\chapter{Quantum Gravity (II)}
We begin with a very important identity which will be use to obtain
the $2-3$ Pachner move in $3D$ quantum gravity.
\begin{exercise}[Biedenharn--Elliot Identity]
The $6j$ symbols are defined by
$$
\xy
(-8,11)*{};(-4,4)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{a}};
(0,11)*{};(-4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{b}};
(8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{d}};
(-4,4)*{};(0,-3)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{j}};
\endxy
=\sum_i\left\{\matrix{a&b&i\cr
c&d&j\cr}\right\}
\xy
(-8,11)*{};(0,-3)**\dir{-} ?(.5)+(2,1)*{\scriptstyle\mathbf{a}};
(0,11)*{};(4,4)**\dir{-} ?(.5)+(2,1)*{\scriptstyle\mathbf{b}};
(8,11)*{};(4,4)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(0,-3)*{};(0,-12)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{d}};
(4,4)*{};(0,-3)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{i}};
\endxy
$$
Prove that they satisfy the Biedenharn--Elliot identity:
$$
\sum_h
\left\{\matrix{a&b&h\cr
c&f&e\cr}\right\}
\left\{\matrix{a&h&i\cr
d&g&f\cr}\right\}
\left\{\matrix{b&c&j\cr
d&i&h\cr}\right\}
=
\left\{\matrix{e&c&j\cr
d&g&f\cr}\right\}
\left\{\matrix{a&b&i\cr
j&g&e\cr}\right\}
$$
[Hint: Use the fact that
$$
\xy
(-12,15)*{};(0,-6)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{a}};
(-4,15)*{};(4,1)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{b}};
(4,15)*{};(8,8)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{c}};
(12,15)*{};(8,8)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{d}};
(8,8)*{};(4,1)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{j}};
(4,1)*{};(0,-6)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{i}};
(0,-6)*{};(0,-15)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{g}};
\endxy
$$
form a basis of spin networks from $\mathbf{a}\otimes\mathbf{b}\otimes\mathbf{c}\otimes\mathbf{d}$
to $\mathbf{g}$, and express
$$
\xy
(12,15)*{};(0,-6)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{d}};
(4,15)*{};(-4,1)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{c}};
(-4,15)*{};(-8,8)**\dir{-} ?(.5)+(2,-1)*{\scriptstyle\mathbf{b}};
(-12,15)*{};(-8,8)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{a}};
(-8,8)*{};(-4,1)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{e}};
(-4,1)*{};(0,-6)**\dir{-} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{f}};
(0,-6)*{};(0,-15)**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{g}};
\endxy
$$
as a linear combination of the former in two different ways; then
equate coefficients.] [Big hint:
$$
\xymatrix{
&&
\xy
(-3,4)*{};(1,-4)**\dir{-};
(-1,4)*{};(-2,2)**\dir{-};
(1,4)*{};(2,2)**\dir{-};
(3,4)*{};(0,-2)**\dir{-};
\endxy
\ar[drr]
\\
\xy
(-3,4)*{};(1,-4)**\dir{-};
(-1,4)*{};(-2,2)**\dir{-};
(1,4)*{};(-1,0)**\dir{-};
(3,4)*{};(0,-3)**\dir{-};
\endxy
\ar[urr]\ar[ddr]
&&&&
\xy
(-3,4)*{};(1,-4)**\dir{-};
(1,4)*{};(2,2)**\dir{-};
(-1,4)*{};(1,0)**\dir{-};
(3,4)*{};(0,-3)**\dir{-};
\endxy
\\
\\
&
\xy
(-3,4)*{};(1,-4)**\dir{-};
(-1,4)*{};(0,2)**\dir{-};
(1,4)*{};(-1,0)**\dir{-};
(3,4)*{};(0,-3)**\dir{-};
\endxy
\ar[rr]
&&
\xy
(-3,4)*{};(1,-4)**\dir{-};
(-1,4)*{};(1,0)**\dir{-};
(1,4)*{};(0,2)**\dir{-};
(3,4)*{};(0,-3)**\dir{-};
\endxy
\ar[uur]
\\}
$$
Follow the steps in the above diagram.]
\end{exercise}
\section{Topological Quantum Field Theories (II)}
We have seen that $n\Cob$ and $\Hilb$ have lots of things in common,
and topological quantum field theories exploit that fact.
