% * observables are all real; then use that to require self-adjt oprs.
% * prob'ty function section
% * einstein indexing note
% * intuition for q and p:  weighted pos'n and change of prob'ty
% * F.T. is sensitive to phase ... it isn't ignorable.  need another example
%   on importance of phase.
% * def square-summable.
% * RA appx:
%   - Riesz rep thm
%   - Riesz-Fischer
% * bounded operators are always defined on all of $\msH$.
% * appx: singular measures?
% * P < 1:  early on, xref fwd to fock spaces.  Note we *don't* always
%   assume vectors are normalized.

% Notation:  \oplus on elements to replace ordered pairs.  Note that this
%   avoids confusion (if not ambiguity) with inner product.

% * sth about hermite fcns ... even if only as wikipedia ref
% * harmonic oscillator:  classical and QM Hamiltonians.

% algebra, func analysis, etc. into bib

%% ================================================================
\section{Introduction}

%% ----------------------------------------------------------------
\subsection{Reason for the course}

Math 541 is a course in mathematical physics --- offered through a mathematics
department, although it is cross-listed with Physics 541. 

%The main intended audience is graduate students in mathematics with a solid
%grounding in analysis (including $L^1$ and $L^2$ spaces and Fourier transforms)
%and linear algebra.

Why offer such a course?
\begin{itemize}
\item It is an essential course for students going on in mathematical physics.
\item It is important for students of mathematics to have at least an
appreciation for where certain mathematical structures came from.
\item It is important for students of physics to have at least an appreciation
for a rigorous approach to mathematical physics.
\item For the sake of the forward progress of mathematics and physics,
students of each discipline must learn to speak one another's languages.
\end{itemize}

Certain mathematical structures and theories come from physics.  For example:
\begin{itemize}
\item Symplectic geometry originates with the Hamiltonian approach to classical
mechanics.
\item Probability theory originates in part from statistical mechanics.
(It also owes a huge debt to gambling!)  In particular, we mention central
limit theorems and the theory of large deviations, which is an active area of
current research.
\item The spectral theory of operators in Hilbert spaces, along with
von Neumann and $C^*$ algebras, originates in quantum theory.
\end{itemize}

%% ----------------------------------------------------------------
\subsection{About these notes}

Dr. Wehr is taking a best-of approach from several texts.  It seems worthwhile
to me to record his systematic treatment in writing, in a way (searchable and
indexed, as handwritten notes can never be) that will be useful for future
reference.

%Throughout this document, the first person singular (``I'') means me; the
%first person plural (``we'') means Dr. Wehr.  Although I may slip from time to
%time.

A disclaimer:  lectures and written material are not the same.  The former is,
as it should be, bracketed by class sessions; one reviews material from the
previous class, and summarizes at the end.  One omits certain details due to
time constraints (for example, ``Now check that the axioms for a metric are
satisfied'').  Writing notes is then more than a mere exercise in transcription
--- it includes removal of segu\'es and insertion of details.  This is a work
in progress --- as of the current writing, these notes retain their
conversational tone (which I would like to retain) with some details still not
fully fleshed out.

The course is being taught with the point of view that the students have seen
elementary quantum mechanics; in fact this is true for most of the students in
the class.  In the main body of the paper we move from spectral theory directly
to many-body theory.  For my own reference, however, I want to create a handy,
short, example-driven list of basic prerequisite facts:  the essentials of what
I need to know to be conversant with quantum theory, assuming first-year
graduate coursework as a given.  These constitute the appendices.  In fact, the
appendices represent my attempt to document, if not fill in, the gaps in my own
knowledge.

My longer-term goal is to create a handy reference for myself containing what I
need to know about quantum mechanics for my graduate study --- given the
pure-math core-course sequence in algebra, analysis, and geometry-topology as
prerequisite material.  These notes will be more and less than a text --- more
examples, less filler and formality.  Also, I will summarize, in this one
location, key concepts from various disciplines on which quantum mechanics
depends:  linear and multilinear algebra, probability, statistics, measure
theory, and functional analysis.

My attitude toward rigor in reference papers I write for myself (such as this
one) is the following:  If a proof is short, and if I can remember it after
reading it, and if the proof contributes something to my understanding, I will
write it out.  Else I will give a reference.  In fact, papers such as this one
are primarily \emph{not} mathematical; they contain far more intuition than
rigor.  So be it.

% In particular I have no patience for \emph{ex nihilo} cleverness; there is
% enough of that afloat in the world already and I won't generate any more.

%% ----------------------------------------------------------------
\subsection{Overview of the course}

The plan is as follows:
\begin{itemize}

\item Give some motivation for quantum theory.

\item Go through spectral theory in as much rigor as possible, while leaving
time for further topics.  (We may need to abbreviate or skip some proofs.)  It
is possible to delve so deeply into spectral theory that at the end of the
semester there is time only for study of the infinite well and the simple
harmonic oscillator.  In fact, the course has been taught this way before.
This semester, we will make time for some topics of current interest in
physics.

\item Extend/specialize the spectral theory to systems of many (usually
identical) particles.  We will use the tools of second quantization and Fock
spaces.

\item Discuss Bose-Einstein condensates and the BCS theory of
superconductivity.  Last, mention superfluids if time permits.

\end{itemize}

The main text for the course is \cite{bib:mr}, with two important deviations:
(1) we will take some time at the beginning to develop quantum-theoretical
concepts from scratch, and (2) we will omit chapter 4 of the text (on the
electron gas) entirely.

%% ----------------------------------------------------------------
\subsection{Notation}

The integers, reals, and complexes are denoted by $\Z$, $\R$, and $\C$
respectively.

For $z=x+iy \in \C$, the \plainidx{complex conjugate} $x-iy$ of $z$
is written either $\overline{z}$ or $z^*$.
The \plainidx{modulus} or \plainidx{absolute value} of $z$, which
is $\sqrt{x^2+y^2} = \sqrt{z\overline{z}}$, is written $|z|$.
(Notice in particular that $z\overline{z}=|z|^2$.)
The real part $x$ and the imaginary part $y$ of $z$ are written
$\Im(z)$ and $\Re(z)$, respectively.

For a map $A$ from an object $V$ to an object $W$, the kernel of $A$ is written
$\ker(A)$; the image of $A$ is written either $\im(A)$ or $A(V)$.
Injective\index{injective mappings} and \plainidx{surjective mappings} are
written $A \into B$ and $A \onto B$, respectively.  Actions of mappings on
elements are written $a \mapsto b$.

The \plainidx{indicator function} of a subset $A$ of a set $X$ is written
$1_A$.  It has values
$$
	1_A(x) = \begin{cases}
		1, & x \in A \\
		0, & x \not\in A.
	\end{cases}
$$

We write
$$
	A := B
$$
when we are defining what $A$ is.  We say ``$A$ is defined to be $B$.''
For example,
$$
	\C^+ := \{z\in\C : \Im(z) > 0\}.
$$

Proofs are ended by $\Box$; examples are bracketed between $\rhd$ and $\lhd$.
Definitions, on the other hand, terminate with the end of a paragraph.

%% ================================================================
\newpage
\section{Main ideas of quantum theory}

In this section, we sketch some ideas in order to fix intuition, before diving
headlong into technical details.  You may think of it as a cartoon version.

%% ----------------------------------------------------------------
\subsection{Classical vs. quantum mechanics}

Consider a one-particle system.  This may conjure up a frightening view of an
all-but-empty universe, but consider that the particle may be acted on by
forces originating with fields or other particles.  For now, we are simply
focusing our spotlight on the behavior of a single particle.

The classical-mechanical view of the state of a single-particle system at time
$t$ is
$$
	(\vecq(t), \vecp(t)) \in \R^3 \times \R^3,
$$
where $\vecq$ is the \emphidx{position} of the particle, and $\vecp = m
\dot\vecq = m\vecv$ is the \emphidx{momentum} of the particle, where $m$ is its
mass.  The \emphidx{time evolution} of the particle is given by the
\emphidx{initial conditions} $(\vecq(0), \vecp(0))$ and a \emphidx{law of
evolution} describing $\dot\vecq$ and $\dot\vecp$.  This is a system of
\emphidx{ordinary differential equations}, and they maybe (wildly) non-linear.
Symplectic geometry addresses this subject further.

