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

The purpose of my comprehensive exam is to exposit the papers \cite{bib:bu} and
\cite{bib:qmath}.  The two works overlap significantly:  the latter is a
conference proceeding which presents and summarizes the former; yet, the latter
also includes some computations which were omitted by the former.  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.

\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.

\end{itemize}

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

Bose and Einstein.  Condensation of the ideal (non-interacting) Bose gas.
Critical temperature.

Effects of interactions ... biblio page from rcm1?

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

What is the order parameter --- ?

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

\Suto{} and Ueltschi.

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

path-integral MC studies --- what ref?

%% ================================================================
\newpage
\section{Model of spatial permutations}

xxx more bibents!!!

%% ----------------------------------------------------------------
\subsection{Definition}

%% ----------------------------------------------------------------
\subsection{Lattice case}

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

Description of main result about occurrence of infinite cycles.

%% ================================================================
\newpage
\section{Existence of infinite cycles}

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{Relation to the quantum Bose gas}

Again, you should know the results, without necessarily understanding all the
steps.  But it would be great if you can precisely state which steps you are
still missing.

%% ----------------------------------------------------------------
\subsection{Step \ldots}

%% ----------------------------------------------------------------
\subsection{Step \ldots}

%% ----------------------------------------------------------------
\subsection{Step \ldots}

%% ----------------------------------------------------------------
\subsection{Step \ldots}

%% ================================================================
\newpage
\section{Simple model with two-cycle interactions}

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

%% ----------------------------------------------------------------
\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).

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

$N_2$ simulations with xref.

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.

%% ================================================================
\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}
	\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}

\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}
	\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}

\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*}
	\Tr(e^{-\beta H})
	= \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}
For $\vecx \in \R^d$, let
$$
	g_t(\vecx) = \frac{1}{(2\pi t)^{d/2}}\, e^{-|\vecx|^2/2t}.
$$
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 Chapman-Kolmogorov equation:
\label{prop:g_iter_2}
$$
	\int_\R g_s(\vecy-\vecx)g_{t-s}(\vecz-\vecy)\,d\vecy = g_{t}(\vecz-\vecx).
$$
\begin{proof}
This follows by completing the square.
\end{proof}
\end{prop}

xxx iteration \ldots.

xxx do I need this?  Not yet \ldots.

%% ----------------------------------------------------------------
\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 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 p.d.f.
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 $\Tr(e^{-\beta H})$ (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
$\Tr(e^{-\beta H})$.  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}

%% ================================================================
\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?

$\Tr(e^{-\beta H})$:  motivate as
$$
	\Tr(e^{-\beta H}) = \sum_j e^{-\beta E_j}
$$
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}),
$$
	Z = \Tr(e^{-\beta H})
$$
so
$$
	\rho = \frac{e^{-\beta H}}{\Tr(e^{-\beta H})}.
$$
Then property (3) of section \ref{sec:mixed_dm} becomes
$$
	\expn{A} = \frac{\Tr(A\,e^{-\beta H})}{\Tr(e^{-\beta H})}.
$$

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

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

%% ================================================================
\newpage
\section{Bose-Einstein condensation in the ideal Bose gas}

critical temperature

critical density

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

%% ================================================================
\newpage
\section{Stochastic calculus and the Feynman-Kac formula}

Here I recall some basic results, all in preparation for a discussion of the
Feynman-Kac formula.  For full information, see \cite{bib:gs},
\cite{bib:lawler}, or \cite{bib:oksendal}.

%% ----------------------------------------------------------------
\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 in each
component
$$
	\bbE[\vecb_s, \vecb_t] = \min\{s,t\}.
$$
Specifically, the probability distribution function (\plainidx{p.d.f.}) 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}

