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

My comprehensive examination exposits the paper \cite{bib:qmath}.  The problem
at hand is to determine the effects of interparticle interactions on the
critical temperature of Bose-Einstein condensation; a so-called random-cycle
model is employed to this end.

A large fraction of this paper is occupied by the appendices.  This reflects
the fact that much of my work on this project has been in filling the gap
between my coursework and the research papers being presented.  The experienced
reader may wish to skip the appendices entirely.

The plan of this paper is as follows:

\begin{itemize}

\item A non-technical discussion of the historical context of the project; a
mention of alternative approaches.

\item The random-cycle approach to the BEC problem requires a model of spatial
permutations.  This model is developed mathematically --- it is of
probabilistic interest on its own --- without reference to the Bose gas.  (Main
results about the latter are summarized in the appendices.)

\item The model of spatial permutations is connected to the Bose gas:

	\begin{itemize}

	\item One begins with a Hamiltonian for particles with two-body
	interactions.

	\item Using a multi-body Feynman-Kac approach involving permutation
	symmetry of bosonic wave functions, one obtains a Hamiltonian in which
	permutation jumps rather than particles interact.

	\item A cluster expansion, to first order in the scattering length
	of the particles, yields a Hamiltonian with only two-jump interactions.  At
	this point, one may employ the random-cycle model of spatial permutations.

	\end{itemize}

\item A simplified random-cycle model, called the two-cycle model, is
described.  In particular, this simplified model is amenable to Monte Carlo
simulation and will play a key role in my dissertation work.  Brownian bridges
are the theoretical workhorse of this paper, but are computationally expensive
to simulate.  The two-cycle model bypasses Brownian bridges entirely.

\end{itemize}

%% ================================================================
\newpage
\section{Historical context}

An excellent summary of the history of the study of Bose-Einstein condensation
may be found in Lieb et al. \cite{bib:lssy}.  Here, we enumerate the main
points.

Study of the interacting Bose gas is a particular instance of a general concern
for interacting multi-body systems in statistical mechanics.

In 1924, the Indian physicist Satyendra Nath Bose realized that the statistics
governing photons are described by the symmetric subspace of the Hilbert space.
[xxx when] shortly thereafter, Albert Einstein realized that the same could be
done with massive particles.  xxx discovered BEC.  xxx describe that in a
sentence of two.  xxx Condensation of the ideal (non-interacting) Bose gas.
xxx Critical temperature.

xxx curiosity [xxx avoid weasel words] ...

xxx Liquid helium is strongly interacting.  Onnes 1908+?  Outside Bose-Einstein
theory.

xxx Effects of interactions on critical temperature ... biblio page from rcm1?

1990's CU-Boulder experiment:  weakly interacting.

xxx ``lowest quantum state'' makes sense in the non-interacting case but
is undefined in the interacting case ... how to describe it?

............

The primary long-term goal of the project is to determine the slope of the
first-order term of the reduced $T_c(a)$.

Order parameter:  Penrose-Onsager $\sigma_{\vecx\vecy}$ and Feynman's
$\rho_\infty$.  xref to DU paper about the connection between the two order
parameters.

BU p. 15:  Feynman.  FK rep; pmtns; ODLRO.

\Suto{} and Ueltschi.

xxx two historical narratives:  (1) Lieb et al.; (2) DU's.

%% ================================================================
\newpage
\section{Alternative approaches}

$O(2) \phi^4$ field theory:
Arnold and Moore 2001 \cite{bib:arnoldmoore}

Worm algorithm (?): Kashurnikov et al. \cite{bib:kps}.

PIMC:
Ceperly, Nho, Landau --- \cite{bib:ceperley,bib:nholandau}

Say something about CPU time etc. ... this is perhaps our main selling point.

%% ================================================================
\newpage
\section{Models of spatial permutations}
\label{sec:rcm}

Here we define and describe two models of spatial permutations from a
mathematical point of view.  These models will be related to the physics of the
Bose gas in section \ref{sec:cluexp}.  The exposition here follows section 2 of
\cite{bib:qmath}, with additional insights drawn from section 2 of
\cite{bib:bu}.

%% ----------------------------------------------------------------
\subsection{Definitions}
\label{sec:model_defns}

\textbf{State space:}
Let $\Lambda \subset \R^d$ be a cube of width $L$ and volume $V=L^d$.  Let $N
\in \Z^+$ and $\vecX = (\vecx_1, \ldots, \vecx_N)$ for $\vecx_1, \ldots,
\vecx_N \in \Lambda$.  The \plainidx{state space} of the model of infinite
permutations is
$$
	\Omega_{\Lambda,N} = \Lambda^N \times \SN
$$
where $\SN$ is the group of permutations of $N$ points.  (See figure
\ref{fig:piham}.)

\begin{figure}[!htb]
\begin{center}
\psfragscanon
\includegraphics{figures/piham.eps}
\caption{A configuration of $\vecX$ and $\pi$ with $N=8$.
	\label{fig:piham}}
\end{center}
\end{figure}

\textbf{Hamiltonian:}
The \plainidx{probability measure} on this state space will be constructed
through a Hamiltonian.  The background probability measure is discrete
(uniform) in $\pi$ and continuous (Lebesgue) in $\vecX$.  The Hamiltonian is as
follows:
$$
	H(\vecX,\pi) = \sum_{i=1}^N \frac{1}{4\beta}\|\vecx_i-\vecx_{\pi(i)}\|
		+ \sum_{1\le i<j<\le N}
			V(\vecx_i, \vecx_{\pi(i)}, \vecx_j, \vecx_{\pi(j)}).
$$
(The $1/4\beta$ factor is the reciprocal of what is more commonly encountered
in statistical mechanics.  For now, think of this as an ansatz.  In section
\ref{sec:U0}, this choice of scale factor will be found to be apprropriate.)

There are two contributions to the energy of a configuration $(\vecX,\pi)$:
\begin{itemize}

\item The first\footnote{The papers \cite{bib:bu} and \cite{bib:qmath}
generalize from $\|\vecx_i-\vecx_{\pi(i)}\|$ to $\xi(\|\vecx_i,\vecx_{\pi(i)})$
where $\xi$ is a spherically symmetric non-negative-valued function on $\R^d$
having integrable $e^{-\xi}$.  This generalization is not of interest in the
current paper, nor will it be of interest in my dissertation to follow.}
contribution to the energy is the sum of squares of permutation jump lengths.
(One jump is marked with a heavy arrow in figure \ref{fig:piham}.)  This makes
permutations with long jumps disfavored; permutations with many short jumps
will be less strongly disfavored.

\item The second contribution to the energy is the double sum over interactions
between permutation jumps.  (One such two-jump interaction is marked with wavy
grey lines in figure \ref{fig:piham}.)

\end{itemize}

\textbf{Jump-interaction potentials:}
The specific form of the two-jump interaction is left unspecified at present,
but we will require the following properties: $V$ is translation-invariant,
i.e.
$$
	V(x,y,x',y') = V(x+a,y+a,x'+a,y'+a)
$$
for all $a \in \Lambda$, and
$$
	V(x,y,x',y') = V(x',y',x,y)
$$
for all $x,y \in \Lambda$.

We will consider three different \plainidx{jump-interaction potentials}:
\begin{itemize}
\item One may certainly take the jump interaction $V$ to be zero.
\item In section \ref{sec:tji}, the jump interaction $V$ will involve the
probability of the intersection of Feynman paths.
\item In section \ref{sec:n2}, we will treat all jump pairs as non-interacting
unless they participate in a two-cycle.
\end{itemize}

\textbf{Partition functions:}
We consider two \plainidx{partition functions}, for
a fixed point configuration $\vecX$ and for an average over point
configurations, respectively:
$$
	Y(\Lambda,\vecX) = \sum_{\sigma\in\SN} e^{-H(\vecX,\sigma)}
	\qand
	Z(\Lambda,N) = \frac{1}{N!} \int_{\Lambda_N} Y(\Lambda,\vecX)\,d\vecX.
$$

\textbf{Probability measures:}
Then we have two corresponding discrete \plainidx{probability measures} on the
finite set $\SN$, for a fixed point configuration $\vecX$ and for an average
over point configurations, respectively:
$$
	P_{\Lambda,\vecX}(\pi)
	= \frac{e^{-H(\vecX,\pi)}}{\sum_{\sigma\in\SN} e^{-H(\vecX,\sigma)}}
	\qand
	P_{\Lambda,N}(\pi)
	= \frac
		{\frac{1}{N!} \int_{\Lambda^N}d\vecX e^{-H(\vecX,\pi)}}
		{\frac{1}{N!} \int_{\Lambda^N}d\vecX \sum_{\sigma\in\SN}
			e^{-H(\vecX,\sigma)}}.
$$
These may also be written in terms of the partition functions as
$$
	P_{\Lambda,\vecX}(\pi)
	= \frac{e^{-H(\vecX,\pi)}}{Y(\Lambda,\vecX)}
	\qand
	P_{\Lambda,N}(\pi)
	= \frac{\int_{\Lambda^N}d\vecX e^{-H(\vecX,\pi)}}{Z(\Lambda,N)N!}.
$$

Note that for the non-interacting $V=0$ case, we have the following heuristic:
\begin{itemize}
\item As $\beta \to 0$, the probability measure becomes supported only on
the identity permutation.
\item As $\beta \to \infty$, the probability measure approaches the
uniform distribution on $\SN$.
\end{itemize}

\textbf{Expectations:}
These probability measures give rise to the following \plainidx{expectations}.
For a random variable $\theta(\pi)$ --- in this paper, we will not consider
random variables which depend on $\vecX$ --- we have
$$
	\bbE_{\Lambda,\vecX}(\theta) = \frac{1}{Y(\Lambda,\vecX)}
		\sum_{\pi\in\SN} \theta(\pi) e^{-H(\vecX,\pi)}
	\qand
	\bbE_{\Lambda,N}(\theta) = \frac{1}{Z(\Lambda,N) N!}
		\int_{\Lambda^N} d\vecX \sum_{\pi\in\SN} \theta(\pi) e^{-H(\vecX,\pi)}.
$$

\textbf{Random variables:}
As we will see in section [xxx xref], Feynman's claim [xxx xref] is that
Bose-Einstein condensation [xxx ham ...] occurs iff there are infinite cycles
in an infinite-volume extension of the above probability model.  In this
section we confine ourselves to mathematical statements only; the connection
with the physics will be made in section [xxx xref].  Yet, we have motivated
the choice of the following random variable.

[xxx also:  Feynman's claim about infinite cycles; cycles depend only on the
permutation and are independent of positions; that's the reason for
$\theta(\pi)$.]

The particular \plainidx{random variable} of interest, $\varrho_{m,n}(\pi)$, is
as follows.  First define
$$
	\ell_i(\pi)
$$
to be the length of the permutation cycle containing the point $\vecx_i$.
For example, for the point configuration $\vecX$ and the permutation $\pi$
in figure \ref{fig:piham}, we have
$$
	\ell_1(\pi) =
	\ell_2(\pi) =
	\ell_3(\pi) =
	\ell_4(\pi) = 4,
	\quad
	\ell_5(\pi) =
	\ell_6(\pi) =
	\ell_7(\pi) = 3,
	\qand
	\ell_8(\pi) = 1.
$$
Also let
$$
	\rho = \frac{N}{V},
$$
i.e. $\rho$ is the particle density.  For $1 \le m \le n \le N$, define
$$
	\varrho_{mn}(\pi)
	= \frac{1}{V} \;
	\# \left\{ i=1, \ldots, N: m \le \ell_i(\pi) \le n \right\}
$$
This random variable, taking values between 0 and $\rho$, is the \bfidx{density
of sites} in cycles of specified length.  For figure \ref{fig:piham}, we have
$\varrho_{2,3}(\pi) = 3/8$.

One may also consider the related random variable
$$
	f_{m,n}
	= \frac{1}{N} \;
	\# \left\{ i=1, \ldots, N: m \le \ell_i(\pi) \le n \right\}
$$
which is $\varrho_{m,n}/\rho$.  This runs from 0 to 1 and is the
\bfidx{fraction of sites} in cycles of specified length.

\textbf{Models:}
The probability measure $P_{\Lambda,\vecX}(\pi)$ and the expectation
$\bbE_{\Lambda,\vecX}(\rho_{m,n})$ is treated in the paper \cite{bib:gru} where
$\vecX$ is a cubic unit lattice; $P_{\Lambda,N}(\pi)$ and
$\bbE_{\Lambda,N}(\rho_{m,n})$ are treated in \cite{bib:bu} and
\cite{bib:qmath}.  The former will be referred to as the \bfidx{lattice model};
the latter will be referred to as the \bfidx{point-process model}.

%% ----------------------------------------------------------------
\subsection{Existence of infinite cycles}

Here we describe theorem 1 of \cite{bib:qmath}, deferring [xxx?] a sketch
of the proof to section [xxx].

\textbf{Thermodynamic limit:}  We inquire about the fraction of sites
participating in short and long cycles (as quantified below) in the
\plainidx{infinite-volume limit}.  Namely, we let $V,N \to \infty$ with fixed
ratio $\rho = N/V$, and we ask about the cycle-length distribution as a
function of $\rho$.

xxx inf vol limit not needed here since we just take lim in $\R$?  mention
extension anyway w/ xref to BU?

xxx stts of the form
$$
	\lim_{V \to \infty} \bbE_{\Lambda,N}(\varrho_{m,n})
$$
...
just convergence of real numbers; finite-volume at each $N$.

\textbf{Critical density:}
xxx def $\rho_c$.  Here, just an ansatz.
$$
	\rho_c = \int_{\R^d} \frac{d\veck}{e^{4\beta\pi^2|\veck|^2}-1}.
$$
Note that $4\beta\pi^2|\veck|^2$ is the Fourier transform of
$\frac{1}{4\beta}|\vecx|^2$.

xxx see also wik bec article.

\textbf{Statement of the theorem:}
For any $0<a<b<1$ (nominally, $a$ is just above 0 and $b$ is just below 1)
and any $s \ge 0$,
\begin{align*}
%
	\lim_{V \to \infty} \bbE_{\Lambda,N}(\varrho_{1,N^a}) &=
		\begin{cases}
		\rho,   & \rho \le \rho_c\\
		\rho_c, & \rho_c \le \rho
		\end{cases} \\
%
	\lim_{V \to \infty} \bbE_{\Lambda,N}(\varrho_{N^a,N^b}) &=
		0 \\
%
	\lim_{V \to \infty} \bbE_{\Lambda,N}(\varrho_{N^b,sN}) &=
		\begin{cases}
		0,           & \rho \le \rho_c \\
		\rho-\rho_c, & \rho_c \le \rho \le s+\rho_c \\
		s,           & s+\rho_c \le \rho.
		\end{cases}
\end{align*}
(See figure \ref{fig:limErho}.)

\begin{figure}[!htb]
\begin{center}
\psfragscanon
\includegraphics{figures/limErho.eps}
\caption{Distribution of cycle lengths as function of density in the
	infinite-volume limit.
	\label{fig:limErho}}
\end{center}
\end{figure}

This statement, in terms of the density of sites in cycles of specified
lengths, may be more intuitively rephrased in terms of the fraction of sites in
cycles of specified length:
\begin{align*}
%
	\lim_{V \to \infty} \bbE_{\Lambda,N}(f_{1,N^a}) &=
		\begin{cases}
		1,   & \rho \le \rho_c\\
		\rho_c/\rho, & \rho_c \le \rho
		\end{cases} \\
%
	\lim_{V \to \infty} \bbE_{\Lambda,N}(f_{N^a,N^b}) &=
		0 \\
%
	\lim_{V \to \infty} \bbE_{\Lambda,N}(f_{N^b,sN}) &=
		\begin{cases}
		0,             & \rho \le \rho_c \\
		1-\rho_c/\rho, & \rho_c \le \rho \le s+\rho_c \\
		s/\rho,        & s+\rho_c \le \rho.
		\end{cases}
\end{align*}

The theorem has the following interpretation:
\begin{itemize}
\item At density below $\rho_c$, all sites are in short cycles and
no sites are in long cycles.
\item As density increases past $\rho_c$ and $\rho_c+s$,
fewer sites are in short cycles and more sites are in long cycles.
In particular, a strictly positive fraction of sites are in long cycles.
\item Asymptotically, all sites are in long cycles.
\end{itemize}

% xxx get the goddamned units correct!!  units of s?  rescale factor?!?

%%% ================================================================
%\newpage
%\section{Proof of existence of infinite cycles}
%
%xxx \cite{bib:bu}
%
%Description of the steps for proving the existence of infinite cycles. It
%involves certain lemmas that could be stated without proof.
%
%%% ----------------------------------------------------------------
%\subsection{Step \ldots}
%
%%% ----------------------------------------------------------------
%\subsection{Step \ldots}
%
%%% ----------------------------------------------------------------
%\subsection{Step \ldots}
%
%%% ----------------------------------------------------------------
%\subsection{Step \ldots}


%% ----------------------------------------------------------------
\newpage
\section{Bosonic Feynman-Kac formulas}
\label{sec:fkboson}

In this section, the heart of this paper, we recast a familiar Hamiltonian
involving \textbf{pair-interacting particles} as a new Hamiltonian involving
\textbf{pair-interacting permutations}.  We use the canonical partition
function as the vehicle for this transformation:
$$
	\textrm{Hamiltonian for particles}
	\longto
	\textrm{partition function}
	\longto
	\textrm{Hamiltonian for permuations}.
$$
A Feynman-Kac formula for $N$ interacting bosons allows the partition
function to be transformed in the middle step.

