% Quantization and Cohomology
% Winter 2007 course notes
% lectures by John Baez
% notes by Apoorva Khare
% figures by Christine Dantas
% version as of 2007/05/28
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\usepackage[breaklinks=true]{hyperref}%To get clickable links to papers
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%% WINTER 2007 SYMBOLS %%%
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\begin{document}
\title[Course notes on Quantization and Cohomology]{Course notes on
Quantization and Cohomology, Fall 2006}
\author[John Baez]{John Baez\\
Notes by Apoorva Khare \\
Figures by Christine Dantas \\
\\
\\
Department of Mathematics, University of California
\\ Riverside, CA 92521, USA}
\maketitle
\tableofcontents
\newpage
\addtocontents{toc}{\protect\vspace{2ex}}
\section{Preface}
These are lecture notes taken at UC Riverside, in the Tuesday lectures of
John Baez's Quantum Gravity Seminar, Winter 2007.
The notes were taken by Apoorva Khare. Figures were prepared by
Christine Dantas based on handwritten notes by Derek Wise.
You can find the most up-to-date version of all this material here:
\begin{center}
\href{http://math.ucr.edu/home/baez/qg-winter2007/index.html#quantization}{\tt
http://math.ucr.edu/home/baez/qg-winter2007/}
\end{center}
These notes are a continuation of the Fall 2006 notes, available
here:
\begin{center}
\href{http://math.ucr.edu/home/baez/qg-fall2006/index.html#quantization}{\tt
http://math.ucr.edu/home/baez/qg-fall2006/}
\end{center}
\noindent
If you see typos or other problems with any of these notes, please let
John Baez know ({\tt baez@math.ucr.edu}).
\addtocontents{toc}{\protect\vspace{2ex}}
\section{Jan 16, 2007: \schr's equation}
\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{1ex}\\
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
\includegraphics[scale=0.5]{AllFigsHere/QABE-FinalFig.pdf}
\end{center}
&
\begin{center}
\includegraphics[scale=0.5]{AllFigsHere/QABE2-FinalFig.pdf}
\end{center}
\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%[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
\[ \ppx := \{ \g : [t_0, t_1] \to Q : \g(t_i) = q_i \} \]
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/QABE3-FinalFig.pdf}
% \caption{ \label{QABE3-FinalFig}}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
%$\ $
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Cusp-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\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{1ex}
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:
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/ParabShapes2-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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:
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.7]{AllFigsHere/QuantumRock-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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:
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.7]{AllFigsHere/QuantumSpring-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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 $\g(t_i) = q_i$.
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/PathGamma-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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 $\g : (t,q) \to
(t',q')$
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/ZigZag-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Linear-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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:
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/WaveFunctions-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Straight-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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,
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Feynman-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Piecewise-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
$\g(t) = x, \g(t') = x'$.
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Gamma-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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 $\g(t_i) = q_i$.
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/OnePath-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/TwoPaths-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
\clearpage
\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
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/BothWays-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\vspace{8ex}
[figure: path that has parts moving both ways in time]
\vspace{2ex}
\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:
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{1ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.8]{AllFigsHere/Lattice1-FinalFig.pdf}
\end{figure}
\vspace{1ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{2ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.9]{AllFigsHere/Lattice2-FinalFig.pdf}
\end{figure}
\vspace{2ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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.
\clearpage
\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{1ex}\\
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
\includegraphics[scale=0.5]{AllFigsHere/GammaPath-FinalFig.pdf}
\end{center}
&
\begin{center}
\includegraphics[scale=0.6]{AllFigsHere/Sigma-FinalFig.pdf}
\end{center}
\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%[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,
$U(1)-\Tor \hookrightarrow \C-\Tor$?)
%inserted by ccdantas%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace{2ex}
\begin{figure}[h]
\centering
\includegraphics[scale=0.6]{AllFigsHere/Circle-FinalFig.pdf}
\end{figure}
\vspace{2ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\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}