\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}(\vecy_1)
		g_{t_2-t_1}(\vecy_2-\vecy_1)
		\cdots
		g_{t_n-t_{n-1}}(\vecy_n-\vecy_{n-1})
		\, d\vecy_1 \,d\vecy_2 \cdots d\vecy_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 p.d.f. of the random variables
$\vecb_{t_1}, \ldots, \vecb_{t_n}$ is
$$
	g_{t_1}(\vecy_1)
	g_{t_2-t_1}(\vecy_2-\vecy_1)
	\cdots
	g_{t_n-t_{n-1}}(\vecy_n-\vecy_{n-1}).
$$
\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}(\vecy_1)
		&
		g_{t_2-t_1}(\vecy_2-\vecy_1)
		\cdots
		g_{t_n-t_{n-1}}(\vecy_n-\vecy_{n-1})
		\\
	&
		f(\vecy_1, \ldots, \vecy_n)
		\; d\vecy_1 \,d\vecy_2 \cdots d\vecy_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}

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 $\vecb_s = \vecy$, is
$$
	g_{t-s}(\vecz-\vecy).
$$
\begin{proof}
This is nothing more than $f_{Z\mid Y}(\vecz\mid \vecy)$ from elementary
probability (see e.g. \cite{bib:gs}), where $Z=\vecb_{t}$ and $Y=\vecb_{s}$.
From equation \ref{eqn:bmpn_bm}, with $n=2$, the joint density of $Y$ and $Z$
is
$$
	g_{t-s}(\vecz-\vecy) g_s(\vecy-\vecx).
$$
We compute
\begin{align*}
	f_{Z\mid Y}(\vecz \mid \vecy) &= \frac{f_{Z,Y}(\vecz,\vecy)} {f_Y(\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 p.d.f. $g_{t-s}(z,y)$ satisfies the following PDEs:
\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*}
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 p.d.f. 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}

\begin{notn}
We write
$$
	\vbb{c}{\vecx}{t} := \vecx + \sqrt{c} \, \vecb_t.
$$
This \bfidx{scaled Brownian motion} $\vbb{c}{\vecx}{t}$ (really, scaled and
shifted Brownian motion) has mean $\vecx$ and variance $ct$ in each component.
\end{notn}

Then 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
	\left(\bigcap_{k=1}^n \vbb{c}{\vecx}{t_k} \in D_k
	\right)
	=
		\int_{D_1}
	&
		\int_{D_2}
		\cdots
		\int_{D_n}
		\\
	&
		g_{ct_1}(\vecy_1-\vecx)
		g_{c(t_2-t_1)}(\vecy_2-\vecy_1)
		\cdots
		g_{c(t_n-t_{n-1})}(\vecy_n-\vecy_{n-1})
		\, d\vecy_1 \,d\vecy_2 \cdots d\vecy_n.
\end{aligned}
\end{equation}
\end{prop}

\begin{rem}
This is equivalent to saying that the joint p.d.f. of the random variables
$\vbb{c}{\vecx}{t_1}, \ldots, \vbb{c}{\vecx}{t_n}$ is
$$
	g_{ct_1}(\vecy_1-\vecx)
	g_{c(t_2-t_1)}(\vecy_2-\vecy_1)
	\cdots
	g_{c(t_n-t_{n-1})}(\vecy_n-\vecy_{n-1}).
$$
\end{rem}

\begin{rem}
\label{rem:ExFomega}
Also, equation \ref{eqn:bmpnE_bm} becomes
\begin{equation}
\label{eqn:bmpnE}
\begin{aligned}
	\bbE[f(\vbb{c}{\vecx}{t_1}, \ldots, \vbb{c}{\vecx}{t_n})] =
		\int_{\R^d}
		\int_{\R^d}
		\cdots
		\int_{\R^d}
		g_{ct_1}(\vecy_1-\vecx)
		&
		g_{c(t_2-t_1)}(\vecy_2-\vecy_1)
		\cdots
		g_{c(t_n-t_{n-1})}(\vecy_n-\vecy_{n-1})
		\\
	&
		f(\vecy_1, \ldots, \vecy_n)
		\; d\vecy_1 \,d\vecy_2 \cdots d\vecy_n.
\end{aligned}
\end{equation}
\end{rem}

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

xxx cite \cite{bib:faris} yet again ... .

xxx xref forward to section(s) where this is used.

\begin{defn}
The (scaled) \bfidx{Brownian bridge} running from $\vecx$ at time 0 to $\vecy$
at time $T$ is
$$
	\vbbb{c}{\vecx}{\vecy}{t} = \vbb{c}{\vecx}{t}
		- \frac{t}{T} (\vbb{c}{\vecx}{T} - \vecy).
$$
\end{defn}

