% Lecture notes for Physical Mathematics I
% Ewan Stewart
% KAIST
% 2011/12/13

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\begin{document}

\section{Integral and differential operators}

\subsection{Integral operators}

An operator $L$ acting on a vector $\ket\phi$ is expressed in components as
\begin{equation}
\bra\alpha L \ket\phi = \int \d\beta \fn{g}{\beta} \fn{L}{\alpha,\beta} \fn{\phi}{\beta}
\end{equation}
The right hand side of this equation is an \keyword{integral operator} $\int \d\beta \fn{g}{\beta} \fn{L}{\alpha,\beta}$ acting on a function $\fn{\phi}{\beta}$.
An integral operator is Hermitian if
\begin{equation}
\fn{L}{\alpha,\beta} = \fn{L^*}{\beta,\alpha}
\end{equation}

\subsection{Differential operators}

Physical operators are often local operators in which case they can be expressed as
\begin{equation} \label{diffop}
\bra\alpha L \ket\phi = L_\alpha \fn{\phi}{\alpha}
\end{equation}
where the right hand side is a \keyword{differential operator} $L_\alpha$ acting on a function $\fn{\phi}{\alpha}$.
For example
\begin{equation}
\bra{x} \hat{p} \ket\psi = - i \hbar \frac{d}{dx} \fn{\psi}{x}
\end{equation}
Canceling $\ket\phi$ from both sides of \eq{diffop} gives
\begin{equation}
\bra\alpha L = L_\alpha \bra\alpha
\end{equation}
Therefore
\begin{equation}
L = \int \d\alpha \fn{g}{\alpha} \ket\alpha L_\alpha \bra\alpha
\end{equation}
and
\begin{equation}
\fn{L}{\alpha,\beta} = \frac{1}{\fn{g}{\beta}} L_\alpha \fn{\delta}{\alpha-\beta}
\end{equation}

\subsection{Hermitian differential operators}

Reexpressing Eq.~(3.2.1) in component form, the Hermitian conjugate of a differential operator
\footnote{Note that $L^\dagger_\alpha \equiv (L^\dagger)_\alpha \neq (L_\alpha)^\dagger = (L_\alpha)^* \equiv L_\alpha^*$ since $L_\alpha$ is a scalar operator.}
is defined by
\begin{equation}
\int \d\alpha \fn{g}{\alpha} \fn{\phi^*}{\alpha} L^\dagger_\alpha \fn{\psi}{\alpha}
= \int \d\alpha \fn{g}{\alpha} \fn{\psi}{\alpha} L_\alpha^* \fn{\phi^*}{\alpha}
\end{equation}
for all $\ket\phi$ and $\ket\psi$ in the Hilbert space.
Therefore
\begin{equation} \label{hcdiffop}
\fn{\phi^*}{\alpha} L^\dagger_\alpha \fn{\psi}{\alpha} - \fn{\psi}{\alpha} L_\alpha^* \fn{\phi^*}{\alpha}
= \frac{1}{\fn{g}{\alpha}} \frac{d}{d\alpha} \fn{W}{\phi,\psi}
\end{equation}
with
\begin{equation} \label{hbc}
\left[ \fn{W}{\phi,\psi} \right]_\mathrm{boundary} = 0
\end{equation}

Note that if $L$ is Hermitian and $\ket\phi$ and $\ket\psi$ are in the same eigenspace, then \eq{hcdiffop} implies
\begin{equation}
\fn{W}{\phi,\psi} = \textrm{constant}
\end{equation}

A second order differential operator has the general form
\begin{equation}
\fn{g}{\alpha} L_\alpha = \fn{a}{\alpha} \frac{d^2}{d\alpha^2} + \fn{b}{\alpha} \frac{d}{d\alpha} + \fn{c}{\alpha}
\end{equation}
\eq{hcdiffop} gives
\begin{equation}
\fn{g}{\alpha} L^\dagger_\alpha = \fn{a^*}{\alpha} \frac{d^2}{d\alpha^2} + \left[ 2 \fn{{a^*}'}{\alpha} - \fn{b^*}{\alpha} \right] \frac{d}{d\alpha} + \fn{{a^*}''}{\alpha} - \fn{{b^*}'}{\alpha} + \fn{c^*}{\alpha}
\end{equation}
with
\begin{equation}
\fn{W}{\phi,\psi} = \fn{a^*}{\alpha} \left[ \fn{\phi^*}{\alpha} \fn{\psi'}{\alpha} - \fn{{\phi^*}'}{\alpha} \fn{\psi}{\alpha} \right] + \left[ \fn{{a^*}'}{\alpha} - \fn{b^*}{\alpha} \right] \fn{\phi^*}{\alpha} \fn{\psi}{\alpha}
\end{equation}
satisfying the boundary condition \eq{hbc}.
If $L$ is Hermitian then
\begin{equation}
\fn{a}{\alpha} = \fn{p}{\alpha}
\quad , \quad
\fn{b}{\alpha} = \fn{p'}{\alpha} + i \fn{r}{\alpha}
\quad , \quad
\fn{c}{\alpha} = \fn{q}{\alpha} + \frac{i}{2} \fn{r'}{\alpha}
\end{equation}
where $p(\alpha)$, $q(\alpha)$ and $r(\alpha)$ are real functions.
Therefore a \keyword{Hermitian second order differential operator} has the general form
\begin{equation}
\fn{g}{\alpha} L_\alpha = \frac{d}{d\alpha} \fn{p}{\alpha} \frac{d}{d\alpha}
+ \frac{i}{2} \left[ \fn{r}{\alpha} \frac{d}{d\alpha} + \frac{d}{d\alpha} \fn{r}{\alpha} \right]
+ \fn{q}{\alpha}
\end{equation}
with
\begin{equation}
\fn{W}{\phi,\psi} = \fn{p}{\alpha} \left[ \fn{\phi^*}{\alpha} \fn{\psi'}{\alpha} - \fn{{\phi^*}'}{\alpha} \fn{\psi}{\alpha} \right]
+ i \fn{r}{\alpha} \fn{\phi^*}{\alpha} \fn{\psi}{\alpha}
\end{equation}
satisfying the boundary condition \eq{hbc}.