To get it over with, let us just state the definition of a topological
quantum field theory and explain all the terms we need for it to make
sense. First, $n\Cob$ and $\Hilb$ are symmetric monoidal
$*$-categories.
\begin{definition}
A {\bf unitary topological quantum field
theory\/} is a symmetric, monoidal, $*$-functor
$$
Z\colon n\Cob\leftarrow \Hilb
$$
\end{definition}
\subsection{Category Theory (II)}
\begin{definition}
We say that a
category $\cal C$ {\bf has adjoints\/} or {\bf duals for morphisms\/}
or {\bf is a $*$-category\/} if there is a contravariant functor
$*\colon{\cal C}\rightarrow{\cal C}$ which takes objects to themselves
and such that $*^2=1$ (the identity functor). For any object $x$ or
morphism $f$, the dual is denoted $*(x)=x^*$ or $*(f)=f^*$.
\end{definition}
Spelling out the definition, $*$ has to satisfy the following
properties:
\begin{itemize}
\item $x^*=x$ for any $x\in\cal C$,
\item for any $f\colon x\rightarrow y$ there is a morphism $f^*\colon
y\rightarrow x$ (this is what ``contravariant'' means),
\item for any $x\in\cal C$, $(1_x)^*=1_{x^*}=1_x$,
\item for any morphisms $f\colon x\rightarrow y$ and $g\colon
y\rightarrow z$, we have $(gf)^*=f^*g^*$, and
\item $(f^*)^*=f$ for any morphism $f$.
\end{itemize}
Examples of $*$-categories are:
\begin{itemize}
\item $n\Cob$, where $M^*$ is obtained by exchanging the roles of input
and output. If the cobordism is imbedded, this can be represented as
reflection along the ``time'' direction.
$$
\xy
(-7,7)*\xycircle(3,2){};
(7,7)*\xycircle(3,2){};
(0,-7)*\xycircle(3,2){};
(-10,7)*{};(-3,-7)**\crv{(-11,3)&(-2,-3)};
(10,7)*{};(3,-7)**\crv{(11,3)&(2,-3)};
(-4,7)*{};(4,7)**\crv{(0,-1)};
\endxy
{\buildrel *\over\longrightarrow}
\xy
(7,-7)*\xycircle(3,2){};
(-7,-7)*\xycircle(3,2){};
(0,7)*\xycircle(3,2){};
(10,-7)*{};(3,7)**\crv{(11,-3)&(2,3)};
(-10,-7)*{};(-3,7)**\crv{(-11,-3)&(-2,3)};
(4,-7)*{};(-4,-7)**\crv{(0,1)};
\endxy
$$
\item $\Hilb$, where the adjoint $T^*$ of a linear operator $T\colon
{\cal H}\rightarrow{\cal H}'$ is defined by
${\bracket{T^*\phi}\psi}_{\cal H}={\bracket\phi{T\psi}}_{{\cal
H}'}$.
\item any {\bf groupoid\/} (a category where every morphism is
invertible), as then the inverse has the properties required of $*$.
\end{itemize}
\begin{definition}
We say that $F\colon{\cal C}\rightarrow{\cal
D}$ is a $*$-functor if, and only if, given $f\colon x\rightarrow y$,
we have $F(f^*)=F(f)^*\colon F(x)\rightarrow F(y)$.
\end{definition}
\begin{definition} A category $\cal C$ is {\bf monoidal\/} if it is
equipped with an operation $\otimes$ with the following properties:
\begin{itemize}
\item for any $x,y\in\cal C$, there is an object $x\otimes y\in\cal C$;
\item for any $f\colon x\rightarrow x'$ and $g\colon y\rightarrow y'$,
there is a morphism $f\otimes g\colon x\otimes y\rightarrow x'\otimes
y'$.
