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\theoremstyle{plain}
\newtheorem{theoremeq}{Theorem}[section]
\newtheorem{prop}[theoremeq]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theoremeq]{Definition}
%\newtheorem{stand}[theoremeq]{Standing Assumption}
\newtheorem{remark}[theoremeq]{Remark}

\begin{document}
\title[Course notes on Quantization and Cohomology]{Course notes on
Quantization and Cohomology, Winter 2007}

\author[John Baez]{John Baez*\\ Notes by Apoorva Khare*}
\footnotetext{$^*$Department of Mathematics, University of California at
Riverside, USA}
\maketitle

\tableofcontents
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Jan 16, 2007: \schr's equation}

These are lecture notes taken at UC Riverside, in the Tuesday lectures of
Professor John Baez's Quantum Gravity Seminar, Winter 2007.
\vspace{3ex}

\noindent{\small
\begin{tabular}{p{2.15in}|p{2.5in}}
\hline
Classical ($p=1$) & Quantum ($p \geq 0$)\\
\hline
\hline
\vspace{0.3ex}
(1) {\bf Lagrangian Mechanics.} The path between two points $(t_i, q_i)$
is $\g \in \pq = \pq Q$ satisfying $\delta S(\g) = 0$. Here,\hfill\break
\noindent $\bullet\ Q$ is the configuration space,\hfill\break
$\bullet\ \pq$ is the path space,\hfill\break
$\bullet\ \{ \g : [t_0, t_1] \to Q : \g(t_i) = q_i \}$,\hfill\break
$\bullet\ L : TQ \to \R$ is the Lagrangian, and\hfill\break
$\bullet\ S(\g) := \int_{t_0}^{t_1} L(\g(t), \gd(t))\ dt$ is the
action.
&
\vspace{0.3ex}
(1) {\bf Lagrangian Mechanics.} The position of a particle at time $t$ is
given by a {\it wavefunction} $\psi_t : Q \to \R$. Moreover, the
amplitudes $|\psi_t|$ form probability densities, so $\psi_t \in \ltq$
for all $t$.

To consider ``paths", we now have to integrate over all initial points
and all paths ending at $q_1$. Thus, $\psi_{t_1}(q_1)$ now
equals\hfill\break
$\displaystyle \int_{q_0 \in Q} \int_{\pq} \eish{\g} \psi_{t_0}(q_0) (\D
\g)\ dq_0$\hfill\break
\vspace{10ex}\\


[figure: $Q$ at both ends + path $\g \in \pq$]

&
[figure: $Q$ at both ends + endpoint wavefunctions $\psi_{t_i}$ + lots of
paths]
\vspace{10ex}\\


(2) {\bf Hamiltonian Mechanics.} We now have the phase space $X = T^*Q$
(or any symplectic manifold), with $H : X \to \R$ the Hamiltonian
function. (In local coordinates $(p_i, q^i)$, and using the Legendre
transform $\la : TQ \to T^*Q$, we have $H = p_i \qd^i - L \circ \la$.)

Then $\gt(t) := (q(t), p(t)) \in X$ satisfies Hamilton's
equations:\hfill\break
$\displaystyle \ddt \gt(t) = v_H(\gt(t))$

\noindent where $v_H$ is the Hamiltonian vector field, with $dH =
\omega(v_H, -)$, where $\omega = -d\aaa$ is the symplectic structure.
&
(2) {\bf Hamiltonian Mechanics.} Each $\psi_t \in \ltq$ satisfies {\it
\schr's equation}:\hfill\break
$\displaystyle \ddt \psi_t = \frac{1}{i\hbar} \hh(\psi_t)$

\noindent (look at it as $\psi : \R \to \ltq$, say,) where $\hh : \ltq
\to \ltq$ is a linear operator obtained (somehow!) by ``quantizing"
$H$.\\
\hline
\end{tabular}
}
\vspace{2ex}


\subsection{Questions}
This chart raises lots of questions. (We mention a couple of them
here.)
\begin{enumerate}
\item How do you do ``path-integrals" $\int_{\pq} \D \g$ over the path
space?

Apparently, there is no meaning to the ``measure" $\D \g$ (or it has not
yet been found!), but there {\it is}, to $\eish{\g} \D \g$, at least in
well-behaved cases. One such has been extensively studied by physicists:
given a smooth finite-dimensional manifold $Q$, define
\[ L(q, \qd) = \frac{m}{2} ||\qd||^2 - V(q) \]

\noindent where we have the obvious kinetic and potential components. (If
we replace $i$ by $-1$ in the case of the {\it harmonic oscillator} $V(q)
= |q|^2$, then the expression above, namely, $e^{-S(\g) / \hbar} \D \g$
is the {\it Wiener measure}.)\medskip


\noindent {\it Digression on complete Riemannian manifolds:}
Note that to define the above Lagrangian, we need additional assumptions
on $Q,V$. Namely, the kinetic component needs a metric, so we assume that
the manifold is Riemannian. Moreover, we want ``closed" manifolds, so
that particles ``don't fall off the edge" - so we assume that $Q$ is
complete as a metric space. Here, we have

\begin{definition}
Given a connected Riemannian manifold $M$,
\begin{enumerate}
\item $M$ is a metric space if we set the distance between points $m_0,
m_1$ $\in M$ to be the {\it Riemannian distance}:
\[ d(m_0, m_1) := \inf_{\g \in \ppm} |\g| \]

\noindent where $\ppm$ is the set of all {\it piecewise regular} curves
(i.e. $\gd(t)$ is zero or undefined at most at finitely many points) from
$m_0$ to $m_1$, and given any (smooth) parametrization $\g : [t_0, t_1]
\to M$, we have its (parametrization-independent) {\it length}
\[ |\g| := \int_{t_0}^{t_1} |\gd(t)|\ dt \]

\noindent Moreover, the metric topology is the same as the manifold
topology.

\item $M$ is {\it geodesically complete} if every maximal geodesic is
defined for all $t \in \R$.
\end{enumerate}
\end{definition}

Then (for ``completeness' sake" $\smiley$!) we have the following result,
that tells us a consequence of completeness:

\begin{theoremeq}[Hopf-Rinow Theorem]
A connected Riemannian manifold is complete (as a metric space) if and
only if it is geodesically complete.
\end{theoremeq}

Moreover, this is if and only if any two points can be joined by a
geodesic - which is why this is relevant to us.
\vspace{3ex}


\noindent {\it Back to our discussion:} We thus define $L(q,\qd)$ as
above, using the assumption that $Q$ is a connected complete Riemannian
manifold (to keep our particle from ``falling off the edge"), and $V : Q
\to \R$ should be smooth and bounded below (again to keep our particle
from acquiring a lot of kinetic energy, and ``shooting off to infinity"
in finite time).\medskip

\noindent {\bf References.} For the basic ideas, try Feynman and Hibbs,
{\it Quantum Mechanics and Path Integrals}.

For mathematical rigor, try Barry Simons' {\it Functional Integration and
Quantum Physics}.
\vspace{3ex}


\item How do we get the Hamiltonian operator $\hh : \ltq \to \ltq$ from
the Hamiltonian function $H : T^*Q \to \R$?

In some cases, it is easy to write down $\hh$, e.g. under the same
assumptions we made while discussing the previous question:
\[ H(q,p) = \frac{|p|^2}{2m} + V(q) \]

\noindent (where $H$ is a connected complete (finite-dimensional)
Riemannian manifold, and $V$ is smooth and bounded below). In this
situation, \schr wrote:
\[ \hh = -\frac{\hbar^2}{2m} \nabla^2 + {\rm mult}_V \]

\noindent where $\nabla^2 := \grad \cdot \grad$ is the {\it Laplacian}
(and $\grad$ the gradient). This sends a square-integrable function $f$
on $Q$ to
\[ f \mapsto -\frac{\hbar^2}{2m} (\grad f, \grad f) + f \cdot V \]

\noindent (where we compute the first term using the Riemannian metric).
This is often written simply as $-\frac{\hbar^2}{2m} \nabla^2 + V$. \schr
got this by guessing the quantization rule: $p \mapsto \frac{\hbar}{i}
\nabla$. (There were physical motivations for such a guess - namely,
results for waves.) This yields:
\[ \frac{|p|^2}{2m} = \frac{(p,p)}{2m} \mapsto -\frac{\hbar^2}{2m}
\nabla^2 \]\smallskip
\end{enumerate}


\subsection{Motivating geometric quantization}
Ideally, we would like
\begin{itemize}
\item a method to get $\hh$ from more general $H$.
\item to use our assumptions (on $Q$ and $V$) to show ``good" properties
of $\hh$.
\end{itemize}

For instance, if $A : K \to K$ is a self-adjoint operator on a Hilbert
space $K$, then $e^{iAt} : K \to K$ is well-defined and unitary
(preserves the inner product), and defining $\psi_t := e^{iAt}(\psi_0)$,
we get
\[ \ddt \psi_t = iA \psi_t \]

This is why we need the \schr operator to be a self-adjoint operator in a
Hilbert space - primarily for obtaining solutions to \schr's equation!
Which is what motivated Von Neumann and others to come up with the theory
of Hilbert spaces and self-adjoint operators on them in the 20th century.
Eventually, Kato and Rellich showed that the $\hh$ in our setup above, is
indeed self-adjoint.
\vspace{3ex}


But we would like a much more systematic theory of ``quantizing"
functions $H : T^*Q \to \R$ and getting operators $\hh : \ltq \to \ltq$.