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

xxx properties (e.g. covariance).

xxx Wiener measure above, for Brownian motion and $dW$ notation.
% int f dW
% = E[f]
% = lim n to infty int...int  gt's... f(y's) dy's... ?

xxx bridge measure and $dW$ notation.

\begin{prop}
The Brownian bridge $\vbbb{c}{\vecx}{\vecy}{t}$ has mean
$$
	\bbE[\vbbb{c}{\vecx}{\vecy}{t}]
	= \vecx + \frac{t}{T}\, (\vecy - \vecx).
$$
\begin{proof}
\begin{align*}
	\bbE\left[ \vbbb{c}{\vecx}{\vecy}{t}\right]
	&= \bbE \left[ \vbb{c}{\vecx}{t}
		- \frac{t}{T} (\vbb{c}{\vecx}{T} - \vecy) \right]
	= \bbE\left[\vbb{c}{\vecx}{t}\right]
		- \frac{t}{T}\, \bbE\left[ \vbb{c}{\vecx}{T}\right]
		+ \frac{t}{T}\, \bbE\left[ \vecy \right] \\
	&= \vecx + \frac{t}{T}\, (\vecy - \vecx)
	= \left(\frac{T-t}{T}\right) \,\vecx - \frac{t}{T} \,\vecy.
\end{align*}
\end{proof}
\end{prop}

%%% ----------------------------------------------------------------
%\subsection{temp}
%
%xxx
%$$
%		\int
%		\exp \left\{\int_0^\beta U(\omega)\,ds \right\}
%		\,dW^{2\beta}_{\vecx\vecx}
%		:=
%		g_\beta(0)
%		\bbE\left[ \exp \left\{\int_0^\beta U(\vbbb{c}{\vecx}{\vecx}{s})\,ds
%		\right\} \right] ?
%$$
%
%xxx
%$$
%	\bbE[f(\vecb)] =
%		\int
%		\exp \left\{\int_0^\beta U(\omega)\,ds \right\}
%		\,dW^{2\beta}_{\vecx\vecx}
%		:=
%		g_\beta(0)
%		\bbE\left[ \exp \left\{\int_0^\beta U(\vbbb{c}{\vecx}{\vecx}{s})\,ds
%		\right\} \right] ?
%$$
%
%\begin{align*}
%	\bbE[f(\vbb{c}{\vecx}{t_1}, \ldots, \vbb{c}{\vecx}{t_n})] =
%		\int_{\R^d}
%		\int_{\R^d}
%		\cdots
%		\int_{\R^d}
%		g_{ct_1}(\vecy_1-\vecx)
%		&
%		g_{c(t_2-t_1)}(\vecy_2-\vecy_1)
%		\cdots
%		g_{c(t_n-t_{n-1})}(\vecy_n-\vecy_{n-1})
%		\\
%	&
%		f(\vecy_1, \ldots, \vecy_n)
%		\; d\vecy_1 \,d\vecy_2 \cdots d\vecy_n.
%\end{align*}
%
%???
%\begin{align*}
%	\bbE[f(\vbb{c}{\vecx}{[0,t]})]
%	&=
%	\lim_{n\to\infty}
%		\int_{\R^d}
%		\cdots
%		\int_{\R^d}
%		g_{ct/n}(\vecy_1-\vecx)
%		g_{ct/n}(\vecy_2-\vecy_1)
%		\cdots
%		g_{ct/n}(\vecy_n-\vecy_{n-1})
%		\\
%	&
%		\qquad \qquad \qquad \qquad
%		f(\vecy_1, \ldots, \vecy_n)
%		\; d\vecy_1 \,d\vecy_2 \cdots d\vecy_n \\
%	%
%	&?:= \int_{\R^d} f(\vbb{c}{\vecx}{[0,t]}) \,dW^{??}_{??}.
%\end{align*}
%
%Q:  limit of Gaussians, vs. discrete $\Delta W$ approximation (It\=o $db_t$).
%
%Do that in 1D first, before anything else.

