% Lecture notes for Physical Mathematics I
% Ewan Stewart
% KAIST
% 2010/12/10

\documentclass[a4paper,12pt]{report}

\usepackage{pm1}

\setcounter{chapter}{3}
\setcounter{section}{4}
\setcounter{page}{22}

\begin{document}

\section{Operator equations}

Consider the operator equation
\begin{equation} \label{opeq}
L \ket\phi = \ket{f}
\end{equation}
Let $\hilbert{L}_\lambda$ and $\hilbert{L}^\dagger_\lambda$ be the eigenspaces of $L$ and $L^\dagger$, and let $P_{\hilbert{L}_\lambda}$ and $P_{\hilbert{L}^\dagger_\lambda}$ be their projection operators.
Then $L P_{\hilbert{L}_0} = 0$ and $L^\dagger P_{\hilbert{L}^\dagger_0} = 0$, and hence $P_{\hilbert{L}^\dagger_0} L = 0$.
Therefore, applying $P_{\hilbert{L}^\dagger_0}$ to \eq{opeq}, we require
\begin{equation} \label{consistency}
P_{\hilbert{L}^\dagger_0} \ket{f} = 0
\end{equation}
for a solution $\ket\phi$ to exist.
If a solution exists, we can decompose it as
\begin{equation}
\ket\phi = P_{\hilbert{L}_0} \ket\phi + \left( 1 - P_{\hilbert{L}_0} \right) \ket\phi
\end{equation}
where $P_{\hilbert{L}_0} \ket\phi$ is undetermined and $\left( 1 - P_{\hilbert{L}_0} \right) \ket\phi$ is proportional to $\ket{f}$, and hence can be written as $G \ket{f}$ for some linear operator G.
Thus the formal solution of \eq{opeq} is
\begin{equation} \label{opeqsol}
\ket\phi = P_{\hilbert{L}_0} \ket\phi + G \ket{f}
\end{equation}
The ambiguity in $G$ coming from \eq{consistency} can be fixed by setting $G P_{\hilbert{L}^\dagger_0} = 0$, in which case applying $P_{\hilbert{L}_0}$ and $L$ to \eq{opeqsol} give $P_{\hilbert{L}_0} G = 0$ and
\begin{equation} \label{Geq}
L G = 1 - P_{\hilbert{L}^\dagger_0}
\end{equation}
respectively.
We can make this more explicit in two important cases.


\subsection{Hermitian operator equations}

If $L$ is Hermitian then its eigenspaces are orthogonal and complete
\begin{equation}
L = \sum_\lambda \lambda P_{\hilbert{L}_\lambda}
\quad \textrm{and} \quad
1 = \sum_\lambda P_{\hilbert{L}_\lambda}
\end{equation}
\eq{opeq} decomposes to
\begin{equation}
\lambda P_{\hilbert{L}_\lambda} \ket\phi = P_{\hilbert{L}_\lambda} \ket{f}
\end{equation}
and inverting for $\lambda \neq 0$ and recombining gives \eq{opeqsol} with
\begin{equation}
G = \sum_{\lambda \neq 0} \lambda^{-1} P_{\hilbert{L}_\lambda}
\end{equation}
which is the inverse of $L$ on $\hilbert{L}_0^\perp$.


\subsection{Green's functions}

If
\begin{equation} \label{Ld0}
\hilbert{L}^\dagger_0 = 0
\end{equation}
then \eq{Geq} reduces to
\begin{equation}
L G = 1
\end{equation}
and if $L$ can be expressed as a differential operator then the Green's function $\fn{G}{\alpha,\beta}$ satisfies
\begin{equation} \label{inhomogreen}
L_\alpha \fn{G}{\alpha,\beta} = \frac{1}{\fn{g}{\alpha}} \fn{\delta}{\alpha-\beta}
\end{equation}
and the homogeneous boundary conditions associated with the Hilbert space.
\eq{inhomogreen} is easy to solve since the delta function is zero apart from at $\alpha = \beta$
\begin{equation} \label{lhomogreen}
L_\alpha \fn{G}{\alpha,\beta} = 0
\qquad \textrm{for $\alpha < \beta$}
\end{equation}
and
\begin{equation} \label{rhomogreen}
L_\alpha \fn{G}{\alpha,\beta} = 0
\qquad \textrm{for $\alpha > \beta$}
\end{equation}
The Green's function is a solution of \eq{lhomogreen} satisfying the boundary condition at $\alpha < \beta$, patched together with a solution of \eq{rhomogreen} satisfying the boundary condition at $\alpha > \beta$, with the matching conditions at $\alpha = \beta$ fixed by the right hand side of \eq{inhomogreen}.