Even better, can we handle the case when the phase space $X$ isn't
$T^*Q$? (So we do not have $Q$ - what would we replace $\ltq$
by?)\medskip

\noindent This leads us to ``geometric quantization". For more on this,
try

\href{http://www.math.ucr.edu/home/baez/quantization.html}{\tt
http://www.math.ucr.edu/home/baez/quantization.html}

\noindent Then try Sniatycki's book.\medskip

A lot of cohomology comes into the game - starting with the fact that
$[\omega] \in H^2(X,\R)$ must come from an {\it integral} cohomology
class, i.e. $[\omega] \in \im \varphi$, where $\varphi : H^2(X, \Z) \to
H^2(X, \R)$.
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Jan 23, 2007: Categorification}

Besides the ``obvious" questions raised by the chart presented last time,
there's a bigger question: {\bf What's really going on?}
Is the ``quantization" some arbitrary trick (that nature has settled
on), or does it have some deeper meaning? Let's try to dig
deeper!\medskip


\subsection{A secret functor}
What sort of entity is the action? Recall: in one approach, we have a
configuration space $Q$, and then the action is a function $S : P_{(t_0,
q_0) \to (t_1, q_1)} \to \R$, where if we denote $x_i := (t_i, q_i) \in
\R \times Q$, then the domain is more precisely written as
\[ \ppx := \{ \g : [t_0, t_1] \to Q : \g(t_i) = q_i \} \]

\vspace{8ex}
[figure: $Q$ at both ends + path $\g \in \ppx$]
\vspace{8ex}

But it's much deeper than this! Note that the action respects {\it
composition of paths}: given $\g_i \in P_{x_{i-1} \to x_i}$, we can
concatenate them to get a path $\g_1 \g_2 \in P_{x_0 \to x_2}$, and then
\[ S(\g_1 \g_2) = S(\g_1) + S(\g_2) \]

What this secretly means is that {\it the action is a functor}, from some
category $\scrp$ with
\begin{itemize}
\item $x = (t,q) \in \R \times Q$ as objects

\item given objects $x_0, x_1$, paths $\g \in \ppx$ as morphisms (so each
$\ppx$ is a Hom-space in $\scrp$)

\item composition of morphisms given by concatenation of paths (this is
indeed associative since we already have $t$ in the ``$x$-data", and
don't need to reparametrize from $[0,1] \to Q$);
\end{itemize}

\noindent to some category $\R$ with
\begin{itemize}
\item one object, $*$

\item all real numbers as morphisms

\item composition of morphisms is just addition (a group is just a one
object category, with all morphisms invertible).\medskip
\end{itemize}


\noindent {\bf Technical remarks}:
\begin{enumerate}
\item We cannot use all possible paths in defining $\scrp$ (since the
action and Lagrangian involved the derivative $\gd$), nor only smooth
paths (since composing paths might result in ``corners"):
\newpage

$\ $

\vspace{8ex}
\noindent [figure: a concatenation of paths $\g_1, \g_2$ with a
``cusp"-like point of non-differentiability at the common point $x_1 =
(t_1, q_1)$]
\vspace{8ex}

We could use piecewise smooth paths, though.\medskip

\item Alternatively, we could have specified both the position and
velocity (to avoid ``corners"), but note that the calculus of variations
involved in picking out an ``extremal" path automatically chooses these
by itself! So if we specified both postiion and velocity, then an
extremal path might not exist. (Physically, we don't want our
particle/path to have ``too much inherent data".)\medskip

\item Note that the above just talks of the action; we'll now see how to
{\it use} this categorical approach for classical and quantum mechanics.
Thus, (in quantum mechanics) a path does not represent a particle's
(time) evolution, only (the integral over) the entire path space does!
So, even though we do not know the positiion and momentum at the same
time for a {\it particle}, we do know it for each specific path!
\end{enumerate}\medskip


\subsection{Bringing in arbitrary categories}
So, can we do classical and quantum mechanics starting with any functor
$S : \calc \to \R$ (where now $\calc$ is {\it any} category), with
\begin{itemize}
\item ``configurations" as objects, and
\item ``paths" as morphisms?
\end{itemize}

\noindent To see this, we ask: how did we use $S : \calc \to \R$ in the
chart last time?\medskip

\noindent {\bf Classically}, we can do one of two things:

\noindent (1) We ``criticize" it - i.e. for each $x, y \in \calc$, we
look at $S : \hhom(x,y) = \hhom_{\calc}(x,y) \to \R$ and seek critical
points, i.e. $\g \in \hhom(x,y)$ with $dS(\g) = 0$.

\vspace{6ex}
[figure: $Q$ (or $\calc$) on both sides, with $\g : x \to y$ and $\delta
S(\g) = 0$]
\vspace{6ex}

This only makes sense if each set $\hhom(x,y)$ is a manifold or a more
general ``infinite-dimensional manifold" (e.g. a space of piecewise
smooth paths in a manifold $Q$), and $S$ is differentiable.

This is addressed by the theory of smooth categories and smooth functors,
cf. \href{http://www.math.ucr.edu/home/baez/2conn.pdf}{\tt
http://www.math.ucr.edu/home/baez/2conn.pdf}.\medskip

\noindent ($1'$) We ``minimize" it - i.e. for each $x,y \in \calc$, we
seek $\g \in \hhom(x,y)$ that minimizes $S(\g)$. (We might need each
$\hhom$-space to be a topological space (possibly compact!), and $S$
``continuous".)\medskip


\noindent {\bf Remarks}:
\begin{enumerate}
\item For both (1) and ($1'$), the issues of {\it existence} and {\it
uniqueness} of a $\g$ criticizing/minimizing the action, are very
important.

\item Case ($1'$) is closer to quantum mechanics, which we can see if we
study {\it Hamilton's principal function}: given $x,y \in \calc$, define
\[ Z(x,y) := \inf_{\g \in \hhom(x,y)} S(\g) \]

\noindent (assuming this infimum exists). In classical mechanics, this is
very important; we get the {\it Hamilton-Jacobi equations} by
differentiating $Z(x,y)$ with respect to $x$ or $y$, and working out the
answer. These are the classical analogue(s) of \schr's equation.
\end{enumerate}\medskip

\noindent Now consider the {\bf quantum case}.

\noindent (2) We integrate its exponential - i.e. for each $x,y$, we
compute a {\it (transition) amplitude}
\[ \zh(x,y) := \int_{\g \in \hhom(x,y)} \eish{\g}\ \D \g \]

For this, we want each set $\hhom(x,y)$ to be a measure space (or
generalized measure space), and $\eish{\g}$ must be integrable.

In this case, we can get \schr's equation by fixing $x$ and
differentiating $\zh(x,y)$ with respect to $y$ (or vice versa), and
working out the answer.\medskip

Now compare cases $(1')$ (i.e. $Z(x,y)$) and (2) (i.e. $\zh(x,y)$): the
classical and quantum cases.\medskip

\noindent{\small
\begin{tabular}{p{2.15in}|p{2.5in}}
\hline
Classical & Quantum\\
\hline
\hline
\vspace{0.3ex}
(a) $S(\g) \in \R$.
&
\vspace{0.3ex}
(a) $\eish{\g} \in U(1) \subset \C$ (because integrating this gives us a
number in $\C$, not $U(1)$).\\

(b) We take the infimum (i.e. minimum).
&
(b) We take the integral (i.e. sum).\\

(c) The group morphism is addition:\hfill\break
$S(\g_1 \g_2) = S(\g_1) + S(\g_2)$.
\vspace*{1ex}
&
(c) The group morphism is multiplication:\hfill\break
$\eish{\g_1 \g_2} = \eish{\g_1} \cdot \eish{\g_2}$.\\
\hline
\end{tabular}
}
\vspace{2ex}

Thus, we now ask: How are both of these, special cases of some
``prescription" for getting physics out of the action?
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Jan 30, 2007: Physics is rigged!}

\subsection{The analogous viewpoints}
From last time, we've seen a ``big analogy" between classical and quantum
mechanics of point particles. As we saw, when we minimize/integrate, the
identity for this operation in $\R$/$\C$ (respectively) is $+\infty$/0.
Thus, we need to replace the $\R$ used above, by $\rmin := \R \cup \{
+\infty \}$ throughout.

Moreover, since we add the actions of paths whenever we compose (in the
classical case), we need to expand the addition operation to all of
$\rmin$. This is done by: $+\infty + x = +\infty$ for all $x$, and
$(\rmin, +, 0)$ is a commutative monoid.\smallskip

We can now say more about the above analogy:\medskip

\noindent{\small
\begin{tabular}{p{1.5in}|p{1.7in}|p{1.3in}}
\hline
&
Classical mechanics of point particles
&
Quantum mechanics of point particles\\
\hline
\hline
\vspace{0.05ex}
(a) We start with the
&
\vspace{0.05ex}
action $S(\g) \in \rmin$.
&
\vspace{0.05ex}
amplitude $\eish{\g} \in \C$.\\

\vspace{0.05ex}
(b) What we do:
&
\vspace{0.05ex}
take the infimum/minimum.
&
\vspace{0.05ex}
take the integral/sum.\\

\vspace{0.05ex}
(c) Operational identity:
&
\vspace{0.05ex}
$+\infty$ - so $(\rmin, \min, +\infty)$ is a commutative monoid.
&
\vspace{0.05ex}
$0$ - so $(\C, +, 0)$ is an abelian group.\\

\vspace{0.05ex}
(d) When we compose paths, we
&
\vspace{0.05ex}
add actions.
&
\vspace{0.05ex}
multiply amplitudes.\\

\vspace{0.05ex}
(e) Operational identity:
&
\vspace{0.05ex}
0.
&
\vspace{0.05ex}
1.\\

\vspace{0.05ex}
(f) Overall:
&
\vspace{0.05ex}
$(\rmin, \min, +\infty, +, 0)$ is a commutative {\it rig}.
&
\vspace{0.05ex}
$(\C, +, 0, \cdot, 1)$ is a commutative ring.\\
\hline
\end{tabular}
}
\vspace{2ex}

\begin{remark}\hfill
\begin{enumerate}
\item We'll see later, how the picture on the right is really a
one-parameter family (one for every $\hbar$), and we ``go to the
classical case" as $\hbar \to 0$.

\item A \underline{rig} is a ``\underline{ring} without
\underline{n}egatives", i.e. a commutative monoid under addition, and a
monoid under multiplication, satisfying left/right distributive laws.
(In particular, every ring is a rig.)