%% ----------------------------------------------------------------
\subsection{Outline}

% xxx decide on notation for old & new H's ...

As in section \ref{sec:model_defns}, we write
$
	\vecX = (\vecx_1, \ldots, \vecx_N)
$
for $\vecx_1, \ldots, \vecx_N$ in a $d$-dimensional cube $\Lambda$ of width
$L$.  The Hamiltonian for $N$ pair-interacting particles is
\begin{align}
\label{eqn:particle_Hamiltonian}
	H(\vecX) &=
	-\sum_{i=1}^N \nabla_i^2 + \sum_{1\le i,j \le N} U(\vecx_i - \vecx_j).
\end{align}
The $U$ considered in this paper is either identically zero (for the
non-interacting case), or a \bfidx{hard-core potential} with radius $a$, i.e.
$U(\vecx_i-\vecx_j)$ is infinite for $|\vecx_i-\vecx_j| \le a$ and $0$ for
$|\vecr| > a$.  (This is an approximation to the true pair potential between
helium atoms.  See figure \ref{fig:ceperley_potential} \cite{bib:ceperley}.)

\begin{figure}[!htb]
\begin{center}
%\psfragscanon
\includegraphics[scale=0.7]{figures/ceperley_potential.eps}
\caption{Pair potential between helium atoms (Ceperley, 1995).
	\label{fig:ceperley_potential}}
\end{center}
\end{figure}

As discussed in appendix \ref{sec:qsm}, the partition function for $N$
distinguishable particles\footnote{For a particle Hamiltonian, the $\beta$
factor is in the expected place.  This is in contrast to the permutation
expression in sections \ref{sec:model_defns} and xxx, where the $\beta$ factor
is, surprisingly, reciprocated.  We will see in section xxx why the
reciprocated $\beta$ is correct for the permutation Hamiltonian.} is
$\Tr(e^{-\beta H})$.  Symmetrizing the partitition function, since our
particles are bosons, the trace is
$$
	\Trsym(e^{-\beta H})
	= \Tr_{L^2} \left( P_+ e^{-\beta H} \right)
	= \Tr_{L^2} \left( e^{-\beta H} P_+ \right)
$$
where
$$
	P_+\, f(\vecx_1, \ldots, \vecx_N)
	:= \frac{1}{N!} \sum_{\pi \in \Sn} M_\pi f(\vecx_1, \ldots, \vecx_N)
$$
and
$$
	M_\pi(f\vecx_1, \ldots, \vecx_N)
	:= f(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)}).
$$
That is,
$$
	\Trsym(e^{-\beta H})
	= \frac{1}{N!}\;
	\sum_{\pi\in\SN} \Tr_{L^2} \left( e^{-\beta H} M_\pi \right).
$$

The following steps remain to develop a bosonic Feynman-Kac formula.  The first
three steps closely parallel the steps used to construct the familiar
single-particle Feynman-Kac formula in appendix \ref{sec:fk1}.

\begin{itemize}

\item
Section \ref{sec:fkboson_exp}:
Interpret $e^{-\beta H} M_\pi$ as an expectation over Brownian motions, as in
proposition \ref{prop:ebHG} for the single-particle case.

\item
Section \ref{sec:fkboson_ker}:
Write $e^{-\beta H} M_\pi$ as an integral operator, and find the kernel.

\item
Section \ref{sec:fkboson_tr}:
Compute $\Tr(e^{-\beta H} M_\pi)$ in terms of Brownian bridges.

\item
Section \ref{sec:fkboson_sum}:
Sum over $\pi \in \SN$ to obtain $Z=\Trsym(e^{-\beta H})$.  Importantly, we
will express $Z$ as sum over permutations $\pi$ of $e^{H_\pi'(\vecX,\pi)}$,
where this new $H_\pi'$ will be viewed a Hamiltonian for a single permuation
$\pi$.  At this point, the permutation Hamiltonian is found inside
$e^{H_\pi'(\vecX,\pi)}$; we lack an expression for its logarithm.

\item
Sections \ref{sec:U0} and \ref{sec:Unon0} decouple
the non-interacting from the interacting terms in the permutation Hamiltonian,
so that we may write $e^{H_\pi^0(\vecX,\pi)+H_\pi^1(\vecX,\pi)}$.

\item
The bosonic Feynman-Kac formula now contains terms for 2-jump interactions,
3-jump interactions, and so on.  In section \ref{sec:cluexp}, we discuss the
cluster expansion which allows us to drop all but 2-jump interactions.  The
cluster expansion furthermore allows us to take the logarithm of
$e^{-H_\pi'(\vecX,\pi)}$, with an explicit expression for $H_\pi'(\vecX,\pi)$.
We recognize the random-cycle model from section \ref{sec:rcm}, with an
explicit 2-jump interaction $V$.

%\item
%The result is equation xxx on page xxx.

\end{itemize}

%% ----------------------------------------------------------------
\subsection{$e^{-\beta H} M_\pi$ as expectation}
\label{sec:fkboson_exp}

Here we simply parallel the development for the single-particle case
in section \ref{sec:fk1_exp}.

\begin{prop}
\label{prop:diffusion_FK_bosonic}
$$
	e^{-\beta H} f(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)})
	= \bbE_{0}^{\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)}} \left[
		\exp\left\{
			\sum_{i<j}
				\int_0^{2\beta} U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
		\right\}
		f \left( \vecw_{c,t}^{(1)}, \ldots, \vecw_{c,t}^{(N)} \right)
	\right].
$$
\begin{proof}
Using the Trotter product formula (equation \ref{eqn:trotter_AB}), as in the
single-particle proposition \ref{prop:diffusion_FK}, we have
%	e^{(A+B)t} &= \lim_{n\to\infty} \left(e^{At/n} e^{Bt/n}\right)^n.
\begin{align*}
	&e^{-\beta H} f(x_{\pi(1)},\ldots,x_{\pi(N)}) \\
	&= \lim_{n\to\infty} \left(
		e^{\frac{t}{n} \sum_i \nabla_i^2}
		e^{-\frac{t}{n} \sum_{i<j} U(\vecx_i-\vecx_j)}
		\right)^n
		f(x_{\pi(1)},\ldots,x_{\pi(N)}) \\
	&= \lim_{n\to\infty}
		e^{\frac{t}{n} \sum_i \nabla_i^2}
		e^{-\frac{t}{n} \sum_{i<j} U(\vecx_i-\vecx_j)}
	\left(
		e^{\frac{t}{n} \sum_i \nabla_i^2}
		e^{-\frac{t}{n} \sum_{i<j} U(\vecx_i-\vecx_j)}
		\right)^{n-1}
		f(x_{\pi(1)},\ldots,x_{\pi(N)}).
\end{align*}
Recall that $e^{-\frac{t}{n} \sum_{i<j} U(\vecx_i-\vecx_j)}$ is simply a
scalar.  Using proposition \ref{prop:diffusion_convolution} to write
$e^{\frac{t}{n} \sum_i \nabla_i^2}$ as an integral operator (since the sum
of $N$ Laplacians, each on $d$ dimensions, is a single $Nd$-dimensional
Laplacian), we have
\begin{align*}
	&e^{-\beta H} f(x_{\pi(1)},\ldots,x_{\pi(N)}) \\
%
	&= \lim_{n\to\infty}
		\int_{\R^{Nd}}
		g_{t/n}\left(
			\colcvecthree{\vecx_1}{\vdots}{\vecx_N} -
			\colcvecthree{\vecz^{(1)}_1}{\vdots}{\vecz^{(1)}_N}
			\right)
		e^{-\frac{t}{n} \sum_{i<j} U(\vecz^{(1)}_i-\vecz^{(1)}_j)} \\
	&\qquad \left(
		e^{\frac{t}{n} \sum_i \nabla_i^2}
		e^{-\frac{t}{n} \sum_{i<j} U(\vecx_i-\vecx_j)}
		\right)^{n-1}
		f(x_{\pi(1)},\ldots,x_{\pi(N)})
		\;d\vecz^{(1)}_1 \cdots d\vecz^{(1)}_N.
\end{align*}
Repeating $n-1$ more times yields
\begin{align*}
	& \lim_{n\to\infty} \int_{\R^{Nnd}}
		g_{t/n}\left(
			\colcvecthree{\vecx_1}{\vdots}{\vecx_N} -
			\colcvecthree{\vecz^{(1)}_1}{\vdots}{\vecz^{(1)}_N}
			\right)
		\cdots
		g_{t/n}\left(
			\colcvecthree{\vecz^{(n-1)}_1}{\vdots}{\vecz^{(n-1)}_N} -
			\colcvecthree{\vecz^{(n)}_1}{\vdots}{\vecz^{(n)}_N}
			\right) \\
	& \qquad \left(\prod_{k=1}^n
		e^{-\frac{t}{n} \sum_{i<j} U(\vecz^{(k)}_i-\vecz^{(k)}_j)}
	\right)
	f(x_{\pi(1)},\ldots,x_{\pi(N)})
		\;d\vecz^{(1)}_1 \cdots d\vecz^{(1)}_N.
\end{align*}
We have an integrand in the form of remark \ref{rem:ExFomega}, with
$t_k = kt/n$, so we can write
\begin{align*}
%
	& \lim_{n\to\infty} \int_{\R^{Nnd}}
		g_{t/n}\left(
			\colcvecthree{\vecx_1}{\vdots}{\vecx_N} -
			\colcvecthree{\vecz^{(1)}_1}{\vdots}{\vecz^{(1)}_N}
			\right)
		\cdots
		g_{t/n}\left(
			\colcvecthree{\vecz^{(n-1)}_1}{\vdots}{\vecz^{(n-1)}_N} -
			\colcvecthree{\vecz^{(n)}_1}{\vdots}{\vecz^{(n)}_N}
			\right) \\
	& \qquad \exp\left\{
		-\frac{t}{n} \sum_{i<j} \sum_{k=1}^n U(\vecz^{(k)}_i-\vecz^{(k)}_j)
		\right\}
%
	f(x_{\pi(1)},\ldots,x_{\pi(N)})
		\;d\vecz^{(1)}_1 \cdots d\vecz^{(1)}_N \\
%
	&= \bbE_0^{\vecx_{\pi(1)},\ldots,\vecx_{\pi(N)}} \left[ \exp\left\{
			\sum_{i<j} \int_0^{2\beta}
				U(\vecw_{c,s}^{(i)} - \vecw_{c,s}^{(j)}) \,ds
		\right\}
	f\left(\vecw_{c,t}^{(1)},\ldots,\vecw_{c,t}^{(N)}\right)
	\right].
\end{align*}
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{$e^{-\beta H} M_\pi$ as an integral operator}
\label{sec:fkboson_ker}

%Section \ref{sec:operator_trace} showed how to compute $\TrebH$ if
%$e^{-\beta H}$ can be represented as an integral operator.  Proposition
%\ref{prop:diffusion_FK} constructed the operator's kernel for the $U=0$ case.
%Here we construct the kernel for the general case, then apply it to find an
%expression for the trace.

\begin{prop}
\label{prop:ebHG_bosonic}
If
\begin{equation*}
	H = -\sum_i \nabla_i^2 + \sum_{i<j} U(\vecx_i-\vecx_j),
\end{equation*}
then
\begin{equation}
\label{eqn:temp1_bosonic}
\begin{aligned}
	e^{-\beta H} f(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)})
	= \int G_{t,U}&(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)},
		\vecy_1, \ldots, \vecy_N) \\
		&f(\vecy_1, \ldots, \vecy_N) \; d\vecy_1 \cdots d\vecy_N
\end{aligned}
\end{equation}
where
\begin{equation}
\label{eqn:temp2_bosonic}
\begin{aligned}
	&G_{t,U}(\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)},
		\vecy_1, \ldots, \vecy_N) = \\
	& \bbE_0^{\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)}} \left[
		\exp\left\{
			\sum_{i<j}
				\int_0^{2\beta} U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
		\right\}
		\prod_{i=1}^N
		\delta\left(\vecw^{(i)}_{c,t}-\vecy^{(i)}\right)
	\right].
\end{aligned}
\end{equation}
\begin{proof}
Insert equation \ref{eqn:temp2_bosonic} into the right-hand side of
\ref{eqn:temp1_bosonic}, interchange expectation and integral, and integrate
out the delta function as in proposition \ref{prop:ebHG}.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{$\Tr(e^{-\beta H} M_\pi)$ using Brownian bridges}
\label{sec:fkboson_tr}

% xxx missing factors of g_t(0)?

\begin{prop}
\label{prop:TrebH_bridge_bosonic}
The trace may be computed using Brownian bridges as follows:
\begin{align*}
	\Tr(e^{-\beta H} M_\pi)
	&=
%	g_t(0) \int
%	\bbE_{0,t}^{\vecx,\vecx} \left[
%			\exp\left\{
%				-\int_0^t U(\vecw_{c,s})\,ds
%			\right\}
%		\right]
%	d\vecx.
	\int \,d\vecX
	\int
		\left(
			\prod_{k=1}^N
			d\vecWhat^{\vecx_k,\vecx_{\pi(k)}}_{0,T}(\vecw_{c,\cdot})
		\right)
		\left[
			\exp\left\{
				\sum_{i<j} \int_0^{2\beta}
					U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
			\right\}
		\right].
\end{align*}
\begin{proof}
Using proposition \ref{prop:operator_trace}, we have
$$
	\Tr(e^{-\beta H} M_\pi)
	=
	\int G_{t,U}(
		\vecx_{\pi(1)},\ldots,\vecx_{\pi(N)},
		\vecx_1,\ldots, \vecx_N)
	\,d\vecX.
$$
Equation \ref{eqn:temp2_bosonic} of proposition \ref{prop:ebHG_bosonic} gives
us an expression for $G$.  Then
\begin{align*}
	\Tr(e^{-\beta H} M_\pi)
	=
	\int
	& \bbE_0^{\vecx_{\pi(1)}, \ldots, \vecx_{\pi(N)}} \left[
		\exp\left\{
			\sum_{i<j}
				\int_0^{2\beta} U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
		\right\}
		\prod_{i=1}^N
		\delta\left(\vecw^{(i)}_{c,t}-\vecx^{(i)}\right)
	\right]
	\,d\vecX.
\end{align*}
Using proposition \ref{prop:Ex_Exy}, we may convert this expectation over
Brownian motion into an expectation over Brownian bridges to obtain
\begin{align*}
	\Tr&(e^{-\beta H} M_\pi) = \\
	&\prod_{i=1}^{N} g_{cT}\left(\vecx_i-\vecx_{\pi(i)}\right)
	\int
		\bbE_{0,t}^{\vecx_1,\vecx_{\pi(1)}; \ldots; \vecx_N,\vecx_{\pi(N)}}
		\left[
			\exp\left\{
				\sum_{i<j} \int_0^{2\beta}
					U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
			\right\}
		\right]
	\,d\vecX.
\end{align*}
Using definition \ref{defn:E_exp_U_bb}, we may write this using
$d\vecWhat$ notation:
\begin{align*}
	\Tr(e^{-\beta H} M_\pi) &=
	\int \,d\vecX
	\int
		\left(
			\prod_{k=1}^N
			d\vecWhat^{\vecx_k,\vecx_{\pi(k)}}_{0,T}(\vecw_{c,\cdot})
		\right)
		\left[
			\exp\left\{
				\sum_{i<j} \int_0^{2\beta}
					U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
			\right\}
		\right].
\end{align*}
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Sum over $\pi \in \SN$}
\label{sec:fkboson_sum}

Applying proposition \ref{prop:TrebH_bridge_bosonic}, we continue the plan laid
out in section \ref{sec:fkboson}, namely:
\begin{equation*}
\begin{aligned}
	\Trsym(e^{-\beta H})
	&= \frac{1}{N!}\;
		\sum_{\pi\in\SN} \Tr_{L^2} \left( e^{-\beta H} M_\pi \right) \\
	&= \frac{1}{N!}\;
		\int \,d\vecX
		\sum_{\pi\in\SN}
			\left[
				\prod_{k=1}^N
				\int
				d\vecWhat^{\vecx_k,\vecx_{\pi(k)}}_{0,T}(\vecw_{c,\cdot})
			\right]
			\left[
				\exp\left\{
					\sum_{i<j} \int_0^{2\beta}
						U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
				\right\}
			\right].
\end{aligned}
\end{equation*}
Notationally, we may split this up as
\begin{equation}
\label{eqn:Tr_L2_ebH}
\begin{aligned}
	\Trsym(e^{-\beta H})
	&= \frac{1}{N!}\;
		\int \,d\vecX
		\sum_{\pi\in\SN}
		e^{-H_\pi'(\vecX, \pi)} \\
	e^{-H_\pi'(\vecX, \pi)}
	&=
		\left[
			\prod_{k=1}^N
			\int
			d\vecWhat^{\vecx_k,\vecx_{\pi(k)}}_{0,T}(\vecw_{c,\cdot})
		\right]
		\left[
			\exp\left\{
				\sum_{i<j} \int_0^{2\beta}
					U(\vecw^{(i)}_{c,s} - \vecw^{(j)}_{c,s})\,ds
			\right\}
		\right].
\end{aligned}
\end{equation}
Here is the pivotal point of this paper --- the original partition function
appears as a sum over $\pi$ of an $X$-averaged quantity.  That quantity is
non-negative so we may write it as the exponential of something which we call
$H_\pi'$.  The sum over permutations of $e^{H_\pi'}$ is precisely what we would
want for a partition function involving energies, not of particles, but of
individual permutations.  The remaining steps of this section involve
separating out the non-interacting terms from the interacting terms in
$e^{H_\pi'}$, and finding an expression for the logarithm of $e^{H_\pi'}$ i.e.
$H_\pi'$.

