\chapter{Quantum Gravity (III)}
\begin{exercise}[$2-3$ Pachner Move]
Using the formula relating the tetrahedron
$$
\xy
(0,0)*\dir{*}="Y"; (-7,4)*\dir{*}="a"; **\dir{-}
?(.5)+(-1,-1)*{\scriptstyle\mathbf{i}}; (0,-9)*\dir{*}="b"; "Y"**\dir{-}
?(.5)+(2,0)*{\scriptstyle\mathbf{q}}; (7,4)*\dir{*}="c"; "Y"**\dir{-}
?(.5)+(-1,2)*{\scriptstyle\mathbf{m}}; "a";"b"**\crv{(-7,-4)}
?(.5)+(-2,-1)*{\scriptstyle\mathbf{p}}; "c";"b"**\crv{(7,-4)}
?(.5)+(2,-1)*{\scriptstyle\mathbf{k}}; "c";"a"**\crv{(0,9)}
?(.5)+(0,2)*{\scriptstyle\mathbf{j}};
\endxy
$$
to the $6j$-symbols and the Biedenharn--Elliot identity, prove the
following identity, which is the $2-3$ Pachner move up to ``fudge
factors'' for lower-dimensional simplices (triangles and edges).
$$
{\xy
(0,18)*\dir{*}="b";(14,8)*\dir{*}="c"**\dir{-};
(14,-8)*\dir{*}="d"**\dir2{-};
(0,-18)*\dir{*}="e"**\dir{-};
(-14,0)*\dir{*}="a"**\dir{-};
"b"**\dir{-};
"d"**\dir{-};
"a"**\dir2{-};
"c"**\dir2{.};
"e"**\dir{.};
\endxy
\over
\xy
(-5,0)*\dir{*}="*";
(5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
"*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
"*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
\endxy}
\quad=\sum_k\quad
{\xy
(0,18)*\dir{*}="b";(14,8)*\dir{*}="c"**\dir2{-};
(14,-8)*\dir{*}="d"**\dir{-};
(0,-18)*\dir{*}="e"**\dir2{-};
(-14,0)*\dir{*}="a"**\dir2{-};
"b"**\dir2{-};
"d"**\dir2{-};
"a"**\dir{-};
"c"**\dir{.};
"e"**\dir2{.};
"b"**\dir3{.} ?(.5)+(2,0)*{\scriptstyle\mathbf{k}};
\endxy
\xy
(-2,0)*\cir<8pt>{};
(3,0)*{\scriptstyle\mathbf{j}};
\endxy
\over
\xy
(-5,0)*\dir{*}="*";
(5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
"*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
"*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
\endxy
\xy
(-5,0)*\dir{*}="*";
(5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
"*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
"*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
\endxy
\xy
(-5,0)*\dir{*}="*";
(5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
"*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
"*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
\endxy}
$$
[Hint: Each side has a factor of
$$
\xy
(0,0)*\dir{*}="Y";
(-7,4)*\dir{*}="a";
**\dir{-} ?(.5)+(-1,-1)*{\scriptstyle\mathbf{i}};
(0,-9)*\dir{*}="b";
"Y"**\dir{-} ?(.5)+(2,0)*{\scriptstyle\mathbf{q}};
(7,4)*\dir{*}="c";
"Y"**\dir{-} ?(.5)+(-1,2)*{\scriptstyle\mathbf{m}};
"a";"b"**\crv{(-7,-4)} ?(.5)+(-2,-1)*{\scriptstyle\mathbf{p}};
"c";"b"**\crv{(7,-4)} ?(.5)+(2,-1)*{\scriptstyle\mathbf{k}};
"c";"a"**\crv{(0,9)} ?(.5)+(0,2)*{\scriptstyle\mathbf{j}};
\endxy
$$
for each tetrahedron, a factor of
$$
{1\over
\xy
(-5,0)*\dir{*}="*";
(5,0)*\dir{*}="o";**\dir{-} ?(.5)+(0,-2)*{\scriptstyle\mathbf{j}};
"*";"o";**\crv{(0,10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{a}};
"*";"o";**\crv{(0,-10)} ?(.5)+(0,-2)*{\scriptstyle\mathbf{b}};
\endxy}
$$
for each triangular face, and a factor of
$$
\xy
(-2,0)*\cir<8pt>{};
(3,0)*{\scriptstyle\mathbf{j}};
\endxy
$$
for each edge. Factors appearing on both sides have been cancelled.]
\end{exercise}
\section{Topological Quantum Field Theories (III)}
\subsection{Category Theory(III)}
\begin{definition}
Given monoidal categories $\cal C$ and $\cal D$, a functor
$F\colon{\cal C}\rightarrow{\cal D}$ is {\bf strictly monoidal} if,
and only if,
\begin{itemize}
\item $F(x\otimes y)=F(x)\otimes F(y)$ for all objects $x,y\in\cal C$,
\item $F(f\otimes g)=F(f)\otimes F(g)$ for all morphisms $f,g\in\cal C$,
and
\item $F(1_{\cal C})=1_{\cal D}$, where $1_{\cal C}$ and $1_{\cal D}$
are the identity objects for the $\otimes$ operation.