\item $\rmin$ is a rig that satisfies $x \min x = x$ for all $x$. Thus,
it satisfies an ``idempotence" property, sort of the opposite of the
usual ``cancellational" addition that we see in a group ($a*b = a*c \To
b=c$.)
\end{enumerate}
\end{remark}\medskip


\subsection{Switching between the classical and quantum viewpoints}
In the rest of this lecture, we'll go back and forth between the
classical and quantum viewpoints. For instance,
\begin{enumerate}
\item In classical mechanics, action is a {\it functor} $S : \calc \to
\rmin$, where $\calc$ is any category (whose objects - resp. morphisms -
are called ``configurations" - resp. ``paths"), and $\rmin$ is a category
with one object whose morphisms are $x \in \rmin$ and composition is
addition (i.e. the ``multiplication" in the rig): $S(\g_1 \g_2) = S(\g_1)
+ S(\g_2)$.\medskip

\noindent Let's now ``quantize" this statement!\medskip

In quantum mechanics, amplitude is a functor $\eish{\cdot} : \calc \to
\C$, where $\calc$ is as above, and $\C$ is a category with one object
whose morphisms are $x \in \C$ and composition is $\cdot$: $\eish{\g_1
\g_2} = \eish{\g_1} \cdot \eish{\g_2}$.\medskip

\begin{remark}
The amplitude functor is called $\eish{\cdot}$ more for sentimental
reasons here, than out of any connection with $S$ itself.
\end{remark}\medskip


\item We now present an {\bf example}, wherein we ``translate" a concept
(over to the ``other side") using this analogy. However, unlike earlier,
we now start on the ``quantum side"!

Say $\calc = \scrp$ (as in the last class), with
\begin{itemize}
\item objects as points $x = (t,q) \in \R \times Q$, where $Q$ is some
``configuration space" (manifold),
\item morphisms $\g : x_0 = (t_0, q_0) \to x_1 = (t_1, q_1)$ are paths $:
[t_0, t_1] \to Q$ so that $\g(t_i) = q_i$.
\end{itemize}\medskip

In the {\bf quantum case}, a wavefunction $\psi : Q \to \C$ tells us the
amplitude for a particle to be at $q \in Q$. Of course, this is not a
``good" thing because we have $Q$ (and not $\R \times Q$), and this is
not really among the objects or morphisms of our category $\calc$.

But now, we can bring in $\calc$ as follows: We describe the
time-evolution of $\psi$ by
\[ \psi(t_1, q_1) = \int_{q_0 \in Q} \int_{\pq} \eish{\g} \psi_{t_0}(q_0)
(\D \g)\ dq_0 \]

The {\bf classical} analogue of a wavefunction is a known entity in
physics; let's call it $\psi_c$ for short. (Just as $S(\cdot) \mapsto
\eish{\cdot}$, we really should have $\psi$ in the quantum case coming
from $-i \hbar \ln \psi$, but we just use $\psi_c$.)

Thus, classically, $\psi_c : Q \to \rmin$ tells us the action for a
particle to be at $q \in Q$. By our analogy, it should evolve in time as
follows:
\[ \psi_c(t_1, q_1) = \inf_{q_0 \in Q} \inf_{\g : [t_0, q_0] \to [t_1,
q_1]} (S(\g) + \psi_c(t_0, q_0)) \]

\begin{remark}\hfill
\begin{enumerate}
\item Note that $t_0$ is fixed.

\item The $\D \g \cdot d q_0$ {\it now} just gives information /
identifies the space over which we integrate / minimize.

\item If we imagine $\psi_c(t_0, q_0)$ as the cost to ``start a trip" at
$q_0 \in Q_0$ at time $t_0$, and $S(\g)$ as the cost of the trip $\g$,
this formula tells us that $\psi_c(t_1, q_1)$ is the cheapest price to be
at $q_1 \in Q$ at time $t_1$. (Note that there may be many different
``equally cheap" ways to get there!)
\end{enumerate}
\end{remark}\medskip


\item In the quantum case, you can go ahead and use the path integral to
compute $\displaystyle \ddt{\psi}$ - you get \schr's equation.

In the classical case, you get the {\it Hamilton-Jacobi equations}.
\end{enumerate}\medskip


\subsection{Wick rotation and a spring in imaginary time (revisited)}
Last quarter, we saw that the {\it dynamics of point particles} is
analogous to the {\it statics of strings}. This analogy involves {\it
Wick rotation}, namely, the substitution $t \mapsto -it$. (Think of how
``rotating" clockwise by $90^\circ$ amounts to multiplying by $-i$ in the
complex plane.)\medskip

For example, consider a rock and a spring in a gravitational field:

\vspace{8ex}
\noindent [figure: rock and spring in gravitational fields - give
parabolic shapes oriented oppositely]
\vspace{8ex}

The action for the rock is
\[ S = \int^{t_1}_{t_0} \left[ \frac{m}{2}\ \qd \cdot \qd - V(q(t))
\right]\ dt \]

\noindent which, under Wick rotation, becomes (note that $\qd = \ddt q$,
so we have to bring in $-1$ from the $\qd \cdot \qd$ now)
\[ \int^{-it_1}_{-it_0} \left[ -\frac{m}{2}\ \qd \cdot \qd - V(q(-it))
\right]\ d(-it) \]

\noindent for the spring. This was what we referred to as the ``spring in
imaginary time" early last quarter! In this (second/spring) case, we
write it as the {\it energy} (cancelling various powers of $(-i)$ and
renaming variables):
\[ E = -iS = \int^{t_1}_{t_0} \left[ \frac{m}{2}\ \qd \cdot \qd + V(q(t))
\right]\ dt \]

\noindent since this is the energy of the spring, where $m$ is now the
{\it tension} (spring constant), $\qd$ refers to how ``stretched" the
spring is, and the energy is the sum of the ``tension energy" and the
gravitational (potential) energy.\medskip

\noindent {\it Next time}: The rig $\R$ (not $\rmin$) comes into play!
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Jan 23, 2007: Statistical mechanics and deformation of rigs}

We saw last time that the classical mechanics (dynamics) of particles
becomes the classical statics of strings by doing the substitutions
\[ t \mapsto -it, \qquad S \mapsto iE \]

\noindent Minimizing action now becomes minimizing energy.\medskip

What does the quantum mechanics (dynamics) of particles become when we do
these substitutions?


\subsection{Statistical mechanics ``quantizes" strings}
In quantum mechanics, the relative amplitude for a particle to trace out
a path is $\eish{}$. In ``statistical" {\it mechanics} (really {\it
thermal statics} - classical statics but with nonzero temperature $T$) -
and let us not even talk of thermo{\it dynamics} now! - the {\it
relative} probability for a system to be in configuration of energy $E$
is 
\[ e^{-E/kT} \]
where $k$ is {\it Boltzmann's constant} (a conversion factor between
energy and temperature).

Note that we have \[ e^{iS/\hbar} \mapsto e^{-E/kT} \] if we do the
substitutions
\[ S \mapsto iE, \qquad \hbar \mapsto kT \]

\noindent (or should we really have $S \mapsto iE/T$, because $\hbar$ is
a constant? But we also want to get to classical mechanics by $\hbar \to
0$, or $T \mapsto 0$!).\medskip

This makes some sense since $\hbar$ measures how big ``quantum
fluctuations" are:

\vspace{8ex}
\noindent [figure: quantum (``wiggling") version of a thrown rock]
\vspace{8ex}

\noindent (Thus, all paths ``far" from the path of least action cancel
one another out, and only the ``nearby" paths contribute.)\medskip

\noindent while $kT$ measures how big ``thermal fluctuations" are:

\vspace{8ex}
\noindent [figure: hanging spring at nonzero (positive!) temperature]
\vspace{8ex}

%(Thus, both systems are ``ensembles" of the same setup, but probabilistic
%in nature. That's how we get the ``wiggles": they represent ``all"
%possible paths/configurations. Since we cannot observe more than one
%possible state at one time point, we observe it at numerous time points
%- to collect ``statistics" about the probabilities.)

\noindent (Moreover, the relative quantities need to be normalized to
give the amplitude/probability.)\medskip


\subsection{A family of rigs via the Boltzmann map}
Henceforth I'll set $k=1$ and use the substitution $\hbar \mapsto T$.
Note that
\begin{itemize}
\item $e^{iS/\hbar} \in \C$, the rig of relative amplitudes, and
\item $e^{-E/T} \in \R^+$, the rig of relative probabilities.\medskip
\end{itemize}

\noindent (Here, $\R^+ = ([0, \infty), +, 0, \cdot, 1)$.)
In short, we have\medskip


\noindent{\small
\begin{tabular}{p{2.3in}|p{2.3in}}
\hline
Particles ($p=1$) & Strings ($p \geq 0$)\\
\hline
\hline
\vspace{0.05ex}
(1) Classical dynamics - we deal with\hfill\break
action $S \in \rmin$
&
\vspace{0.05ex}
(1) Classical statics - we deal with\hfill\break
energy $E \in \rmin$ (a ``different/imaginary" $\rmin$)\\

\vspace{0.05ex}
(2) Quantum dynamics - we deal with\hfill\break
relative amplitude $e^{iS/\hbar} \in \C$
&
\vspace{0.05ex}
(2) Thermal statics - we deal with\hfill\break
relative probability $e^{-E/T} \in \R^+$\\
\hline
\end{tabular}
}\medskip

We note that to go from the first column to the second, we use Wick
rotation:
\[ t \mapsto -it, \qquad S \mapsto iE, \qquad \hbar \mapsto T \]

We'd like to understand how quantum mechanics reduces to classical
mechanics as $\hbar \to 0$, but it's easier to understand how thermal
statics reduces to classical statics as $T \to 0$.

To do this, we'll formulate thermal statics using $E$ instead of
$e^{-E/T}$: for any $T>0$, we consider the {\it Boltzmann map} $\beta_T :
\rmin \to \R^+$:
\[ E \mapsto e^{-E/T} \qquad (+\infty \mapsto 0) \]

This isn't a rig homomorphism, just a one-to-one and onto function. So,
we'll pull back the rig structure on $\R^+$ to (the {\it set}) $\rmin$
via $\beta_T$, and get a rig $\R^T$.