By contrast, the quantum-mechanical view of the state of a single-particle
system is a function $\psi(\vecx)$, for $\vecx \in \R^3$, from which position
and momentum are derived: $$ \vecx \mapsto (\vecq,\vecp).  $$ We again want to
know the time evolution of the system.  (We will intentionally mix between
writing $\psi(\vecx)$ and $\psi(\vecx,t)$.  The reason for this will be given
in section [write and xref].) The \emphidx{initial conditions} are
$\psi(\vecx,0)$; the \emphidx{law of evolution} is the \emphidx{Schr\"odinger
equation}, which is a PDE (since $\psi$ is a function of both $\vecx$ and $t$)
but is (happily) a linear PDE\footnote{This is not to say that nonlinear
equations don't appear in quantum theory; they do.  The basic underlying PDEs
\emph{are} linear, but may (especially with multi-particle systems) become
quite complicated.  There exist approximation techniques which (perhaps
drastically) reduce the number of variables, but which do so at the price of
introducing nonlinearity in the approximation equations.}.

Why do we care about only $\vecq$ and $\vecp$?  It turns out that many
physical quantities are functions of position and momentum.  For example,
the classical kinetic energy is
$$
	T = \frac{m\|\vecv\|^2}{2} = \frac{\|\vecp\|^2}{2m}.
$$
More generally, we can look at
$$
	F(\vecq(t), \vecp(t))
$$
evolving in time.  These functions of the \plainidx{state variables} $\vecq$
and $\vecp$ are called \emphidx{observables}.  They are not, in fact, arbitrary
functions of state; rather, we will view them as \emphidx{linear operators}
acting on the state.  In particular, we will view $\vecq$ and $\vecp$
themselves as linear operators acting on $\psi$.  (There are all sorts of
linear operators; the ones in this course will all be multiplication or
differentiation operators, or some combination thereof.)

%The \emphidx{position operator} $\vecq$ is the vector-times-scalar product
%$$
%	\phi(\vecx) \mapsto \vecx \phi(\vecx);
%$$
%we will define the position operator soon.

The possible functions $\psi$ will form a \emphidx{Hilbert space} $\msH$;
observables will be \emphidx{unbounded linear operators} in $\msH$.  We say
``in'' rather than ``on'' because these operators may not be defined on all of
$\msH$.  This is not merely an annoyance --- rather, it is key to the power of
the theory.

%% ----------------------------------------------------------------
\subsection{Hilbert space and linear operators}
\label{sec:cartoon_HS_and_lin_ops}

The particle state is encapsulated in a non-zero complex-valued function $\psi
\in L^2(\R^3, \C)$.

\begin{defn}
The $L^2$ condition, or \emph{square-integrability
condition}\index{square integrable}, means\footnote{The use of a superscript
asterisk for complex conjugation is a physics convention which we will follow.}
that
$$
	\int_{\R^3} |\psi(\vecx)|^2\,d\vecx =
	\int_{\R^3} \psi^*(\vecx) \psi(\vecx)\,d\vecx < +\infty.
$$
(When time dependence is included, i.e. $\psi:\R ^4 \to \C$, we require
square-integrability only in the spatial coordinates.)

We will treat any two such functions differing only a complex constant, i.e.
$\psi_1(\vecx) = c\psi_2(\vecx)$ for all $\vecx \in \R^3$, as being the same.
We \emphidx{normalize} so that we assume
$$
	\|\psi\|_{L^2}^2 = 1
$$
where
$$
	\|\psi\|_{L^2}^2 = \int_{\R^3} |\psi(\vecx)|^2\,d\vecx.
$$
\end{defn}

%\begin{interp}

The physical interpretation of $\psi$ is that $|\psi|^2$ --- which is
real-valued --- plays the role of a probability-density function as discussed
in appendix \ref{sec:prb}.  [xxx make sure to mention, here and/or in the
appendix, griffith's caveat:  probability in terms of repeated identical
experiments.  Also foreshadow (with xref) the collapse of the wave function
and make it clear we can't immediately repeat the experiment on the \emph{same} system.]

%\end{interp}

\begin{defn}
We define\footnote{Conjugating the first function in the integral rather than
the second is another physics convention which we will follow.} a
\emphidx{scalar product} or \emphidx{inner product} on $\msH=L^2(\R^3,\C)$

$$
	(\phi,\psi) := \int_{\R^3} \phi^*(\vecx)\psi(\vecx)d\vecx.
$$
(For the physical interpretation, see [write and xref].) With this inner
product,
$$
	\msH=(L^2(\R^3,\C),(\cdot,\cdot))
$$
is a \emphidx{Hilbert space}.
\end{defn}

\begin{notn}
The standard mathematical notation for the inner product is $(\phi,\psi)$ as
written above.  The physics notation (called \emphidx{Dirac notation} or
\emphidx{bra-ket notation}) for the inner product $(\phi,\psi)$ is
$$
	\braket{\phi}{\psi}.
$$
We will later [write and xref] define what the symbols $\bra\phi$ and
$\ket\psi$ mean individually.
\end{notn}

So far, this is a poor man's quantum theory:  we only have a single particle,
and we have no \plainidx{spin} --- without which (as we will see below [write
and xref]) there is no superconductivity.  Later, we will modify and extend the
theory to include multiple particles as well as spin.

An \emphidx{observable quantity} will be represented by a \emphidx{self-adjoint
operator} in $\msH$.  We say ``in'' rather than ``on'' since there will
be \plainidx{domain issues}: these operators may not be defined on all of
the functions in $\msH$.  (We will see why in a few paragraphs.)

\begin{defn}
Observable position\index{position operator} $\vecq$ has an operator
representation.  It is defined by the vector-times-scalar product
$$
	\psi \mapsto \vecx \psi(\vecx).
$$
I.e. if $\vecx = (x^1, x^2, x^3)$ then we have
$$
	\psi \stackrel{q^\alpha}{\longmapsto} x^\alpha \psi
$$
being an operator $q^\alpha:\msH \to \msH$.
\end{defn}

The domain issue is that even if $\psi$ is square integrable, $\vecx\psi$ might
not be:  multiplication by $\vecx$ is an \emphidx{unbounded linear operator}.
We will leave this issue at present; for now, just be aware of difficulties to
come.

% [xxx heuristic:  x times psi is just probty-weighted position!]

\begin{defn}
Similarly, the observable momentum\index{momentum operator} $\vecp$ has an
operator representation.  This is $p^\alpha:\msH \to\msH$ given by
$$
	\psi \stackrel{p^\alpha}{\longmapsto} -i\hbar \frac{\D\psi}{\Dx^\alpha}.
$$
\end{defn}

The value $\hbar$ is called \emphidx{Planck's constant}; its precise value and
units are unimportant at present.  The key thing to know for now is that it is
small.  This is why quantum effects are typically visible only microscopically.
Again, there are domain issues:  functions $\psi$ which are square-integrable
certainly do not need to be differentiable.

\begin{defn}
The \emphidx{commutation relation} is
$$
	[\vecq,\vecp]=\vecq\vecp-\vecp\vecq
$$
where the multiplication is composition of operators, which we recall is
non-commutative in general.
\end{defn}

This is also called the \emphidx{canonical commutation relation} or
\emphidx{Heisenberg's commutation relation}.  This is a misnomer; it describes
\emph{lack} of commutativity.

\begin{prop}
We have
$$
	[\vecq,\vecp]=i\hbar
$$
i.e.
$$
	[q^\alpha,p^\beta] := q^\alpha p^\beta - p^\beta q^\alpha
	= i\hbar \delta_{\alpha\beta}.
$$
\begin{proof}
Using $\psi$ as a test function, we have (using the product rule)
\begin{align*}
	[q^\alpha,p^\beta]\,\psi(\vecx)
		&= q^\alpha p^\beta\psi - p^\beta q^\alpha\psi \\
	&= -i\hbar x^\alpha\frac{\D\psi}{\D x^\beta} +
		i\hbar \frac{\D}{\Dx^\beta}\left(x^\alpha \psi \right) \\
	&= -i\hbar x^\alpha\frac{\D\psi}{\D x^\beta}
		+i\hbar x^\alpha\frac{\D\psi}{\D x^\beta}
		+i\hbar \frac{\D x^\alpha}{\D x^\beta} \psi \\
	&= i\hbar \delta_{\alpha\beta} \psi.
\end{align*}
Removing the test function and collecting the coordinate functions into column
vectors, we have
$$
	[\vecq,\vecp] = i\hbar.
$$
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Fourier transforms}

\begin{defn}
The \emphidx{Fourier transform} of $\psi(\vecx)$, written
$$
	\psitilde(\veck) = \mcF(\psi(\vecx)),
$$
is
$$
	\psitilde(\veck) := (2\pi)^{-3/2} \int_{\R^3}
		e^{-i\veck\cdot\vecx} \psi(\vecx)\,d\vecx.
$$
\end{defn}

\begin{rem}
The $3$ in $(2\pi)^{-3/2}$ comes from the dimension of $\R^3$; in $d$
dimensions it would be $(2\pi)^{-d/2}$.
\end{rem}

\begin{rem}
There are numerous definitions in the literature involving the scale
factor out front and the $-i$ in the integrand.  We have carefully chosen
the scale factor to make the Fourier transform an \emphidx{isometry},
so that
$$
	\int_{\R^3} |\psitilde(\veck)|^2\,d\veck
$$
if $\psi$ is normalized to 1.  More generally,
$$
	\int_{\R^3} |\psitilde(\veck)|^2\,d\veck =
	\int_{\R^3} |\psi(\vecx)|^2\,d\vecx.
$$
\end{rem}

\begin{rem}
For the Fourier transform to be easily defined, we actually require $\psi \in
L^1(\R^3,\C) \cap L^2(\R^3,\C)$.  We gloss this detail for now.
\end{rem}

\begin{defn}
We will soon need the inverse transform, so we'll define it now:
$$
	\psi(\vecx) =
	\mcF^{-1}(\psitilde(\veck)) = 
	(2\pi)^{-3/2} \int_{\R^3} e^{+i\veck\cdot\vecx} \psitilde(\veck)\,d\veck.
$$
\end{defn}