%% ----------------------------------------------------------------
\subsection{Extraction of the non-interacting terms}
\label{sec:U0}

Let $U \equiv 0$ in equation \ref{eqn:Tr_L2_ebH}.  Then we have
\begin{align*}
	e^{-H_\pi'(\vecX, \pi)}
	&= \left[
		\prod_{k=1}^N
		\int
		d\vecWhat^{\vecx_k,\vecx_{\pi(k)}}_{0,T}(\vecw_{c,\cdot})(1)
	\right].
\end{align*}
Recall from proposition \ref{prop:dW_normalization} and notation
\ref{notn:g_t} that
$$
	\int d\vecWhat^{\vecx_k,\vecx_{\pi(k)}}_{0,T}(\vecw_{c,\cdot})(1)
	= g_{cT}(\vecx_k-\vecx_{\pi(k)})
	= \frac{1}{(2\pi c T)^{d/2}}
		\exp\left\{ -\frac{1}{2cT} \|\vecx_k-\vecx_{\pi(k)}\|^2 \right\}
$$
Then
\begin{align*}
	e^{-H_\pi'(\vecX, \pi)}
	&= \left[
		\frac{1}{(2\pi c T)^{dN/2}}
		\prod_{k=1}^N
		\exp\left\{ -\frac{1}{2cT} \|\vecx_k-\vecx_{\pi(k)}\|^2 \right\}
	\right] \\
	&= \left[
		\frac{1}{(2\pi c T)^{dN/2}}
		\exp\left\{ -\frac{1}{2cT}
			\sum_{k=1}^N
			\|\vecx_k-\vecx_{\pi(k)}\|^2 \right\}
	\right].
\end{align*}
We write
\begin{align*}
	e^{-H_\pi'(\vecX, \pi)}
	&= \frac{1}{(2\pi c T)^{dN/2}} e^{-H^{(0)}(\vecX,\pi)}
\end{align*}
where
\begin{align*}
	H^{(0)}(\vecX,\pi) &= \frac{1}{2cT} \sum_{k=1}^N
		\|\vecx_k-\vecx_{\pi(k)}\|^2.
\end{align*}
Here is the key point where we discover that the $\beta$ in a permutation
Hamiltonian is indeed reciprocated --- in contrast to our experience with
particle Hamiltonians.

%% ----------------------------------------------------------------
\subsection{$U \not\equiv 0$ \ldots}
\label{sec:Unon0}


%% ----------------------------------------------------------------
\subsection{Cluster expansion and two-jump interactions; relation to
spatial permutations}
\label{sec:cluexp}

xxx section 4 of \cite{bib:qmath}:  cluster expansion.  Two-body interactions
$\longmapsto$ many-jump interactions; cluexp for two-jump interactions.

xxx xref fwd to $N_2$ interactions.

%% ================================================================
\newpage
\section{Model with full two-jump interactions}
\label{sec:tji}

%% ================================================================
\newpage
\section{Simple model with two-cycle interactions}
\label{sec:n2}

xxx section 5 of \cite{bib:qmath}

%% ----------------------------------------------------------------
\subsection{Motivation}

xxx strongest interactions are two-cycles.  xxx closed-form formula of Daniel's
...

%% ----------------------------------------------------------------
\subsection{$\alpha$ ...}

xxx
$$
	\alpha = \left(\frac{8}{\pi\beta}\right)^{1/2} + O(a^2).
$$

%% ----------------------------------------------------------------
\subsection{Pressure ...}

pressure and chemical potential.

xxx
$$
	p^{(\alpha)}(\beta, \mu) = -\frac{1}{\beta}
		\int_{\R^3} \log\left(1-e^{-\beta(|2\pi k^2|^2-\mu)}\right) dk
	-
	\frac{e^{2\beta\mu}}{2^{11/2}\pi^{3/2}\beta^{5/2}}(1-e^{-\alpha}).
$$

xxx free energy (leg xform):
$$
	f^{(\alpha)}(\beta,\rho) = \sup_\mu\left[
		\rho \mu - p^{(\alpha)}(\beta,\mu)
		\right].
$$

%% ----------------------------------------------------------------
\subsection{Conjecture for the critical density}

Conjecture for the critical density (recall that it can be computed, but it is
not proved that it corresponds to infinite cycles).

xxx thm 2 of \cite{bib:qmath} section 5 --- ?

xxx
$$
	\rho_c^{(\alpha)} = \frac{\partial p^{(\alpha)}}{\partial\mu}
		\Bigg|_{\mu=0-}
	= \rho_c^{(0)} - \frac{1}{2^{9/2}\pi^{3/2}\beta^{3/2}} (1-e^{-\alpha}).
$$
Using
$$
	\alpha = \left(\frac{8}{\pi\beta}\right)^{1/2} a,
$$
xxx
$$
	\frac{T_c^{(\alpha)}-T_c^{(0)}}{T_c^{(0)}}
	= 0.37 \rho^{1/2}a + o(a).
$$

%% ================================================================
\newpage
\section{Future directions}

$N_2$ simulations with xref.

lattice and point-process models with xrefs back to section \ref{sec:rcm}.

Fully interacting model \ldots appears intractable.  Maybe with large-scale
computing, or a closed-form expression for $V$, or some form of virial approach
\ldots .

%% ================================================================
\newpage
\section{Acknowledgements}

Bill Faris' notes on the Feynman-Kac formula have proved invaluable, as has
Janek Wehr's spring 2008 course in stochastic differential equations.  Tom
Kennedy made it possible for me to participate in this research project, and
has been a patient and learned sounding board throughout.  Daniel Ueltschi has
shared with me his extensive knowledge and his enthusiasm, and has created an
environment of great mutual benefit.  Each of these gentlemen deserves my
gratitude; this work would not have been possible without any of them.

Additionally, much of the work here was done during the Summer School for
Current Topics in Mathematical Physics at the Erwin Schr\"odinger Institute in
July 2008.  The Institute provided a productive and fruitful atmosphere; the
University of Arizona Department of Mathematics helped make the trip possible.

%% ================================================================
\appendix
\newpage
\section{Quantum mechanics}

For information on quantum mechanics, see \cite{bib:griffiths}, \cite{bib:nc},
and \cite{bib:sakurai}; see \cite{bib:huang}, \cite{bib:widom}, and
\cite{bib:pathria} for statistical mechanics.  The topics presented here are a
digestion of these sources, with unified vocabulary and notation, serving as a
single point of reference for the main body of the paper.

%% ----------------------------------------------------------------
\subsection{Postulates of quantum mechanics}
\label{sec:postulates}

I take elementary quantum mechanics as given; in particular, I will use bra-ket
notation without explanation.  (I should point out that I take the complex
inner product to be linear on the right, in accordance with
mathematical-physics convention.) However, I will recall the basic postulates
of quantum mechanics in order to smoothly elucidate the usefulness of the
otherwise mysterious density matrix.  In turn, density matrices are needed to
formulate quantum statistical mechanics in section \ref{sec:qsm}.

\begin{itemize}

\item[(1)] To any isolated physical system is associated a separable complex
Hilbert space $\msH$, called a \emphidx{state space}, with inner product
$\braket{\phi}{\psi}$.  States (pure states; see section
\ref{sec:pure_and_mixed}) of the physical system are described by unit vectors
(often called \emphidx{state vectors} or \emphidx{wave functions}) in $\msH$,
i.e. vectors $\psi$ such that $\braket{\psi}{\psi}=1$.  Vectors that are scalar
multiples of one another are considered to be equivalent.  (If $\psi_1 =
c\psi_2$, then, since $\psi_1$ and $\psi_2$ have norm 1, $|c|=1$.  That is,
$c=e^{i\theta}$ for some real $\theta$.  Thus arises the saying that state
vectors are distinct up to a \emphidx{phase factor}.) The state space of a
composite system is the \plainidx{tensor product} of the state spaces of the
component systems.

\item[(2)] State vectors evolve in time via \emphidx{unitary transformations}.
Specifically, the unitary operator is described by the \emphidx{Schr\"odinger
equation}:
$$
	i \hbar \frac{\D\psi(t,\vecx)}{\Dt}
	= \frac{-\hbar^2}{2m} \nabla^2 \psi(t,\vecx) + V(t,\vecx) \psi(t,\vecx)
	= H \psi(t,\vecx)
$$
i.e.
$$
	\frac{\D\psi}{\Dt}
	= \frac{-i H}{\hbar} \psi
$$
where $H$ is the Hermitian operator
$$
	H = \frac{-\hbar^2}{2m} \nabla^2 + V.
$$
The wave function varies in position and time; the \emphidx{potential
function}, conventionally written with the letter $V$, may also vary in
position and time.  The potential represents external interactions with the
particle.

We can write down a solution
\begin{align}
\label{eqn:SE}
	\frac{d\psi}{dt} &= \frac{-i}{\hbar} \, H \psi \\
	\psi(t) &= e^{-iHt/\hbar} \psi(0).
\end{align}
One checks this solution by differentiating:
\begin{align*}
	\frac{d\psi}{dt} &= \frac{-iH}{\hbar} \, e^{-iHt/\hbar} \psi(0)
	= \frac{-iH}{\hbar} \, \psi(t)
	= \frac{-i}{\hbar} \, H \psi(t).
\end{align*}
As is well known, the imaginary exponential of a hermitian matrix is
\plainidx{unitary}.  The importance of this is that unitary matrices $U$ are
\plainidx{norm-preserving}.  That is, if $\|\psi\|=1$ then $\|U\psi\|$ is still
1; probability functions remain probability functions as they evolve in time.
This fact is called the \emphidx{conservation of probability}.

%Real spectrum.  Existence of orthonormal basis of eigenvectors.
%Why they \emph{must} be orthonormal.

\item[(3)] An observable $\mcO$ corresponds to a densely defined Hermitian
operator $A:\msH\to \msH$.
Since $A$ is Hermitian, $\msH$ has an orthonormal basis $\{\phi_j\}$ of
eigenvectors of $A$.  Since $\msH$ is separable, the basis is a countable set.
In the case that $A$ has discrete spectrum, the eigenstates of $A$ have
respective (necessarily real) eigenvalues $a_j$ and $A$ has a
\emphidx{spectral decomposition}
$$
	A = \sum a_j \ketbra{\phi_j}
$$
(The operators $P_j = \ketbra{\phi_j}$ are known as
\emphidx{projection operators}.) A (necessarily square-summable) wave function
$\psi$ is a linear combination of the eigenbasis:
$$
	\psi = \sum_{j} c_j \phi_j
	\qqwith
	c_j = \braket{\phi_j}{\psi}
	\qqand
	\sum_j |c_j|^2 = 1.
$$
A measurement of $\mcO$ has as its possible outcomes the numbers $a_j$ with
probabilities
$$
	\bbP(\mcO=a_j) = |\braket{\phi_j}{\psi}|^2.
$$
The expected value of $A$ is
\begin{align*}
	\expn{A} &= \sum_j a_j \bbP(\mcO=a_j)
	= \sum_j a_j |\braket{\phi_j}{\psi}|^2
	= \sum_j a_j \braket{\psi}{\phi_j} \braket{\phi_j}{\psi} \\
	&= \sum_j a_j
		\bra{\psi}\; \bigg(\ketbra{\phi_j}\bigg) \; \ket{\psi}
	= \sum_j \bra{\psi}\;
		\bigg(a_j \ketbra{\phi_j}\bigg)
		\; \ket{\psi} \\
	&= \braCket{\psi}{A}{\psi}.
\end{align*}
This is the \plainidx{average value} or \plainidx{mean} of all the
observations, averaged over repeated measurements on identical systems.

\item[(4)] Immediately after measurement of an observable $\mcO$, the state of
the system is described \emph{only} by an eigenstate $\phi_j$ of $A$.  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{itemize}

%% ----------------------------------------------------------------
\subsection{Pure and mixed states}
\label{sec:pure_and_mixed}

The terms \emphidx{pure state} and \emphidx{mixed state} are horribly
misleading.  Nonetheless, they are entrenched in the literature and cannot be
dispensed with.  A pure state is simply a state of a quantum-mechanical system:
the adjective \emph{pure} is superfluous.  A mixed state not a single state at
all:  it is a statistical description of a \emph{set} of many states.

\begin{defn}
An \bfidx{ensemble} is a list of $n$ distict state vectors $\{\psi_1, \ldots,
\psi_n\}$ along with respective probabilities $\{p_1, \ldots, p_n\}$ with $0
\le p_j \le 1$ and $\sum_{j=1}^n p_j = 1$.  (One may think of an ensemble as a
probability mass function.)
\end{defn}

\begin{defn}
A \bfidx{mixed state} is an ensemble.  A \bfidx{pure state} is (with abuse of
notation) either a single state, or an ensemble with $n=1$, or an ensemble with
$n \ge 1$ but only one $p_k=1$ and the remaining $p_j=0$ for $j\ne k$.
\end{defn}

\begin{rem*}
Pure and mixed states are not to be confused with the following:
\begin{itemize}
\item A quantum-mechanical state is a linear combination of eigenstates.
	Such states are pure, not mixed.
\item Entangled and separable states arised in the theory of multi-particle
	systems, e.g. the tensor product $\mcH_1 \tensor \mcH_2$ of Hilbert
	spaces $\mcH_1$ and $\mcH_2$.  An entangled state is an indecomposable
	tensor in $\mcH_1 \tensor \mcH_2$; a separable state is a decomposable
	tensor.
\end{itemize}
\end{rem*}

%% ----------------------------------------------------------------
\subsection{Density matrices}

Density matrices are a notational convenience for working with ensembles (mixed
states) of quantum systems.  I will develop\footnote{This development follows,
in part, that in \texttt{http://electron6.phys.utk.edu/QM1}, retrieved on June
12, 2008.} density matrices in three steps:  (1)  In this section, I will show
how to construct a density matrix from an ensemble.  (2) In section
\ref{sec:pure_dm}, I will show how density matrices serve to notationally
encapsulate the postulates of quantum mechanics as described in section
\ref{sec:postulates}.  (3) In section \ref{sec:mixed_dm} I will show how this
notation permits one to extend those postulates to the mixed-state case.

\begin{defn}
A \bfidx{density matrix} is a positive-definite matrix with trace 1.
\end{defn}

\begin{prop}
\label{prop:rho_from_ens}
A density matrix may be obtained from an ensemble by
$$
	\rho = \sum_{j=1}^n p_j \ketbra{\psi_j}.
$$
\begin{proof}
Let $\rho$ be such a matrix.
Fix an orthonormal basis $\{\phi_i\}$ for $\msH$.
I will use the characterization that a matrix $A$ is positive-definite iff it
is Hermitian and $\braCket\xi A \xi > 0$ for all non-zero $\xi$.  To see that
$\rho$ is Hermitian, write
\begin{align*}
	\rho^* &= \left(\sum_{j=1}^n p_j \ketbra{\psi_j}\right)^*
	= \sum_{j=1}^n p_j \left(\ketbra{\psi_j} \right)^*
	= \sum_{j=1}^n p_j \bra{\psi_j}^* \ket{\psi_j}^*
	= \sum_{j=1}^n p_j \ketbra{\psi_j}
	= \rho.
\end{align*}
Now let $0 \ne \xi \in \msH$.  Then
\begin{align*}
	\braCket \xi \rho \xi
	&= \BraCKet
			{\xi}
			{\sum_{j=1}^n p_j \ketbra{\psi_j}}
			{\xi}
	= \sum_{j=1}^n p_j \braket{\xi}{\psi_j} \braket{\psi_j}{\xi}
	= \sum_{j=1}^n p_j \braket{\xi}{\psi_j} \braket{\xi}{\psi_j}^*
	= \sum_{j=1}^n p_j | \braket{\xi}{\psi_j} |^2.
\end{align*}
Since this is a sum of non-negative real numbers, it is non-negative.  But by
positive-definiteness of the inner product, it is zero iff either $\xi$ is
zero, which it is not by hypothesis, or all of the $\psi_j$'s are zero, which
they are not since they all have norm 1.