\item for any objects $x,y,z\in\cal C$ there is an isomorphism
$a_{xyz}\colon (x\otimes y)\otimes z\rightarrow x\otimes(y\otimes z)$ called {\bf associator\/} and satisfying the {\bf pentagon
identity\/}:
$$
\xymatrix{
&&\scriptstyle{(w\otimes x)\otimes(y\otimes z)}
\ar[drr]^{a_{wx(yz)}}\\
\scriptstyle{\bigl((w\otimes x)\otimes y\bigr)}
\otimes z\ar[urr]^{a_{(wx)yz}}\ar[ddr]^{a_{wxy}\otimes 1_z}
&&&&\scriptstyle{w\otimes\bigl(x\otimes(y\otimes z)\bigr)}\\
\\
&\scriptstyle{\bigl(w\otimes(x\otimes y)\bigr)\otimes z}
\ar[rr]^{a_{w(xy)z}}
&&\scriptstyle{w\otimes\bigl((x\otimes y)\otimes z\bigr)}
\ar[uur]^{1_w\otimes a_{xyz}}
\\}
$$
\item there is an object $1$ such that, for any object $x\in C$, there
are isomorphisms $l_x\colon 1\otimes x\rightarrow x$ and $r_x\colon
x\otimes 1\rightarrow x$ called {\bf units\/} satisfying the {\bf
other identity\/}:
$$
\xymatrix{&&x\otimes 1\ar[drr]^{r_x}\\
(1\otimes x)\otimes 1\ar[urr]^{l_x\otimes 1_1}
\ar[ddr]^{a_{1x1}}
&&&&x\\
\\
&1\otimes (x\otimes 1)\ar[rr]^{1_1\otimes r_x}&&1\otimes
x\ar[uur]^{l_x}\\}
$$
\item finally, given $f\colon x\rightarrow y$, $g\colon
y\rightarrow z$, $f'\colon x'\rightarrow y'$ and $g'\colon
y'\rightarrow z'$, we require that $(g\otimes g')(f\otimes
f')=(gf)\otimes(g'f')$, which just says that the following diagram
is unambiguous:
$$
\xy
(-5,8)*++{f}*\cir{}="f";
(-5,-8)*++{g}*\cir{}="g";**\dir{-} ?(.5)+(-2,0)*{\scriptstyle y};
"f";(-5,20)**\dir{-} ?(.5)+(-2,0)*{\scriptstyle x};
"g";(-5,-20)**\dir{-} ?(.5)+(-2,0)*{\scriptstyle z};
(5,8)*+{f'}*\cir{}="f'";
(5,-8)*++{g'}*\cir{}="g'";**\dir{-} ?(.5)+(-2,0)*{\scriptstyle y'};
"f'";(5,20)**\dir{-} ?(.5)+(-2,0)*{\scriptstyle x'};
"g'";(5,-20)**\dir{-} ?(.5)+(-2,0)*{\scriptstyle z'};
\endxy
$$
\end{itemize}
\end{definition}
MacLane's theorem guarantees that if the above two diagrams commute,
then any diagram that can be constructed from the associator and the
units commutes.
Examples of monoidal categories are
\begin{itemize}
\item $\Group$: objects are groups, morphisms are group homomorphisms and
$\otimes$ is the direct product of groups.
\item $n\Cob$: the $\otimes$, both for objects and for morphisms, is the
disjoint union of manifolds.
\item $\Vect$ or $\Hilb$: the $\otimes$ is the tensor product. This is
how, in quantum mechanics, two things are put together.
\item $\Elect$: it has just one object, morphisms are electrical circuit
elements, composition is serial combination of components, and
$\otimes$ is parallel or shunted combination of components.
\end{itemize}
Notice that parenthesised expressions can be drawn as trees. This is
how we start to see that category theory can be useful for physics,
and also we get a hint of what the Biendenharn-Elliot identity means.
Note also that we are promoting all the theorems we proved in the
first quarter to axioms. We can draw morphisms in any monoidal
category as two-dimensional diagrams just like we did in $\Vect$.
The moral of this story is that categorification---the process of
replacing equalities by isomorphisms---is a way to understand
processes (that is, time) at a deeper level.
%\proclaim Definition. The category $n\Cob$ is defined as follows:
%\unorderedlist
%\li {\bf objects\/} are closed $(n-1)$-dimensional unoriented manifolds,
%\li {\bf morphisms\/} are $n$-dimensional unoriented manifolds with boundary,
%\li {\bf products\/} are disjoint unions,
%\li it has an obvious {\bf symmetric braiding\/}, and
%\li the {\bf dual\/} of an $(n-1)$-dimensional manifold is itself, and the
% {\bf dual\/} of an $n$-cobordism is itself, exchanging the roles
% of initial and final object.
