
\subsection{\bf $i\frac{d}{dx}$ as a Hermitean Operator}

\begin{slide}

\begin{itemize}
\item Suppose we consider periodic functions
\begin{eqnarray*}
f(b) & = & f(a) \\
g(b) & = & g(a) \\
\end{eqnarray*}

\begin{description}
\item[---] i.e., impose particular boundary conditions
\end{description}

\item Then surface term vanishes and 
\begin{displaymath}
\int^b_a g(x) \frac{d}{dx} f(x)\, dx = - \int^b_a f(x) \frac{d}{dx} g(x)\, dx
\end{displaymath}

\item The operator $L = i \frac{d}{dx}$ with $\underline{\rm
boundary\ conditions}$ $f(a) = f(b)$ is Hermitean.  The $i$ in $L$
cancels minus sign.

\item Must include boundary conditions when discussing operators in
infinite dimensional space
\end{itemize}

\end{slide}