To see that $\rho$ has trace 1, write
\begin{align}
	\Tr(\rho) &= \Tr\left(\sum_{k=1}^n p_k \ketbra{\psi_k}\right) \\
	&= \sum_{k=1}^n p_k \Tr\left(\ketbra{\psi_k}\right).
\end{align}
Write each $\psi_k$ in terms of the orthonormal basis as
$$
	\psi_k = \sum_\ell c_{k\ell} \phi_\ell.
$$
Note that $\ketbra{\psi_k}$ has $ij$th matrix element
$$
	\left( \ketbra{\psi_k} \right)_{ij} = c_{ki}\,c_{kj}^*.
$$
Then
$$
	\Tr\left( \ketbra{\psi_k} \right) =
		\sum_i c_{ki}\,c_{ki}^* =
		\sum_i | c_{ki} |^2 = 1,
$$
from which
$$
	\Tr(\rho) = \sum_{k=1}^n p_k = 1.
$$
\end{proof}
\end{prop}

Furthermore, we have a criterion for whether a density matrix corresponds
to a pure or mixed state.

\begin{prop}
If $\rho$ is a density matrix for an ensemble $\{\psi_1, \ldots, \psi_n\}$,
then $\Tr(\rho^2) \le 1$, with equality if and only if the ensemble is a pure
state.
\begin{proof}
First note that for a matrix $A$, we have
$$
	\Tr(A^2) =
		\sum_i
		\sum_j
		A_{ij}
		A_{ji}.
$$
Then
\begin{align*}
	\Tr(\rho^2) &=
		\sum_i
		\sum_j
		\rho_{ij}
		\rho_{ji} \\
	&=
		\sum_i
		\sum_j
		\left(
		\sum_{k=1}^n p_k
			c_{ki}
			c_{kj}^*
		\right)
		\left(
		\sum_{\ell=1}^n p_\ell
			c_{\ell j}
			c_{\ell i}^*
		\right)
		\\
	&=
		\sum_{k=1}^n p_k
		\sum_{\ell=1}^n p_\ell \;
		\sum_i
		\left(
			c_{\ell i}^*
			c_{ki}
		\right)
		\sum_j
		\left(
			c_{kj}^*
			c_{\ell j}
		\right)
		\\
	&=
		\sum_{k=1}^n p_k
		\sum_{\ell=1}^n p_\ell \;
		\braket{\psi_\ell}{\psi_k}
		\braket{\psi_k}{\psi_\ell}
		\\
	&=
		\sum_{k=1}^n p_k
		\sum_{\ell=1}^n p_\ell \;
		\braket{\psi_k}{\psi_\ell}^*
		\braket{\psi_k}{\psi_\ell}
		\\
	&=
		\sum_{k=1}^n p_k
		\sum_{\ell=1}^n p_\ell \;
		| \braket{\psi_k}{\psi_\ell} |^2.
\end{align*}
First suppose that one $p_{k_0}$ is 1 and the rest are zero.  (This is the pure
case.)  Then
\begin{align*}
	\Tr(\rho^2) &=
		| \braket{\psi_{k_0}}{\psi_{k_0}} |^2 = 1.
\end{align*}
Now suppose that at least two $p_k$'s are non-zero.  Decompose the sum as
\begin{align*}
	\Tr(\rho^2) &=
		\sum_{k=1}^n p_k^2
		| \braket{\psi_k}{\psi_k} |^2
	+
		\sum_{k=1}^n p_k
		\sum_{\ell \ne k} p_\ell \;
		| \braket{\psi_k}{\psi_\ell} |^2 \\
	&=
		\sum_{k=1}^n p_k^2
	+
		\sum_{k=1}^n
		\sum_{\ell \ne k}
		p_k
		p_\ell \;
		| \braket{\psi_k}{\psi_\ell} |^2.
\end{align*}
To finish the proof, note that $\sum_{k=1}^n p_k^2$ is the diagonal part of
$(\sum_{k=1}^n p_k)^2$.  We have
\begin{align*}
	1 = (1)^2 &= \left(\sum_{k=1}^n p_k\right)^2
	= \sum_{k=1}^n \sum_{\ell=1}^n
		p_k p_\ell
	= \sum_{k=1}^n \sum_{\ell \ne k}
		p_k p_\ell
		+
		\sum_{k=1}^n
		p_k^2
	\\
	\sum_{k=1}^n p_k^2 &= 1 - \sum_{k=1}^n \sum_{\ell \ne k} p_k p_\ell.
\end{align*}
Then
\begin{align*}
	\Tr(\rho^2) &=
	1 - \sum_{k=1}^n \sum_{\ell \ne k} p_k p_\ell
		(1 - \cos^2\theta_{k\ell})
\end{align*}
where
$$
	\braket{\psi_k}{\psi_\ell} = \|\psi_k\| \;\|\psi_\ell\|
		\;\cos\theta_{k\ell}.
$$
Note that all the terms in the sum are non-negative.  If two $p_k$'s are
non-zero --- without loss of generality, say $p_1$ and $p_2$ --- then the
product $p_1p_2$ is non-zero.  Furthermore, $\cos^2\theta_{1,2}$ is strictly
less than 1.  Otherwise, $\phi_1$ and $\phi_2$ (which are unit vectors)
would be identical up to global phase, contradicting the distinctness of the
members of the ensemble.  Thus, $\Tr(\rho^2) < 1$.
\end{proof}
\end{prop}

\begin{rem}
Two different ensembles can give the same density matrix.  In \cite{bib:nc} is
a theorem characterizing the conditions under which this can happen.  [xxx type
this up and include an example.]
\end{rem}

Lastly, we have two lemmas which will be used in section \ref{sec:pure_dm}.

\begin{lem}
\label{lem:rho_herm_idempotent}
The operator $\rho$ is Hermitian.  If $\rho=\ket\psi$,
then $\rho$ satisfies $\rho^2=\rho$.
\begin{proof}
For the first claim,
$$
	\rho^* =
	\left(\sum_k p_k \ketbra{\psi_k} \right)^* =
	\sum_k p_k \bra{\psi_k}^* \ket{\psi_k}^* =
	\sum_k p_k \ketbra{\psi_k} = \rho.
$$
For the second claim,
\begin{align*}
	\rho^2 = \ket\psi \braket\psi\psi \ket\psi = \ket\psi \bra\psi = \rho
\end{align*}
since $\braket \psi\psi = 1$.
\end{proof}
\end{lem}

\begin{lem}
\label{lem:expn_trace}
For a Hermitian operator $A$,
$$
	\braCket{\psi}{A}{\psi} = \Tr(A \ket\psi \bra\psi).
$$
\begin{proof}
The operator $\rho=\ketbra\psi$ is Hermitian by lemma
\ref{lem:rho_herm_idempotent}, so it makes sense to write $\braCket \psi A
\psi$.  Expand $\psi$ in terms of an orthonormal basis $\{\phi_j\}$ for $\msH$
as
$$
	\psi = \sum_j c_j \phi_j.
$$
Recall that
$$
	\braCket \psi A \psi = \sum_{ik} c_i^* A_{ik} c_k.
$$
On the other hand, for matrices $C$ and $D$,
\begin{align*}
	(CD)_{ij} &= \sum_k C_{ik} D_{kj}
	\qqand
	\Tr(CD)   = \sum_{ik} C_{ik} D_{ki},
\end{align*}
so
\begin{align*}
	\Tr(A \ketbra\psi) &= \sum_{ik} A_{ik} c_k c_i^*
	= \braCket \psi A \psi.
\end{align*}
\end{proof}
\end{lem}

%% ----------------------------------------------------------------
\subsection{Pure-state density matrices}
\label{sec:pure_dm}

For this section, fix a state $\psi$ and let
$$
	\rho = \ketbra\psi.
$$
Here, four key properties of the density matrix $\rho$ are described,
paralleling the four postulates of quantum mechanics listed in section
\ref{sec:postulates}.

\begin{itemize}

\item[(1)]  The state vector $\psi$ has norm 1.  The density matrix $\rho$
has trace 1, as proved in proposition \ref{prop:rho_from_ens}.

\item[(2)]  The state vector $\psi$ evolves in time according to
the \plainidx{Schr\"odinger equation}
$$
	\frac{d\psi}{dt} = \frac{-i}{\hbar} \, H \psi;
$$
the density matrix $\rho$ evolves in time according to
\begin{align}
\label{eqn:pure_state_drho_dt}
	\frac{d\rho}{dt} = \frac{-i}{\hbar} \, [H, \rho].
\end{align}

\begin{proof}
Using the product rule,
\begin{align*}
	\frac{d\rho}{dt}
		&= \frac{d\ket\psi}{dt} \bra\psi + \ket\psi \frac{d\bra\psi}{dt}.
\end{align*}
To find $d\bra\psi/dt$, take the adjoint of the Schr\"odinger equation and
recall that the Hamiltonian is Hermitian (so $H^*=H$):
\begin{align*}
	\left(\frac{d\ket\psi}{dt}\right)^*
		&= \left(\frac{-i}{\hbar} H \ket\psi \right)^* \\
	\frac{d\bra\psi}{dt}
		&= \frac{i}{\hbar} \bra\psi H^*
		= \frac{i}{\hbar} \bra\psi H.
\end{align*}
Then
\begin{align*}
	\frac{d\rho}{dt}
	&= \left(\frac{-i}{\hbar}\,H\ket\psi \right) \bra\psi
		+ \ket\psi \left( \frac{i}{\hbar}\,\bra\psi H \right)
	= \frac{-i}{\hbar}\,H \rho + \frac{i}{\hbar} \rho H
	= \frac{-i}{\hbar}\, [H,\rho].
\end{align*}
\end{proof}

In section \ref{sec:mixed_dm} we will find a solution for this equation.

\item[(3)]  Given an observable $\mcO$ with operator $A$, eigenstates
$\phi_j$, and corresponding eigenvalues $a_j$, we have
$$
	\bbP(\mcO = a_j) = |\braket{\phi_j}{\psi}|^2
	\qqand
	\expn{A} = \braCket{\psi}{A}{\psi};
$$
in terms of $\rho$, we have
$$
	\bbP(\mcO = a_j) = \Tr(\rho P_j)
	\qqand
	\expn{A} = \Tr(\rho A)
$$
where
$$
	P_j = \ketbra{\phi_j}.
$$

\begin{proof}
For the first claim,
\begin{align*}
	\bbP(\mcO = a_j) &= |\braket{\phi_j}{\psi}|^2
	= \braket{\psi}{\phi_j} \braket{\phi_j}{\psi}
	= \braCket{\psi}{P_j}{\psi}
	= \Tr(\rho P_j),
\end{align*}
where the last step follows from the centrality of trace, along with
lemma \ref{lem:expn_trace}.  The second claim also follows from the lemma.
\end{proof}

\item[(4)]  After measurement of the observable $\mcO$ with operator $A$, the
state vector $\psi$ is replaced by an eigenstate $\phi_j$.  The density matrix
$\ketbra\psi$ is replaced by $\ket{\phi_j}\bra{\phi_j}$.

\end{itemize}

%% ----------------------------------------------------------------
\subsection{Mixed-state density matrices}
\label{sec:mixed_dm}

Now we suppose
$$
	\rho = \sum_k p_k \ketbra{\psi_k},
$$
and we enumerate the postulates of quantum mechanics in terms of the density
matrix $\rho$.

\begin{itemize}

\item[(1)]  The density matrix $\rho$ has trace 1, as proved in proposition
\ref{prop:rho_from_ens}.

\item[(2)]  The density matrix $\rho$ evolves in time according to
\begin{align}
\label{eqn:mixed_state_drho_dt}
	\frac{d\rho}{dt} = \frac{-i}{\hbar} \, [H, \rho].
\end{align}

\begin{proof}
Using the product rule,
\begin{align*}
	\frac{d\rho}{dt}
		&= \sum_k p_k \frac{d\ket{\psi_k}}{dt} \bra{\psi_k}
		+ \sum_k p_k \ket{\psi_k} \frac{d\bra{\psi_k}}{dt}.
\end{align*}
As in section \ref{sec:pure_dm},
$$
	\left(\frac{d\ket{\psi_k}}{dt}\right)^* = \frac{i}{\hbar} \bra{\psi_k} H,
$$
so
\begin{align*}
	\frac{d\rho}{dt}
	&= \sum_k p_k \left(\frac{-i}{\hbar}\,H\ket{\psi_k} \right) \bra{\psi_k}
	+ \sum_k p_k \ket{\psi_k} \left( \frac{i}{\hbar}\,\bra{\psi_k} H \right) \\
%
	&= \frac{-i}{\hbar} \sum_k H p_k \ketbra{\psi_k}
	+ \frac{ i}{\hbar} \sum_k p_k \ketbra{\psi_k} H \\
%
	&= \frac{-i}{\hbar}\,H \rho + \frac{i}{\hbar} \rho H
	= \frac{-i}{\hbar}\, [H,\rho].
\end{align*}
\end{proof}

Furthermore, if wave functions evolve by a unitary operator $U_t$ via
$$
	\psi_t = U_t \psi_0,
$$
then the density matrix evolves via
$$
	\rho_t = U_t \rho_0 U_t^*.
$$
In particular, if the Hamiltonian is time-independent, we have
\begin{align}
\label{eqn:rho_evolution}
	\rho_t = e^{-iHt/\hbar} \rho_0 e^{iHt/\hbar}.
\end{align}
One verifies by differentiation that \ref{eqn:rho_evolution} solves
\ref{eqn:mixed_state_drho_dt}.

\item[(3)]  Given an observable $\mcO$ with operator $A$, eigenstates
$\phi_j$, and corresponding eigenvalues $a_j$,
$$
	\bbP(\mcO = a_j) = \Tr(\rho P_j)
	\qqand
	\expn{A} = \Tr(\rho A)
$$
where
$$
	P_j = \ketbra{\phi_j}.
$$

\begin{proof}
For the first claim, let $\mcO_k$ be the observation of $\psi_k$.  Using
conditional probability, recalling result 3 from section \ref{sec:pure_dm}, and
using linearity of trace, we have
\begin{align*}
	\bbP(\mcO = a_j) &=
		\sum_k p_k \, \bbP(\mcO_k = a_j) \\
	&= \sum_k p_k \, \Tr\left( \ketbra{\psi_k} P_j \right) \\
	&= \Tr\left( \sum_k p_k \, \ketbra{\psi_k} P_j\right) \\
	&= \Tr\left( \left(\sum_k p_k \, \ketbra{\psi_k} \right)
		P_j\right) \\
	&= \Tr(\rho P_j).
\end{align*}
Similarly, for the second claim we have
\begin{align*}
	\expn{A} &= \sum_k p_k \expn{A_k} \\
	&= \sum_k p_k \Tr\left( \ketbra{\psi_k} A \right) \\
	&= \Tr\left( \sum_k p_k \ketbra{\psi_k} A \right) \\
	&= \Tr(\rho A).
\end{align*}
\end{proof}

\item[(4)]  After measurement of the observable $\mcO$ with operator $A$, the
density matrix $\rho$ is replaced by
$$
	\sum_j P_j \rho P_j.
$$
\begin{proof}
First,
\begin{align*}
	\sum_j P_j \rho P_j
	&= \sum_j
		\left(\ketbra{\phi_j}\right)
		\left(\sum_k p_k \ketbra{\psi_k}\right)
		\left(\ketbra{\phi_j}\right) \\
	&= \sum_{jk}
		p_k
		\ket{\phi_j}
		\braket{\phi_j}{\psi_k}
		\braket{\psi}{\phi_j}
		\bra{\phi_j} \\
	&= \sum_{jk}
		p_k
		\ket{\phi_j}
		| \braket{\phi_j}{\psi} |^2
		\bra{\phi_j} \\
	&= \sum_{jk}
		p_k
		|c_{kj}|^2
		\ketbra{\phi_j}.
\end{align*}
On the other hand, from result 4 in section \ref{sec:pure_dm} we know that each
state $\psi_k$ collapses to eigenstate $\phi_j$ with probability
$$
	|c_{kj}|^2 = |\braket{\phi_j}{\phi_k}|^2.
$$
Thus, again using conditional probability, the post-measurement ensemble should
be described by pure-state density matrices $\ketbra{\phi_j}$ with
probabilities $p_k|c_{kj}|^2$, i.e.
$$
	\sum_{jk} p_k |c_{kj}|^2 \ketbra{\phi_j} = \sum_j P_j \rho P_j.
$$
\end{proof}
\end{itemize}

%% ----------------------------------------------------------------
\subsection{Additional properties of density matrices}

\begin{lem}
\label{lem:sum_ketbra_phi_j_is_I}
If $\{\phi_j\}$ is an orthornomal basis for $\msH$, then
$$
	\sum_j \ketbra{\phi_j} = I.
$$
\begin{proof}
Take an arbitrary vector $\psi \in \msH$.  Expanded in terms of the basis,
this is
$$
	\psi = \sum_j c_j \phi_j
$$
where
$$
	c_j = \braket{\phi_j}{\psi}.
$$
Then
\begin{align*}
	\left( \sum_j \ketbra{\phi_j} \right) \ket\psi
	= \sum_j \ket{\phi_j} \braket{\phi_j}{\psi}
	= \sum_j \ket{\phi_j} c_j
	= \psi = I \psi.
\end{align*}
\end{proof}
\end{lem}

\begin{lem}
\label{lem:ebH_eig}
Eigenvectors of a self-adjoint operator $H$ are eigenvectors of $e^{-\beta H}$.
That is, if $H\phi_j = E_j\phi_j$ then
$$
	e^{-\beta H} \phi_j = e^{-\beta E_j} \phi_j.
$$
\begin{proof}
This may be justified formally by using the Taylor expansion for the
operator exponential:
\begin{align*}
	e^{-\beta H} \phi_j
	&= \sum_{k=0}^\infty \frac{(-\beta)^k H^k \phi_j}{k!}
	= \sum_{k=0}^\infty \frac{(-\beta)^k E_j^k \phi_j}{k!}
	= e^{-\beta E_j} \phi_j.
\end{align*}
\end{proof}
\end{lem}

%% ----------------------------------------------------------------
\subsection{Trace in coordinates}

With respect to a basis $\{\phi_i\}$, $A$ has matrix element $A_{ij}$
$$
	A_{ij} = \braCket{\phi_i}{A}{\phi_j}.
$$
Then
\begin{align}
\label{eqn:Tr_bracket}
	\Tr(A) &= \sum_i A_{ii}
	= \sum_i \braCket{\phi_i}{A}{\phi_i}.
\end{align}
If $\{\phi_i\}$ is an orthonormal eigenbasis for $A$, with respective
eigenvalues $\{\lambda_i\}$, then
\begin{align*}
	\Tr(A)
	= \sum_i \braCket{\phi_i}{\lambda_i}{\phi_i}
	= \sum_i \lambda_i \braket{\phi_i}{\phi_i}
	= \sum_i \lambda_i.
\end{align*}
Likewise, using lemma \ref{lem:ebH_eig},
\begin{align*}
	\TrebH
	= \sum_i \braCket{\phi_i}{e^{-\beta H}}{\phi_i}
	= \sum_i \braCket{\phi_i}{e^{-\beta E_i}}{\phi_i}
	= \sum_i e^{-\beta E_i} \braket{\phi_i}{\phi_i}
	= \sum_i e^{-\beta E_i}.
\end{align*}

%% ================================================================
\newpage
\section{Analysis results}

The results here are standard --- see, for example, \cite{bib:HN}.  They are
collected here for ready reference, using notation which is consistent with the
rest of the paper.