As a set, $\R^T$ is just $[0, \infty)$, but now it's a rig with
\begin{eqnarray*}
a +_T b      & := &  \beta_T^{-1}(\beta_T(a) + \beta_T(b))\\
0_T          & := &  \beta_T^{-1}(0)\\
a \cdot_T b  & := &  \beta_T^{-1}(\beta_T(a) \beta_T(b))\\
1_T          & := &  \beta_T^{-1}(1)
\end{eqnarray*}

\noindent {\bf Homework.} Work out $+_T, 0_T, \cdot_T, 1_T$ explicitly,
and show that
\[ \lim_{T \to 0} \beta_T^{-1}(\R^+) = \rmin \]

\noindent or, in other words, that
\[ \lim_{T \to 0} (+_T, 0_T, \cdot_T, 1_T) = (\min, +\infty, +, 0) \]

So, the ``topological rig" $\R^T$ converges to the topological rig
$\rmin$ as $T \to 0$.\medskip

Also note that as $T \to +\infty$, all elements $\beta_T(a)$ converge to
either 0 (if $a \in \R$) or 1 (if $ a = +\infty$) - ``impossible" or
``possible" events respectively. (That is, when the going gets hot,
everything that is possible appears equally likely!)

(We'd be very happy if this were to be in a different rig - the logic rig
of truth values ($\{ 0, 1 \}, \vee, \wedge$). But this is not
so.)\medskip


\subsection{The analogous situation for quantization}
The moral of the above analysis is that ``thermal statics reduces to
classical statics as $T \to 0$"; in both cases we're really doing linear
algebra over some rig, and $\R^T \to \rmin$ as $T \to 0$.

Alas, seeing classical mechanics as an $\hbar \to 0$ limit of quantum
mechanics is harder, since
\[ \beta_\hbar : \rmin \to \C \mbox{ sending } S \mapsto e^{iS/\hbar},\
+\infty \mapsto 0 \]

\noindent is neither one-to-one nor onto, and its image is not a subrig
(though it's closed under multiplication). So we can't pull the rig
structure on $\C$ back to $\rmin$.\medskip

However, people do study quantization {\it indirectly} using the
$\displaystyle \lim_{T \to 0} \R^T = \rmin$ idea, which is called
\begin{itemize}
\item {\it tropical mathematics} (a really stupid term for the work of
Brazilian mathematicians - ``Arctic mathematics" would be better for $T
\to 0$ math!)

\item {\it idempotent analysis} (since $a \min a = a$ in $\rmin$)

\item {\it Maslov dequantization} (in reference to how $T \to 0$ limit
lets us study the $\hbar \to 0$ limit).
\end{itemize}\medskip


\noindent {\bf Solution to the homework.} It is easy to see the
following:
\begin{eqnarray*}
\beta_T^{-1}(a)  & := &  -T \ln(a)\\
a +_T b          & := &  -T \ln(e^{-a/T} + e^{-b/T})\\
0_T              & := &  -T \ln 0 = +\infty\\
a \cdot_T b      & := &  -T \ln(e^{-(a+b)/T}) = a+b\\
1_T              & := &  -T \ln 1 = 0
\end{eqnarray*}

This means that there is only one limit left to verify: that of $a +_T b$
as $T \to 0$. So say $a \leq b$. Then we compute:
\[ a +_T b = -T \ln(e^{-a/T}(1 + e^{(a-b)/T})) = -T \ln(e^{-a/T}) - T
\ln(1 + e^{(a-b)/T}) \]

The first term clearly equals $a$. Now denote $\aaa = e^{a-b}$. Then
$\aaa \in (0,1]$. As $(1 >) T \to 0^+$, we see that $\aaa^{1/T} \in
(0,1]$, whence $\ln (1 + \aaa^{1/T})$ is bounded. Therefore, by the
Pinching Theorem,
\[ \lim_{T \to 0} a +_T b = a + \lim_{T \to 0^+} T \cdot \ln (1 +
\aaa^{1/T}) = a + 0 \]

\noindent and we are done, since $a \min b = a$. \qed
\newpage

\addtocontents{toc}{\protect\vspace{2ex}}
\section{Feb 13, 2007: An example of path integral quantization - I}

We have a strategy for quantization, given any category $\calc$ (of
``configurations" and ``paths") and a functor
\[ S : \calc \to (\R, +) \]

\noindent (the ``action"). This gives a functor
\[ e^{iS/\hbar} : \calc \to (\C, \cdot) \]

\noindent (the ``amplitude"), and we compute the ``transition amplitude"
from any object $x \in \calc$ to $y \in \calc$ via
\[ \zh(x,y) = \int_{\g : x \to y} \eish{\g} \D \g \]

\noindent which requires also that we have measures on $\hom$-sets
$\hom_\calc(x,y)$.\medskip


\subsection{Example: free particle on the real line}

Let's do an example - the free particle on $\R$. Here, the objects of
$\calc$ form the set $\R^2 \ni (t,q)$, and morphisms $\g : (t_0, q_0) \to
(t_1, q_1)$ are paths $\g : [t_0, t_1] \to \R$, so that $\g(t_i) = q_i$.

\vspace{8ex}
[figure: path $\g$ from $x_0 = (t_0, q_0)$ to $x_1 = (t_1, q_1)$]
\vspace{8ex}

\noindent (This is the category $\scrp$ - with $Q = \R$ - that we
introduced earlier, and then the morphism spaces $\{ \g : x_0 \to x_1 \}$
were called $\pq$.) Thus,
\[ \zh((t_0, q_0), (t_1, q_1)) = \int_{\pq} \eish{\g}\ \D \g \]


\noindent where $S$ is the action for a free (i.e. no potential) particle
of mass $m$:
\[ S(\g) = \int_{t_0}^{t_1} L(\g(t), \gd(t))\ dt \]

\noindent with Lagrangian given by
\[ L(q, \qd) = \frac{m}{2} \qd^2 \]

\noindent since there's no potential.\medskip

To do the path integral over {\it all} paths $\g$, we first integrate
only over piecewise linear paths (and worry about ``taking the limit"
later). However, we first need a change of notation!
\[ (t_0, q_0) \leftrightarrow (t,q), \qquad (t_1, q_1) \leftrightarrow
(t',q') \]

\noindent We thus consider piecewise linear paths $\g : (t,q) \to
(t',q')$

\vspace{8ex}
[figure: zigzagging path with corners at $(t_j, x_j)$ for $j=2, 3, \dots,
N$]
\vspace{8ex}

\noindent for some chosen times $t_1, \dots, t_{N+1}$, where
\[ \pt := \{ t = t_1 < t_2 < \dots < t_N < t_{N+1} = t' \} \]

\noindent is a (not necessarily regular) partition of $[t,t']$ (also
called a {\it mesh}), and $\g$ is piecewise linear on each subinterval.

To integrate over all these piecewise linear paths, we just integrate
over $x_2, \dots, x_N \in \R$, where $x_j = \g(t_j)$.

{\it Then} we'll try to show that these integrals over (the space of)
piecewise linear paths converge as the {\it norm} $||\pt||$ of the
partition goes to zero. To use another name for the norm, we'll see what
happens as the {\it mesh spacing}
\[ \max_j (t_{j+1} - t_j) \]

\noindent goes to zero.\medskip


\subsection{Doing the math}

But first, let's see what these integrals look like - let's compute one:
\[ A_{\pt} = A_{\pt}((t,q), (t',q')) = \int_{\R^{N-1}} \exp \left(
\frac{i}{\hbar} \int_t^{t'} \frac{m}{2} \gd(s)^2\ ds \right)\ dx_2 \dots
dx_N \]

\noindent Actually, we need to rescale the Lebesgue measure by
{\it normalizing factors}:
\[ d x_j \mapsto \frac{dx_j}{c_j},\ c_j \in \R \]

\noindent where $c_j$ depends on $t_{j+1} - t_j$ (for all $j$). We need
these to get convergence as the mesh spacing goes to zero.\medskip

\noindent {\bf Question.} The $c_j$'s are chosen to make the math work
out fine. (These are what one calls {\it (re?)normalizations}.) But what
is the {\it physics} behind choosing them? (E.g. is it just analogous to
rescaling in order to get the total probability to equal 1?) Many people
would be very happy to know...\medskip

But $\g$ is piecewise-linear, so on the $j$th piece $[t_j, t_{j+1}]$, we
have
\[ \gd(s) \equiv \frac{x_{j+1} - x_j}{t_{j+1} - t_j} = \diff{j} \]

\noindent where $\dd x_j = x_{j+1} - x_j,\ \dd t_j = t_{j+1} - t_j$ for
all $j$. Hence,
\[ A_{\pt} = \int_{\R^{N-1}} \exp \left( \frac{im}{2\hbar} \sum_{j=1}^N
\left( \diff{j} \right)^2 \dd t_j \right)\ \frac{dx_2}{c_2} \dots
\frac{dx_N}{c_N} \]

\noindent The crucial thing is that if we choose the $c_j$'s correctly,
$A_{\pt}$ is actually {\it independent} of the mesh/partition $\pt$ - so
convergence is trivial!

In other words, we can actually compute $\zh((t,q),(t',q'))$ as an
integral over {\it linear} paths

\vspace{8ex}
[figure: a straight line path between $(t,q)$ and $(t',q')$]
\vspace{8ex}

\noindent of which there is only one.\medskip


To prove that $A_{\pt}$ is independent of the mesh $\pt$, let's think
instead about the rule for evolving a wavefunction $\psi$ in time:

\vspace{8ex}
\noindent [figure: wavefunctions $\psi_t, \psi_{t'} : Q \to \C$, and a
possible path from $(t,q)$ to $(t',q')$]
\vspace{8ex}

\[ \psi_{t'}(q') = \int_{\R} \int_{\g : (t,q) \to (t',q')} \eish{\g}
\psi_t(q)\ \D \g\ dq = \int_{\R} \zh((t,q), (t',q')) \psi_t(q)\ dq \]

When we approximate $\zh$ by integrating over piecewise linear paths, we
get ($\widetilde{\psi_t}$ is the approximation, and we write out the
exponentials in the order in which they are applied to the original
wavefunction $\psi_t(q)$):
\begin{eqnarray*}
&& \widetilde{\psi}_{t'}(q')\\
& = & \int_{\R^N} 
\exp \left( \frac{im}{2 \hbar} \frac{(\dd x_N)^2}{\dd t_N} \right)
\cdots
\exp \left( \frac{im}{2 \hbar} \frac{(\dd x_1)^2}{\dd t_1} \right)
\psi_t(q)\ dq\ \frac{dx_2}{c_N} \cdots\ \frac{dx_N}{c_N}\\
& = & \int_{\R^N} K(\dd t_N, \dd x_N) \cdots  K(\dd t_1, \dd x_1)
\psi_t(q)\ dq\ dx_2 \cdots dx_N
\end{eqnarray*}