%% ----------------------------------------------------------------
\subsection{The Feynman-Kac formula for diffusion with potential}

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.  The
discussion in this section is again an elaboration on \cite{bib:faris}.

\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
$$
	\psi(t, \vecx)
	= e^{(c\nabla^2 + U(\vecx))t} \psi(0, \vecx)
	= \bbE \left[
		\exp\left\{
		%e^{
			\int_0^t U(\vbb{c}{\vecx}{s})\,ds
		%}
		\right\}
		\psi(0, \vbb{c}{\vecx}{t})
	\right],
$$
where $\vbb{c}{\vecx}{t}$ is scaled Brownian motion with mean $\vecx$ and
variance $ct$.
\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-\vecy_1) 
		e^{U(\vecy_1)t/n}
		\left(e^{c\nabla^2 t/n} e^{U(\vecy_1)t/n}\right)^{n-1}
		\psi(0, \vecy_1)
		\,d\vecy_1.
\end{align*}
Repeating yields
\begin{align*}
	\psi(t,\vecx)
	&= \lim_{n\to\infty}
		\int_{\R^d}
		\int_{\R^d}
		g_{ct/n}(\vecx-\vecy_1) 
		g_{ct/n}(\vecy_1-\vecy_2) 
		e^{U(\vecy_1)t/n}
		e^{U(\vecy_2)t/n}
		\left(e^{c\nabla^2 t/n} e^{U(\vecy_2)t/n}\right)^{n-2}
		\psi(0, \vecy_2)
		\,d\vecy_1
		\,d\vecy_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-\vecy_1) 
		g_{ct/n}(\vecy_1-\vecy_2) 
		\cdots
		g_{ct/n}(\vecy_{n-1}-\vecy_n)  \\
	&
		e^{U(\vecy_1)t/n}
		e^{U(\vecy_2)t/n}
		\cdots
		e^{U(\vecy_n)t/n}
		\psi(0, \vecy_n)
		\,d\vecy_1
		\,d\vecy_2 \dots
		\,d\vecy_n
	\\
%
	= \lim_{n\to\infty}
		\int_{\R^d}
		\int_{\R^d}
		\cdots
		\int_{\R^d}
	&
		g_{ct/n}(\vecx-\vecy_1) 
		g_{ct/n}(\vecy_1-\vecy_2) 
		\cdots
		g_{ct/n}(\vecy_{n-1}-\vecy_n)  \\
	&
		e^{\sum_{k=1}^n U(\vecy_k)t/n}
		\psi(0, \vecy_n)
		\,d\vecy_1
		\,d\vecy_2 \dots
		\,d\vecy_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\left[
		%e^{\sum_{k=1}^n U(\vbb{c}{\vecx}{kt/n})t/n}
		\exp\left\{ \sum_{k=1}^n U(\vbb{c}{\vecx}{kt/n}) \frac{t}{n} \right\}
		\psi(0, \vbb{c}{\vecx}{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\left[
		\exp\left\{ \int_0^t U(\vbb{c}{\vecx}{s}) \,ds \right\}
		\psi(0, \vbb{c}{\vecx}{t})
	\right].
\end{align*}
\end{proof}
\end{prop}

\begin{rem}
Note that in the $U=0$ case, we have
$$
	\psi(t, \vecx)
	= e^{c\nabla^2t} \psi(0, \vecx)
	= \bbE \left[ \psi(0, \vbb{c}{\vecx}{t}) \right].
$$
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 p.d.f. for $Y$.  Here, for each fixed $t$ and $\vecx$,
$Y$ is $\vbb{c}{\vecx}{t}$, $h(Y)$ is $\psi(0,\vbb{c}{\vecx}{t})$, and the
p.d.f. of $\vbb{c}{\vecx}{t}$ is $g_{ct}(\vecy-\vecx)$ as a function of
$\vecy$.
\end{rem}

%% ----------------------------------------------------------------
\subsection{Feynman-Kac formula for single-particle $\Tr(e^{-\beta H})$}
\label{sec:fk1}