Any real second order differential operator
\begin{equation}
L_\alpha = \fn{A}{\alpha} \frac{d^2}{d\alpha^2} + \fn{B}{\alpha} \frac{d}{d\alpha} + \fn{C}{\alpha}
\end{equation}
with $\left|\fn{A}{\alpha}\right| > 0$ can be expressed as a Hermitian operator
\begin{equation}
\fn{g}{\alpha} L_\alpha = \frac{d}{d\alpha} \fn{p}{\alpha} \frac{d}{d\alpha} + \fn{q}{\alpha}
\end{equation}
with
\begin{equation}
\frac{\fn{p'}{\alpha}}{\fn{p}{\alpha}} = \frac{\fn{B}{\alpha}}{\fn{A}{\alpha}}
\end{equation}
and
\begin{equation}
\fn{g}{\alpha} = \frac{\fn{p}{\alpha}}{\fn{A}{\alpha}}
\quad , \quad
\fn{q}{\alpha} = \frac{\fn{C}{\alpha} \fn{p}{\alpha}}{\fn{A}{\alpha}}
\end{equation}
if the boundary condition \eq{hbc} is satisfied.

\subsection{Eigenfunctions}

All of our formalism for the eigenvectors of linear operators translates directly to the eigenfunctions of differential operators simply by taking components with respect to a continuous basis $\ket\alpha$.

\subsubsection{Discrete set of eigenvectors}

In the case of a discrete set of eigenvectors $\ket{\psi_a}$, the eigenvector, orthonormality, completeness and eigenvector expansion equations become
\begin{eqnarray}
L \ket{\psi_a} = \lambda_a \ket{\psi_a}
\quad & \to & \quad
L_\alpha \fn{\psi_a}{\alpha} = \lambda_a \fn{\psi_a}{\alpha}
\\
\braket{\psi_a}{\psi_b} = \delta_{ab}
\quad & \to & \quad
\int \d\alpha \fn{g}{\alpha} \fn{\psi_a^*}{\alpha} \fn{\psi_b}{\alpha} = \delta_{ab}
\\
\sum_a \ket{\psi_a} \bra{\psi_a} = 1
\quad & \to & \quad
\sum_a \fn{\psi_a}{\alpha} \fn{\psi_a^*}{\beta} = \frac{1}{\fn{g}{\alpha}} \fn{\delta}{\alpha-\beta}
\\
\ket\phi = \sum_a \phi_a \ket{\psi_a}
\quad & \to & \quad
\fn{\phi}{\alpha} = \sum_a \phi_a \fn{\psi_a}{\alpha}
\\
\phi_a = \braket{\psi_a}{\phi}
\quad & \to & \quad
\phi_a = \int \d\alpha \fn{g}{\alpha} \fn{\psi_a^*}{\alpha} \fn{\phi}{\alpha}
\end{eqnarray}

\subsubsection{Continuous set of eigenvectors}

Similarly, in the case of a continuous set of eigenvectors $\ket{\fn{\psi}{a}}$, we have
\begin{eqnarray}
L \ket{\fn{\psi}{a}} = \fn{\lambda}{a} \ket{\fn{\psi}{a}}
\quad & \to & \quad
L_\alpha \fn{\psi}{a,\alpha} = \fn{\lambda}{a} \fn{\psi}{a,\alpha}
\\
\braket{\fn{\psi}{a}}{\fn{\psi}{b}} = \frac{1}{\fn{h}{a}} \fn{\delta}{a-b}
\quad & \to & \quad
\int \d\alpha \fn{g}{\alpha} \fn{\psi^*}{a,\alpha} \fn{\psi}{b,\alpha} = \frac{1}{\fn{h}{a}} \fn{\delta}{a-b}
\nonumber \\ \\
\int \d{a} \fn{h}{a} \ket{\fn{\psi}{a}} \bra{\fn{\psi}{a}} = 1
\quad & \to & \quad
\int \d{a} \fn{h}{a} \fn{\psi}{a,\alpha} \fn{\psi^*}{a,\beta} = \frac{1}{\fn{g}{\alpha}} \fn{\delta}{\alpha-\beta}
\nonumber \\ \\
\ket\phi = \int \d{a} \fn{h}{a} \fn{\phi}{a} \ket{\fn{\psi}{a}}
\quad & \to & \quad
\fn{\phi}{\alpha} = \int \d{a} \fn{h}{a} \fn{\phi}{a} \fn{\psi}{a,\alpha}
\\
\fn{\phi}{a} = \braket{\fn{\psi}{a}}{\phi}
\quad & \to & \quad
\fn{\phi}{a} = \int \d\alpha \fn{g}{\alpha} \fn{\psi^*}{a,\alpha} \fn{\phi}{\alpha}
\end{eqnarray}

\end{document}