%\begin{interp}

The interpretation is that $|\psitilde|^2$ is the density of distribution of
momenta.  [xxx similar to caveats for distribution of $\vecq$: write and
xref.]  That is, the probability that $\vecp$ is in a (Borel) subset $B$
of $\R^3$ is
$$
	\int_B |\psitilde(\veck)|^2\,d\veck.
$$

%\end{interp}

A crucial fact is how $\vecp$ acts in $\veck$-space.  In $\vecx$-space it was a
derivative operator --- what about here?  To find out, let's carefully define
it.
\begin{defn}
The momentum operator in $\veck$-space is really shorthand for the
compostion of three operators:
$$
	\vecp\psitilde(\veck) := \mcF \circ \vecp \circ \mcF^{-1} \psitilde.
$$
\end{defn}

% xxx write out the steps

Writing out the steps (which are easy, but require justification) we obtain
$$
	\vecp\psitilde(\veck) = \hbar\veck\psitilde(\veck).
$$
That is, $\vecp$ is a multiplication operator in $\veck$-space, just as $\vecq$
is a multiplication operator in $\vecx$-space.  We call $\psitilde(\veck)$ the
\emphidx{wave function} of the particle in the \plainidx{momentum
representation}; $\psi(\vecx)$ is called the wave function of the particle in
the \plainidx{position representation}.  (When we discuss Bose-Einstein
condensation in section [write and xref], we will see that the condensation
appears explicitly only in the momentum representation.)

We emphasize that the Fourier transform turned a differentiation operator in
the position representation into a multiplication operator in the momentum
representation.  Also, we will see in section \ref{sec:cartoon_SE}, the
potential-energy operator will be a multiplication operator and the
kinetic-energy operator will be a differentiation operator.  It is (generally
speaking) easier to solve problems involving position or potential energy in
the position representation, and to solve problems involving momentum or
kinetic energy in the momentum representation:  in each case, the operators of
interest become multiplication operators.  One of the challenges of quantum
mechanics is that there is no obvious representation which simplifies position
\emph{and} momentum, or potential \emph{and} kinetic energy, simultaneously.

%% ----------------------------------------------------------------
\subsection{The measurement theorem}

Suppose that $A$ is a linear operator corresponding to an observable quantity.
Assume self-adjointness of $A$.  The linear-algebra point of view of
self-adjointness (glossing details which occur in infinite-dimensional vector
spaces, and ignoring domain issues whereby $A$ is not defined on all of $\msH$)
is that for all $\phi,\psi\in\msH$,
$$
	(A\phi,\psi) = (\phi,A\psi).
$$
It is known that this implies that the \plainidx{spectrum}\index{real spectrum}
of $A$ is real.  In the finite-dimensional case, the spectrum consists only of
the (real) eigenvalues $a_n$, and there exists a spanning set of orthonormal
eigenvectors $\phi_n$ such that
$$
	A\phi_n=a_n\phi_n.
$$

\begin{interp}[\plainidx{First postulate of quantum mechanics}]
The interpretation is as follows.  Suppose that $\psi$ is the wave function of
our one-particle system.  Then measurement of the quantity to which $A$
corresponds has as its possible outcomes the numbers $a_n$ with probabilities
$|(\phi_n,\psi)|^2$.  If this is to work, we would want the probabilities
to be non-negative (which we have by positive-definiteness of the inner
product), and we would want them to sum to 1.  In fact, they do.
\end{interp}

\begin{prop}
We have
$$
	\sum_n |(\phi_n,\psi)|^2 = 1.
$$
\begin{proof}
First, we know we've normalized $\psi$ so that
$$
	\|\psi\|^2 = (\psi,\psi) = 1.
$$
Since the eigenfunctions $\phi_n$ are a spanning set, we have
$$
	\psi=\sum_n c_n \phi_n
$$
with
$$
	c_n = (\phi_n,\psi).
$$
Then
\begin{align*}
	1 &= (\psi,\psi) = \left( \sum_m c_m \,\phi_m, \sum_n c_n \,\phi_n \right)
		= \sum_m \sum_n \left(c_m \phi_m, c_n \phi_n \right) \\
	&= \sum_m \sum_n c_m^* c_n \delta_{mn} \\
	&= \sum_n |c_n|^2
\end{align*}
by the orthonormality of the $\phi_n$'s.  But the $c_n$'s were my shorthand for
$(\phi_n,\psi)$ so we have
$$
	\sum_n |(\phi_n,\psi)|^2 = 1
$$
which is what we wanted to show.
\end{proof}
\end{prop}

Next we can calculate the \emphidx{mean} of $A$.  [write and xref to
appendix.]  This is the sum over each possible measurement times the
probability of that measurement.  We have
\begin{align*}
	(\psi, A\psi)
	&= \left( \sum_m (\phi_m,\psi)\,\phi_m, \sum_n (\phi_n,\psi)\, A \,\phi_n
		\right) \\
	&= \left( \sum_m (\phi_m,\psi)\,\phi_m, \sum_n (\phi_n,\psi)\, a_n \,\phi_n
		\right) \\
	&= \sum_n a_n |(\phi_n,\psi)|^2
\end{align*}
which is the mean of the value measured.

\begin{interp}[\plainidx{Second postulate of quantum mechanics}]
Immediately after measurement, the state of the system is described by $\phi_n$
\emph{only}.  This is called the \emphidx{collapse of the wave function}.
The evolution of the single-particle system thereafter is described by
the Schr\"odinger equation, with new initial conditions.
\end{interp}

%% ----------------------------------------------------------------
\subsection{The Schr\"odinger equation}
\label{sec:cartoon_SE}

The\index{Schr\"odinger equation} wave function $\psi(\vecx,t)$ satisfies (as
verified by experiment)
$$
	i\hbar\frac{\D\psi}{\Dt} = H\psi
$$
with $H$ being a self-adjoint operator in $\msH$.  This $H$ is called the
\emphidx{energy operator} or \emphidx{Hamiltonian operator} for the system.  It
corresponds to classical energy just as the $\vecq$ and $\vecp$ operators
correspond to the classical notions of position and momentum.  A common set-up
for a single particle is
$$
	H = T+V
$$
where $T$ is the \emphidx{kinetic energy} and $V$ is the \emphidx{potential
energy}.  Classically, potential energy comes from such things as gravity,
spring tension, electrical fields, etc.  This is often position-dependent,
although it could also be time-dependent (e.g. an electrical field which
varies sinusoidally).  The classical kinetic energy is
$$
	H = \frac{m\|\vecv\|^2}{2} = \frac{\|\vecp\|^2}{2m}
$$
where $m$ is the mass of the particle.

We want to mimic this as much as possible.  We have
$$
	H = T+V = \frac{\vecp^2}{2m} + V(\vecq).
$$
Classically and quantum-mechanically, we write
$$
	\vecp^2 = \sum_{\alpha=1}^3 p^\alpha \cdot p^\alpha.
$$
Classically, the $p^\alpha$'s are numbers; quantum-mechanically they are
differentiation operators.  It is key to note that $\vecp^2$ is still a
linear operator, even though it looks quadratic.  Writing out the second
derivatives, we have (making note of various notations in common use)
$$
	T = -\frac{\hbar^2}{2m} \frac{\D}{\Dx_j^2}
	= -\frac{\hbar^2}{2m} \Delta
	= -\frac{\hbar^2}{2m} \nabla^2,
$$
i.e. $-\hbar^2/2m$ times the \plainidx{Laplacian operator}.
We also set $V$ to be the operator which multiplies by $V(\vecx)$.