\noindent where
\[ K(\dd t_j, \dd x_j) := \frac{1}{c_j}
\exp \left( \frac{im}{2 \hbar} \frac{(\dd x_j)^2}{\dd t_j} \right) \]

\noindent is the ``kernel".\medskip

\begin{remark}\hfill
\begin{enumerate}
\item We have to justify the third equality in
\begin{eqnarray*}
\psi_{t'}(q') & = & \int_{\R^N} \zh \cdot \psi_t(q) = \int_{\R^N}
\lim_{||\pt|| \to 0} A_{\pt} \cdot \psi_t(q)\\
& = & \lim_{||\pt|| \to 0} \int_{\R^N} A_{\pt} \cdot \psi_t(q) =
\lim_{||\pt|| \to 0} \widetilde{\psi}_{t'}(q')
\end{eqnarray*}

\noindent But if $A_{\pt}$ is independent of $\pt$, then the equality
becomes obvious! Moreover, we shall see next time that this {\it does},
indeed, hold.\medskip

\item To show that $A_{\pt}$ is independent of $\pt$, we just need to
show that $\widetilde{\psi}_{t'}(q')$ is independent of $\pt$ for all
$\psi_t$. (This is exactly like showing that $\int fg = 0\ \forall g \To
f = 0$, for then we can choose various delta-functions for $\psi_t$, and
get that $\widetilde{\psi}_{t'}(q) = 0$ for all $q$.)\medskip

\item To show independence from $\pt$, it is enough to show that if we
refine the mesh/partition by one point, we get the {\it same} kernel.
This is because if we are then given $A_P, A_Q$ for partitions $P,Q$,
then they are both equal to $A_{P \cup Q}$.

Thus, it is now enough to show that
\[ K(t_3 - t_1, x_3 - x_1) = \int_{x_2 \in \R} K(t_3 - t_2, x_3 - x_2)
K(t_2 - t_1, x_2 - x_1)\ dx_2 \]

\vspace{8ex}
\noindent [figure: $\g$ is a line, but we introduce an intermediate time
$t_2$ and a non-linear, piecewise-linear curve from $t_1$ to $t_3$]
\vspace{8ex}

\item Note that this is a proof where the $c_j$'s are unknown! So we will
in fact use the proof to determine the $c_j$'s (so that the above
condition holds). Moreover, we need $c_j$ to be a function of at most the
time - and as we will see next time, $c_j$ actually depends only on $\dd
t_j$!
\end{enumerate}
\end{remark}
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Feb 20, 2007: An example of path integral quantization - II}

Last time, we considered path integral quantization for a free particle
on a line. (This is ``exactly solvable".) We claimed that the exact
answer could be found by considering just {\it one} path - the straight
line path

\vspace{8ex}
[figure: straight line from $(t,x)$ to $(t',x')$]
\vspace{8ex}

\noindent - and essentially evaluating $e^{iS/\hbar}$ to go from $(t,x)$
to $(t',x')$:
\[ K(\dd t, \dd x) = \frac{1}{c(\dd t)}\exp \left( \frac{im}{2\hbar}
\frac{(\dd x)^2}{\dd t} \right) \]

\noindent where $c = c(\dd t)$ is a normalizing factor (that we have yet
to determine). We saw that to prove this, we just need to check
\begin{equation}\label{Ekernel}\tag{$\bigstar$}
K(t_3 - t_1, x_3 - x_1) = \int_{x_2 \in \R} K(t_3 - t_2, x_3 - x_2) K(t_2
- t_1, x_2 - x_1)\ dx_2
\end{equation}

\noindent i.e., in pictures,

\vspace{8ex}
\noindent [figure: Feynman diagram line = $\displaystyle \int \bigg($
Feynman diagram piecewise line broken at $(t_2, x_2) \bigg)\ dx_2$]
\vspace{8ex}


\subsection{Time-evolution operators}

To prove \eqref{Ekernel} (for some $c(\dd t)$), we could just do the
integral - but this is too annoying (for JB!). So instead, we'll take a
more conceptual route:

Consider the operator $U(t)$ which describes one step of time evolution
(via straight-line paths only):

\vspace{8ex}
\noindent [figure: wavefunctions at times $t_1, t_2$; the endpoint is
fixed at $(t_2, x_2)$, but there are lots of lines leading to it, all
from potential starting points $(t_1, x_1)$]
\vspace{8ex}

\[ \left( U(t_2 - t_1) \psi_{t_1} \right)(x_2) = \int_{\R} K(t_2 - t_1,
x_2 - x_1) \psi_{t_1}\ dx_1 \]

\noindent This tells us the wavefunction at time $t_2$ in terms of the
wavefunction at time $t_1$ (i.e. $\psi_{t_1}$), as an integral over
straight-line paths from $(t_1, x_1)$ to $(t_2, x_2)$.\medskip

\begin{remark}
Thus, {\it kernels} really are functions of two variables that are like
matrices, but with entries indexed by a {\it continuous} set. Integrating
against a kernel is like matrix multiplication - see the last two
equations above!
\end{remark}\medskip


In these (above) terms, \eqref{Ekernel} simply says
\[ U(t_3 - t_1) \psi_{t_1} \equiv U(t_3 - t_1) U(t_2 - t_1) \psi_{t_1}\
\forall \psi_{t_1},\ \forall t_i \in \R \]

\noindent or equivalently,
\[ U(t+s) \equiv U(t) U(s)\ \forall t,s \in \R \]

\noindent (One way is clear: integrate the second identity against any
$\psi_{t_1}$ to get the first. Conversely, the familiar principle of
$\int fg = 0\ \forall g \To f \equiv 0$ suggests that we use
delta-functions in place of $\psi_{t_1}$ to get the second equation at
all points.)\medskip


\subsection{Bringing in the Hamiltonian}

We would know this if we could write
\[ U(t) \equiv \exp \left( \frac{-itH}{\hbar} \right) = \exp \left(
\frac{t}{i\hbar} H \right) \]

\noindent for some operator $H$, since we then get
\[ \exp \left( \frac{-i(t+s)H}{\hbar} \right) \equiv \exp \left(
\frac{-itH}{\hbar} \right) \cdot \exp \left( \frac{-isH}{\hbar} \right)
\]

So we will show that $U(t) = \exp((t/i \hbar)H)$ for
\[ H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \]

\noindent (as \schr had obtained). Here, $H$ is the {\it Hamiltonian}
for the free particle.

When we show this, we'll see that if $\psi_t = U(t) \psi_0$, then
$\psi_t$ will satisfy \schr's equation:
\[ \ddt{\psi_t} = \ddt \left[ \exp \left( \frac{t}{i \hbar} H \right)
\psi_0 \right] = \frac{1}{i \hbar} H \left( e^{(t/i \hbar)H} \psi_0
\right) = \frac{1}{i \hbar} H \psi_t \]

So we just need to check that $(e^{(t/i \hbar)H} \psi)(x)$ is the same as
\[ (U(t) \psi)(x) = \int_{y \in \R} K(t,x-y) \psi(y)\ dy \]

Since both sides depend linearly on $\psi$, and both sides are
translation-invariant, it suffices to check this for $\psi = \delta$, the
Dirac-delta at the origin. In other words,
\[ \left( e^{(t/i \hbar)H} \delta \right)(x) = K(t,x) = K(t-0, x-0) \]

\begin{remark}\hfill
\begin{enumerate}
\item To make this rigorous, we need to
\begin{itemize}
\item introduce some topology on the space of wavefunctions,

\item show that in this space, the set of finite linear combinations of
delta functions is ``dense" in this topology, and

\item show that $\exp(\frac{t}{i \hbar}H), U(t)$ are ``continuous" on
this space in this topology.
\end{itemize}\medskip


\item Note that in this last equation,
\begin{itemize}
\item the left side is the Hamiltonian way of computing the amplitude for
a particle to end up at position $x$ at time $t$ if it (definitely!)
starts at the origin at time 0; and\medskip

\item the right side is the Lagrangian way to compute the same thing -
but integrating only over straight-line paths!
\end{itemize}
\end{enumerate}
\end{remark}\medskip


\subsection{Computing the normalizing factors}

Since we (almost) know one side - $K(t,x)$ - we'll just compute the other
side of the above equation. We'll use the {\it Fourier Transform}
\[ \widehat{\psi}(k) = \twopi \int_{\R} e^{-ikx} \psi(x)\
dx \qquad (k \in \R) \]

\noindent and its {\it inverse}
\[ \psi(x) = \twopi \int_{\R} e^{ikx} \widehat{\psi}(k)\ dk \]

\noindent (Here, ``$k$" stands for momentum.)\medskip

Next, we need a couple of small computations. Firstly, how do
differentiation and the Fourier transform interact? We use integration by
parts to compute:
\[ \widehat{\psi'}(k) = \twopi \int_{\R} e^{-ikx} \frac{d}{dx} \psi(x)\
dx = ik \twopi \int_{\R} e^{-ikx} \psi(x)\ dx = ik \widehat{\psi}(k) \]

\noindent so that for $\displaystyle H = -\frac{\hbar^2}{2m}
\frac{d^2}{dx^2}$, we get (using the Taylor series expansion)
\begin{eqnarray*}
\widehat{e^{(t/i \hbar)H} \psi}(k) & = & \widehat{e^{\frac{-t}{i \hbar}
\frac{\hbar^2}{2m} \frac{d^2}{dx^2}}\psi}(k) = \sum_{n \geq 0}
\frac{1}{n!} \left( \frac{it \hbar}{2m} \right)^n
\widehat{\frac{d^{2n}}{dx^{2n}} \psi}(k)\\
& = & \sum_{n \geq 0} \frac{1}{n!} \left( \frac{it \hbar \cdot i^2
k^2}{2m} \right)^n \widehat{\psi}(k) = e^{-itk^2 \hbar / 2m}
\widehat{\psi}(k)
\end{eqnarray*}

To simplify future computations, let's pick units where $\hbar = 1$ and
$m=1$, to lessen the mess. So we now know that $e^{(t/i \hbar)H}$ now
becomes $e^{-itH}$, and we compute:
\[ \widehat{e^{-itH} \delta}(k) = e^{-itk^2/2} \widehat{\delta}(k) =
e^{-itk^2/2} \twopi \int_{\R} e^{-ikx} \delta(x)\ dx =
\frac{e^{-itk^2/2}}{\sqrt{2 \pi}} \]