Section \ref{sec:operator_trace} showed how to compute $\Tr(e^{-\beta H})$ 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_{\beta,U}(\vecx, \vecy) f(\vecy) \,d\vecy
\end{equation}
where
\begin{equation}
\label{eqn:temp2}
	G_{\beta,U}(\vecx,\vecy) = \bbE\left[
		\exp\left\{
			-\int_0^\beta U(\vbb{c}{\vecx}{s})\,ds
		\right\}
		\,\delta(\vbb{c}{\vecx}{\beta}-\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_{\beta,U}(\vecx, \vecy) f(\vecy) \,d\vecy
	&= \int
	\bbE\left[
		\exp\left\{
			-\int_0^\beta U(\vbb{c}{\vecx}{s})\,ds
		\right\}
		\,\delta(\vbb{c}{\vecx}{\beta}-\vecy)
	\right]
	\,f(\vecy) \,d\vecy
\end{align*}
Interchange expectation and integral [xxx justify this] and integrate out the
delta function:
\begin{align*}
	\int G_{\beta,U}(\vecx, \vecy) f(\vecy) \,d\vecy
	&= \bbE\left[
		\int
		\exp\left\{
			-\int_0^\beta U(\vbb{c}{\vecx}{s})\,ds
		\right\}
		\,\delta(\vbb{c}{\vecx}{\beta}-\vecy)
	\,f(\vecy) \,d\vecy
	\right] \\
%
	&= \bbE\left[
		\exp\left\{
			-\int_0^\beta U(\vbb{c}{\vecx}{s})\,ds
		\right\}
	\,f(\vbb{c}{\vecx}{\beta})
	\right] \\
%
	&= e^{-\beta H} f(\vecx).
\end{align*}
where the last step follows from proposition \ref{prop:diffusion_FK}.
\end{proof}
\end{prop}

\begin{prop}
The trace may be computed using Brownian bridges as follows:
\begin{align*}
	\Tr(e^{-\beta H})
	&=
	\bbE\left[
			\exp\left\{
				-\int_0^\beta U(\vbbb{c}{\vecx}{\vecx}{s})\,ds
			\right\}
			\vbbb{c}{\vecx}{\vecx}{\beta}
		\right] \\
\end{align*}
\begin{proof}
Using proposition \ref{prop:operator_trace}, we have
$$
	\Tr(e^{-\beta H})
	=
	\int G_{\beta,U}(\vecx,\vecx)\,d\vecx;
$$
proposition \ref{prop:ebHG} gives us an expression for $G(\vecx,\vecy)$.
Setting $\vecy=\vecx$ in equation \ref{eqn:temp2}, we have
\begin{align*}
	\Tr(e^{-\beta H})
	&= \int
		\bbE\left[
			\exp\left\{
				-\int_0^\beta U(\vbb{c}{\vecx}{s})\,ds
			\right\}
			\,\delta(\vbb{c}{\vecx}{\beta}-\vecx)
		\right]
	\,d\vecx.
\end{align*}
xxx:
\begin{align*}
	\Tr(e^{-\beta H})
	&= g_\beta(0)\, \int
		\bbE\left[
			\exp\left\{
				-\int_0^\beta U(\vbbb{c}{\vecx}{\vecx}{s})\,ds
			\right\}
		\right]
	\,d\vecx.
\end{align*}
\end{proof}
\end{prop}

xxx DU notation:
\begin{prop}
$$
	\Tr \left(e^{-\beta H} \right) =
		\int
		\int
		\exp \left\{- \int_0^\beta U(\omega)\,ds \right\}
		\,dW^{2\beta}_{\vecx\vecx} \,d\vecx.
$$
\begin{proof}
xxx what about DU's definition in terms of limit of Gaussian integrals.

xxx is that definable only for exp's of integrals?

xxx find and fix the factors of 1/2 which appear in U/BU \ldots.
\end{proof}
\end{prop}

%% ----------------------------------------------------------------
\subsection{Feynman-Kac formula for bosonic $\Tr(e^{-\beta H})$}
\label{sec:fkboson}

%% ================================================================
\newpage
\section{The It\=o perspective on the Feynman-Kac formula}