With these definitions, the Schr\"odinger equation becomes
\begin{align*}
	i\hbar\frac{\D\psi}{\Dt}
	&= H\psi \\
	&= T\psi+V\psi \\
	&= -\frac{\hbar^2}{2m} \nabla^2\psi + V(\vecx)\psi.
\end{align*}

The next task is to attempt to write down some solutions.
We will soon [write and xref] in some detail the \bfidx{harmonic oscillator},
where
$$
	V(\vecq) = \frac{1}{2}\gamma \|\vecq\|^2.
$$
Here, in this very special case, the Schr\"odinger equation can be solved in
explicit detail.

%% ----------------------------------------------------------------
\subsection{The ODE perspective and the evolution operator}

We can think of the Schr\"odinger equation as follows.
Consider the mapping
$$
	t \mapsto \psi_t \in \msH.
$$
Then the Schr\"odinger equation is an $\msH$-valued ODE
$$
	i\hbar\frac{d\psi_t}{dt} = H\psi_t.
$$
(This is merely formal manipulation of symbols; think of it as a device for
guessing solutions.)  If $\msH$ were a finite-dimensional vector space, with
$H$ represented by a matrix, then we would write
$$
	\psi_t = e^{-\frac{i}{\hbar} Ht} \psi_0 = U(t)\psi_0
$$
where we define the \bfidx{evolution operator} to be
$$
	U(t) = e^{-\frac{i}{\hbar} Ht}.
$$
In an ODE course, we would define the \bfidx{operator exponential} to be
$$
	e^{-\frac{i}{\hbar} Ht} = \sum_{n=0}^\infty \frac{\left(
		-\frac{i}{\hbar} Ht \right)^n}{n!}.
$$
However, for an unbounded operator $H$, the series does not converge.

We may use a different procedure to define the operator exponential.
First, diagonalize:  find a family $\psi_n$ of eigenfunctions, which we
hope spans $\msH$, with corresponding eigenvalues $\veps_n$ such that
$$
	H\psi_n = \veps_n \psi_n.
$$
Define an action on eigenfunctions
$$
	e^{-\frac{i}{\hbar}Ht}\psi_n
	e^{-\frac{i}{\hbar}\veps_n t}\psi_n
$$
which we hope extends by linearity.  One reason for concern is the following.
If
$$
	\psi=\sum_n c_n \psi_n
$$
with
$$
	\sum_n |c_n|^2=1
$$
and we try to write
$$
	U(t)\psi=\sum_n d_n \psi_n
$$
then there is no guarantee that
$$
	\sum_n |d_n|^2=1.
$$
A concrete example [write and xref: SHO] is
$$
	H\psi_n = \hbar(n+1/2)\psi_n.
$$
The coefficients here are not square-summable.

We do want to be able to define $U(t)$ on all of $\msH$, but this is not
automatic in the infinite-dimensional setting.  This is called the
\bfidx{extension problem}.  We will see in [write and xref] that there is in
fact a one-to-one correspondence between groups of evolution operators and
[missed it --- xxx type up].

What properties do we expect of an evolution operator?  Here is a list:
\begin{itemize}

\item[(1)] Observables are real, so operators are self-adjoint.  (There exists
complex observables, but their real and imaginary parts correspond to
self-adjoint operators.)  We sometimes write $H^*=H$.  Using mathematicians'
notation, this means that for all $\phi,\psi\in\msH$,
$$
	(H\phi,\psi) = (\phi,H\psi).
$$
The Dirac notation reads simply $\braCket{\phi}{H}{\psi}$ in either --- this
notation was designed for self-adjoint operators and so it does not permit us
to communicate about non-self-adjoint operators.

(A remark:  the class of self-adjoint operators is not closed under linear
combinations.  In particular, if $A$ and $B$ are self-adjoint, $A+B$ may not
be.  We will see below [write and xref] that $a$ and $a^\dagger$ are
non-self-adjoint linear combinations of self-adjoint operators.)

\item[(2)] From this it follows that
$$
	U(t)^* = U(t)^{-1}.
$$
Here is an informal proof, using formal manipulation of symbols:
$$
	U^* = (e^{-\frac{i}{\hbar} Ht})^*
	= e^{+\frac{i}{\hbar} Ht}
	= e^{-\frac{i}{\hbar} H\cdot(-t)}
	= (e^{-\frac{i}{\hbar} H\cdot(t)})^{-1}
$$
where we think of backward evolution undoing forward evolution.

\item[(3)] Preservation of normalization of $\psi$ under $U$:  i.e. if $\int
|\psi(\vecx)|^2 \,d\vecx=1$, then $\int |U\psi(\vecx)|^2 \,d\vecx=1$.  If
$|\psi|^2$ is a probability density function, time evolution shouldn't change
that.  This is easy to show using unitarity of $U$, since \bfidx{unitary
operators} preserve norms in Hilbert spaces.  (A consequence is that
eigenvalues of $U$ will lie on the unit circle in the complex plane.)

\item[(4)] $U(t_1+t_2)=U(t_1)\circ U(t_2)$:  A wave function should be able to
evolve forward by $t_1$ time units, starting from initial conditions
$\psi(\vecx,0)$, then evolve further with new initial conditions
$\psi(\vecx,t_1)$.  That is, the $U(t)$'s form a \bfidx{unitary group of
operators}\footnote{One would rather say \emph{group of unitary operators}, but
the name has stuck.}

One in fact may make the following grand summary: \emph{Quantum mechanics says
that the energy operator generates the group of time-evolution operators.}

This is based mathematically on the theory of passing back and forth between
$H$ and $U(t)$, which we will see [write and xref] in \bfidx{Stone's theorem}
and the \bfidx{spectral theorem}.

\end{itemize}

%% ----------------------------------------------------------------
\subsection{A few words about many-body theory}

For a while now we will be either restricting our attention to single-particle
systems, or talking about abstract Hilbert spaces.  We will not handle
multi-particle systems until [write and xref].  However, it is important to at
least sketch at this point what some of these other Hilbert spaces are --- lest
we make the mistake early on of thinking of $\msH$ as $L^2(\R^3,\C)$ only.

A \plainidx{wave function} of a system of $N$ particles is, in the
\plainidx{position representation},
$$
	\psi(\vecx_1, \ldots, \vecx_N),\quad \vecx_j \in \R^3
$$
with the \plainidx{normalization}
$$
	\int_{\R^{3N}} |\psi(\vecx_1,\ldots,\vecx_N)|^2\,
		d\vecx_1\cdots d\vecx_N = 1.
$$
The \plainidx{momentum representation}\index{Fourier transform} for the wave
function is
$$
	\psitilde(\veck_1, \ldots, \veck_N)
	= (2\pi)^{-N/2} \int_{\R^{3N}}
	\exp\left({-i\sum_{j=1}^N \veck_j\cdot \vecx_j}\right)
	\psi(\vecx_1,\ldots,\vecx_N)\,d\vecx_1\cdots d\vecx_N.
$$

The physical interpretation is as before:
$|\psi(\vecx_1, \ldots, \vecx_N)|^2$ is the probability density function
for the positions of the particles, and
$|\psi(\veck_1, \ldots, \vec,_N)|^2$ is the PDF for their momenta.
In particular, for $N=2$ and $A \subseteq \R^3$, we have
$$
	\int_{A\times A} |\psi(\vecx_1, \vecx_2)|^2\,d\vecx_1 \,d\vecx_2
	= P(\vecx_1 \in A, \vecx_2 \in A).
$$
The key question is, can we \emphidx{label} particles 1 and 2?  Sometimes we
cannot distinguish them, and these symmetry requirements drastically reduce the
number of possible wave functions.  For example, if the two particles
are \emphidx{bosons} (e.g. certain atomic nuclei) then
$$
	\psi(\vecx_1, \vecx_2) = \psi(\vecx_2, \vecx_1)
$$
and more generally for any permutation $\pi\in\mcS_N$,
$$
	\psi(\vecx_1, \ldots, \vecx_N)
	= \psi(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)}).
$$
Likewise, for \emphidx{fermions},
$$
	\psi(\vecx_1, \vecx_2) = -\, \psi(\vecx_2, \vecx_1)
$$
and more generally
$$
	\psi(\vecx_1, \ldots, \vecx_N)
	= \sgn(\pi)\psi(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)}).
$$

Keep in mind that a Hilbert space may be $L^2(\R^{3N},\C)$ --- or perhaps
a (highly) proper subspace, or a quotient space.

%% ================================================================
\newpage
\section{Hilbert-space theory}

Having completed our sketch of quantum theory, we now take a more careful
approach to the underlying mathematics.