\subsubsection{Boundary Green's function}

For $L_\alpha$ of the form of Eq.~(3.4.12), and given $\fn{\zeta}{\alpha}$ satisfying
\footnote{Note that \eq{ll0} is possible despite \eq{Ld0} if $\fn{\zeta}{\alpha}$ does not satisfy the boundary condition at $\alpha > \beta$ and so is not in the Hilbert space.}
\begin{equation} \label{ll0}
L_\alpha \fn{\zeta}{\alpha} = 0
\quad \textrm{for} \quad \alpha < \beta
\end{equation}
including the boundary condition at $\alpha < \beta$,  and $\fn{\xi}{\alpha}$ satisfying
\begin{equation}\label{rl0}
L_\alpha \fn{\xi}{\alpha} = 0
\quad \textrm{for} \quad \alpha > \beta
\end{equation}
including the boundary condition at $\alpha > \beta$, the Green's function is
\begin{equation}
\fn{G_\mathrm{b}}{\alpha,\beta} = \left\{
\begin{array}{ccc}
\displaystyle
\frac{ \fn{\zeta}{\alpha} \fn{\xi}{\beta} }{ \fn{a}{\beta} \left[ \fn{\zeta}{\beta} \fn{\xi'}{\beta} - \fn{\zeta'}{\beta} \fn{\xi}{\beta} \right] }
& \textrm{for} & \alpha < \beta
\\ \\
\displaystyle
\frac{ \fn{\zeta}{\beta} \fn{\xi}{\alpha} }{ \fn{a}{\beta} \left[ \fn{\zeta}{\beta} \fn{\xi'}{\beta} - \fn{\zeta'}{\beta} \fn{\xi}{\beta} \right] }
& \textrm{for} & \alpha > \beta
\end{array}
\right.
\end{equation}
Note that for $L_\alpha$ of the form of Eq.~(3.4.19)
\begin{equation}
\fn{a}{\beta} \left[ \fn{\zeta}{\beta} \fn{\xi'}{\beta} - \fn{\zeta'}{\beta} \fn{\xi}{\beta} \right]
= \fn{W}{\zeta^*,\xi}
= \textrm{constant}
\end{equation}

\begin{figure}[h]
\centering
\begin{tikzpicture}[x=0.5\textwidth,y=0.5\textwidth]
% axes
\draw[->] (0,0) -- (1,0) -- ++ (10pt,0pt) node[right]{$\alpha$};
\draw[->] (0,-0.25) -- (0,0.25) node[above](delta){$\fn{G_\mathrm{b}}{\alpha,\beta}$};
\draw (0,0) -- ++(-2pt,0pt) ++(0pt,-2pt) node[below left]{$0$};
\draw (0.667,0) -- ++(0pt,-2pt) node[below]{$\beta$};
\draw (1,0) -- ++(0pt,-2pt) node[below]{$1$};
% Green's function
\draw[red] (0,0) -- (0.667,-0.222);
\draw[blue] (0.667,-0.222) -- (1,0);
\end{tikzpicture}
\caption{ \label{figGb}
$\frac{d^2}{d\alpha^2} \fn{G_\mathrm{b}}{\alpha,\beta} = \fn{\delta}{\alpha-\beta}$ with $\fn{G_\mathrm{b}}{0,\beta} = \fn{G_\mathrm{b}}{1,\beta} = 0$.
}
\end{figure}
For example, for $L_\alpha = d^2/d\alpha^2$ with boundary condition $\fn{\phi}{0} = \fn{\phi}{1} = 0$, \eq{ll0} has solution $\fn{\zeta}{\alpha} \propto \alpha$ and \eq{rl0} has solution $\fn{\xi}{\alpha} \propto 1 - \alpha$ giving the Green's function
\begin{equation} \label{Gbex}
\fn{G_\mathrm{b}}{\alpha,\beta} = \left\{
\begin{array}{ccc}
- \left( 1 - \beta \right) \alpha
& \textrm{for} & \alpha < \beta
\\
- \beta \left( 1 - \alpha \right)
& \textrm{for} & \alpha > \beta
\end{array}
\right.
\end{equation}
shown in Figure~\ref{figGb}.