%% ----------------------------------------------------------------
\subsection{The It\=o formula in one dimension}

\begin{defn}
Let $b_t$ be one-dimensional Brownian motion.  Let $e(t,x)$ and $f(t,x)$ be
$\mcF_t$-measurable where $\mcF_t$ is the filtration of Brownian motion.  Then
$$
	dX_t = e(t,X_t)\,dt + f(t,X_t)\,db_t
$$
is called an \emphidx{It\=o process}.
\end{defn}

\begin{prop}
\label{prop:ito_1d}
If
\begin{align}
\label{eqn:dX_edt_fdb}
	dX_t = e(t,X_t)\,dt + f(t,X_t)\,db_t
\end{align}
is an It\=o process and
$$
	y = g(t,x)
$$
is a $C^2$ map from $[0,+\infty) \times \R$ to $\R$, then
$$
	Y_t = g(t, X_t)
$$
is again an It\=o process with
\begin{align}
\label{eqn:ito_1d_1}
	dY_t = \fracDgt(t,X_t)\,dt + \fracDgx(t,X_t)\,dX_t
		+ \frac{1}{2} \fracDgxx(t,X_t)\,dX_t^2.
\end{align}
\begin{rem*}
Recall that we use the It\=o calculus rules
$$
	db_t^2 = dt, \qquad
	dt^2 = 0, \qquad
	dt\,db_t = 0
$$
when expanding $dX_t^2$.
\end{rem*}
\begin{rem*}
If we expand $dX_t$ (equation \ref{eqn:dX_edt_fdb}) into \ref{eqn:ito_1d_1},
we obtain an alternate expression
\begin{align}
\label{eqn:ito_1d_2}
	dX_t^2 &= e^2\,dt^2 + 2ef\,dt\,db_t + f^2 db_t^2 = f^2 dt \\
	dY_t &= \fracDgt(t,X_t)\,dt + \fracDgx(t,X_t)\,(e(t,X_t)\,dt
		+ f(t,X_t)\,db_t) + \frac{f(t,X_t)^2}{2} \fracDgxx(t,X_t)\,dt \\
	&= \left(\fracDgt(t,X_t)
		+ e(t,X_t)\,\fracDgx(t,X_t)
		+ \frac{f(t,X_t)^2}{2} \fracDgxx(t,X_t)
		\right)\,dt
		+ f(t,X_t)\,\fracDgx(t,X_t)\,db_t.
\end{align}
\end{rem*}
\end{prop}

%% ----------------------------------------------------------------
\subsection{The It\=o formula in higher dimensions}

\begin{prop}
\label{prop:ito_nd}
Let
$$
	\colvecthree{dX^{(1)}_t}{\vdots}{dX^{(n)}_t}
	=
	\colvecthree{u_1\,dt}{\vdots}{u_n\,dt}
	+
	\prmatrix{
		v_{1,1} & \cdots & v_{1,m} \\
		\vdots  &        & \vdots  \\
		v_{n,1} & \cdots & v_{n,m} \\
	}
	\colvecthree{db^{(1)}_t}{\vdots}{db^{(m)}_t},
$$
i.e. in matrix notation
$$
	d\vecX_t = \vecu\,dt + \vecV \,d\vecb_t.
$$
(This is called am $n$-dimensional It\=o process.)
Let $\vecg:[0,+\infty) \times \R^n \to \R^p$ be a $C^2$ map:
$$
	\colvecthree{y_1}{\vdots}{y_p}
	= \colvecthree
		{g_1(t, \vecx)}
		{\vdots}
		{g_p(t, \vecx)}
	= \colvecthree
		{g_1(t, x_1, \ldots, x_n)}
		{\vdots}
		{g_p(t, x_1, \ldots, x_n)}.
$$
Then
$$
	\vecY_t = \vecg(t, \vecX_t)
$$
is again an It\=o process with $k$th component (for $k = 1, \ldots, p$)
$$
	dY^{(k)}_t
		= \fracDgt(t,\vecX_t)\,dt
		+ \sum_{i=1}^n \frac{\Dg_k}{\Dx_i} (t,\vecX_t)\, dX^{(i)}_t
		+ \frac{1}{2} \sum_{i,j=1}^n \frac{\D^2 g_k}{\Dx_i \Dx_j}
			(t,\vecX_t)\, dX^{(i)}_t \,dX^{(j)}_t.
$$
\begin{rem*}
Recall that we use the It\=o calculus rules
$$
	db^{(i)}_t \, db^{(j)}_t = \delta_{ij}, \qquad
	dt^2 = 0, \qquad
	dt\,db^{(i)}_t = 0
$$
when expanding $dX^{(i)}_t \, dX^{(j)}_t$.
\end{rem*}
\end{prop}