First, a caveat: In the introductory version above, we've tacitly assumed that
operators in $\msH$ are not unlike matrices acting on a finite-dimensional
vector space.  However, this notion cannot be overused.  First, the
finite-dimensional viewpoint excludes the case of continuous spectra.  Second,
consider the following commutation relation from section
\ref{sec:cartoon_HS_and_lin_ops}:
$$
	[q^\alpha,p^\beta] = i\hbar \delta_{\alpha\beta}I.
$$
Can one have matrices $P$ and $Q$ on a finite-dimensional vector space
such that
$$
	[P,Q]=i\hbar I?
$$
No, and here is one reason:  take traces.  Trace is real-linear and moves
through the difference, so
$$
	\tr(QP-PQ)=\tr(PQ)-\tr(QP).
$$
Also, trace is central and so, even when $PQ\ne QP$, $\tr(PQ)=\tr(QP)$.  On the
right-hand side, we have $\tr(i\hbar I)=i\hbar n$ where $n$ is the dimension of
the vector space.  These facts yield the absurdity $0=i\hbar n$.  In the
infinite-dimensional setting, some operators have a trace defined on them (we
say they are \emph{of trace class}\index{trace class}); some do not.

A note about notation: As mentioned above, there are mathematicians' and
physicists' notations for things.  In this section we will use the former,
eschewing Dirac notation, in case the student consults mathematical reference
texts on this material.  However we will keep the convention throughout this
course that we conjugate the first argument, not the second, in complex inner
products.

%% ----------------------------------------------------------------
\subsection{Axioms}

\begin{defn}
A \bfidx{Hilbert space} $\msH$ is an infinite-dimensional complex inner-product
space where the inner product $(\cdot,\cdot)$ satisfies:
\begin{itemize}
\item Conjugate symmetry\index{conjugate symmetry}:  for all $\phi,\psi\in\msH$,
	$(\phi,\psi)=\overline{(\psi,\phi)}$.  (Note that as a consequence,
	$(\psi,\psi)$ is real.)
\item Right linearity\index{right linearity}:  for all $\phi,\psi\in\msH$
	and all $a,b \in \C$, $(\phi, a\psi_1+b\psi_2) = a(\phi, \psi_1) +b(\phi,
	\psi_2)$.
\item Left antilinearity\index{left antilinearity}\index{antilinearity}:
	$(a\phi_1+b\phi_2,\psi_2) = \abar(\phi_1, \psi) + \bbar(\phi_2, \psi)$.
\item Positive definiteness:  $(\psi,\psi) \ge 0$ for all $\psi \in \msH$,
	with $(\psi,\psi)=0$ iff $\psi=0$.
\end{itemize}
\end{defn}

\begin{prop}
The inner product \plainidx{induces} a \bfidx{norm} on $\msH$:
$$
	\|\psi\| := (\psi,\psi)^{1/2}.
$$
\begin{proof}
Exercise.  Make use of the \plainidx{Cauchy-Schwartz} inequality.
\end{proof}
\end{prop}

\begin{prop}
The norm induces a \bfidx{metric} on $\msH$:
$$
	d(\phi,\psi) = \|\phi-\psi\|.
$$
\begin{proof}
Exercise.
\end{proof}
\end{prop}

\begin{prop}
$\msH$ is \bfidx{complete} in this metric.
\begin{proof}[Proof outline]
One needs to show that a \plainidx{Cauchy sequence} in $\msH$ converges to
something in $\msH$.  That is, if $\{\phi_n\}$ is a sequence in $\msH$ such
that $\|\phi_m-\phi_n\|\to 0$ as $m,n\to\infty$, then there exists
$\phi\in\msH$ such that $\|\phi_n-\phi\|\to 0$ as $n \to\infty$.  (This type of
convergence is called \emphidx{convergence in norm}.)
\end{proof}
\end{prop}

[xxx projection theorem and the need for spectral analysis?  This is why we
have Lebesgue measure --- not just for esoteric constructions such as
integrating the indicator function of the rationals on the unit interval.]

[xxx Riesz-Fischer theorem is non-trivial?  What is the context of this
comment?  The thm is that $\psi$ is square-integrable if its Fourier
series converges in $\ell^2$ \ldots .]

%% ----------------------------------------------------------------
%\subsection{Spectral theory of unbounded operators in Hilbert spaces}
\subsection{Operators in Hilbert spaces}

There are many definitions here.  Where we are headed is the spectral theorem
which says that \plainidx{self-adjoint} operators have a so-called
\emph{spectral resolution property} which makes them manageable.

\begin{defn}
Let $\msH_1, \msH_2$ be two Hilbert spaces.  A \bfidx{linear operator} from
$\msH_1$ to $\msH_2$ consists of a \bfidx{domain of definition} $\DA$, which
is a linear subspace of $\msH_1$, and a \bfidx{linear map} $A:\DA\to\msH_2$.
\end{defn}

Often, but not necessarily, $\DA$ is dense in $\msH_1$.  If it is, we say that
$A$ is \bfidx{densely defined}.  The domains of definition are usually
described by elements of an $L^2$ space having ``enough'' decay at infinity, or
``enough'' regularity (degree of differentiability).

\beginex
The position operator $\vecq$ is, in each coordinate, multiplication by the
unbounded function $x^\alpha$.  Let $\msH_1 = \msH_2 = L^2(\R,\C)$.  Recalling
examples \ref{ex:Cauchy_PDF} and \ref{ex:standard_normal_PDF}, and recalling
that $\psi\psibar$ is a probability density function, let
$$
	\psi(x) = \sqrt{ \frac{1}{\pi(x^2+1)} }.
$$
Then
$$
	\int_\R \psi(x)^*\psi(x)\,dx = \frac{1}{\pi} \int_\R \frac{1}{x^2+1}\,dx = 1.
$$
However,
$$
	\int_\R \psi(x)^*\,x\, \psi(x)\,dx
	= \frac{1}{\pi} \int_\R \frac{x}{x^2+1}\,dx
$$
diverges because the integrals on each side of the origin are infinite (we do
not have absolute integrability).  On the other hand, let
$$
	\phi(x) = \sqrt{ \frac{1}{\sqrt{2\pi}} e^{-x^2/2} }
$$
Then
$$
	\int_\R \phi(x)^*\,x\, \phi(x)\,dx = \frac{1}{\sqrt{2\pi}}
		\int_\R x\, e^{-x^2/2}\,dx = 0.
$$
Thus, the position operator leaves only one of these two wavefunctions in
$\msH$.  Note in particular that as $|x|$ increases, $\phi$ goes to zero more
quickly than $\psi$ does.
\mendex

\begin{defn}
The \bfidx{operator norm} (or sometimes \bfidx{uniform norm}) of a linear operator $A$ is
$$
	\|A\|_\op := \sup_{f\in\DA, \|f\|_{\msH_1}=1} \|Af\|_{\msH_2}
$$
or equivalently
$$
	\|A\|_\op
	= \sup_{0 \ne f\in\DA} \frac{\|Af\|_{\msH_2}}{\|f\|_{\msH_1}}.
$$
\end{defn}

\begin{defn}
A linear operator $A$ is \bfidx{bounded} (this is bad terminology but
universally used) if
$$
	\|A\|_\op < \infty.
$$
\end{defn}

\begin{exer}
Show that $A$ is bounded iff it is continuous as a mapping $A:\msH_1\to\msH_2$.
\end{exer}

One can always extend $A$ from $\DA$ to $\DAbar$ by continuity.  This means
that if $\DA$ is dense in $\msH_1$ then a bounded operator can be uniquely
extended to all of $\msH_1$.

We have the usual concepts of \bfidx{kernel} and \bfidx{image}.  If
$\ker(A)=\{0\}$ then $A^{-1}$ is defined as a linear map from $\im(A)$ to
$\DA$.  The sum $A_1+A_2$ is defined on the domain $\mcD_{A_1}\cap\mcD_{A_2}$.
Likewise, if $A:\msH_1\to\msH_2$ and $B:\msH_2\to\msH_3$ then $B \circ A$ has
domain of definition $\{f\in\DA: Af\in\mcD_B\}$.