%% ----------------------------------------------------------------
\subsection{Gaussians}

\begin{notn}
\label{notn:g_t}
For $\vecx \in \R^d$, let
$$
	g_t(\vecx) = \frac{1}{(2\pi t)^{d/2}}\, e^{-|\vecx|^2/2\beta}.
$$
Then $g_t(\vecx-\veca)$ is a Gaussian, with integral 1, having mean $\veca$ and
variance $t$ in each component.
\end{notn}

\begin{prop}
The family of $g_t$'s satisfies a \plainidx{Chapman-Kolmogorov equation}:
\label{prop:g_iter_2}
$$
	\int_{\R^d} g_s(\vecx-\vecz)g_t(\vecz-\vecy)\,d\vecz = g_{s+t}(\vecx-\vecy).
$$
\begin{proof}
This follows (albeit messily) by completing the square.
\end{proof}
\end{prop}

\begin{prop}
\label{prop:g_iter}
\begin{align*}
	\int_{\R^d} \cdots \int_{\R^d}
	&
	g_{t_1}(\vecx-\vecz_1)
	g_{t_2}(\vecz_1-\vecz_2)
	\cdots
	g_{t_{n-1}}(\vecz_{n-2}-\vecz_{n-1})
	g_{t_{n}}  (\vecz_{n-1}-\vecy)
	\\
%
	&
	\,d\vecz_1
	\,d\vecz_2
	\cdots
	\,d\vecz_{n-2}
	\,d\vecz_{n-1} \\
%
	&
	= g_{t_1 + \cdots + t_n}(\vecx-\vecy).
\end{align*}
\begin{proof}
Use induction, iterating the integrals and applying proposition
\ref{prop:g_iter_2}.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{The diffusion equation}
\label{sec:diffusion_equation_exp}

We begin to solve the \plainidx{diffusion equation}, namely,
$$
	\frac{\D\psi(t,\vecx)}{\Dt}
	= c \nabla^2 \psi(t,\vecx) + U(\vecx)\psi(t,\vecx).
$$
Note that $c=1/2$ corresponds to the forward diffusion equation (equation
\ref{eqn:forward_diffusion}).

%Note that $c=1/2$ corresponds to the forward diffusion equation
%and $c=-1/2$ corresponds to the backward diffusion equation (equations
%\ref{eqn:forward_diffusion} and \ref{eqn:backward_diffusion}, respectively).

Solving the diffusion equation seems an odd thing to do --- this is not quite
the Schr\"odinger equation, where the left-hand side has $i\hbar$ prepended,
and where the solution uses an operator of the form $e^{-iHt}$ rather than
$e^{Ht}$.  Our interest in the diffusion equation arises because (1) the PDF
of Brownian motion evolves according to the diffusion equation, as will be seen
in section \ref{sec:bm}, and (2) computation of the quantum-mechanical
partition function $\TrebH$ (sections xxx, xxx, and xxx) also involves an
exponentiated operator without the $i$.

\begin{prop}
\label{prop:diffusion_exp}
The solution to
$$
	\frac{\D\psi(t,\vecx)}{\Dt} = c \nabla^2 \psi(t,\vecx)
		+ U(\vecx) \psi(t,\vecx)
$$
is
$$
	\psi(t, \vecx) = e^{(c\nabla^2 + U(\vecx))t} \psi(0, \vecx).
$$
\begin{proof}
Differentiate:
\begin{align*}
	\frac{\D\psi(t,\vecx)}{\Dt}
	&= (c\nabla^2 \, + U(\vecx)) e^{(c\nabla^2 + U(\vecx)) t} \psi(0,\vecx)
	= c\nabla^2 \, \psi(t,\vecx).
\end{align*}
\end{proof}
\end{prop}

It remains to interpret the exponentiated operator $e^{(c\nabla^2 +
U(\vecx))t}$.  This is done in section [xxx] for the $U=0$ case, and in section
[xxx] for the general case.

%% ----------------------------------------------------------------
\subsection{Operator trace}
\label{sec:operator_trace}

As described in section \ref{sec:canon_ens_rho}, we will want to compute
$\TrebH$.  Here is a partial result, assuming that $e^{-\beta H}$
can be written as an integral operator.  The construction of that integral
operator will need to wait until section [xxx].

\begin{prop}
\label{prop:operator_trace}
If an operator $A$ has a $G(\vecx, \vecy)$ such that
$$
	A\, f(\vecx) = \int G(\vecx,\vecy) f(\vecy) \,d\vecy,
$$
then
$$
	\Tr(A) = \int G(\vecx, \vecx) \,d\vecx.
$$
\begin{proof}
As shown in equation \ref{eqn:Tr_bracket}, with $\{\phi_j\}$ being a
(countable) basis for the (separable) Hilbert space $\msH$,
$$
	\Tr(A) = \sum_j \braCket{\phi_j}{A}{\phi_j}.
$$
Then
\begin{align*}
	\Tr(A)
	&= \sum_j \braCket{\phi_j}{A}{\phi_j} \\
	%&= \sum_j \braket{\phi_j} {\int G(\vecx, \vecy) \,\phi_j(\vecy)\,d\vecy} \\
	&= \sum_j \int \int
		\phi_j^*(\vecx) \,G(\vecx, \vecy) \,\phi_j(\vecy)\,d\vecy\,d\vecx \\
	&= \int \int \,G(\vecx, \vecy)
		\left(\sum_j \phi_j^*(\vecx) \,\phi_j(\vecy)\right) \,d\vecy\,d\vecx \\
	&= \int \int \,G(\vecx, \vecy)
		\delta(\vecx-\vecy) \,d\vecy\,d\vecx \\
	&= \int \,G(\vecx, \vecx) \,d\vecx.
\end{align*}
\end{proof}
\end{prop}

xxx prove the completeness relation.

\begin{rem}
Note that $G(\vecx,\vecy)$ only appears in an integral, and thus may be a
distribution rather than a function.  This will in fact be the case in section
\ref{sec:fk1}, where the $G(\vecx, \vecy)$ we construct will contain a
delta-function term.
\end{rem}

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

\begin{notn}
Write
$$
	\psihat(\veck) = (\mcF\psi)(\vecx)
	\qqor
	\psihat = \mcF(\psi)
$$
for the Fourier transform of $\psi$.
\end{notn}

\begin{defn}
$$
	\ghat_t(\veck) = e^{-|\veck|^2 t}.
$$
\end{defn}

\begin{prop}
$$
	\mcF(g_t)(\vecx) = \ghat_t(\veck).
$$
\begin{proof}
xxx prove it.
\end{proof}
\end{prop}

\begin{rem}
xxx $t$ vs. $1/t$; reciprocal bell widths.
\end{rem}

\begin{defn}
\label{def:operator_transform}
The \bfidx{Fourier transform} or \bfidx{operator transform} of an operator $A$
is
$$
	\Ahat = \mcF A \mcF^{-1}.
$$
Note this this is equivalent to
$$
	\Ahat \mcF = \mcF A.
$$
\end{defn}

That is, this is the frequency-domain operator which takes a $\veck$ function,
back-transforms to an $\vecx$ function, applies $A$, and forward-transforms:

\setlength{\unitlength}{1.0in}
\begin{picture}(6.5,1.2)
%\put(0.0,0.0){\framebox(6.5,1.2)}

\put(1.00,0.50){Spatial domain}
\put(5.00,0.50){Frequency domain}
\put(2.70,0.50){$A$}
\put(4.20,0.50){$\Ahat$}
\put(3.40,1.05){$\mcF^{-1}$}
\put(3.40,0.20){$\mcF$}

\put(3.00,0.90){\vector(0,-1){.75}}
\put(4.00,0.90){\vector(0,-1){.75}}

\put(3.90,0.95){\vector(-1,0){.80}}
\put(3.10,0.10){\vector(1,0){.80}}

\end{picture}

\begin{lem}
\label{lem:FT_of_deriv}
$$
	\mcF\left( \frac{d^n \psi(x)}{dx^n} \right)
	= (ik)^n \hat{\psi}(k).
$$
\begin{proof}
Integrate by parts.
\end{proof}
\end{lem}

xxx do that for $\R^d$.

\begin{prop}
The Fourier transform of the scaled Laplacian is a multiplication operator:
$$
	(\mcF c\nabla^2)(\psi(\vecx))
	= -c |\veck|^2 \psihat(\veck).
$$
\begin{proof}
Apply the lemma with $n=2$.  xxx do some algebra here if necessary.
$$
	(\mcF c\nabla^2)(\psi(\vecx))
	= \mcF\left(c \nabla^2 \psi(\vecx)\right)
	= \mcF\left(c \sum_{j=1}^d \frac{\D^2\psi(\vecx)}{\Dx_j^2}\right)
	%= c \sum_{j=1}^d (-k_j^2 \psihat(\veck))
	= -c k^2 \mcF(\psi(\vecx))
	= -c k^2 \psihat(\veck).
$$
\end{proof}
\end{prop}

\begin{prop}
The Fourier transform of the exponentiated scaled Laplacian is a
convolution operator:
$$
	\mcF\left(e^{c\nabla^2 t} \right) f = g_{ct} * f.
$$
\begin{proof}
For brevity, write
$$
	A_c = e^{c\nabla^2 t}.
$$
Applying linearity of the Fourier transform to the power-series expansion
of $e^{c\nabla^2 t}$, and assuming convergence, one has
$$
	\Ahat_c
	= \mcF(A_c)
	= \mcF\left(\sum_{j=0}^\infty \frac{(c\nabla^2 t)^j}{j!}\right)
	= \sum_{j=0}^\infty \frac{\mcF((c\nabla^2)^j) t^j}{j!}
	= \ldots
	= e^{-ck^2 t}.
$$
xxx fix me by using convolution on powers of $c\nabla^2$.

xxx track down the scale factor.

Since $\ghat_{ct}(k) = e^{-ck^2 t}$, we have
$$
	\Ahat_c(\fhat) = \ghat_{ct} \fhat.
$$
Then
\begin{align*}
	\Ahat_c(\fhat) &= \ghat_{ct} \fhat \\
	\Ahat_c \mcF f &= \ghat_{ct} \fhat \\
	\mcF A_c f &= \ghat_{ct} \fhat \\
	\mcF A_c f &= \mcF(g_{ct} * f) \\
	A_c f &= g_{ct} * f \\
	e^{c\nabla^2 t}f &= g_{ct} * f.
\end{align*}
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{The diffusion equation, continued}
\label{sec:diffusion_equation}

Following \cite{bib:faris}, we show how to solve the diffusion equation
using Fourier transforms and Gaussian functions.

\begin{prop}
\label{prop:diffusion_convolution}
The exponentiated Laplacian may be written as an integral operator:
$$
	e^{c\nabla^2 t} f(\vecx) = \int_{\R^d} g_{ct}(\vecx-\vecy) f(\vecy)\,d\vecy.
$$

\begin{proof}
xxx cite above.
\end{proof}
\end{prop}

\begin{cor}
The solution to
$$
	\frac{\D\psi(t,\vecx)}{\Dt}
	= c \nabla^2 \psi(t,\vecx)
$$
is
$$
	\psi(t, \vecx)
	= e^{c\nabla^2 t} \psi(0,\vecx)
	= \int_{\R^d} g_{ct}(\vecx-\vecy) \psi(0, \vecy)\,d\vecy.
$$
\end{cor}



%%% ================================================================
%\newpage
%\section{Thermodynamics}
%
%intensive vs. extensive
%
%$E$
%$U$
%$A$
%
%$T$
%$V$
%$N$
%
%$S$
%$P$
%$\mu$
%
%pairs ...
%
%3 laws ...
%
%\begin{itemize}
%
%\item[(1)]
%\begin{align*}
%	dU &= T\,dS - P\,dV + \mu\,dN \\
%	dA &= -S\,dT - P\,dV + \mu\,dN \\
%	U  &= A - TS.
%\end{align*}
%
%Interp:
%	\begin{itemize}
%
%	\item Entropy describes how the free energy changes as the temperature
%	changes.  The product $T\,dS$ is \bfidx{heat} added to the system.
%
%	\item Pressure describes how the internal energy changes as volume changes.
%	The product $P\,dV$ is \bfidx{work} done on the system.
%
%	\item The chemical potential descries how the internal energy changes as
%	particles are added to the system.  The product $\mu\,dN$ is called [xxx?].
%
%	\end{itemize}
%
%\item[(2)]
%$$
%	\frac{\D S}{\D T} \ge 0
%$$
%and describe when.
%
%\item[(3)]
%$$
%	S = k \ln \Omega
%$$
%so $S$ is minimum when $\Omega=1$ which (xxx xref) happens at $T=0$.
%
%xxx This definition of $S$ is microscopic.  For this section, we need
%a macroscopic version \ldots .
%
%% How to reconcile this with degenerate ground state in the 1D Ising model?
%
%\end{itemize}



%% ================================================================
\newpage
\section{Statistical mechanics}

xxx mention mechanics of $N$ classical or quantum-mechanical particles, when
$N$ can be quite large (cf. $\sim10^{23}$ for a typical laboratory experiment).
Average out the microscopic details and obtain simpler macroscopic quantities.

N V E T and entropy ...

CAREFUL (!) definitions of system, phase space, state, ensemble, distribution

%% ----------------------------------------------------------------
\subsection{Microcanonical ensemble}

$N V E$

fixed-E subspaces as foliations (compact submanifolds) of phase space?
unif prob distr on cpct?

key point is uniform probability.

See section \ref{sec:1D_ising} for an example.

%% ----------------------------------------------------------------
\subsection{Canonical ensemble}

$N V T$

xxx microstates and macrostates

The \bfidx{Boltzmann distribution law}:  Probability of occurrence of
microstate with energy $E_i$ is proportional to
$$
	e^{-\beta E_i}.
$$
Probability of occurrence of macrostate with energy $E_i$ is proportional to
$$
	g(E_i) e^{-\beta E_i}
$$
where $g(E_i)$ is the number of microstates with common energy $E_i$.  One also
calls $g(E_i)$ the \emphidx{degeneracy} of the macrostate with energy $E_i$.

See section \ref{sec:1D_ising} for an example.

xxx motivate in two ways:  (1) Widom's independence argument; (2) Huang?
Pathria?  Find it and type it up.

The \bfidx{partition function} or \bfidx{canonical partition function}
is obtained by \ldots

\begin{align*}
	\bbP(E = E_i) &= \frac{e^{-\beta E_i}}{\sum_j e^{-\beta E_j}}
\end{align*}
so
\begin{align*}
	Z &= \sum_j e^{-\beta E_j}.
\end{align*}

xxx $Z$ function of $T$, $V$, and $N$.

xxx partials ... \bfidx{potentials} ... \cite{bib:widom}:

Using conditional probability:
\begin{align*}
	U = \overline{E} &=
		\frac{\sum_i E_i e^{-\beta E_i}}{\sum_j E_j e^{-\beta E_j}}.
\end{align*}
Using the chain rule one immediately obtains
\begin{align*}
	U &= -\left( \frac{\D}{\D\beta} \ln Z \right).
\end{align*}
xxx:
\begin{align*}
	S &= k \ln \Omega \\
	S &= - \sum_i P_i \ln(P_i)
\end{align*}
xxx:
\begin{align*}
	U &= A + TS \\
	A &= -\ln Z/\beta \\
	S &= -\D A / \D T \\
	P &= -\D A / \D V \\
	\mu &= \D A / \D N.
\end{align*}

xxx degeneracy of $E$ and $g(E)$ \ldots.