\begin{cor}
\label{prop:stochastic_product_rule}
If $X_t$ and $Y_t$ are stochastic processes then
$$
	d(X_t Y_t) = X_t\,dY_t + Y_t\,dX_t + dX_t\,dY_t.
$$
\begin{proof}
Use the proposition with $g(x,y) = xy$.
\end{proof}
\end{cor}

%% ----------------------------------------------------------------
\subsection{The Feynman-Kac formula}

The following is adapted from \cite{bib:lawler}.  See \cite{bib:oksendal}
for a uniqueness proof.

\begin{prop}
The solution to the partial differential equation
\begin{align}
\label{eqn:fk1_stop}
	\fracDphit(t,x)
		&= \frac{f(x)^2}{2} \fracDphixx(t,x)
		+ e(x) \fracDphix(t,x) + g(x) \phi(t,x)
\end{align}
with initial condition
$$
	\phi(0,x) = h(x)
$$
is
\begin{align}
\label{eqn:fk1_start}
	\phi(t,x) &= \bbE^x\left[
		h(X_t) \exp\left\{ \int_0^t g(X_u) \,du \right\}
	\right],
\end{align}
where $X_t$ satisfies
$$
	dX_t = e(X_t)\,dt + f(X_t)\,db_t
$$
and $\bbE^x[Y]$ denotes $\bbE[Y \mid X_0 = x]$.
(We assume that this expectation exists for all $t,x$.)

\begin{rem*}
We call $X_t$ a \emphidx{time homogeneous diffusion}, or simply a
\emphidx{diffusion}, since $e$ and $f$ depend on $X_t$ but not on $t$ directly.
Note that $X_t$ is Markovian.

\end{rem*}
\begin{proof}
Let $\phi(t,x)$ be given by equation \ref{eqn:fk1_start}.  We will show that
it satisfies \ref{eqn:fk1_stop}.
For brevity, write
$$
	J_t = \exp\left\{\int_0^t g(X_u)\,du \right\}.
$$
Note that for $s<t$,
\begin{align*}
	J_t &= \exp\left\{\int_0^t g(X_u)\,du \right\}
	= \exp\left\{\int_0^s g(X_u)\,du \right\}
		\exp\left\{\int_s^t g(X_u)\,du \right\}
	= J_s
		\exp\left\{\int_s^t g(X_u)\,du \right\}.
\end{align*}
If $s<t$ then, since $J_s$ is $\mcF$-measurable,
\begin{align*}
	\bbE[h(X_t)J_t\mid \mcF_s]
	&= J_s \bbE\left[
		h(X_t) \exp\left\{ \int_s^t g(X_u) \,du \right\}
		\mid \mcF_s
		\right] \\
	%
	&= J_s \phi(t-s,X_s).
\end{align*}
Call this quantity $M_s$.  We claim that it is a martingale for $0 \le s < t$.
To see this, suppose $r<s$.  Then
\begin{align*}
	\bbE\Bigg[\bbE\bigg[h(X_t) J_t \mid \mcF_s\bigg] \mid \mcF_r\Bigg]
		&= \bbE[h(X_t) J_t \mid \mcF_r].
\end{align*}

