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$H^{-1}$


$\displaystyle G$ $\textstyle =$ $\displaystyle \left( \begin{array}{cc} E-H_A & -H_{AB} \\ -H_{BA} & E-H_B \end{array} \right)^{-1}$ (262)
  $\textstyle =$ $\displaystyle \left( \begin{array}{cc} \frac{1}{E-H_A - H_{AB} \frac{1}{E-H_B} ... ... H_{BA} } & \frac{1}{E-H_B - H_{BA} \frac{1}{E-H_A} H_{AB} } \end{array}\right)$ (263)


kino 平成17年12月15日