%% ----------------------------------------------------------------
\subsection{1D Ising-model example}
\label{sec:1D_ising}

xxx aligning nearest-neighbor energy, periodic boundary conditions:
$$
	E = -\sum_{i=1}^N \sigma_i \sigma_{i+1}
$$
with
$$
	\sigma_{N+1} \equiv \sigma_{1}.
$$

xxx $N=10$, $\beta=1$:

% E=10
% -+-+-+-+-+
% +-+-+-+-+-
%
% E=6
%
% ---+-+-+-+
% --+--+-+-+
% +++-+-+-+-
%
% E=2
%
% ----+-+--+
% -+-----+-+
% +++-+---+-
%
% E=-2
%
% -----++-+-
% +++--+++--
% ++++++-++-
%
% E=-6
%
% -----++---
% -+--------
% +++--+++++
%
% E=-10
%
% ----------
% ++++++++++

$$
\begin{tabular}{|r|r|r|r|r|r|r|}
\hline
	\multicolumn{1}{|c|}{$i$} &
	\multicolumn{1}{|c|}{Representative spins} &
	\multicolumn{1}{|c|}{$g(E_i)$} &
	\multicolumn{1}{|c|}{$E_i$} &
	\multicolumn{1}{|c|}{$e^{-E_i}$} &
	\multicolumn{1}{|c|}{$\bbP(E=E_i)$}
	\\
\hline

\hline 1 & $+-+-+-+-+-$ &   2 & $ 10$ &     0.0000454 &  $1.1\cdot 10^{-9}$ \\
\hline 2 & $+++-+-+-+-$ &  90 & $  6$ &     0.0024788 &  $2.7\cdot 10^{-6}$ \\
\hline 3 & $+++++-+-+-$ & 420 & $  2$ &     0.1353353 &  $6.8\cdot 10^{-4}$ \\
\hline 4 & $+++++++-+-$ & 420 & $- 2$ &     7.3890561 &  $0.037$ \\
\hline 5 & $+++++++++-$ &  90 & $- 6$ &   403.4287935 &  $0.434$ \\
\hline 6 & $++++++++++$ &   2 & $-10$ & 22026.4657948 &  $0.527$ \\

\hline
\end{tabular}
$$

$U = -7.95$

xxx note that for high $\beta$, $U=-10$; for low $\beta$, $U=0$.

%% ----------------------------------------------------------------
\subsection{Classical ideal gas}

xxx two idealizations:  (1) classical/quantum; (2) ideal/real.

xxx boltzmann statistics

xxx here is where Widom gets into chemical complications I don't need ...

xxx maxwell-boltzmann statistics

xxx $N$-fold partition function decomposes as product (independence of
events ...)

%% ----------------------------------------------------------------
\subsection{Classical real gas}

xxx $N$-fold partition function no longer decomposes as product
(non-independence of events ...)

%% ----------------------------------------------------------------
\subsection{Grand canonical ensemble}

$$
	\Xi(T,V,\mu) = \sum_{N=0}^\infty e^{\mu\beta N} Z(T,V,N).
$$

xxx grand partition function

%% ----------------------------------------------------------------
\subsection{Order parameter}

sth non-analytic at the phase xitn \ldots.

%% ================================================================
\newpage
\section{Quantum statistical mechanics}
\label{sec:qsm}

xxx $d\rho/dt=0$:  stationary distribution.  Forces $\rho$ to commute with
$H$.  Also makes the Hamiltonian time-independent --- ?

pathria pp. 108-109:  random phases / incoherence?

$\TrebH$:  motivate as
$$
	\TrebH = \sum_j e^{-\beta E_j}
$$
for the diagonal case and then use invariance under change of basis?

%% ----------------------------------------------------------------
\subsection{Density matrix for the canonical ensemble}
\label{sec:canon_ens_rho}

Here we work out the density matrix for the canonical ensemble.  The $p_j$'s of
\ref{prop:rho_from_ens} are
$$
	p_j = \frac{e^{-\beta E_j}}{Z}.
$$
with $Z$ being a to-be-determined normalization factor that makes $\sum_j
p_j=1$.  The $\phi_j$'s are eigenstates of the Hamiltonian with respective
eigenvalues $E_j$.  Then
\begin{align*}
	\rho
	&= \frac{1}{Z} \sum_j e^{-\beta E_j} \ketbra{\phi_j}
	= \frac{1}{Z} \sum_j e^{-\beta H} \ketbra{\phi_j}
	= \frac{1}{Z} e^{-\beta H} \sum_j \ketbra{\phi_j}
	= \frac{1}{Z} e^{-\beta H}.
\end{align*}
by lemmas
\ref{lem:sum_ketbra_phi_j_is_I}
and
\ref{lem:ebH_eig}.

It is now clear what the partition function $Z$ is:  since trace is linear, for
a trace-class operator $A$ we have
$$
	\Tr\left(\frac{A}{\Tr(A)}\right) = \frac{\Tr(A)}{\Tr(A)} = 1.
$$
Thus, to normalize $\rho$ (see property 1 in section \ref{sec:mixed_dm}),
$$
	Z = \TrebH
$$
so
$$
	\rho = \frac{e^{-\beta H}}{\TrebH}.
$$
Then property (3) of section \ref{sec:mixed_dm} becomes
$$
	\expn{A} = \frac{\TrXebH{A})}{\TrebH}.
$$

%% ----------------------------------------------------------------
\subsection{Fermi and Bose ideal gases}

%% ----------------------------------------------------------------
\subsection{The real Bose gas}

%% ================================================================
\newpage
\section{Bose-Einstein condensation in the ideal Bose gas}
\label{sec:ideal_bec}

critical temperature

critical density

fraction of points in the condensation as function of $T$.

%% ================================================================
\newpage
\section{Stochastic calculus}

Here I recall some basic results, all in preparation for a discussion of the
Feynman-Kac formula.  The presentation here is almost entirely an elaboration
on \cite{bib:faris}.  For full information, see \cite{bib:gs},
\cite{bib:lawler}, or \cite{bib:oksendal}.

%% ----------------------------------------------------------------
\subsection{Expectations and covariance}
\label{sec:expn_covar}
\newcommand{\fooa}{{\vecx,\vecx}}
\newcommand{\foob}{{0,T}}

\begin{notn}
\label{notn:cov}
The covariance of $\R^d$-valued random variables
$$
	\vecX = \colvecthree{X_1}{\vdots}{X_d}
	\qand
	\vecY = \colvecthree{Y_1}{\vdots}{Y_d},
$$
having respective means $\vecmu$ and $\vecnu$, is written
$$
	\Cov(\vecX,\vecY) = \bbE[(\vecX-\vecmu)(\vecY-\vecnu)']
	= \bbE[\vecX\vecY'] - \vecmu\vecnu'.
$$
We write $\vecX'$ for the transpose of $\vecX$ from column to row vector.  Then
$\vecX\vecY'$ and $\vecmu\vecnu'$ denote the $d \times d$ outer product of
$\vecX$ and $\vecY$ and $\vecmu$ and $\vecnu$, respectively, and the
expectations are taken componentwise:
\begin{align*}
	\bbE[\vecX\vecY']
	&= \bbE\left[
		\colvecthree{X_1}{\vdots}{X_d}
		\rowvecthree{Y_1}{\vdots}{Y_d}
		\right]
	= \pcmatrix{
		\bbE[X_1 Y_1] & \cdots & \bbE[X_1 Y_d] \\
		\vdots        &        & \vdots        \\
		\bbE[X_d Y_1] & \cdots & \bbE[X_d Y_d] \\
		}.
\end{align*}
In case the covariance matrix has constant $c$ along the diagonal and zeroes
elsewhere, we write $\bbE[\vecX\vecY] = c I$.
\end{notn}

\begin{notn}
For
\index{$\mathbb{E}$, $\mathbb E_0^\vecx$, $\mathbb E_\fooa^\foob$}
\index{$\mathbb{P}$, $\mathbb P_0^\vecx$, $\mathbb P_\fooa^\foob$}
a stochastic process $\vecX_t$, defined on, say, $t \in [0, \infty)$ or $t \in
[0, T]$ for some positive real $T$, and for some Borel set $D \subset \R^d$, we
write the conditional probabilities \index{conditional probability}
$$
	\bbP_0^\veca(\vecX_t \in D) := \bbP(\vecX_t \in D \mid \vecX_0 = \veca)
$$
and
$$
	\bbP_{0,T}^{\veca,\vecb}(\vecX_t \in D)
	:= \bbP(\vecX_t \in D \mid \vecX_0 = \veca, \vecX_T = \vecb).
$$
Likewise, we write the conditional expectations\index{conditional expectation}
$$
	\bbE_0^\veca[F(\vecX_t)] := \bbE[F(\vecX_t) \mid \vecX_0 = \veca]
$$
and
$$
	\bbE_{0,T}^{\veca,\vecb}[F(\vecX_t)]
	:= \bbE[F(\vecX_t) \mid \vecX_0 = \veca, \vecX_T = \vecb].
$$
Similarly, conditional covariances\index{conditional covariance} are
$$
	\Cov_0^\veca(\vecX_t,\vecY_t)
	:= \bbE[(\vecX_t -\vecmu_t)(\vecY_t-\vecnu_t)
		\mid \vecX_0 = \veca, \vecY_0 = \veca]
$$
and
$$
	\Cov_{0,T}^{\veca,\vecb}(\vecX_t,\vecY_t)
	:= \bbE[(\vecX_t -\vecmu_t)(\vecY_t-\vecnu_t)
		\mid \vecX_0 = \veca, \vecY_0 = \veca, \vecX_T=\vecb, \vecY_T=\vecb].
$$
\end{notn}

\begin{notn}
\label{notn:expn_x1_xN}
For $N$ stochastic processes $\vecX^{(1)}, \ldots, \vecX^{(N)}$, we write the
conditional expectations\index{conditional expectation}
$$
	\bbE_{0}^{\veca^{(1)},\ldots,\veca^{(N)}}
	[F(\vecX^{(1)}_t, \ldots, \vecX^{(N)}_t)]
	:= \bbE\left[F(\vecX^{(1)}_t,\ldots,\vecX^{(N)})
		\mid
			\vecX^{(1)}_0 = \veca^{(1)}, \ldots, \vecX^{(N)}_0 = \veca^{(N)}
		\right]
$$
and
\begin{align*}
	\bbE_{0,T}^{\veca^{(1)},\vecb^{(1)}; \ldots;\veca^{(N)},\vecb^{(N)}}
	&[F(\vecX^{(1)}_t, \ldots, \vecX^{(N)}_t)] \\
	&:= \bbE\left[F(\vecX^{(1)}_t, \ldots, \vecX^{(N)}_t)
		\mid
			\vecX^{(1)}_0 = \veca^{(1)},
			\vecX^{(1)}_T = \vecb^{(1)},
			\ldots,
			\vecX^{(N)}_0 = \veca^{(N)},
			\vecX^{(N)}_T = \vecb^{(N)}
		\right].
\end{align*}
\end{notn}

%% ----------------------------------------------------------------
\subsection{Brownian motion}
\label{sec:bm}

In this section we review fundamental properties of Brownian motion,
highlighting its connections with the Gaussian function and the diffusion
equation.

Brownian motion (see \cite{bib:oksendal,bib:lawler,bib:gs} for more careful
treatments) is defined for $t \in [0, \infty) \mapsto \vecb_t \in \R^d$.  It
has the following properties:

\begin{itemize}

\item $\vecb_t$ is a stochastic process, or random function of $t$, which is
almost surely continuous (with respect to the probability measure defined
below).

\item $\vecb_0 = 0$.

\item $\vecb_t$ has Gaussian distribution with mean $0$ and variance $t$ in
each of its $d$ components for all $t>0$, and it has covariance
$$
	\Cov(\vecb_s, \vecb_t) = \bbE[\vecb_s \vecb_t] = (s \wedge t) I
$$
where $s \wedge t$ denotes\index{wedge notation $s \wedge t$} $\min\{s, t\}$.
Specifically, the probability distribution function (\plainidx{PDF}) of
$\vecb_t$ at time $t$ is Gaussian with mean $0$ and
component variances $t$.  For a box
$$
	D = [\ell^{(1)}, u^{(1)}] \times \cdots \times [\ell^{(d)}, u^{(d)}],
$$
we have
\begin{align}
\label{eqn:bmp1}
	\bbP(\vecb_t \in D) &= \int_D g_t(\vecy) \,d\vecy.
\end{align}
where $g_t$ is notation \ref{notn:g_t} for the Gaussian with variance $t$.

\item $\vecb_t$ has \emphidx{independent increments}:  For all $s<t<u<v$,
$\vecb_t-\vecb_s$ is independent of $\vecb_v-\vecb_u$.

\end{itemize}

It turns out (see also \cite{bib:faris}) that Brownian motion is uniquely
characterized by an extension of equation \ref{eqn:bmp1}.
\begin{prop}
\label{prop:bm_chzn}
Brownian motion is uniquely characterized by the following:  for all $n \ge 1$,
all times $0 < t_1 < \ldots < t_n$, and all boxes
$$
	D_k = 
	[\ell_k^{(1)}, u_k^{(1)}]
	\times \cdots \times
	[\ell_k^{(d)}, u_k^{(d)}],
$$
\begin{equation}
\label{eqn:bmpn_bm}
\begin{aligned}
	\bbP
	\left(\bigcap_{k=1}^n \vecb_{t_k} \in D_k
	\right)
	=
		\int_{D_1}
	&
		\int_{D_2}
		\cdots
		\int_{D_n}
		\\
	&
		g_{t_1}(\vecz_1)
		g_{t_2-t_1}(\vecz_1-\vecz_2)
		\cdots
		g_{t_n-t_{n-1}}(\vecz_{n-1}-\vecz_{n})
		\, d\vecz_1 \,d\vecz_2 \cdots d\vecz_n.
\end{aligned}
\end{equation}
\end{prop}

\begin{rem}
That is, we set up some hoops and calculate the probability that a given
realization of the Brownian motion will jump through all of them (figure
\ref{fig:bm_hoops}).  The only family of functions which jump through such sets
of hoops with these normally distributed rates of success is the family of
realizations of Brownian motion.  This defines a probability measure on
continuous functions $f:[0,+\infty) \to \R^d$ with $f(0)=0$.
\end{rem}

\begin{figure}[!htb]
\begin{center}
%\psfragscanon
\includegraphics{figures/comp_bm_hoops.eps}
\caption{Ten realizations of Brownian motion, moving past three boxes.
	\label{fig:bm_hoops}}
\end{center}
\end{figure}

\begin{rem*}
This is equivalent to saying that the joint PDF of the random variables
$\vecb_{t_1}, \ldots, \vecb_{t_n}$ is
$$
	g_{t_1}(\vecz_1)
	g_{t_2-t_1}(\vecz_1-\vecz_2)
	\cdots
	g_{t_n-t_{n-1}}(\vecz_{n-1}-\vecz_n).
$$
\end{rem*}

\begin{rem}
\label{rem:ExFomega_bm}
Faris points out that one may generalize equation \ref{eqn:bmpn_bm}.
Rephrase it by writing
\begin{align*}
	f(\vecb_{t_1}, \ldots, \vecb_{t_n})
	&= \chi_{\cap_{k=1}^n \vecb_{t_k} \in D_k }
		(\vecb_{t_1}, \ldots, \vecb_{t_n}).
\end{align*}
Then equation \ref{eqn:bmpn_bm} becomes
\begin{equation}
\label{eqn:bmpnE_bm}
\begin{aligned}
	\bbE[f(\vecb_{t_1}, \ldots, \vecb_{t_n})] =
		\int_{\R^d}
		\int_{\R^d}
		\cdots
		\int_{\R^d}
		g_{t_1}(\vecz_1)
		&
		g_{t_2-t_1}(\vecz_1-\vecz_2)
		\cdots
		g_{t_n-t_{n-1}}(\vecz_{n-1}-\vecz_n)
		\\
	&
		f(\vecz_1, \ldots, \vecz_n)
		\; d\vecz_1 \,d\vecz_2 \cdots d\vecz_n.
\end{aligned}
\end{equation}
This holds not only for the characteristic function $\chi$ which appears in
equation \ref{eqn:bmpn_bm}, but for any Borel-measurable $f$.
\end{rem}

\begin{defn}
\label{defn:E_exp_U_bm}
We extend this to exponentiated integrals by defining
\begin{equation}
\begin{aligned}
	\bbE\left[\exp\left\{\int_0^T f(\vecb_t)\,dt\right\} \right]
	:= \lim_{n\to\infty}
		\int_{\R^{d}}
		\cdots
		\int_{\R^{d}}
		&
		g_{t/n}(\vecz_1)
		g_{t/n}(\vecz_1-\vecz_2)
		\cdots
		g_{t/n}(\vecz_{n-1}-\vecz_n)
		\\
	&
		\exp\left\{\frac{T}{n} \sum_{j=1}^n f(\vecz_i) \right\}
		\; d\vecz_1 \cdots d\vecz_n.
\end{aligned}
\end{equation}
\end{defn}

Now consider a realization of Brownian motion moving forward in time, having
already passed time $s$ with $\vecb_s = \vecy$ (figure \ref{fig:bm_cond}).  We
want to find the conditional density of $\vecb_t$.