\subsubsection{Causal Green's function}

For initial condition problems, the solution of \eq{ll0} is zero but \eq{rl0} has two independent solutions, giving the causal Green's function
\begin{equation}
\fn{G_\mathrm{c}}{\alpha,\beta} = \frac{ \fn{\xi_1}{\beta} \fn{\xi_2}{\alpha} - \fn{\xi_1}{\alpha} \fn{\xi_2}{\beta} }{ \fn{a}{\beta} \left[ \fn{\xi_1}{\beta} \fn{\xi_2'}{\beta} - \fn{\xi_1'}{\beta} \fn{\xi_2}{\beta} \right] } \fn{\theta}{\alpha-\beta}
\end{equation}

\begin{figure}[h]
\centering
\begin{tikzpicture}[x=0.5\textwidth,y=0.5\textwidth]
% axes
\draw[->] (0,0) -- (1,0) -- ++ (10pt,0pt) node[right]{$\alpha$};
\draw[->] (0,0) -- (0,0.5) node[above](delta){$\fn{G_\mathrm{c}}{\alpha,\beta}$};
\draw (0,0) -- ++(-2pt,0pt) (0,0) -- ++(0pt,-2pt) ++(-2pt,0pt) node[below left]{$0$};
\draw (0.667,0) -- ++(0pt,-2pt) node[below]{$\beta$};
% Green's function
\draw[red] (0,0) -- (0.667,0);
\draw[blue] (0.667,0) -- (1,0.333);
\end{tikzpicture}
\caption{ \label{figGc}
$\frac{d^2}{d\alpha^2} \fn{G_\mathrm{c}}{\alpha,\beta} = \fn{\delta}{\alpha-\beta}$ with $\fn{G_\mathrm{c}}{0,\beta} = \fn{\frac{d G_\mathrm{c}}{d\alpha}}{0,\beta} = 0$.
}
\end{figure}
For example, for $L_\alpha = d^2/d\alpha^2$ with initial condition $\fn{\phi}{0} = \fn{\phi'}{0} = 0$, \eq{rl0} has solutions $\fn{\xi_1}{\alpha} \propto 1$ and $\fn{\xi_2}{\alpha} \propto \alpha$ giving the Green's function
\begin{equation} \label{Gcex}
\fn{G_\mathrm{c}}{\alpha,\beta} = \left( \alpha - \beta \right) \fn{\theta}{\alpha-\beta}
\end{equation}
shown in Figure~\ref{figGc}.


\subsection{Inhomogeneous differential equations}

If $L$ can be expressed as a differential operator, then \eq{opeq} takes the component form
\begin{equation} \label{copeq}
L_\alpha \fn{\phi}{\alpha} = \fn{f}{\alpha}
\end{equation}
and the Green's function solution \eq{opeqsol} takes the component form
\begin{equation} \label{copeqsol}
\fn{\phi}{\alpha} = \fn{\phi_0}{\alpha} + \int \d\beta \fn{g}{\beta} \fn{G}{\alpha,\beta} \fn{f}{\beta}
\end{equation}
where $\ket{\phi_0} \in \hilbert{L}_0$.

For example, in the case of \eq{Gbex}, \eq{copeq} has solution
\begin{equation}
\fn{\phi}{\alpha} = - \alpha \int_\alpha^1 \d\beta \left( 1 - \beta \right) \fn{f}{\beta}
- \left( 1 - \alpha \right) \int_0^\alpha \d\beta \beta \fn{f}{\beta}
\end{equation}
while in the case of \eq{Gcex}
\begin{equation}
\fn{\phi}{\alpha} = \int_0^\alpha \d\beta \left( \alpha - \beta \right) \fn{f}{\beta}
\end{equation}

However, one often needs to solve a differential equation of the form
\begin{equation} \label{prob}
L_\alpha \fn{\psi}{\alpha} = \fn{h}{\alpha}
\end{equation}
where $\fn{\psi}{\alpha}$ satisfies inhomogeneous boundary conditions, i.e.\ boundary conditions which are not satisfied by $\fn{\psi}{\alpha} = 0$, which cannot define a vector space.
So first we set
\begin{equation} \label{vectorize}
\fn{\psi}{\alpha} = \fn{\psi_\mathrm{b}}{\alpha} + \fn{\phi}{\alpha}
\end{equation}
where $\fn{\psi_\mathrm{b}}{\alpha}$ satisfies the boundary conditions.
Then $\fn{\phi}{\alpha}$ satisfies homogeneous boundary conditions, consistent with a vector space structure.
For example, if $\fn{\psi}{a} = A$ and $\fn{\psi}{b} = B$, then $\fn{\psi_\mathrm{b}}{a} = A$ and $\fn{\psi_\mathrm{b}}{b} = B$, and $\fn{\phi}{a} = \fn{\phi}{b} = 0$.
Substituting \eq{vectorize} into \eq{prob} gives \eq{copeq} with
\begin{equation}
\fn{f}{\alpha} = \fn{h}{\alpha} - L_\alpha \fn{\psi_\mathrm{b}}{\alpha}
\end{equation}
and \eq{copeqsol} gives
\begin{equation}
\fn{\psi}{\alpha} = \fn{\psi_\mathrm{b}}{\alpha} + \fn{\phi_0}{\alpha}
+ \int \d\beta \fn{g}{\beta} \fn{G}{\alpha,\beta} \left[ \fn{h}{\beta} - L_\beta \fn{\psi_\mathrm{b}}{\beta} \right]
\end{equation}
Note that if $L$ is Hermitian then \eq{Ld0} implies $\hilbert{L}_0 = 0$ and hence $\fn{\phi_0}{\alpha} = 0$, but $\fn{\psi_\mathrm{b}}{\alpha}$ can still be chosen to satisfy $L_\alpha \fn{\psi_\mathrm{b}}{\alpha} = 0$ since it is not in the Hilbert space.
\footnote{Alternatively, we can think in terms of a larger Hilbert space in which $L$ is not Hermitian and choose $\fn{\psi_\mathrm{b}}{\alpha} \in \hilbert{L}_0 \neq \hilbert{L}^\dagger_0$.}