\noindent where we compute $\widehat{\delta}(k)$ from first
principles.\medskip

Now take the inverse Fourier transform:
\begin{eqnarray*}
(e^{-itH} \delta)(x) & = & \frac{1}{2 \pi} \int_{\R} e^{ikx}
e^{-itk^2/2}\ dk\\
& = & \frac{1}{2 \pi} \int \exp \left[ -\left( \frac{i}{2} \left(tk^2 -
2xk + \frac{x^2}{t} \right) - \frac{i}{2} \frac{x^2}{t} \right) \right]\
dk\\
& = & \frac{e^{ix^2/2t}}{2 \pi} \int \exp \left[ \frac{-i}{2} \left(
\sqrt{t} k - \frac{x}{\sqrt{t}} \right)^2 \right]\ dk\\
& = & \frac{e^{ix^2/2t}}{2 \pi \sqrt{t}} \int \exp \left( \frac{-i}{2}
u^2 \right)\ du
\end{eqnarray*}

\noindent where we use the $u$-substitution $u = \sqrt{t} k -
\frac{x}{\sqrt{t}}$ in the last step - and the last integral is just a
number.\medskip

So just as desired, $(e^{-itH} \delta)(x) = K(t,x)$, where $\displaystyle
K(t,x) = \frac{e^{\frac{i}{2} \frac{x^2}{t}}}{c(t)}$, and the normalizing
factor is
\[ \frac{1}{c(t)} = \frac{1}{2 \pi} \frac{1}{\sqrt{t}} \int_{\R} \exp
\left( \frac{-i}{2} u^2 \right)\ du \]

\noindent {\bf Note} that this integral is not always absolutely
convergent (this is similar, for instance, to the fact that a series does
not converge if its summand terms do not go to zero). However, to
evaluate it, we can consider $\lim_{R \to \infty} \int^R_{-R}$, or
alternatively, consider
\[ \int e^{-\tau u^2/2}\ du \]

\noindent where $\tau \in \C$ is close to $i$, but has a small positive
real part.

\noindent This then converges absolutely; we then take the limit as $\tau
\to i$. Let's do it:
\begin{eqnarray*}
\int_{\R} e^{-\tau u^2/2}\ du & = & \sqrt{ \int_{\R} e^{-\tau x^2/2}\ dx
\cdot \int_{\R} e^{-\tau y^2/2}\ dy}\\
& = & \sqrt{ \iint_{\R^2} e^{-\tau (x^2 + y^2)/2}\ dx\ dy}
\end{eqnarray*}

Now convert to polar coordinates, and the Jacobian is $r$ (this procedure
is very well-known), and compute
\[ = \sqrt{\int_0^\infty \int_0^{2 \pi} e^{-\tau r^2/2} r\ d\theta\ dr} =
(v = \frac{r^2}{2}) = \sqrt{2\pi \int_0^\infty e^{-\tau v}\ dv} =
\sqrt{\frac{2 \pi}{\tau}} \]

\noindent We now take the limit as $\tau \to i$. Thus,
\[ K(t,x) = \frac{e^{ix^2/2t}}{2 \pi \sqrt{t}} \cdot \sqrt{\frac{2
\pi}{i}} = \frac{e^{ix^2/2t}}{\sqrt{2 \pi i t}} \]

This is how to do the ``simplest path-integral in the world"!\medskip

\noindent {\it Next time:} We throw in a potential - then it may not be
exactly solvable, but we can still go some of the distance.
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Feb 27, 2007: More examples of path integrals}

\subsection{A potential problem}

Now let's consider a quantum particle of mass $m$ on the real line, but
in a potential
\[ V : \R \to \R \]

\noindent Let's compute its time evolution using a path integral, using
the fact that we already did (this for) the free particle. Our particle
can trace out any path
\[ \g : [0,T] \to \R \]

\noindent (where now, without loss of generality, we start the clock
ticking at 0), and its action is
\[ S(\g) = \int^T_0 \left( \frac{m}{2} \gd(t)^2 - V(\g(t)) \right) dt \]

\noindent so the path integral philosophy tells us:\medskip

Given a wavefunction $\psi_0 \in L^2(\R)$ at time 0, it will evolve to
$\psi_T \in L^2(\R)$ at time $T$, where
\[ \psi_T(x') = \int_{x \in \R} \int_{\g \in P_{x \to x'}} \eish{\g}
\psi_0(x)\ \D \g\ dx \]

\noindent (Here, $P_{x \to x'} = \{ \g : [0,T] \to \R, \g$ piecewise
regular, $\g(0) = x, \g(T) = x' \}$.)\medskip

To do this, we first integrate over piecewise linear paths like

\vspace{8ex}
\noindent [figure: piecewise linear path with corners at $x = x_0, x_1,
\dots, x_n = x'$, but with timepoints $t_i$ in a {\it regular partition}
of $[0,T]$]
\vspace{8ex}

\noindent (Here, $\dd t = T/n$, and we only consider {\it regular}
partitions of $[0,T]$. We now set $x_k = \g(k \dd t),\ \dd x_k = x_k -
x_{k-1}$ whenever defined.)\medskip

We now take the limit $n \to \infty$, if possible. So we hope
\[ \psi_T(x') = \lim_{n \to \infty} \int_{\R^n} \eish{\g} \psi_0(x_0)
\frac{dx_0}{c(\dd t)}\cdots \frac{dx_{n-1}}{c(\dd t)} \]

\noindent where
\[ S(\g) = \int^T_0 \left( \frac{m}{2} \gd(t)^2 - V(\g(t)) \right) dt
\sim \sum_{k=1}^n \left( \frac{m}{2} \left( \frac{\dd x_k}{\dd t}
\right)^2 \dd t - V(x_{k-1}) \right) \dd t \]

\noindent where the last expression merely an approximation, not an
equality. Thus, we have decided to approximate $\displaystyle
\int_{(k-1)\dd t}^{k \dd t} V(\g(t))$ by $V(x_{k-1})$. (Note that as $n
\to \infty$, this approximation might not matter!)\medskip

So, we hope:
\[ \psi_T(x') = \lim_{n \to \infty} \int_{\R^n} \prod_{k=1}^n
e^{\frac{im}{2\hbar} \frac{(\dd x_k)^2}{\dd t}} e^{\frac{-i}{\hbar}
V(x_{k-1}) \dd t} \cdot \psi_0(x_0) \frac{dx_0}{c(\dd t)}\cdots
\frac{dx_{n-1}}{c(\dd t)} \]


Hence we start with our wavefunction $\psi_0$, and repeatedly applying
the following two types of operators in alternation:
\begin{enumerate}
\item multiplication by $\displaystyle \exp \left( \frac{-i}{\hbar} V \dd
t \right)$, and

\item evolving it for a time $\dd t$ as if it were a free particle.
\end{enumerate}\medskip

The latter step, we've seen, amounts to the operator
\[ \psi \mapsto e^{-i H \dd t / \hbar} \psi \]

\noindent where $H$ is the {\it Hamiltonian} for a free particle:
\[ H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \]

\noindent So $\displaystyle \psi_T = \lim_{n \to \infty} \left( e^{-iH
\dd t/\hbar} e^{-iV \dd t/\hbar} \right)^n \psi_0$.\medskip


\subsection{The Lie-Trotter Theorem and self-adjoint operators}

We now need the following theorem.

\begin{theoremeq}[Lie-Trotter]
If $A$ and $B$ are (possibly unbounded) self-adjoint operators defined on
a Hilbert space $\mathscr{H}$ (with dense domains $D(A)$, $D(B)$
respectively), so that $A+B$ is essentially self-adjoint on $D(A) \cap
D(B)$, then
\[ e^{i (A+B)t} \psi = \lim_{n \to \infty} (e^{iAt/n} e^{iBt/n})^n \psi
\]

\noindent for all $\psi \in \mathscr{H}$.
\end{theoremeq}

\begin{remark}\hfill
\begin{enumerate}
\item Unbounded self-adjoint operators are only ``densely defined" on
$\mathscr{H}$.
Thus, $A+B$ is only necessarily defined on $D(A) \cap D(B)$.

\item Now, if $A$ is such an operator, then for each $t$, $e^{iAt}$ is
unitary. We can thus define $e^{iAt/n} e^{iBt/n}$ on $D(A) \cap D(B)$,
and the definition can be extended (for {\it this} operator) to all of
$\mathscr{H}$.

\item For details, see Volume 1, {\it Methods in Modern Mathematical
Physics}, by Reed and Simon.
\end{enumerate}
\end{remark}\medskip


In fact, $H$ and $V$ (i.e. mult$_V$) are self-adjoint on $L^2(\R)$, and
$H+V$ is essentially self-adjoint on $D(H) \cap D(V)$, if $V$ is
reasonably nice - e.g. continuous and bounded below. So in this case,
\[ \psi_T = \lim_{n \to \infty} \left( e^{-iH \dd t/\hbar} e^{-iV \dd
t/\hbar} \right)^n \psi_0 \]

\noindent exists and equals $\displaystyle e^{-i(H+V)T/\hbar} \psi_0$.
So we get that $\displaystyle \psi_T = e^{-i(H+V)T/\hbar} \psi_0$, and if
$\psi_0 \in D(H+V)$, we can differentiate this and get (set $T
\leftrightarrow t$):
\[ \frac{d}{dt} \psi_t = \frac{-i}{\hbar} (H+V) \psi_t \]

\noindent which is {\it \schr's equation}.\medskip


\subsection{Generalization to complete Riemannian manifolds}

We can also handle the case of a particle on a complete (connected)
Riemannian manifold $Q$ (see the lecture on January 16, 2007). Here
again,
\[ H = -\frac{\hbar^2}{2m} \nabla^2 \]

\noindent and ``$V$" = mult$_V$ ($V : Q \to \R$) are self-adjoint
operators on $L^2(Q)$,
and if $V$ is continuous and bounded below, $H+V$ is essentially
self-adjoint on $D(H) \cap D(V)$.\medskip

\begin{remark}\hfill
\begin{enumerate}
\item The notion of piecewise linear paths does not make sense here. But
we can still go for ``piecewise geodesic" paths, each piece in a small
enough coordinate-patch.