\begin{figure}[!htb]
\begin{center}
%\psfragscanon
\includegraphics{figures/comp_bm_cond.eps}
\caption{Ten realizations of Brownian motion, conditioned on $\vecb_s=\vecy$.
	\label{fig:bm_cond}}
\end{center}
\end{figure}

\begin{lem}
The conditional density of $\vecb_t$ given $\vecb_s = \vecy$, for $t > s$,
is
$$
	g_{t-s}(\vecz-\vecy).
$$
\begin{proof}
This is nothing more than $f_{\vecZ\mid \vecY}(\vecz\mid \vecy)$ from
elementary probability (see e.g. \cite{bib:gs}), where $\vecZ=\vecb_{t}$ and
$\vecY=\vecb_{s}$.  From equation \ref{eqn:bmpn_bm}, with $n=2$, the joint
density of $\vecY$ and $\vecZ$ is
$$
	g_{t-s}(\vecz-\vecy) g_s(\vecy-\vecx).
$$
We compute
\begin{align*}
	f_{\vecZ\mid \vecY}(\vecz \mid \vecy)
	&= \frac{f_{\vecZ,\vecY}(\vecz,\vecy)} {f_\vecY(\vecy)}
	= \frac{ g_{t-s}(\vecz-\vecy) g_s(\vecy-\vecx) } {g_s(\vecy-\vecx)}
	= g_{t-s}(\vecz-\vecy).
\end{align*}
\end{proof}
\end{lem}

\begin{prop}
The conditional PDF $g_{t-s}(z,y)$ satisfies the following PDE, known
as the \emphidx{forward diffusion equation}:
\begin{align}
\label{eqn:forward_diffusion}
	\frac{\D}{\Dt} g_{t-s}(\vecz-\vecy)
		&=  \frac{1}{2} \nabla^2 g_{t-s}(\vecz-\vecy)
%		&& \textrm{(Forward diffusion equation)}
%		\\
%\label{eqn:backward_diffusion}
%	\frac{\D}{\Ds} g_{t-s}(\vecz-\vecy)
%		&= -\frac{1}{2} \nabla^2 g_{t-s}(\vecz-\vecy)
%		&& \textrm{(Backward diffusion equation)}.
\end{align}
\begin{proof}
Compute the partial derivatives and simplify.
\end{proof}
\end{prop}

\begin{rem*}
The intuition is as follows.  A large number of realizations of Brownian
motion, all having passed through $\vecy$ at time $s$, form a cloud of points
initially concentrated at $\vecy$, diffusing away from $\vecy$ as $t$
increases.  The PDF plays the role of $\psi$ in section
\ref{sec:diffusion_equation_exp}, with the initial condition being a point mass
at 0.
\end{rem*}

%% ----------------------------------------------------------------
\subsection{Scaled Brownian motion}
\label{sec:scaled_bm}

Here we generalize Brownian motion (what \cite{bib:lawler} calls
\emphidx{standard Brownian motion}) in two ways:  (1) by allowing a
\emphidx{diffusion constant} to scale the variance, and (2) by removing the
requirement that the motion start at 0.

\begin{notn}
For fixed $\vecx$, we write
\begin{equation}
\label{eqn:scaled_bm}
	\vecw^\vecx_{c,t} := \vecx + \sqrt{c} \, \vecb_t.
\end{equation}
If $\vecx$ is allowed to vary, we simply write $\vecw_{c,t}$.  However, this is
not well defined as a stochastic process unless we either set a probability
distribution for $\vecx$, or write $\vecw_{c,t}$ inside a conditional
expectation.  We will take the latter approach in this paper.  Also, we take
$\bbE^\vecx[\vecw_{c,t}]$ and $\bbE[\vecw^\vecx_{c,t}]$ to be equivalent.
\end{notn}

\begin{prop}
Scaled Brownian motion has conditional mean and conditional covariance
$$
	\bbE_0^\vecx[\vecw_{c,t}] = \vecx
	\qqand
	\Cov_0^\vecx[\vecw_{c,s} \vecw_{c,t}] = c\, (s \wedge t) I.
$$
\begin{proof}
These follow immediately from equation \ref{eqn:scaled_bm}, along with the mean
and covariance for Brownian motion given in section \ref{sec:bm}.
\end{proof}
\end{prop}

Proposition \ref{prop:bm_chzn} becomes the following.

\begin{prop}
For all $n \ge 1$, all times $0 < t_1 < \ldots < t_n$, and all boxes
$$
	D_k = 
	[\ell_k^{(1)}, u_k^{(1)}]
	\times \cdots \times
	[\ell_k^{(d)}, u_k^{(d)}],
$$
we have
\begin{equation}
\label{eqn:bmpn}
\begin{aligned}
	\bbP_0^\vecx
	\left(\bigcap_{k=1}^n \vecw_{c,t_k} \in D_k
	\right)
	=
		\int_{D_1}
	&
		\int_{D_2}
		\cdots
		\int_{D_n}
		\\
	&
		g_{ct_1}(\vecz_1-\vecx)
		g_{c(t_2-t_1)}(\vecz_1-\vecz_2)
		\cdots
		g_{c(t_n-t_{n-1})}(\vecz_{n-1}-\vecz_n)
		\, d\vecz_1 \,d\vecz_2 \cdots d\vecz_n.
\end{aligned}
\end{equation}
\end{prop}

\begin{rem}
This is equivalent to saying that the joint PDF of the random variables
$\vecw_{c,t_1}, \ldots, \vecw_{c,t_n}$, conditioned on $\vecw_0=\vecx$, is
$$
	g_{ct_1}(\vecz_1-\vecx)
	g_{c(t_2-t_1)}(\vecz_1-\vecz_2)
	\cdots
	g_{c(t_n-t_{n-1})}(\vecz_{n-1}-\vecz_n).
$$
\end{rem}

\begin{rem}
\label{rem:ExFomega}
Also, equation \ref{eqn:bmpnE_bm} becomes
\begin{equation}
\label{eqn:bmpnE}
\begin{aligned}
	\bbE_0^\vecx[f(\vecw_{c,t_1}, \ldots, \vecw_{c,t_n})] =
		\int_{\R^d}
		\int_{\R^d}
		\cdots
		\int_{\R^d}
		g_{ct_1}(\vecz_1-\vecx)
		&
		g_{c(t_2-t_1)}(\vecz_1-\vecz_2)
		\cdots
		g_{c(t_n-t_{n-1})}(\vecz_{n-1}-\vecz_n)
		\\
	&
		f(\vecz_1, \ldots, \vecz_n)
		\; d\vecz_1 \,d\vecz_2 \cdots d\vecz_n.
\end{aligned}
\end{equation}
\end{rem}

\begin{defn}
\label{defn:E_exp_U_wm}
Definition \ref{defn:E_exp_U_bm} becomes
\begin{equation}
\begin{aligned}
	\bbE_0^\vecx \left[\exp\left\{\int_0^T f(\vecw_{c,t})\,dt\right\} \right]
	:= \lim_{n\to\infty}
		\int_{\R^{d}}
		\cdots
		\int_{\R^{d}}
		&
		g_{cT/n}(\vecz_1-\vecx)
		g_{cT/n}(\vecz_1-\vecz_2)
		\cdots
		g_{cT/n}(\vecz_{n-1}-\vecz_n)
		\\
	&
		\exp\left\{\frac{T}{n} \sum_{j=1}^n f(\vecz_i) \right\}
		\; d\vecz_1 \cdots d\vecz_n.
\end{aligned}
\end{equation}
\end{defn}

%% ----------------------------------------------------------------
\subsection{Brownian bridges}

\cite{bib:faris} treats only bridges from $\vecx$ to $\vecx$; here, we treat
the more general case of bridges from $\vecx$ to $\vecy$.  This is needed for
\cite{bib:bu}, \cite{bib:qmath}, and in particular for section
\ref{sec:fkboson} of this paper.

\begin{defn}
The (scaled) \bfidx{Brownian bridge} running from $\vecx$ at time 0 to $\vecy$
at time $T$ is
\begin{align}
\label{eqn:bridge_process}
	\vecw_{c,T,t}^{\vecx,\vecy}
	&= \vecx + \sqrt{c}\, \vecb_t
		+ \frac{t}{T} \left(\vecy - \vecx - \sqrt{c}\, \vecb_T\right).
\end{align}
We may also write this as
\begin{align}
\label{eqn:bridge_process_aux}
	\vecw_{c,T,t}^{\vecx,\vecy}
	&= \frac{(T-t)\vecx + t\vecy}{T}
	+ \sqrt{c}\, \frac{T\vecb_t - t\vecb_T}{T}.
\end{align}
This looks ungainly, but it has the advantage that the mean is precisely the
first term.
\end{defn}

\begin{notn}
As in section \ref{sec:scaled_bm}, if $\vecx$ and $\vecy$ are allowed to vary,
then we again have simply $\vecw_{c,t}$.  We will write $\vecw_{c,t}$
inside a conditional expectation in order to make this notation well defined as
a stochastic process.  As well, we take $\bbE^{\vecx,\vecy}_{0,T}[\vecw_{c,t}]$
and $\bbE[\vecw^{\vecx,\vecy}_{c,T,t}]$ to be equivalent.
\end{notn}

%This could be defined for all $t\ge 0$ --- after all, we are simply adding a
%linear drift to the scaled Brownian motion $\sqrt{c} \vecb_t$, with offset
%$\vecx$ and slope $(\vecy-\vecx)/T$ --- but it is conventionally taken to be
%defined only for $t \in [0, T]$.

\begin{defn}
\label{defn:E_exp_U_bb}
As in definition \ref{defn:E_exp_U_wm}, we extend this to exponentiated
integrals by defining
\begin{equation}
\begin{aligned}
	\bbE_{0,T}^{\vecx,\vecy}
		\left[\exp\left\{\int_0^T f(\vecw_{c,t})\,dt\right\} \right]
	&:= \lim_{n\to\infty}
		\int_{\R^{d}}
		\cdots
		\int_{\R^{d}} \\
%
	&
		g_{cT/n}(\vecx-\vecz_1)
		g_{cT/n}(\vecz_1-\vecz_2)
		\cdots
		g_{cT/n}(\vecz_{n-2}-\vecz_{n-1})
		g_{cT/n}(\vecz_{n-1}-\vecy)
		\\
%
	&
		\exp\left\{\frac{T}{n} \left(
			f(\vecx) + \sum_{j=1}^{n-1} f(\vecz_i) \right)
		\right\}
		\; d\vecz_1 \cdots d\vecz_{n-1} \\
%
	&:= \int \exp\left\{ \int_0^{T} f(\vecw_{c,t})\,dt \right\}
		\,d\vecW_{0,T}^{\vecx,\vecy}(\vecw_{c}).
\end{aligned}
\end{equation}
\end{defn}

\begin{notn}
\label{notn:bridge_measure}
In the last line we have defined\index{$d\mathbf{W}$} the notation for the
bridge measure as used in \cite{bib:bu} and \cite{bib:qmath}.
\end{notn}

\begin{prop}
The bridge process $\vecw^{\vecx,\vecy}_{c,T,t}$ has mean
$$
	\vecx + \frac{t}{T} \left(\vecy-\vecx\right)
$$
and covariance
$$
	c I \left( s \wedge t - \frac{st}{T} \right).
$$
\begin{proof}
The mean follows immediately from taking the expectation of equation
\ref{eqn:bridge_process}.  For the covariance, we may take advantage of
equation \ref{eqn:bridge_process_aux} where the mean is readily subtractible,
and recall notation \ref{notn:cov}.  Then, since $\bbE[\vecb_s \vecb_t'] = (s
\wedge t)I$, we have
\begin{align*}
	\Cov^{\vecx,\vecy}_{0,T}[\vecw_{c,s},\vecw_{c,t}]
	%&= \bbE\left[ 
	%	\left(\sqrt{c}\, \frac{T\vecb_s - s\vecb_T}{T}\right)
	%	\left(\sqrt{c}\, \frac{T\vecb_t - t\vecb_T}{T}\right)'
	%	\right] \\
	&= \frac{c}{T^2} \, \bbE\left[ 
		\left(T\vecb_s - s\vecb_T\right)
		\left(T\vecb_t - t\vecb_T\right)'
		\right] \\
	&= \frac{c}{T^2} \, \left(
		T^2 \bbE[\vecb_s \vecb_t'] -
		s T \bbE[\vecb_T \vecb_t'] -
		t T \bbE[\vecb_s \vecb_T'] +
		s t \bbE[\vecb_S \vecb_T']
		\right) \\
	&= c I \left( s \wedge t - \frac{st}{T} \right).
\end{align*}
\end{proof}
\end{prop}

\begin{rem}
If $s<t$ we may write this (perhaps more memorably) as
$$
	c I \, \frac{s(T-t)}{T}.
$$
\end{rem}

%% ----------------------------------------------------------------
\subsection{Brownian motion and Brownian bridges}

We continue to follow \cite{bib:faris} by computing expectations over delta
functions.  This may seem bizarre, but it is needed in the proof of proposition
\ref{prop:TrebH_bridge}.

%\begin{lem}
%\label{lem:gb0}
%For the scaled Brownian motion starting at $\vecx$, we have
%$$
%	\bbE_0^\vecx[\delta(\vecw_{c,T}-\vecx)] = g_{cT}(0)
%$$
%where $g_t$ is notation \ref{notn:g_t} for the Gaussian with variance $t$.
%\begin{proof}
%Using remark \ref{rem:ExFomega} for the expectation, we have
%\begin{align*}
%	\bbE_0^\vecx[\delta(\vecw_{c,T}-\vecx)]
%	&= \int_{\R^d}
%		g_{cT}(\vecx-\vecz)
%		\delta(\vecz-\vecx) \,d\vecz
%	= g_{cT}(\vecx-\vecx)
%	= g_{cT}(0).
%\end{align*}
%\end{proof}
%\end{lem}

\begin{lem}
\label{lem:gb0}
For the scaled Brownian motion starting at $\vecx$, we have
$$
	\bbE_0^\vecx[\delta(\vecw_{c,T}-\vecy)] = g_{cT}(\vecx-\vecy)
$$
where $g_t$ is notation \ref{notn:g_t} for the Gaussian with variance $t$.
\begin{proof}
Using remark \ref{rem:ExFomega} for the expectation, we have
\begin{align*}
	\bbE_0^\vecx[\delta(\vecw_{c,T}-\vecy)]
	&= \int_{\R^d}
		g_{cT}(\vecx-\vecz)
		\delta(\vecz-\vecy) \,d\vecz
	= g_{cT}(\vecx-\vecy).
\end{align*}
\end{proof}
\end{lem}

\begin{lem}
\label{lem:wmcovar}
The Wiener-measure covariance of the bridge process
$\vecw^{\vecx,\vecy}_{c,T,t}$ and the scaled Brownian motion $\vecw_{c,T}$ is
zero.
\begin{proof}
\begin{align*}
	\Cov(\vecw^{\vecx,\vecy}_{c,T,t}, \vecw_{c,T})
	&= \bbE\left[
		\left(\vecw^{\vecx,\vecy}_{c,T,t} - \vecx\right)\vecw_{c,T}' \right] \\
	&= \bbE\left[ \left(\frac{t}{T}(\vecy-\vecx) + \sqrt{c} \vecb_t
		- \sqrt{c}\frac{t}{T} \vecb_T\right) \sqrt{c} \vecb_T' \right] \\
	&= \frac{t}{T}(\vecy-\vecx) \sqrt{c} \,\bbE[\vecb_T]'
		+ c \,\bbE[\vecb_t\vecb_T'] - c \frac{t}{T} \,\bbE[\vecb_T\vecb_T'] \\
	&= c \left(t - \frac{t}{T} T \right) = 0.
\end{align*}
\end{proof}
\end{lem}

The following proposition is needed for our treatment of the Feynman-Kac
formula in sections \ref{sec:fk1} and \ref{sec:fkboson_tr}.  There,
$F(\vecw_{c,\cdot})$ will be an exponentiated integral in the form of
definition \ref{defn:E_exp_U_bb}.