The \bfidx{graph} of $A$, written $\GA$, is the subspace of
$\msH_1\oplus\msH_2$ consisting of pairs $\{(f,Af):f\in\DA\}$.  (Beware
that the ordered-pair notation looks like an inner product!) Then
$\msH_1\oplus\msH_2$ is a Hilbert space with inner product
$$
	((f_1,f_2),(g_1,g_2))_{\msH_1\oplus\msH_2} :=
	(f_1,f_2)_{\msH_1} +
	(g_1,g_2)_{\msH_2}.
$$

\begin{exer}
Prove this claim.
\end{exer}

This construction will be very useful below.
The above definitions were probably familiar; here is the first new one.

\begin{defn}
A linear operator $A$ is called \bfidx{closed} if its graph $\GA$ is closed
in $\msH_1\oplus\msH_2$.
\end{defn}

This is an elegant definition, but akwward to apply; a more useful
characterization is coming soon.

\begin{defn}
A linear operator $A$ is \bfidx{closable} if $\GAbar$ is a graph of a linear
operator $\Abar$, called the \plainidx{closure} of $A$ (which typically has a
larger domain of definition).
\end{defn}

The more useful characterization of closedness is the following.  It is
presented as an exercise, but a very important one.

\begin{exer}
A linear operator $A$ is closed if whenever there is a sequence $f_n\in\DA$
converging (in norm) to $f\in\msH$ and $Af_n \to g \in \msH_2$, we have
$f\in\DA$ and $Af=g$.  (Think of this as a substitute for continuity.)
\end{exer}

A weaker special case is the following:  $A$ is closable iff $f_n\to 0 \in
\DA$ and $Af_n \to g\in\msH_2$ implies $g=0$.

\begin{rem}
The graph $\GA$ cannot contain pairs $(0,g)$ with $g\ne 0$, since it already
contains the ordered pair $(0,0)$.  This is an obvious fact at face value, but
is often useful in proofs.
\end{rem}

We want a correct definition of \plainidx{self-adjoint operator}
$A:\msH\to\msH$ for use in quantum mechanics.  We want $H$ to be self-adjoint
for
$$
	H = \frac{-\hbar^2}{2m}\nabla^2+V(\vecq),
$$
over a broad class of potentials $V(\vecq)$.  However, if the definition is too
broad, it will include operators which are unmanageable.

A starting point is to require that
$$
	(Af,g)=(f,Ag)
$$
for all $f,g\in\msH$.  If $A$ satisfies this, it is said to be a
\bfidx{symmetric operator}.  This is meaningful but (as von Neumann realized)
this is not enough for a good spectral theorem.  (Note that in
finite-dimensional vector spaces, symmetry is equivalent to self-adjointness.)
In fact, there exist operators which are symmetric but which are terrible to
work with, and which cannot be the Hamiltonian of any reasonable physical
system.

An operator will be self-adjoint if it is equal to its own adjoint, so we need
to define adjoint.  Return to $A:\msH_1\to\msH_2$.

\begin{defn}
\label{defn:adjoint}
Let $A:\msH_1 \to\msH_2$ be a linear operator.  Suppose that $\DAbar=\msH_1$.
Then an \bfidx{adjoint operator} $A^*:\msH_2\to\msH_1$ has domain
$\mcD_{A^*}$ consisting of those $g\in\msH_2$ for which 
there is some $g^*\in\msH_1$ such that for all $f\in\DA$,
$$
	(Af,g) = (f,g^*).
$$
\end{defn}

Note that we have defined the domain, but not the operator itself!
We want to define the action of $A^*$ on $g$.  Hold $g$ fixed.
Consider
$$
	(Af,g)=(f,g^*).
$$
Such a $g^*$ is unique by density.  If the above holds, we define $A^*g$
to be $g^*$.  We can check that this is a linear operator on the
above-specified domain, straight from definitions.  Note that $\mcD_{A^*}$
could be $\{0\}$, but won't be in cases of interest.

Now suppose (this is a strong hypothesis!) that $A$ is bounded and is defined
on all of $\msH_1$.  Then the bounded functional
$$
	f\mapsto(Af,g)
$$
is (by the Riesz representation theorem) always of the form
$$
	(f,g^*)
$$
for a unique $g^*$.  So, we directly define $A^*$ on all of $\msH_2$ by
$$
	(Af,g)=(f,A^*g).
$$
This $A^*$ is a bounded linear operator from all of $\msH_2$ to $\msH_1$:
Note that
$$
	\|(Af,g)\| \le \|A\|_\op \; \|f\|_{\msH_1} \; \|g\|_{\msH_2}
$$
by the Cauchy-Schwartz inequality.  This is the constant
$\|A\|_\op\;\|g\|_{\msH_2}$ times $\|f\|_{\msH_1}$.

If $A$ is not defined on on all of $\msH_1$, then typically it is hard to
explicitly describe $\mcD_{A^*}$.

\begin{defn}
We write $A_1 \subseteq A_2$ if $\mcD_{A_1} \subseteq \mcD_{A_2}$ and
$A_1 = A_2 |_{\mcD_{A_1}}$.
\end{defn}

\begin{exer}
Show that $A_1 \subseteq A_2$ implies $A_2^* \subseteq A_1^*$.
\end{exer}

It can be shown that $A^*$ is a closed operator, whether $A$ is closed or not.

\begin{defn}
Let $M$ be a linear subspace of $\msH$.  (Really, $M$ can be merely a set.)
Then define the \bfidx{perpendicular space} (or \bfidx{orthogonal complement})
of $M$ to be
$$
	M^\perp := \{g\in\msH: (f,g)=0 \textrm{ for all } f\in M\}.
$$
\end{defn}

\begin{prop}
\label{prop:perp_space_is_closed}
$M^\perp$ is always closed, whether $M$ was or not.
\begin{proof}
The condition survives the taking of the limit $g_n \to g$.
\end{proof}
\end{prop}

\begin{exer}
Show that $\ker(A^*)=(\im(A))^\perp.$
\end{exer}

\begin{exer}
Show that $A$ and $A^*$ are both invertible iff $\ker(A)=0$ and $\im(A)$
is dense in $\msH_2$.
\end{exer}

\begin{rem}
\label{rem:close_perp_perp}
Show that $M^{\perp \perp} = \overline{M}$.  In particular, $M^\perp=\{0\}$
implies $\overline{M}=\msH$.
\end{rem}

\begin{prop}
$A^*$ is densely defined iff $A$ is closable.
\begin{rem}
In this case,
$$
	\Abar = A^{**}.
$$
\end{rem}
\begin{proof}
The proof uses the geometric construction of $A^*$, which we have not seen
until now.  First we restate the definition of $A^*$ geometrically.  The pair
$$
	(g,g^*) \in \mcG_{A^*},
$$
for $g\in\msH_2$ and $g^*\in\msH_1$, iff
$$
	(g^*,-g) \perp \GA.
$$
The graph of $A^*$ is $\mcG_{A^*}=\mcG_A^\perp$.  Check this by writing out the
definition of adjoint.  For all $f\in\DA$,
$$
	(f,g^*) + (Af,-g)=0
$$
i.e.
$$
	((f,Af),(g^*,-g))_{\msH_1\oplus\msH_2} = 0.
$$
(Note that we are using parentheses in two different ways here: the inner pairs
are for two-tuples; the outer pair denotes an inner product.) This shows that
$A^*$ is a closed operator, since the graph is perpendicular to something.
Alternatively, vectors of the form $(Ag^*,-g)$ constitute $\mcG_A^\perp$
and consequently $\GAbar$ is the orthogonal complement of these vectors.

The proof is now easy with this setup, and it goes similarly in both directions.
If $\mcD_{A^*}$ is not dense in $\msH_2$ then there is $0\ne h \in\msH_2$ such that
$h \perp \mcD_{A^*}$, i.e.
$$
	(0,h) \perp (A^*g,-g),
$$
for all $g\in\mcD_{A^*}$.  This means that
$$
	(0,h) \in \GAbar
$$
so $\GAbar \ne \GA$.  This means that $A$ is not closable.

Now suppose that $A$ is closable.  Then $A^*$ is densely defined, so $A^{**}$
makes sense, and the graph of $A^{**}$ is $\GA^{\perp\perp}=\GAbar$.
\end{proof}
\end{prop}

\begin{defn}
A linear operator $A:\msH\to\msH$ with dense domain of definition $\DA$
is called \bfidx{symmetric} if
$$
	A \subseteq A^*,
$$
i.e. if for all $f,g \in \DA$,
$$
	(Af,g)=(f,Ag).
$$
\end{defn}

Note that since $\mcD_{A^*} \supseteq \mcD_A$, $A^*$ is also densely defined.
So, by the theorem we just proved, $A$ is closable.  That is to say, \emph{a
symmetric operator is closable}.

\begin{prop}
The expectation
$$
	\pig{A} = (f,Af)
$$
is real if $A$ is symmetric.  That is, observables are real.
\begin{proof}
Obvious.
\end{proof}
\end{prop}

\begin{defn}
A densely defined operator $A: \msH\to\msH$ is \bfidx{self-adjoint} if
$$
	A = A^*.
$$
\end{defn}

Note that adjoints are always closed, so a self-adjoint operator is necessarily
closed.