\item Then, $e^{-i HT/\hbar}$ is not exactly an integral over such paths
with only a few pieces; we need to take the limit as the number $n$ of
pieces goes to infinity.

\item So again, the final answer is as desired, but only in the limit -
so the intermediate steps are only approximations now, unlike the case
$Q=\R$.

\item Some people study this after applying Wick rotation; this leads to
the study of the {\it Heat equation} and of Brownian motion on manifolds.
\end{enumerate}
\end{remark}\medskip


\noindent {\bf Upshot.} So again, skipping lots of steps, we obtain this
formula for the time evolution of a wavefunction for a particle on $Q$
with potential $V$:
\[ \psi_T = \lim_{n \to \infty} \left( e^{-iH T/n \hbar} e^{-iV T /n
\hbar} \right)^n \psi_0 = e^{-i(H+V)T/\hbar} \psi_0 \]

\noindent where $\displaystyle H+V \leftrightarrow -\frac{\hbar^2}{2m}
\nabla^2 +$ mult$_V$. (See the notes from January 16, 2007!) Here, the
operator $\nabla^2$ is defined for general $Q$, and $H+V$ is the {\it
Hamiltonian} for our particle.\medskip


\subsection{Back to the general picture}

Now let's return to our general story. We have a category $\calc$ whose
objects are ``configurations" and whose morphisms are ``paths". In the
example we just saw, objects were points in $\R \times Q$ (spacetime) and
a morphism $\g : (t,x) \to (t',x')$ is a path $\g : [t,t'] \to Q$ so that
$\g(t) = x, \g(t') = x'$.

\vspace{8ex}
[figure: $\g : (t,x) \to (t',x')$]
\vspace{8ex}

We also have a functor $S : \calc \to \R$ (where $\R$ is the category
with one object, reals as morphisms, $+$ as composition) serving as our
action, giving
\[ e^{iS / \hbar} : \calc \to U(1) \subset \C \]

\noindent {\bf Question.} How do we get a Hilbert space from this general
framework? In the examples we just saw, we could use $L^2(Q)$ - but
there's no ``$Q$" in general! $Q$ came from our ability to ``slice" he
set of objects $\R \times Q$ into slices $\{ t =$ constant$\}$, called
{\it Cauchy surfaces} - surfaces on which we can freely specify ``initial
data" (= {\it Cauchy data}) for our wavefunction (and then we can solve
differential equations given such Cauchy data - for example, \schr's
equation).\medskip


\subsection{Digression of the day: Cauchy surfaces}

In Newtonian mechanics, spacetime is $\R \times \R^3$, so Cauchy surfaces
are the level sets for the first coordinate.

\vspace{8ex}
[figure: parallel family of horizontal straight lines]
\vspace{8ex}

In special relativity, spacetime is $\R^4$ with the Minkowski metric. So
Cauchy surfaces can be one of many different families of parallel lines.
{\it Lorentz transformations} take one family of Cauchy surfaces to
another.

\vspace{8ex}
\noindent [figure: parallel families of horizontal straight lines,
cutting one another transversely]
\vspace{8ex}

Finally, in general relativity, spacetime is a 4-manifold with a
Lorentzian metric; then Cauchy surfaces can be very badly behaved. They
may not even exist! For instance, consider $S^1 \times \R^3$. There are
closed timelike loops here.

(In fact, the problem in many (most?) time-traveller paradoxes in science
fiction, is that of a lack of Cauchy surfaces. You {\it cannot} be in two
places at once in the same universe!)
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Mar 6, 2007: Hilbert spaces and operator algebras from
categories}

Suppose we have a category $\calc$ of ``configurations and processes" and
an ``action" functor $S : \calc \to (\R,+)$ giving the phase (set $\hbar =
1$ henceforth) $e^{iS} : \calc \to (U(1), \cdot)$ describing the
amplitude for any process to occur.

How do we get a Hilbert space out of this? (One approach is to try and
get a Cauchy surface; however, we will try something different.) Here's
one avenue of attack:\medskip


\subsection{(Pre-)Hilbert spaces from categories}

First, as a zeroth approximation to our Hilbert space, form a vector
space as follows: let $\obc$ (resp. $\morc$) be the set of all objecs
(resp. morphisms) in $\calc$. Then we have $s,t : \morc \to \obc$,
assigning to any morphism $\g : x \to y$ its {\it source} $s(\g) = x$ and
{\it target} $t(\g) = y$ respectively.

Form the vector space $\fun(\obc)$ of ``nice" complex-valued functions on
$\obc$ - where we'll have to see what ``niceness" is required.

Then define for $\psi, \phi \in \fun(\obc)$ an ``inner product":
\[ \tphipsi := \int_{\g : x \to y} e^{iS(\g)} \overline{\phi(y)} \psi(x)\
\D \g \D x \D y = \int_{\morc} e^{iS(\g)} \overline{\phi(t(\g))}
\psi(s(\g))\ \D \g \]

\begin{remark}\hfill
\begin{enumerate}
\item For this to make sense, we need a measure on $\morc$, and $\psi,
\phi$ should be nice enough so that the integral converges - for
instance, $\psi \circ s, \phi \circ t \in L^2(\morc)$.

\item Under ``nice" conditions, given a measure $\D \g$ on $\morc$, we
can find a measure on $\obc \times \obc$, and for any point $(x,y)$ here,
a measure on Mor$_{\calc}(x,y)$, so that $\D \g \leftrightarrow \D \g\ \D
x\ \D y$.
\end{enumerate}
\end{remark}\medskip


Now we have questions:
\begin{enumerate}
\item Is $\tphipsi$ linear in $\psi$ and antilinear in $\phi$?

\item Is $\overline{\tphipsi} = \tangle{\psi,\phi}$?

\item Is $\tangle{-,-}$ nondegenerate? That is, given $\phi$ so that
$\tphipsi = 0\ \forall \psi$, is $\phi=0$?

\item Is $\tangle{\psi,\psi} \geq 0$ for all $\psi$?
\end{enumerate}\medskip

\noindent Consider these in turn:
\begin{enumerate}
\item is obvious if the integral is well-behaved.\medskip


\item is more interesting:
\begin{enumerate}
\item On the one hand,
\[ \overline{\tphipsi} = \int_{\g : x \to y} e^{-iS(\g)}
\overline{\psi(x)} \phi(y)\ \D y\ \D x\ \D y \]

\noindent whereas
\[ \tangle{\psi,\phi} =  \int_{\g : x \to y} e^{i S(\g)}
\overline{\psi(x)} \phi(y)\ \D \g \D x \D y = \int_{\g : y \to x} e^{i
S(\g)} \overline{\psi(y)} \phi(x)\ \D \g \D x \D y \]

\noindent upon relabelling $x \leftrightarrow y$. Thus, the equality of
the two comes from ``time reversal symmetry". It's easy if $\calc$ is a
groupoid, since then, given $\g : x \to y$, we get $\g^{-1} : y \to x$ -
and since $S$ is a functor, $S(\g^{-1}) = -S(\g)$. So we'll get
$\overline{\tphipsi} = \tangle{\psi,\phi}$ if the measure $\D \g$ on
$\morc$ is preserved by the transformation $^{-1} : \morc \to
\morc$.\medskip

\item But - our favourite example is not a groupoid! Recall - given a
manifold $Q$, we have a category with $\obc = \R \times Q$, and a
morphism $\g : (t_1, q_1) \to (t_2, q_2)$ is a path $\g : [t_1, t_2] \to
Q$ with $\g(t_i) = q_i$.

\vspace{8ex}
[figure: path from $(t_1, q_1)$ to $(t_2, q_2)$]
\vspace{8ex}

Here, we've been assuming $t_1 \leq t_2$, so this is not a groupoid. We
{\it could} adjoin inverses to get a groupoid, but then we'd get
morphisms like

\vspace{8ex}
\noindent [figure: two paths between $(t_1, q_1) \to (t_2, q_2) \to (t_1,
q_3)$ - so there's backwards motion in time!]
\vspace{8ex}

\noindent Such morphismsm do indeed show up in Feynman diagrams involving
antimatter, but would require further thoughts.\medskip

\item Research topics:
\begin{enumerate}
\item Study Feynman's original work on path integrals for a
special-relativistic particle, and see if he allowed paths like

\vspace{8ex}
[figure: path that has parts moving both ways in time]
\vspace{8ex}

\item If so, formalize what he did using some category $\calc$. Is it a
groupoid, or merely a $*$-{\it category}?
\end{enumerate}

\item A $*$-{\it category} is a category $\calc$ with a contravariant
functor $* : \calc \to \calc$ that is the identity on objects, and
satisfies $** = 1_{\calc}$. Equivalently, for any morphism $\g : x \to
y$, there is a morphism $\g^* : y \to x$ so that
\begin{enumerate}
\item $(\g_1 \circ \g_2)^* = \g_2^* \circ \g_1^*$
\item $(\g^*)^* = \g$
\end{enumerate}

\noindent (These imply that $(1_x)^* = 1_x\ \forall x \in \obc$.) This is
also called a {\it category with involution}, an {\it involutive
category}, or in quantum computing, a $\dagger$-{\it category}.\medskip


\item We now make some remarks.
\begin{remark}\hfill
\begin{enumerate}
\item An obvious eample is a groupoid with one object - also called a
group! Then $*$ is the inverse map on morphisms.

\item The main example, however, is the category of Hilbert spaces and
bounded linear operators, denoted by Hilb: given $T : H \to H'$, we get
the {\it adjoint operator} $T^* : H' \to H$, defined by
\[ \tangle{T^* \phi, \psi} := \tangle{\phi, T \psi} \]

\item The requirement that $x^* = x$ for all objects $x$ of $\calc$ is
somewhat ``evil"; we might ask for a ``non-strict" (ala monoidal
categories) version only.
\end{enumerate}
\end{remark}
\end{enumerate}\medskip

\item $\tangle{-,-}$ is usually degenerate, but that's not bad; we can
form the vector subspace $K \subset \fun(\obc)$ by
\[ K = \{ \psi : \tphipsi = 0\ \forall \phi \in \fun(\obc) \} \]

\noindent and form the quotient space
\[ H_0 = \fun(\obc)/K \]

\noindent on which we have $\tangle{-,-}$ defined by
\[ \tangle{[\phi], [\psi]} := \tphipsi \]

\noindent Then this new $\tangle{-,-}$ is nondegenerate on $H_0$.\medskip


\item Is $\tangle{\psi,\psi} \geq 0$?