%\endunorderedlist
%
%\proclaim Definition. The category $\Hilb$ is defined as follows:
%\unorderedlist
%\li {\bf objects\/} are complex, finite-dimensional vector spaces,
%\li {\bf morphisms\/} are linear mappings,
%\li {\bf products\/} are tensor products,
%\li reversing the order of factors in a product is a {\bf symmetric
% braiding\/}, and
%\li the {\bf dual\/} of a vector space is itself (through the
% canonical antiisomorphism induced by the inner product), and the
% dual of a linear map is its {\bf adjoint}, defined by
% $$
% \bracket{T^*\phi}\psi=\bracket\phi{T\psi}.
% $$
%\endunorderedlist
\chapter{Quantum Mechanics form a Category-Theoretic Viewpoint (II)}
Recall that in quantum mechanics, we describe the evolution of the
state vector $\psi\in\cal H$ by a one-parameter, strongly continuous,
unitary group $U$, i.e., $\psi(t)=U(t)\psi(0)$. Then Stone's theorem
guarantees that $U(t)=\exp(-itH)$, where $H$ is self-adjoint and
corresponds to the energy of the system. Schr\"odinger's equation
$$
i{d\over dt}\psi(t)=-iH\psi(t),
$$
which is equivalent to $\psi(t)=U(t)\psi(0)$ if $\psi(t)$ is
differentiable, just states that ``energy is the same as rate of
change with respect to time''. This is one of the most significant
discoveries of the twentieth century, and although there were
indications of this coming from analytical mechanics in the nineteenth
century, it wasn't until the advent of quantum mechanics that the
connection was established so explicitly.
Now the question is, how do we get the Hamiltonian $H$ for a
particular problem? A possible answer, and the first we shall explore
here, is to ``steal it'' from classical mechanics.
\section{A Point particle on a line}
The classical Hamiltonian for a particle on a line is
$$
H={p^2\over 2m}+V(q),\qquad{\rm with}\quad\{q,p\}=1,
$$
where $q$ is an affine coordinate on the line and $(q,p)$ are
coordinates on the cotangent bundle. The Lagrangian is
$$
L=p\dot q-H(q,p),
$$
and the Hamilton equations of motion are
$$
\dot q={\partial H\over\partial p}=p/m\qquad{\rm and}\quad\dot
p=-{\partial H\over\partial q}.
$$
This is translated into quantum mechanics as follows.
The Hilbert space of a point particle on a line is ${\cal H}=L^2(\R)$,
the space of complex functions on the line. States are unit-norm
functions $\psi(x)$, and $\int_A\left|\psi(x)\right|^2dx$ is the
probability to find a particle in $A\subset\R$. The fact that $\psi$
is normalised just means that the probability that the particle is
{\it somewhere\/} is $1$.
To get a Hamiltonian operator, we interpret $(q\psi)(x)=x\psi(x)$ and
$(p\psi)(x)=-i\psi'(x)$. The form of these operators is restricted by
the Dirac quantization condition $[q,p]=i\{q,p\}$ relating the
commutator of the quantum operators and the Poisson bracket of the
classical variables. Then the Hamiltonian is
$$
H={p^2\over 2m}+V(q)={-1\over 2m}{\partial^2\over \partial x^2}+V(q),
$$
where $\bigl(V(q)\psi\bigr)(x)=V(x)\psi(x)$.
The Schr\"odinger equation is
$$
i{\partial\over\partial t}\psi(t,x)={-1\over 2m}{\partial^2\over
\partial x^2}\psi(t,x)+V(x)\psi(x,t)
$$
so, for a free particle,
$$
\Bigl(i{\partial\over\partial t}+{1\over 2m}{\partial^2\over
\partial x^2}\Bigr)\psi(t,x)=0.
$$
We can obtain solutions by guessing. Since the equation is linear we
use exponential trial functions
$$
\psi(t,x)=e^{-i(Et-kx)},
$$
which satisfy Schr\"oedinger's equation if $E={k^2\over 2m}$. In fact,
every solution is of the form
$$
\psi(t,x)=\int{dk\over 2\pi}\exp\Bigl[-i\Bigl({k^2\over 2m}t-kx\Bigr)\Bigr]f(k),
$$
and if $f\in L^2(\R)$, then $\psi(t)\in L^2(\R)$ for all $t$. There
are special solutions generically called ``wavepackets'' that preserve
their form as they evolve, apart from getting gradually more and more
spread out. These can be described as ``Gaussian bumps with a
corkscrew twist''.