%% ----------------------------------------------------------------
\subsection{Self-adjointness of multiplication operators}

Here is a preview of the spectral theorem, involving multiplication operators
by real-valued functions.  Let $\msH=L^2(M,\sigma)$ for some not necessarily
linear space $M$ and some $\sigma$-finite measure $\sigma$, with
$$
	(f,g)=\int_M \fbar g\,d\sigma.
$$
Then $\msH$ is a Hilbert space (this is a restatement of Riesz-Fischer).
Let $a:M\to \R$ be a measurable function.  Then $T_a:\msH\to\msH$
with (as usual)
$$
	\DTa = \{f \in \msH: a \cdot f \in \msH\}
$$
i.e. $f$ such that $a \cdot f$ is still square-integrable.
On this domain of definition, $T_a f := a \cdot f$.

The following result is very important.  It is similar in spirit to
diagonalization of matrices.  The heart of spectral theory is that \emph{all}
self-adjoint operators are one of these in disguise (up to unitary
transformations).  The reason for all the precise definitions in this section
is that they capture operators of this form.

\begin{prop}
$T_a$ is self-adjoint.
\begin{proof}
Recalling definition \ref{defn:adjoint}, we first need to show that the domain
$\DTa$ is dense in $\msH$, i.e. that the closure of $\DTa$ is all of $\msH$.
Using remark \ref{rem:close_perp_perp}, we will show that $\DTa^\perp = \{0\}$.
Suppose $g \perp \DTa$; we want to show $g=0$.  We will start by proving
a weaker result.  Let $f$ be arbitrary in $L^2(M,\sigma)$.  For each positive
integer $N$, let
$$
	M_N = \{m \in \msH : |a(m)| \le N\}.
$$
Then
$$
	\int_\msH \gbar f 1_{M_N} d\sigma = 0
$$
since $f1_{M_N} \in \DTa$ for all $f \in L^2(M,\sigma)$.  We can write this
in an equivalent way as
$$
	\int_{M_N} \gbar f d\sigma = 0
$$
for all $f \in L^2(M,\sigma)$.  This is true in particular for $g$.  So,
$$
	\int_{M_N} \gbar g d\sigma = \int_{M_N} |g|^2d\sigma = 0
$$
which implies $g=0$ $\sigma$-a.e. on $M_N$.  Remember for below that the whole
space $L^2(M,\sigma)$ is the countable union of the $M_N$'s.

Second we show the self-adjointness of $T_a$.  Let $g \in \DTastar$.  This
means, by definition \ref{defn:adjoint}, that there exists a $g^* \in
L^2(M,\sigma)$ such that for all $f \in \DTa$,
$$
	(g,T_af) = (g^*,f).
$$
(Recall that we then define $T_a^*$ by $T_a^* g := g^*$ as we vary $g$.)  In
terms of integrals, this is
$$
	\int_\msH \gbar T_a f d\sigma = \int_\msH \gbar a f d\sigma
	= \int_\msH \gstarbar f d\sigma.
$$
For uniqueness of $g^*$, we need $\DTa$ to be dense in $L^2(M,\sigma)$, which
we were partway through proving above.  We want to conclude
$$
	g^* = ag;
$$
then $g \in \DTa$ will follow automatically.  The issue (and the reason for the
$M_N$'s) is that we don't yet know that $ag$ is square-integrable.
However, we do know that
$$
	\int_\msH \gbar a f d\sigma = \int_\msH \gstarbar f d\sigma
$$
for a dense set of $f$'s.

Use
$$
	\int_{M_N} \gbar a f d\sigma = \int_{M_N} \gstarbar f d\sigma
$$
from which
$$
	\int_{M_N} (\gbar a - \gstarbar) f d\sigma = 0.
$$
Note $\gbar a - \gstarbar \in L^2(M_N,\sigma)$; then take $N \to \infty$.

So $g \in \DTa$, and the action of the operator $T_a^*$ is given by
$$
	T_a^* g = ag.
$$
(Remember $a$ is real-valued.)
Since
$$
	T_a g = ag
$$
and $g$ was arbitrary in $\DTastar$, $T_a$ is self-adjoint.
\end{proof}
\end{prop}

%\label{prop:perp_space_is_closed}

%% ----------------------------------------------------------------
\subsection{Spectrum}

Here we generalize the notion of \plainidx{eigenvalue}.  We define the spectrum
of an operator $A$ in a rather backward way:  first we define what it means for
a complex number \emph{not} to be in the spectrum of $A$.

For motivation, recall that in finite dimensions, $A$ is always representable
by a matrix (which we also call $A$).  Then $z$ is not an eigenvalue of $A$ iff
$\ker(A-zI)=0$, which is the case iff $(A-zI)^{-1}$ exists.  We think of the
spectrum of $A$, written $\sigma(A)$, as the eigenvalues of $A$ for this
finite-dimensional case.

\begin{defn}
Let $A$ be a densely defined closed operator in $\msH$.  (Closedness is not
necessary, but all our operators are closed.)  We say that $z\in\C$ is in the
\bfidx{resolvent set} $\rho(A)$ if there exists a bounded operator
$$
	R_z := (A-zI)^{-1}
$$
defined on all of $\msH$.  We call $R_z$ the \bfidx{resolvent operator} of $A$.
(Note that it depends on $A$ as well as $z$, but this dependence does not
appear in the notation.)
\end{defn}

The term \emph{resolvent} comes about because of the following.  Let $f,g\in
\msH$.  Then $(A-zI)^{-1}g=f$ iff $(A-zi)f=g$ which is to say $Af-zf=g$, and
$z$ \emph{resolves} this equation.  (If $A$ is a multiplication operator, this
is an algebraic equation; if $A$ is a differentiation operator, this is a
differential equation.)

\begin{rem}
Physicists call $R_z$ a \bfidx{Green's function}.  One asks:  A function of
what?  To solve for $z$, one often obtains an integral operator for $R_z$ with
a kernel (in the integral sense of the word) which is a Green's function.
\end{rem}

\begin{defn}
Let $A$ be a densely defined operator in $\msH$.
The \bfidx{spectrum} of $A$, written $\sigma(A)$, is
$$
	\sigma(A) := \C \; \setminus \; \rho(A).
$$
\end{defn}

We want to know the kernel and image of $(A-zI)$ in order to find out the
domain of its inverse.

\begin{prop}
The resolvent $\rho(A)$ is an open set in $\C$, and the operator-valued
function $z \mapsto R_z$ is an analytic function.
\begin{proof}
Easy, but deferred until later.  The proof is important because it
introduces methods of complex analysis.

First we need to define analyticity for operator-valued functions.

\begin{defn}
The operator-valued function $z\mapsto R_z$ is \bfidx{analytic} if for all $z$
sufficiently close to $z_0\in\rho(A)$,
$$
	R_z = \sum_{n=0}^\infty C_n (z-z_0)^n
$$
with the $C_n$'s all bounded operators defined on all of $\msH$, and with
convergence in the metric induced by the operator norm.
\end{defn}

A sketch of the proof is that one tries to form the inverse $1/(A-zI)$
algebraically.  With an eye to using a geometric series we have
$$
	\frac{1}{A - z_0I - (z-z_0)I}.
$$
Factor out and continue.
\end{proof}
\end{prop}

The following corollary is easy.

\begin{cor}
Fix $f \in \msH$ and take the quadratic form $z \mapsto (f,R_z\,f)$.
This complex-valued function is analytic in $\rho(A)$ for all $f \in\msH$.
\end{cor}

This analyticity will be crucial:  it will provide a link between classical
complex analysis and modern spectral theory.

\begin{prop}[Hilbert resolvent identity]
For\index{Hilbert resolvent identity} all $z,z' \in \rho(A)$,
$$
	R_z - R_{z'} = (z-z') R_z R_{z'}.
$$
\begin{proof}
Write $R_z=(1/A-z)$ and expand as a geometric series.  Do the algebra, then
check it works for operators.  Then write $R_z-R_{z'}$ in the following
artificial but useful way:
$$
	R_z - R_{z'} = R_z(A-z'I)R_{z'} - R_{z'}
$$
Note that the first term on the right-hand side is the identity on $\msH$.
This is
\begin{align*}
	R_z - R_{z'} &= R_z(A-zI + (z-z')I)R_{z'} - R_{z'} \\
	&= R_{z'} + R_z(z-z')R_{z'} - R_{z'} \\
	&= (z-z')R_z R_{z'}.
\end{align*}
\end{proof}
\end{prop}

\begin{cor}
Resolvent operators on the same $A$ commute with one another:
$$
	R_z R_{z'} = R_{z'}R_z.
$$
\end{cor}

Of course, this result could have been obtained without making a corollary of
the proposition.

