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Eigenvectors

We diagonalise \bgroup\color{col1}$ H$\egroup according to
$\displaystyle H = SDS^{-1},$     (1.13)

where \bgroup\color{col1}$ D$\egroup is the diagonal matrix of the eigenvalues and \bgroup\color{col1}$ S$\egroup the orthogonal matrix of the eigenvectors. This yields
$\displaystyle e^{\alpha H}= e^{\alpha SDS^{-1}}= S e^{\alpha D}S^{-1}$     (1.14)

which follows from the definition of the power series (Exercise: CHECK)! For \bgroup\color{col1}$ H=T_c \sigma_x$\egroup, we already calculated the EVs in an earlier chapter,
$\displaystyle \left(\begin{array}{cc}
0 & T_c \\
T_c & 0 \\
\end{array}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\e...
...rac{1}{\sqrt{2}}\left(\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{array}\right)$ (1.15)

with \bgroup\color{col1}$ \varepsilon_{\pm}=\pm T_c$\egroup. Thus,
$\displaystyle U(t,t_0)$ $\displaystyle \equiv$ $\displaystyle e^{-i H (t-t_0)}= S e^{-i(t-t_0)D}S^{-1}$  
  $\displaystyle =$ $\displaystyle \frac{1}{2} \left(\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{arr...
...d{array}\right)
\left(\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \frac{1}{2} \left(\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{arr...
...^{-i(t-t_0)T_c} \\
e^{+i(t-t_0)T_c} & -e^{+i(t-t_0)T_c} \\
\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{cc}
\cos [(t-t_0)T_c] & -i \sin [(t-t_0)T_c] \\
-i \sin [(t-t_0)T_c] & \cos [(t-t_0)T_c] \\
\end{array}\right)$  
  $\displaystyle =$ $\displaystyle \cos [(t-t_0)T_c] \hat{1} -i \sin [(t-t_0)T_c] \sigma_x.$ (1.16)


next up previous contents index
Next: Quantum Oscillations in Two-Level Up: Example: Two-Level System Previous: Power Series   Contents   Index
Tobias Brandes 2005-04-26