
\subsection{\bf Orthogonality of Eigenvectors}
\subsection{\bf of Hermitean Matrices}

\begin{slide}

\begin{itemize}
\item Now take any two eigenvectors/values
\begin{eqnarray*}
A | v_1 > & = & \lambda_1 | v_1 > \\
A | v_2 > & = & \lambda_2 | v_2 > \\
\langle v_2 | A | v_1 \rangle & = & \lambda_1 \langle v_2 | v_1
\rangle \hspace{1em} ( \alpha ) \\
\langle v_1 | A | v_2 \rangle & = & \lambda_2 \langle v_1 | v_2
\rangle
\end{eqnarray*}

\item Take complex conjugate of this equation
\begin{eqnarray*}
\langle v_2 | A^+ | v_1 \rangle & = & \lambda_2^* \langle v_2 | v_1
\rangle \\
\langle v_2 | A | v_1 \rangle & = & \lambda_2 \langle v_2 | v_1
\rangle \hspace{1em} ( \beta ) \\
\mbox{as}\ A = A^+ && \lambda_2 = \lambda_2^*
\end{eqnarray*}

\item Subtract $(\alpha )$ and $(\beta )$
\[
\left( \lambda_1 - \lambda_2 \right) \langle v_2 | v_1 \rangle = 0
\]

\item either $\lambda_1 = \lambda_2$ or $\langle v_2 | v_1 \rangle
= 0$ so eigenvectors of different eigenvalues are orthogonal.

\item Note if $\vert v_1>$ and $\vert v_2>$ are distinct eigenvectors
of same degenerate eigenvalue $\lambda_1 = \lambda_2$, then we can
replace $\vert v_2>$ by $\vert v_2^\prime >$ where $\vert v_2^\prime >$ is 
a linear combination of $\vert v_1>$ and $\vert v_2>$, and $< v_1
\vert v_2^\prime > = 0$.
\end{itemize}

\end{slide}

