\gamma\,,\quad
\\[3mm]&\!\!\!\!\!\!\!\!\!\!\mbox{Note que}\quad
\left\vert\begin{array}{l}
\phantom{\sum_{\eta}}{\cal G}_{\eta\eta'} = {\cal G}_{\eta'\eta}&&
\\
\sum_{\eta}{\cal G}_{\eta\eta'} =
\sum_{\eta'}{\cal G}_{\eta\eta'} = 0
\end{array}\right.
\\[3mm]
\gamma &\equiv
{4\pi e^{2} \over h}
\partiald{}{\mu}\int_{-\infty}^{\infty}\dd\omega\,
\rho\pars{\omega + \mu}\fermi\pars{\omega}
\end{align*}
\begin{align*}
\gamma &=
{4\pi e^{2} \over h}\,\partiald{}{\mu}\int_{-\infty}^{\infty}\dd\omega\,
{\Gamma/\pi \over
\pars{\omega + \mu - \epsilon + eV_{G}}^{2} + \Gamma^{2}}\,
\fermi\pars{\omega}
\\[3mm]&=
{4\pi e^{2} \over h}\,\partiald{}{\mu}\braces{\half - {1 \over \pi}\,
\Im\Psi\pars{%
\half + {\Gamma + \bracks{\epsilon - \mu - eV_{G}}\ic \over 2\pi\kb T}}}
\\[3mm]&=
{4\pi e^{2} \over h}\,\pars{-\,{1 \over \pi}}\,
\Im\braces{\Psi'\pars{%
\half + {\Gamma + \bracks{\epsilon - \mu - eV_{G}}\ic \over 2\pi\kb T}}
\pars{-\,{1 \over 2\pi\kb T}\,\ic}}
\end{align*}
$$
\gamma =
{4 e^{2} \over h}\,{1 \over 2\pi\kb T}\,\Re\Psi'\pars{%
\half + {\Gamma + \bracks{\epsilon - \mu - eV_{G}}\ic \over 2\pi\kb T}}
$$
Algunos valores particulares vienen dados por:
$$
\gamma = \left\lbrace\begin{array}{lcl}
{4\pi e^{2} \over h}\,
{\Gamma/\pi \over \pars{\mu - \epsilon + eV_{G}}^{2} + \Gamma^{2}}
={4\pi e^{2} \over h}\,\rho\pars{\mu}
& \quad\mbox{si}\quad &
{\verts{\Gamma + \pars{\mu - \epsilon + eV_{G}}\ic} \over 2\pi\kb T}
\to \infty
\\[3mm]
{\pi e^{2} \over h}\,{1 \over \kb T}
& \quad\mbox{si}\quad &
{\verts{\Gamma + \pars{\mu - \epsilon + eV_{G}}\ic} \over 2\pi\kb T}
\gtrsim 0
\end{array}\right.
$$
Un Ejemplo Particular
En este caso particular solo ilustramos, por simplicidad, cuatro
reservorios. En uno de ellos el flujo de carga se establece
igual
a cero $\pars{~I_{3} = 0~}$ tal que $I_{0} + I_{1} + I_{2} = 0$.
$$
\pars{\begin{array}{c}I_{0} \\ I_{1} \\ I_{2} \\ I_{3}\end{array}}
=
\pars{\begin{array}{cccc}
{\cal G}_{00} & {\cal G}_{01} & {\cal G}_{02} & {\cal G}_{03}
\\
{\cal G}_{10} & {\cal G}_{11} & {\cal G}_{12} & {\cal G}_{13}
\\
{\cal G}_{20} & {\cal G}_{21} & {\cal G}_{22} & {\cal G}_{23}
\\
{\cal G}_{30} & {\cal G}_{31} & {\cal G}_{32} & {\cal G}_{33}
\end{array}}\
\pars{\begin{array}{c}V_{0} \\ V_{1} \\ V_{2} \\ V_{3}\end{array}}
$$
En el ejemplo presente, por simplicidad, escogemos
$\ds{\Gamma_{\eta} = {1 \over 4}\,\Gamma}$ tal que
$\ds{{\cal G}_{\eta\eta} = {3 \over 16}\,\gamma\Gamma}$ y
$\ds{\left.{\cal G}_{\eta\eta'}\right\vert_{\eta \not= \eta'}
= -\,{1 \over 16}\,\gamma\Gamma}$:
$$
\pars{\begin{array}{c}I_{0} \\ I_{1} \\ -I_{0} - I_{1} \\ 0\end{array}}
=
{1 \over 16}\,\gamma\Gamma\,
\pars{\begin{array}{rrrr}
3 & -1 & -1 & -1
\\
-1 & 3 & -1 & -1
\\
-1 & -1 & 3 & -1
\\
-1 & -1 & -1 & 3
\end{array}}\
\pars{\begin{array}{c}V_{0} \\ V_{1} \\ V_{2} \\ V_{3}\end{array}}
$$
$$
\imp\quad
\left\lbrace\begin{array}{cclcl}
I_{0} & = &
{1 \over 16}\,\gamma\Gamma\,\pars{3V_{0} - V_{1} - V_{2} - V_{3}}
& = &\phantom{-\,}
{4 \over 3}\,\gamma\Gamma\,
\bracks{2\pars{V_{0} - V_{2}} - \pars{V_{1} - V_{2}}}
\\[1mm]
I_{1} & = &
{4 \over 3}\,\gamma\Gamma\,\pars{-V_{0} + 3V_{1} - V_{2} - V_{3}}
& = & -\,{4 \over 3}\,\gamma\Gamma\,
\bracks{\pars{V_{0} - V_{2}} -2 \pars{V_{1} - V_{2}}}
\\[1mm]
I_{2} & = & -I_{0} - I_{1}&&
\\[1mm]
0 & = & {1 \over 16}\,\gamma\Gamma\,\pars{-V_{0} - V_{1} - V_{2} + 3V_{3}}
& \imp & V_{3} = {V_{0} + V_{1} + V_{2} \over 3}
\end{array}\right.
$$
Note que $V_{3}$ debe escogerse igual a $\pars{V_{0} + V_{1} + V_{2}}/3$
para garantizar que el flujo de carga desde el $\mbox{reservorio-}3$ se
anule. Escogiendo $V_{2} = 0$ ( origen del potencial
eléctrico ):
\begin{align*}
I_{0} &= \phantom{-\,}{4 \over 3}\,\gamma\Gamma\,\pars{2V_{0} - V_{1}}
\\[3mm]
I_{1} &= -\,{4 \over 3}\,\gamma\Gamma\,\pars{V_{0} - 2V_{1}}
\\[3mm]
I_{2} &= -\,{4 \over 3}\,\gamma\Gamma\,\pars{V_{0} + V_{1}}
\\[5mm]
V_{3} &= \phantom{-\,}{V_{0} + V_{1} \over 3}
\end{align*}