To get this, we need some extra conditions - but we'd need to look at
some examples to find {\it nice} sufficient conditions.

\noindent This is somehow related to ``reflection positivity" in the
Osterwalder-Schrader Theorem.
\end{enumerate}\medskip

\noindent Anyhow, if we get that these properties all hold, then $H_0$ is
called a {\it pre-Hilbert space}; we can then {\it complete} it to get a
Hilbert space.\medskip


\subsection{Operators and multiplying them}
Besides the issue of producing a Hilbert spaces, there's the issue of
operators. How can we get some ``nice" operators in $\fun(\obc)$?

We can get them from elements $F \in \fun(\morc)$, some space of ``nice"
complex-valued functions on $\morc$:
\[ (F \psi)(y) := \int_{\g : x \to y} F(\g) \psi(x)\ \D \g\ \D x \]

\noindent where ``nice" means this converges.

In fact, we get an algebra of such operators with some luck:
\[ GF(\g) = \int_{\mathscr{P}} G(\g_2) F(\g_1)\ \D p \]

\noindent where we integrate over the set
\[ \mathscr{P} := \{ (\g_1, \g_2) \in \morc \times \morc : \g_2 \circ
\g_1 = \g \} \]

\noindent and $\D p$ is a measure on $\mathscr{P}$.\medskip

\begin{remark}\hfill
\begin{enumerate}
\item This is ``convolution"; $\fun(\morc)$ is called the ``category
algebra" of $\calc$.

\item If we're working over a {\it groupoid} $\calc$, the above integral
can be converted to an integral only over morphisms to one point, by a
``change of variables"; moreover, the measure here is one that has
already shown up earlier above.
\end{enumerate}
\end{remark}
\newpage


\addtocontents{toc}{\protect\vspace{2ex}}
\section{Mar 13, 2007: The big picture}


\subsection{The case of finite categories}
Last time, we sketched how to get a Hilbert space from a category $\calc$
(of ``configurations" and ``processes") equipped with an ``amplitude"
functor
\[ A : \calc \to U(1) \]

There are lots of subtleties involving analysis, but these evaporate when
$\calc$ is finite (so all morphism spaces are finite sets, all integrals
are finite sums). Then we form the vector space $\fun(\obc)$ - which now
means {\it all} functions $\psi : \obc \to \C$.

$\fun(\obc)$ is isomorphic to $\C[\obc]$ - the space of formal linear
combinations of objects of $\C$. (This corresponds to allowing {\it
superpositions} in quantum mechanics.)

Then we define a $\C$-sesquilinear map
\[ \tangle{-,-} : \C[\obc] \times \C[\obc] \to \C \]

\noindent by
\[ \tangle{y,x} := \sum_{\g : x \to y} A(\g)\ \forall x,y \in \obc \]

\noindent Recall that $A = e^{iS/\hbar}$ here. (The prefix ``sesqui"
means ``one-and-a-half", and this is particularly appropriate, since it
is antilinear - hence only $\R$-linear - in the first coordinate, but
$\C$-linear in the second coordinate.)

We're doing a ``path integral", but now it's a sum over morphisms - we're
implicitly using counting measure on $\hhom_{\calc}(x,y)$.

Take $\C[\obc]$ and mod out by
\[ \{ \psi \in \C[\obc] : \tangle{\psi,\phi} = 0\ \forall \phi \in
\C[\obc] \} \]

\noindent to get a vector space $H$ with a nondegenerate sesquilinear
form on it; if that's positive definite, then $H$ is a
(finite-dimensional) Hilbert space.\medskip


\subsection{Example: particle on a line}
Alex Hoffnung and I (=JB) have been looking at examples like this:

\vspace{8ex}
[figure: ``a particle on a line" : lattice in $T$ vs. $X$ plot]
\vspace{8ex}

\noindent In this example,
\begin{itemize}
\item $\obc = \{ 1, 2, \dots, T \} \times \{ 1, 2, \dots, X \}$.

\item Morphisms in $\calc$ are freely generated (under the composition /
concatenation operation, and including the identity) by morphisms $\g :
(t,x) \to (t+1,y)$ for all $t \in \{ 1, 2, \dots, T-1 \}$ and $x,y \in \{
1, 2, \dots, X \}$.
\end{itemize}\medskip

So a typical morphism in $\calc$ is like

\vspace{5ex}
\noindent [figure: a zigzagging ``rising" path in the lattice, connected
but not necessarily from the bottom to the top]
\vspace{5ex}

Now $\calc$ is a ``quiver".\medskip

If you choose the amplitude $A : \calc \to U(1)$ to be a discretized
version of the amplitude for a particle on a line, we recover standard
physics in the continuous limit.

\subsection{From particles to strings}
We really want to categorify all this. We now make a table as we have
done earlier; we shall make explanatory remarks later on.\medskip

\noindent{\small
\begin{tabular}{p{1.7in}|p{2.9in}}
\hline
Particles & Strings\\
\hline
\hline
\vspace{0.3ex}
(1) A category $\calc$.
&
\vspace{0.3ex}
(1) A 2-category (or double category) $\calc$.
\vspace{5ex}\\


[figure: path $\g : x \to y$]
&
[figure: $\g_1, \g_2 : x \to y$ and the enclosed area $\Sigma$ - or a
``rectangle" with horizontal edges $\g_i$ and enclosed area $\Sigma$ -
the {\it worldsheet} of a string]
\vspace{5ex}\\

\vspace{0.3ex}
(2) A functor $A : \calc \to U(1) \subset \C$. Here, $U(1)$ is the
1-category with one object $*$, and the morphisms are the elements of
$U(1)$.
&
\vspace{0.3ex}
(2) A 2-functor $: \calc \to U(1)[1]$. (This is explained later.)\\

\vspace{0.3ex}
(3) From $A : \calc \to U(1)$, we try to build a Hilbert space - but
first we form the vector space $\fun(\obc)$, which, if $\calc$ is finite,
is just\hfill\break
$\hhom(\obc,\C) \cong \C[\obc]$
&
\vspace{0.3ex}
(3) From $A : \calc \to U(1)[1]-\Tor \cong U(1)[1]$, we try to build a
2-Hilbert space $FUN(OB(\calc))$, which, if $\calc$ is finite, is
just\hfill\break
$\hhom(OB(\calc), \vectc) \cong$ (we hope)
$\vectc[OB(\calc)]$.\hfill\break

Here, $OB(\calc)$ could be the category formed by discarding the
2-morphisms in our 2-category $\calc$ - but this only works if $\calc$ is
strict. What to do in general? Good questions.\\


\vspace{0.3ex}
(4) We define $\tangle{-,-}$ on $\C[\obc]$ by\hfill\break
$\displaystyle \tangle{y,x} = \sum_{\g : x \to y} A(\g)$.\hfill\break
Here, we use $U(1) \hookrightarrow \C$ to add elements of $U(1)$ and get
elements of $\C$.
&
\vspace{0.3ex}
(4) $\tangle{-,-}$ on $\vectc$ should satisfy\hfill\break
$\displaystyle \tangle{y,x} = \bigoplus_{\g : x \to y} A(\g) \in
\vectc$.\hfill\break
Here we use $U(1)-\Tor \hookrightarrow \vectc$; this is explained
below.\\
\hline
\end{tabular}
}
\vspace{2ex}

\begin{remark}
We present some facts about torsors below; let us first address the other
points in the table above.
\begin{enumerate}
\item For any abelian group $A$ and $n \geq 0$, we can form an
$n$-category $A[n]$: the objects form a singleton set $\{ * \}$, the
1-morphisms are $\{ 1_* \}$, the 2-morphisms are $\{ 1_{1_*} \}$, and so
on, until the $n$-morphisms (between the unique $(n-1)$-morphism and
itself). This set is different - and equals $A$.\medskip

\item While defining $\tangle{-,-}$ in the case of strings, we use that
$U(1)-\Tor \hookrightarrow \vectc$, sending a torsor to its corresponding
1-dimensional vector space (that contains the circle $U(1)$). (In fact,
$U(1)-\Tor \hookrightarrow \C-\Tor$?)

\vspace{8ex}
[figure: circle maps to shaded plane containing circle]
\vspace{8ex}

\noindent Moreover, this vector space is actually a {\it Hilbert} space.
For more on this, see Daniel Freed's {\it Higher algebraic structures and
quantization}.
\end{enumerate}
\end{remark}\medskip

\subsection{Digression on torsors}
Given any group $G$, a $G$-{\it torsor} is a $G$-set isomorphic to $G$
(viewed as a $G$-set via left-multiplication). Denote the set of
$G$-torsors by $G-\Tor$. In other words, ``a torsor is a group that has
forgotten its identity".

Thus, $G$-torsor morphisms are $G$-set maps that are also bijections.

\begin{enumerate}
\item If $G$ is abelian, then $G-\Tor$ is a monoidal category with
\[ X \otimes Y := X \times Y / \{ (xg = gx, y) \sim (x, gy) \} \]

\noindent where $X,Y \in  G-\Tor$ and $g \in G$ acts on the right on $X$
(since $G$ is abelian and acts on $X$).\medskip


\item If $G$ is abelian, then $G-\Tor$ is a 2-category as follows: it has

one object $*$;

$G$-torsors as morphisms, composed using $\otimes$ defined above; and

$G$-torsor morphisms as 2-morphisms.

\item Next, recall the definition of the 2-category $G[1]$ for the
abelian group $G$. We now have that

\noindent $G[1]$ {\it is a skeleton of} $G-\Tor$ - so we have $G[1] \cong
G-\Tor$.\medskip


To see this, we need one ``special" morphism for every object. Thus,
identify one of the torsors with $G$! This is to correspond to the
identity morphism, since we can verify that $G \otimes G = G$ (where
$\otimes$ was defined above).
\end{enumerate}
\end{document}