\subsection{Poisson's equation}

Poisson's equation
\begin{equation} \label{poisson}
- \nabla^2 \fn{\phi}{\vect{r}} = \fn{\rho}{\vect{r}}
\end{equation}
with boundary condition
\begin{equation} \label{pbc}
\fn{\phi}{\vect{r}} \to 0
\quad \textrm{as } |\vect{r}| \to \infty
\end{equation}
can be solved using the Green's function $\fn{G}{\vect{r},\vect{r}'}$ satisfying
\begin{equation} \label{poissongreen}
- \nabla^2 \fn{G}{\vect{r},\vect{r}'} = \fn{\delta}{\vect{r}-\vect{r}'}
\end{equation}
with the boundary condition
\begin{equation}
\fn{G}{\vect{r},\vect{r}'} \to 0 \quad \textrm{as } |\vect{r}| \to \infty
\end{equation}
Using translational symmetry, we can simplify \eq{poissongreen} to
\begin{equation} \label{pgt}
- \nabla^2 \fn{G}{\vect{r}} = \fn{\delta}{\vect{r}}
\end{equation}
Now
\begin{equation}
\int \mathinner{d^3\vect{r}} \fn{\delta}{\vect{r}} = \int 4 \pi r^2 \d{r} \fn{\delta}{\vect{r}} = \int \d{r} \fn{\delta}{r}
\end{equation}
and so
\begin{equation} \label{deltar}
\fn{\delta}{\vect{r}} = \frac{1}{4\pi r^2} \fn{\delta}{r}
\end{equation}
Using spherical symmetry and \eq{deltar}, \eq{pgt} simplifies to
\begin{equation} \label{rge}
- \frac{d}{dr} r^2 \frac{d}{dr} \fn{G}{r} = \frac{1}{4\pi} \fn{\delta}{r}
\end{equation}
The general solution of the homogeneous equation
\begin{equation}
- \frac{d}{dr} r^2 \frac{d}{dr} \fn{\psi}{r} = 0
\end{equation}
is
\begin{equation}
\fn{\psi}{r} = \frac{A}{r} + B
\end{equation}
The boundary condition \eq{pbc} sets $B=0$ and so
\begin{equation}
\fn{G}{r} = \frac{A}{r} \qquad \textrm{for } r > 0
\end{equation}
To fix the normalisation we can integrate \eq{rge}
\begin{eqnarray}
\frac{1}{4\pi} & = & \frac{1}{4\pi} \int_0^\infty \d{r} \fn{\delta}{r}
\\
& = & - \int_0^\infty \d{r} \frac{d}{dr} r^2 \frac{d}{dr} \fn{G}{r}
\\
& = & \left[ - r^2 \frac{d}{dr} \fn{G}{r} \right]_0^\infty
\\
& = & A - 0
\end{eqnarray}
since $\fn{G}{r}$ should be smoothable at $r=0$.
Therefore
\begin{equation}
\fn{G}{r} = \frac{1}{4\pi r}
\end{equation}
\begin{equation}
\fn{G}{\vect{r}} = \frac{1}{4\pi |\vect{r}|}
\end{equation}
and
\begin{equation}
\fn{G}{\vect{r},\vect{r}'} = \frac{1}{4\pi |\vect{r} - \vect{r}'|}
\end{equation}
Therefore the Green's function solution of \eq{poisson} is
\begin{equation} \label{poissonsol}
\fn{\phi}{\vect{r}} = \int \mathinner{d^3\vect{r}'} \fn{G}{\vect{r},\vect{r}'} \fn{\rho}{\vect{r}'}
= \int \mathinner{d^3\vect{r}'} \frac{\fn{\rho}{\vect{r}'}}{4\pi |\vect{r} - \vect{r}'|}
\end{equation}
$\fn{G}{\vect{r},\vect{r}'}$ can be interpreted as the electric potential due to a point charge at $\vect{r} = \vect{r}'$ and the Green's function solution \eq{poissonsol} can be interpreted as summing up these contributions over the charge density field $\fn{\rho}{\vect{r}}$.

\end{document}

