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次へ: 上へ: sg8_6 戻る: use transfer matrix

block matrix, non-orthogonal basis


\begin{displaymath}
(\omega S-H) G= I
\end{displaymath}

consider L,C and R region and rewrite it
\begin{displaymath}
(w S -H) = \left[ \begin{array}{ccc}
\omega S_{L}-H_{L} & \...
...\omega S_{RC}-H_{RC} & \omega S_{R}-H_{R}
\end{array} \right]
\end{displaymath} (186)


$\displaystyle G_{C}$ $\textstyle =$ $\displaystyle \left[ \omega S_C -H_{C}
- (\omega S_{CR}-H_{CR}) (\omega -H_{R})^{-1}(\omega S_{RC}- H_{RC} )\right.$  
    $\displaystyle \left.
- (\omega S_{CL}-H_{CL}) (\omega -H_{L})^{-1}(\omega S_{LC}- H_{LC}) \right]^{-1}$ (187)
  $\textstyle =$ $\displaystyle \{ \omega S_{C} -H_{C} -\Sigma_R -\Sigma_L \}^{-1}$ (188)
$\displaystyle \Sigma_R$ $\textstyle =$ $\displaystyle (\omega S_{CR}-H_{CR}) (\omega S_R-H_{R})^{-1}( \omega S_{RC} - H_{RC})$ (189)
$\displaystyle \Sigma_L$ $\textstyle =$ $\displaystyle (\omega S_{CL}-H_{CL}) (\omega S_L-H_{L})^{-1}( \omega S_{LC} - H_{LC})$ (190)



kino 平成18年4月17日