[xxx make fwd xrefs to here from the SA section.]

\begin{prop}
Suppose $A:\msH\to\msH$ is self-adjoint.  Then:

(1) $\sigma(A) \subseteq \R$.  In particular, $R_z$ is defined on the
open half-planes
$$
	\C^+ := \{\Im(z) > 0\} \qqand \C^- := \{\Im(z) < 0\}.
$$

(2) Recall that $R_z$ is bounded so the adjoint $R_z^*$ exists.
We have
$$
	R_z^*=R_{\zbar}.
$$
Moreover,
$$
	\|R_z\|_\op \le |\Im(z)|^{-1}.
$$

\begin{rem*}
This result is emphatically not true for an arbitrary symmetric operator;
self-adjointness is crucial.
\end{rem*}
\begin{proof}
From the definitions, using the antilineary of the inner product, we have
$$
	(A-zI)^* = A^* - \zbar I.
$$
Since adjoint and inverse commute, it follows that
$$
	R_z^* = R_\zbar.
$$
Now we calculate
$$
	\|(A-zI)f\|^2 = (Af-zf, Af-zf).
$$
Let $z=x+iy$.  Then
\begin{align*}
	\|(A-zI)f\|^2 = (Af,Af) - x(Af,f) - iy(Af,f) - x(f,Af) +iy(f,Af)
		+ |z|^2(f,f).
\end{align*}
Now, $(Af,Af)$ and $(f,f)$ are non-negative and real by the
positive-definiteness of the inner product.  The terms with $x$ in them
are complex conjugates of one another, and $x$ is real, so
$x(Af,f)-x(f,Af)$ is real.  We have some real number, minus
$iy(Af,f)$, plus $iy(f,Af)$, so these latter two terms sum to zero.
We claim that
$$
	\|(A-zI)f\|^2 = \|(A-xI)f\|^2 + y^2\|f\|^2 = \|(A-zI)f\|^2.
$$
We want to show this is not too small because we are interested in the norm
of the inverse operator.  The result is
$$
	\|(A-zI)f\|^2 \ge y^2\|f\|^2.
$$

We did not assume $z\in\rho(A)$; we only assumed $\Im(z)\ne 0$ and we want to
conclude that $z\in\rho(A)$.  We know $\ker(A-zI)=0$ by the inequality.  Look
at the range of the operator $(A-zI)$.  We want $R_z$ defined on all of $\msH$:
if the kernel of $A-zI$ is zero and if it is onto, then it is invertible on all
of $\msH$.  We have
$$
	\im(A-zI)^\perp = \ker((A-zI)^*)
$$
by an exercise from above [xxx xref].  Using self-adjointness of $A$, this
is
$$
	\im(A-zI)^\perp = \ker((A-zI)^*) = \ker(A^*-\zbar I)
	= \ker(A-\zbar I).
$$
Note that $\Im(z)\ne 0$ iff $\Im(\zbar) \ne 0$.  So $\im(A-zI)$ is dense
in $\msH$, but this operator is closed so the image is all of $\msH$.
Thus $(A-zI)^{-1}$ exists.  Also we have
$$
	\|(A-zI)^{-1}\} \le \frac{1}{y} = \Im(z)^{-1}
$$
as desired.
\end{proof}
\end{prop}

Here is another property the resolvent operator satisfies.

\begin{prop}
Fix $0\ne f
\in\msH$.  Consider the function
$$
	\Phi(z) = (f,R_zf).
$$
This is defined on $\rho(A)$ so it is at least defined on $\C^+$ and $\C^-$ (as
well as perhaps parts of the real line).  This function is analytic and maps
$\C^+$ and $\C^-$ into themselves, respectively.

Furthermore, $\Phi(z)$ satisfies
$$
	\Phi(\zbar) = \overline{\Phi(z)}
$$
and
$$
	|\Phi(\zbar)| \le (f,f) \Im(z)^{-1}.
$$
\begin{rem*}
We call $\Phi(z)$ a \bfidx{Herglotz function}.
\end{rem*}
\begin{proof}
The last two points are obvious from the previous theorem; use the
Cauchy-Schwartz inequality.

To see that $\Phi(z)$ maps $\C^\pm$ into $\C^\pm$, let
$$
	R_z f = g.
$$
Then
$$
	f = (A-zI)g.
$$
In particular, $g \ne 0$.  Also,
$$
	(f, R_zf) = ((A-zI)g,g) = (Ag,g) - \zbar(g,g).
$$
Since $A$ is symmetric, $(Ag,g)$ is real.  Therefore the imaginary part
of the left-hand side is $\Im(\zbar(g,g))$.  Chasing signs, we find that
the right-hand side has imaginary part $\Im(z)(g,g)$.
\end{proof}
\end{prop}

There is a \plainidx{representation theorem} for Herglotz functions:
for each $\Phi(z)$ there is a $\sigma(x)$ such that we have the following
Lebesgue-Stieltjes integral:
$$
	\Phi(z) = \int_\R \frac{d\sigma(x)}{x-z}
$$
Here $\sigma(x)$ is right-continuous and non-decreasing, with
$\lim_{x\to\-\infty}(x)=0$ and and $\lim_{x\to+\infty}(x) < +\infty$.  The
measure has point masses at discontinuities of $\sigma$.  As may be seen from
the integral representation, the analyticity of $\Phi(z)$ breaks down as $z$
approaches the real line; the function $\sigma(x)$ will encapsulate the
continuity and/or discreteness of the spectrum $\sigma(A)$.

%% ----------------------------------------------------------------
\subsection{Preview of the spectral theorem}

The following analogy is an illustration of the spectral theorem on a
finite-dimensional vector space.  This makes it easier to follow the proof of
the spectral theorem, in particular the choice of $M$ and $\sigma$ which arise
there.

For this section, let $\msH$ be an $N$-dimensional complex Hilbert space.  Let
$A:\msH\to\msH$ be self-adjoint (which is equivalent to symmetric in finite
dimension).  As is well-known, there exists an orthonormal basis of
eigenvectors $\{f_i\}$ with eigenvalues $a_i$ such that for $i=1,2,\ldots, N$,
$A f_i = a_i f_i$.  Assume for now that $A$ has a \emphidx{simple spectrum},
i.e. all the eigenvalues of $A$ are distinct.  For each $f \in \msH$ there
exists a unique representation
$$
	f = \sum_{i=1}^N c_i f_i.
$$
Define an operator $U$ trading $f$ for the $N$-tuple $(c_1, \ldots, c_N) \in
\C^N$.  Since the $f_i$'s form an orthonormal basis, by writing out the inner
product one may show that
$$
	\|f\|_\msH^2 = \sum_{i=1}^N |c_i|^2.
$$
Thus, we may think of $\C^N$ as an $L^2$ space.  However, we do this in a
particular way:  we write
$$
	\vecc = (c_1, \ldots, c_N) \in L^2(\{a_1, \ldots, a_N\}).
$$
That is, we put the $c_i$'s as a graph over the $a_i$'s.  (This is why we
assume the $a_i$'s are distinct.)  Then $U$ is a unitary transformation from
$\msH \to L^2(\{a_1, \ldots, a_N\})$.

This elementary construction achives the same goal as in the spectral theorem.
We define $T_a:L^2 \to L^2$ by
$$
	{}[T_a(\vecc)]_i = a_i c_i
$$
and then we write
$$
	A = U^{-1} T_a U.
$$

Now suppose there are $N/2$ eigenvalues, each of multiplicity 2.
Do the above, but with the space
$$
	L^2(a_1, \ldots, a_{N/2}) \oplus L^2(a_1, \ldots, a_{N/2}).
$$

One may generalize to the case of arbitrary multiplicities.  In all cases, the
$L^2$ case is the spectral theorem's $M$.

%% ----------------------------------------------------------------
\subsection{Spectral theorem}

%\begin{thm}
%Let $A$ be a self-adjoint operator in a Hilbert space $\msH$.  Then $A$ is
%real-diagonalizable, i.e. $\msH$ admits an orthonormal basis of eigenvectors
%of $A$ with real eigenvalues.
%\end{thm}

%% ----------------------------------------------------------------
\newpage
\section{Multi-body systems}

Note as of Sep. 27 2007:  I had some difficulty following the discussion of
spectral theory and will defer typing up any notes until I have some good
examples in hand.  Also, as of about a week ago, the discussion in class has
been closely following the text (\cite{bib:mr}), so there I feel reduced
urgency for these notes.  However I do feel that Martin and Rothen's text could
bear some worked examples, which I will include here as I have them.
