Eigenvectors

We diagonalise $H$ according to

$$H = SDS^{-1},$$ (1.13)

where $D$ is the diagonal matrix of the eigenvalues and $S$ the orthogonal matrix of the eigenvectors. This yields

$$e^{\alpha H}= e^{\alpha SDS^{-1}}= S e^{\alpha D}S^{-1}$$ (1.14)

which follows from the definition of the power series (Exercise: CHECK)! For $H=T_c \sigma_x$, we already calculated the EVs in an earlier chapter,

$$\left(\begin{array}{cc} 0 & T_c \\ T_c & 0 \\ \end{array}\right) = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \e... ...rac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array}\right)$$ (1.15)

with $\varepsilon_{\pm}=\pm T_c$. Thus,

$$U(t,t_0) \equiv e^{-i H (t-t_0)}= S e^{-i(t-t_0)D}S^{-1}$$

$$= \frac{1}{2} \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{arr... ...d{array}\right) \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{array}\right)$$

$$= \frac{1}{2} \left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \\ \end{arr... ...^{-i(t-t_0)T_c} \\ e^{+i(t-t_0)T_c} & -e^{+i(t-t_0)T_c} \\ \end{array}\right)$$

$$= \left(\begin{array}{cc} \cos [(t-t_0)T_c] & -i \sin [(t-t_0)T_c] \\ -i \sin [(t-t_0)T_c] & \cos [(t-t_0)T_c] \\ \end{array}\right)$$

$$= \cos [(t-t_0)T_c] \hat{1} -i \sin [(t-t_0)T_c] \sigma_x.$$ (1.16)