Using the stochastic product rule (proposition
\ref{prop:stochastic_product_rule}), we have
\begin{align*}
	dM_s &=
		J_s\, d\phi(t-s,X_s)
		+ dJ_s\, \phi(t-s,X_s)
		+ dJ_s\, d\phi(t-s,X_s).
\end{align*}
For the first differential, using the chain rule and the second fundamental
theorem of calculus,
\begin{align*}
	dJ_s &= \fracDs\left(\exp\left\{\int_0^s g(X_u)\,du \right\}\right)\,ds
	= \exp\left\{\int_0^s g(X_u)\,du \right\} g(X_s) \,ds
	= J_s \,g(X_s)\,ds.
\end{align*}
For the second differential, applying the It\=o formula (assuming sufficient
differentiability) in the form of equation \ref{eqn:ito_1d_2} to $Y_s =
\phi(t-s, X_s)$, we obtain
\begin{align*}
	d\phi(t-s,X_s)
	&= \left(\fracDphis(t-s,X_s)
		+ e(X_s)\,\fracDphix(t-s,X_s)
		+ \frac{f(X_s)^2}{2} \fracDphixx(t-s,X_s)
		\right)\,ds \\
	&+ f(X_s)\,\fracDphix(t-s,X_s)\,db_s.
\end{align*}
The third product term $dJ_s\,d\phi(t-s,X_s)$ is zero since its differentials
are of the form $ds^2$ and $ds\,db_s$.  Then
\begin{align*}
	dM_s &=
		J_s\,
		\left(\fracDphis(t-s,X_s)
			+ e(X_s)\,\fracDphix(t-s,X_s)
			+ \frac{f(X_s)^2}{2} \fracDphixx(t-s,X_s)
			\right)\,ds \\
	& + J_s \,f(X_s)\,\fracDphix(t-s,X_s)\,db_s
	+ J_s \,g(X_s) \, \phi(t-s,X_s) \,ds.
\end{align*}
Since $M_s$ is a martingale, the $ds$ terms must sum to zero.  There is a
common factor of $J_s$, but the exponential function is everywhere non-zero so
the following four terms must be zero:
\begin{align*}
	0 &=
	\fracDphis(t-s,X_s)
	+ e(X_s) \fracDphix(t-s, X_s)
	+ \frac{f(X_s)^2}{2} \fracDphixx(t-s, X_s)
	+ g(X_s) \phi(t-s,X_s) \\
	%
	-\fracDphis(t-s,X_s)
	&=
	\frac{f(X_s)^2}{2} \fracDphixx(t-s, X_s)
	+ e(X_s) \fracDphis(t-s, X_s)
	+ g(X_s) \phi(t-s,X_s).
\end{align*}
Putting $s=0$ and recalling $X_0=x$, we are done:
\begin{align*}
	\fracDphit(t,x)
	&=
	\frac{f(x)^2}{2} \fracDphixx(t, x)
	+ e(x) \fracDphix(t, x)
	+ g(x) \phi(t,x).
\end{align*}
\end{proof}
\end{prop}

\begin{rem}
In the context of the diffusion equation, where
$$
	\frac{\D\psi(t,\vecx)}{\Dt} = c\nabla^2 \psi(t,\vecx)
		+ U(\vecx)\psi(t,\vecx),
$$
this is
$$
	\psi(t,\vecx) = \bbE\left[
		\psi(0, \vbb{c}{\vecx}{t})
			\exp\left\{\int_0^t U(\vbb{c}{\vecx}{s})\,ds\right\}
		\right].
$$
This matches proposition \ref{prop:diffusion_FK}.
\end{rem}

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

%% ----------------------------------------------------------------
\subsection{Unitary evolution of the density matrix}

xxx $\rho \mapsto U \rho U^*$

% xxx boundedness and convergence ...

%% ----------------------------------------------------------------
\subsection{$\bbE^x$}

xxx
\begin{defn}
$$
	\bbE^\vecx(X_t) := \bbE[X_t \mid X_0 = \vecx].
$$
\end{defn}

xxx??
\begin{defn}
$$
	\bbE^{\vecx,\vecy}_T(X_t) := \bbE[X_t \mid X_0 = \vecx, X_T = \vecy].
$$
\end{defn}

%% ----------------------------------------------------------------
\subsection{Questions}

\begin{itemize}

\item $\vecb_0=0$ vs. $\vecb_0=\vecx$, notationally.  Maybe $\vecomega_t =
\vecx+\vecb_t$?

\item Where to put the scaling $c$?

\end{itemize}