%\begin{prop}
%\label{prop:Ex_Exx}
%Expectations over $\vecx$-to-$\vecx$ bridges and expectations over scaled
%Brownian motion are related by
%$$
%	\bbE_{0,T}^{\vecx,\vecx}[F(\vecw_{c,\cdot})]
%	= \frac
%		{\bbE_0^\vecx[F(\vecw_{c,\cdot}) \delta(\vecw_{c,T}-\vecx)]}
%		{g_{cT}(0)}.
%$$
%\begin{proof}
%We start with the right-hand side.  By construction, at time $T$ the Brownian
%bridge coincides with the Brownian motion conditioned on taking value $\vecx$
%at time $T$:  $\vecw_{c,T} = \vecx = \vecw^{\vecx,\vecx}_{c,T}$.  Thus
%$$
%	\frac
%		{\bbE_0^\vecx[F(\vecw_{c,\cdot}) \delta(\vecw_{c,T}-\vecx)]}
%		{g_T(0)}
%	=
%	\frac
%		{\bbE_0^\vecx[F(\vecw^{\vecx,\vecx}_{c,T}) \delta(\vecw_{c,T}-\vecx)]}
%		{g_T(0)}.
%$$
%By lemma \ref{lem:wmcovar}, the Wiener-measure covariance of
%$\vecw^{\vecx,\vecx}_{c,T,t}$ and $\vecw_{c,T}$ is zero.  Therefore the
%expectation factors and the right-hand side becomes
%$$
%	\frac
%		{
%		\bbE_0^\vecx[F(\vecw^{\vecx,\vecx}_{c,T,\cdot})] \;
%		\bbE_0^\vecx[\delta(\vecw_{c,T}-\vecx)]
%		}
%		{g_T(0)}.
%$$
%By lemma \ref{lem:gb0}, the denominator is equal to the second term in
%the numerator.  Thus the right-hand side equals the left-hand side.
%\end{proof}
%\end{prop}

\begin{prop}
\label{prop:Ex_Exy}
Expectations over Brownian bridges and expectations over scaled Brownian motion
are related by
$$
	\bbE_{0,T}^{\vecx,\vecy}[F(\vecw_{c,\cdot})]
	= \frac
		{\bbE_0^\vecx[F(\vecw_{c,\cdot}) \delta(\vecw_{c,T}-\vecy)]}
		{g_{cT}(\vecx-\vecy)}.
$$
\begin{proof}
We start with the right-hand side.  By construction, at time $T$ the Brownian
bridge coincides with the Brownian motion conditioned on taking value $\vecy$
at time $T$:  $\vecw_{c,T} = \vecy = \vecw^{\vecx,\vecy}_{c,T}$.  Thus
$$
	\frac
		{\bbE_0^\vecx[F(\vecw_{c,\cdot}) \delta(\vecw_{c,T}-\vecy)]}
		{g_T(\vecx-\vecy)}
	=
	\frac
		{\bbE_0^\vecx[F(\vecw^{\vecx,\vecy}_{c,T}) \delta(\vecw_{c,T}-\vecy)]}
		{g_T(\vecx-\vecy)}.
$$
By lemma \ref{lem:wmcovar}, the Wiener-measure covariance of
$\vecw^{\vecx,\vecy}_{c,T,t}$ and $\vecw_{c,T}$ is zero.  Therefore the
expectation factors and the right-hand side becomes
$$
	\frac
		{
		\bbE_0^\vecx[F(\vecw^{\vecx,\vecy}_{c,T,\cdot})] \;
		\bbE_0^\vecx[\delta(\vecw_{c,T}-\vecy)]
		}
		{g_T(\vecx-\vecy)}.
$$
By lemma \ref{lem:gb0}, the denominator is equal to the second term in
the numerator.  Thus the right-hand side equals the left-hand side.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Normalized bridges}

We modify definition \ref{defn:E_exp_U_bb} to obtain a \plainidx{normalized
measure}:  we set a scale factor, then prove that it is chosen correctly.

\begin{defn}
Let\index{$d\hat{\mathbf{W}}$}
\begin{equation}
	\,d\vecWhat_{0,T}^{\vecx,\vecy}(\vecw_{c})
	=
	\frac{1}{g_{ct}(\vecx-\vecy)}
	\,d\vecW_{0,T}^{\vecx,\vecy}(\vecw_{c}).
\end{equation}
\end{defn}

\begin{prop}
\label{prop:dW_normalization}
We have
$$
	\int \,d\vecWhat_{0,T}^{\vecx,\vecy}(\vecw_{c}) = 1.
$$
\begin{proof}
Starting with definition \ref{defn:E_exp_U_bb} for the non-normalized measure,
we have
\begin{align*}
	\int \,d\vecW_{0,T}^{\vecx,\vecy}(\vecw_{c})
	&= \int \exp\left\{ \int_0^T 0 \, dt \right\}
		\,d\vecW_{0,T}^{\vecx,\vecy}(\vecw_{c}) \\
	&:= \lim_{n\to\infty}
		\int_{\R^{d}}
		\cdots
		\int_{\R^{d}}
		g_{cT/n}(\vecx-\vecz_1)
		\cdots
		g_{cT/n}(\vecz_{n-1}-\vecy)
		\; d\vecz_1 \cdots d\vecz_{n-1} \\
	&= g_{ct}(\vecx-\vecy)
\end{align*}
where the last step follows from proposition \ref{prop:g_iter}.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\newpage
\section{Single-particle Feynman-Kac formulas}
\label{sec:fk1}

The material here is familiar from many references.  Here I review it for two
reasons:  for my own understanding, and to fix notation which will be used for
the derivation of the bosonic Feynman-Kac formula in section \ref{sec:fkboson}.

%% ----------------------------------------------------------------
\subsection{$e^{-\beta H}$ as expectation}
\label{sec:fk1_exp}

In proposition \ref{prop:diffusion_convolution} we saw how to interpret the
exponentiated operator $e^{c\nabla^2 t}$ as an integral operator with a
Gaussian kernel.  We now ask how to interpret $e^{(c\nabla^2 + U)t}$ where $U$
is a multiplication operator corresponding to a potential energy.

\begin{prop}
\label{prop:diffusion_FK}
The solution to
$$
	\frac{\D\psi(t,\vecx)}{\Dt} = c \nabla^2 \psi(t,\vecx)
		+ U(\vecx) \psi(t,\vecx)
$$
is
$$
	\psi(t, \vecx)
	= e^{(c\nabla^2 + U(\vecx))t} \psi(0, \vecx)
	= \bbE_0^\vecx \left[
		\exp\left\{
		%e^{
			\int_0^t U(\vecw_{c,s})\,ds
		%}
		\right\}
		\psi(0, \vecw_{c,t})
	\right].
$$
\begin{proof}
As shown in proposition \ref{prop:diffusion_exp},
$$
	\psi(t, \vecx) = e^{(c\nabla^2+U(\vecx))t} \psi(0, \vecx)
$$
is a solution; it remains to find a way to interpret the exponentiated
operator.

The \plainidx{Trotter product formula} (see \cite{bib:simon} for a proof) says
that for self-adjoint operators $A$ and $B$,
\begin{align}
\label{eqn:trotter_AB}
	e^{(A+B)t} &= \lim_{n\to\infty} \left(e^{At/n} e^{Bt/n}\right)^n.
\end{align}
With $A=c\nabla^2$ and $B=U(\vecx)$, we have
\begin{align*}
	e^{(c\nabla^2+U(\vecx))t}\psi(0, \vecx)
	&= \lim_{n\to\infty} \left(e^{c\nabla^2 t/n} e^{U(\vecx)t/n}\right)^n
		\psi(0, \vecx) \\
	&= \lim_{n\to\infty}
		e^{c\nabla^2 t/n} e^{U(\vecx)t/n}
		\left(e^{c\nabla^2 t/n} e^{U(\vecx)t/n}\right)^{n-1}
		\psi(0, \vecx).
\end{align*}
Recall that $e^{U(\vecx)t/n}$ is simply a scalar.  Using proposition
\ref{prop:diffusion_convolution} to write $e^{c\nabla^2 t/n}$ as an integral
operator, we have
\begin{align*}
	\psi(t,\vecx)
	&= \lim_{n\to\infty}
		\int_{\R^d} g_{ct/n}(\vecx-\vecz_1) 
		e^{U(\vecz_1)t/n}
		\left(e^{c\nabla^2 t/n} e^{U(\vecz_1)t/n}\right)^{n-1}
		\psi(0, \vecz_1)
		\,d\vecz_1.
\end{align*}
Repeating yields
\begin{align*}
	\psi(t,\vecx)
	&= \lim_{n\to\infty}
		\int_{\R^d}
		\int_{\R^d}
		g_{ct/n}(\vecx-\vecz_1) 
		g_{ct/n}(\vecz_1-\vecz_2) 
		e^{U(\vecz_1)t/n}
		e^{U(\vecz_2)t/n}
		\left(e^{c\nabla^2 t/n} e^{U(\vecz_2)t/n}\right)^{n-2}
		\psi(0, \vecz_2)
		\,d\vecz_1
		\,d\vecz_2,
\end{align*}
and finally
\begin{align*}
	\psi(t,\vecx)
	= \lim_{n\to\infty}
		\int_{\R^d}
		\int_{\R^d}
		\cdots
		\int_{\R^d}
	&
		g_{ct/n}(\vecx-\vecz_1) 
		g_{ct/n}(\vecz_1-\vecz_2) 
		\cdots
		g_{ct/n}(\vecz_{n-1}-\vecz_n)  \\
	&
		e^{U(\vecz_1)t/n}
		e^{U(\vecz_2)t/n}
		\cdots
		e^{U(\vecz_n)t/n}
		\psi(0, \vecz_n)
		\,d\vecz_1
		\,d\vecz_2 \dots
		\,d\vecz_n
	\\
%
	= \lim_{n\to\infty}
		\int_{\R^d}
		\int_{\R^d}
		\cdots
		\int_{\R^d}
	&
		g_{ct/n}(\vecx-\vecz_1) 
		g_{ct/n}(\vecz_1-\vecz_2) 
		\cdots
		g_{ct/n}(\vecz_{n-1}-\vecz_n)  \\
	&
		e^{\sum_{k=1}^n U(\vecz_k)t/n}
		\psi(0, \vecz_n)
		\,d\vecz_1
		\,d\vecz_2 \dots
		\,d\vecz_n.
\end{align*}

Now we recognize an integrand in the form of remark \ref{rem:ExFomega}, with
$t_k = kt/n$, and we can write
\begin{align*}
	\psi(t,\vecx)
	&= \lim_{n \to \infty}
		\bbE_0^\vecx\left[
		\exp\left\{ \sum_{k=1}^n U(\vecw_{c,kt/n}) \frac{t}{n} \right\}
		\psi(0, \vecw_{c,t})
	\right].
\end{align*}
Interchanging limit and expectation by a dominated-convergence argument
[xxx type it up], and recognizing the limit of the sum as a Riemann integral,
we obtain the desired result:
\begin{align*}
	\psi(t,\vecx)
	&= \bbE_0^\vecx\left[
		\exp\left\{ \int_0^t U(\vecw_{c,s}) \,ds \right\}
		\psi(0, \vecw_{c,t})
	\right].
\end{align*}
\end{proof}
\end{prop}

\begin{rem}
Note that in the $U=0$ case, we have
$$
	\psi(t, \vecx)
	= e^{c\nabla^2\beta} \psi(0, \vecx)
	= \bbE_0^\vecx \left[ \psi(0, \vecw_{c,t}) \right].
$$
Yet in proposition \ref{prop:diffusion_convolution} we also saw that
$$
	\psi(t,\vecx)
	= e^{c\nabla^2 t} \psi(0,\vecx)
	= \int_{\R^d} g_{ct}(\vecx-\vecy) \psi(0,\vecy) \,d\vecy.
$$
These two points of view are reconciled by using the \plainidx{Law of the
Unconscious Statistician}\index{Unconscious Statistician, Law of} from
elementary probability.  Namely, for a measurable function $h$ of a random
variable $Y$,
$$
	\bbE[h(Y)] = \int h(y) f_Y(y) \,dy
$$
where $f_Y(y)$ is the PDF for $Y$.  Here, for each fixed $t$ and $\vecx$,
$Y$ is $\vecw_{c,t}$, $h(Y)$ is $\psi(0,\vecw_{c,t})$, and the PDF of
$\vecw{c,t}$ is $g_{ct}(\vecy-\vecx)$ as a function of $\vecy$.
\end{rem}

%% ----------------------------------------------------------------
\subsection{$e^{-\beta H}$ as an integral operator}
\label{sec:fk1_ker}

Section \ref{sec:operator_trace} showed how to compute $\TrebH$ if
$e^{-\beta H}$ can be represented as an integral operator.  Proposition
\ref{prop:diffusion_FK} constructed the operator's kernel for the $U=0$ case.
Here we construct the kernel for the general case, then apply it to find an
expression for the trace.

\begin{prop}
\label{prop:ebHG}
If
\begin{equation*}
	H = c\nabla^2 + U(\vecx),
\end{equation*}
then
\begin{equation}
\label{eqn:temp1}
	e^{-\beta H} f(\vecx) = \int G_{Ut}(\vecx, \vecy) f(\vecy) \,d\vecy
\end{equation}
where
\begin{equation}
\label{eqn:temp2}
	G_{Ut}(\vecx,\vecy) = \bbE_0^\vecx\left[
		\exp\left\{
			-\int_0^t U(\vecw_{c,s})\,ds
		\right\}
		\,\delta(\vecw_{c,t}-\vecy)
	\right].
\end{equation}
\begin{proof}
Inserting equation \ref{eqn:temp2} into the right-hand side
of \ref{eqn:temp1} gives
\begin{align*}
	\int G_{t,U}(\vecx, \vecy) f(\vecy) \,d\vecy
	&= \int
	\bbE_0^\vecx\left[
		\exp\left\{
			-\int_0^t U(\vecw_{c,s})\,ds
		\right\}
		\,\delta(\vecw_{c,t}-\vecy)
	\right]
	\,f(\vecy) \,d\vecy.
\end{align*}
Now interchange expectation and integral [xxx justify this] and integrate out
the delta function:
\begin{align*}
	\int G_{t,U}(\vecx, \vecy) f(\vecy) \,d\vecy
	&= \bbE_0^\vecx\left[
		\int
		\exp\left\{
			-\int_0^t U(\vecw_{c,s})\,ds
		\right\}
		\,\delta(\vecw_{c,t}-\vecy)
	\,f(\vecy) \,d\vecy
	\right] \\
%
	&= \bbE_0^\vecx \left[
		\exp\left\{
			-\int_0^t U(\vecw_{c,s})\,ds
		\right\}
	\,f(\vecw_{c,t})
	\right] \\
%
	&= e^{-\beta H} f(\vecx).
\end{align*}
where the last step follows from proposition \ref{prop:diffusion_FK}.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{$\Tr(e^{-\beta H})$ using Brownian bridges}
\label{sec:fk1_tr}

\begin{prop}
\label{prop:TrebH_bridge}
The trace may be computed using Brownian bridges as follows:
\begin{align*}
	\TrebH
	&=
	g_t(0) \int
	\bbE_{0,t}^{\vecx,\vecx} \left[
			\exp\left\{
				-\int_0^t U(\vecw_{c,s})\,ds
			\right\}
		\right]
	d\vecx.
\end{align*}
\begin{proof}
Using proposition \ref{prop:operator_trace}, we have
$$
	\TrebH
	=
	\int G_{t,U}(\vecx,\vecx)\,d\vecx;
$$
proposition \ref{prop:ebHG} gives us an expression for $G(\vecx,\vecy)$.
Setting $\vecy=\vecx$ in equation \ref{eqn:temp2}, we have
\begin{align*}
	\TrebH
	&= \int
		\bbE_0^\vecx\left[
			\exp\left\{
				-\int_0^t U(\vecw_{c,s})\,ds
			\right\}
			\,\delta(\vecw_{c,t}-\vecx)
		\right]
	\,d\vecx.
\end{align*}
Using proposition \ref{prop:Ex_Exy}, we may convert this expectation over
Brownian motion into an expectation over Brownian bridges:
\begin{align*}
	\TrebH
	&= g_t(0)\, \int
		\bbE_{0,t}^{\vecx,\vecx} \left[
			\exp\left\{
				-\int_0^t U(\vecw_{c,s})\,ds
			\right\}
		\right]
	\,d\vecx.
\end{align*}
\end{proof}
\end{prop}

\begin{rem}
Using definition \ref{defn:E_exp_U_bb}, we may match the notation of
\cite{bib:bu} and \cite{bib:qmath}:
$$
	\TrebH =
		\int d\vecx
		\int d\vecW^{2\beta}_{\vecx\vecx}(\vecw_{c,\cdot})
		\,
		\exp \left\{- \frac{1}{2} \int_0^{2\beta} U(\vecw_{c,s})\,ds \right\}.
$$
xxx find and fix the factors of 1/2 which appear in U/BU \ldots.
\end{rem}



%%% ================================================================
%\newpage
%\section{Cluster expansions}
%
%Keep it simple.
%
%Do it for 1D or 2D Ising.
%
%Xref to DU papers for the general (abstract!) case.



%% ================================================================
\newpage
\section{To be filed}

%% ----------------------------------------------------------------
\subsection{Misc.}

\begin{itemize}

\item Fokker-Planck vs. generator; write them out w/ cites; they coincide for
scaled Laplacian.  For B.M., $\frac{1}{2}\nabla^2$.

\item Fix $g_{2t}$ vs. $g_t$.

\item xxx really need diffusion eq?  E.g. diffusion-with-potential section.

\item find and fix all scale factors of $1/2$, $-U$'s, $1/g$ scale factors for
$d\vecWhat$'s, and $\beta$ vs. $c$ vs. $T$'s.

\end{itemize}

%% ----------------------------------------------------------------
\subsection{To-do list}

\begin{itemize}

% xxx boundedness and convergence ...

% xxx onpageref macro ... use it.

% xxx forward xrefs for lemmas & props.

%xxx mention other random variables:  $N_2(\pi)$, $\ell_{\textrm{max}}$.

\item $t$ to $\beta$ ... :/

\item somewhere rmk on CDF $\rho_{0,n}$ vs. PMF so that otherwise-invisible
spikes are now easy to see.

\item Pictures for introduction.

	Brownian bridges as a route to eliminate their own necessity.  Final
	computational model doesn't involve them at all.

\item Historical context.  Cite the Seiringer et al. text?

\item Critical density and Fourier transform --- ?!?

\item Write up $1/\rho$ scaling for clarification of theorem 1.  Pictures.

\item (Should be quick.)  Write up F-K rep section.

\item Cluster expansion \ldots needs lots of figures.

\item (Easy.)  Simplified interaction ($dWdW$ to $dW$):  do the algebra.

\item Eliminate some/all thermo-appx subsections.  Eliminate some
	stat-mech-appx and QSM subsections.

\item Annotate the bib.

\item Minimize the diffusion-equation stuff.  Maybe a footnote or some such.

\item Somewhere write down the non-interacting particle hamiltonian and
solve it as per Lieb et al.  Maybe before/in/after fkboson section?

\end{itemize}