\end{itemize}
\end{definition}
\begin{definition}
A monoidal category is {\bf braided\/} if, and only if, for every
objects $x,y\in\cal C$ there is an isomorphism $B_{x,y}\colon x\otimes
y\rightarrow y\otimes x$ such that
\begin{itemize}
\item the {\bf hexagon identity\/} $BaB=aBa$ is satisfied, and
\item for any morphisms $f\colon x\leftarrow x'$ and $g\colon
x'\rightarrow y'$, we have $B(f\otimes g)=(g\otimes 1)B(1\otimes
f)=(g\otimes f)B=(1\otimes f)B(1\otimes g)$.
\end{itemize}
A braided category is {\bf symmetric\/} if $B_{x,y}=B_{y,x}^{-1}$.
\end{definition}
\begin{definition}
A monoidal functor $F\colon {\cal C}\rightarrow{\cal D}$ between
braided categories is {\bf braided\/} if, and only if,
$F(B_{x,y})=B_{F(x),F(y)}$.
\end{definition}
A braided functor between symmetric categories is automatically
symmetric.
\begin{definition}
A monoidal category has {\bf duals for objects\/} if, for every object
$x$ there is an object $x^*$ and morphisms $i_x\colon 1\rightarrow
x\otimes x^*$ and $e_x\colon x^*\otimes x\rightarrow 1$ such that
$(1_x\otimes e_x)(i_x\otimes 1_x)=1_x$ and $(e_x\otimes
1_{x^*})(1_{x^*}\otimes i_x)=1_{x^*}$.
\end{definition}
Any monoidal functor will preserve duals for objects.
Now that all the terms have been defined we can restate the definition
of Topological Quantum Field Theory.
\begin{definition}
A {\bf topological quantum field theory\/} is a symmetric monoidal
functor $Z\colon n\Cob\rightarrow \Vect$.
\end{definition}
\begin{definition}
A {\bf unitary topological quantum field theory\/} is a symmetric,
monoidal $*$-functor $Z\colon n\Cob\rightarrow
\Hilb$.
\end{definition}
Depending on the dimension of space-time, topological quantum field
theories have varying degrees of complexity.
\begin{itemize}
\item TQFTs on $1\Cob$ are extremely simple;
\item TQFTs on $2\Cob$ are related to commutative algebras;
\item TQFTs on $3\Cob$ involve the spin-network technology we have been
developing, and are related to monoidal categories;
\item nobody really understands TQFTs on $4\Cob$. This is unfortunate,
as we have seen that General Relativity is like a TQFT on $4\Cob$ with
some extra features.
\end{itemize}
\subsection{One-dimensional Topological Quantum Field Theories (I)}
To see just how simple one-dimensional TQFTs are, let us characterise
them completely.
A $1$-cobordism is a one-dimensional manifold with boundary. This
means it is a disjoint union of arcs and circles, and the boundary is
formed by the endpoints of the arcs. If we consider oriented
cobordisms, the boundary is also oriented, in such a way that, if
$\gamma\colon p\rightarrow q$ is an oriented arc between $p$ and $q$,
then $p$ is considered to be positively oriented and $q$ is considered
to be negatively oriented. If we represent positive points by $\circ$
and negative points by $\bullet$, then all the $1$-cobordisms are
disjoint unions of the following elementary ones:
$$
\xy
(-15,5)*{\circ};(-15,-5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
(-9,-5)*{\circ};(-9,5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
(-3,5)*{\circ};(3,5)*{\bullet};**\crv{(-0,0)} ?(.6)*\dir{>};
(9,-5)*{\circ};(15,-5)*{\bullet};**\crv{(12,0)} ?(.6)*\dir{>};
\endxy
$$
Cobordisms are composed in such a way that the orientation of the
arrows is preserved, so we need to identify a positive and a negative
endpoints, which cancel due to the opposite orientations. For example,
$$
\xy
(0,0)*\cir<8pt>{};
\endxy
=
\xy
(-3,1)*{\circ};(3,1)*{\bullet};**\crv{(0,6)} ?(.6)*\dir{>};
(3,-1)*{\circ};(-3,-1)*{\bullet};**\crv{(0,-6)} ?(.6)*\dir{>};
\endxy
$$
For our purposes we need identity cobordisms, so we adopt the
convention of reversing the orientation of the output. Then, the
elementary morphisms are
$$
\xy
(-15,5)*{\circ};(-15,-5)*{\circ};**\dir{-} ?(.6)*\dir{>};
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
(-3,5)*{\circ};(3,5)*{\bullet};**\crv{(-0,0)} ?(.6)*\dir{>};
(9,-5)*{\bullet};(15,-5)*{\circ};**\crv{(12,0)} ?(.6)*\dir{>};
\endxy
\qquad
\hbox{so that}
\quad
\xy
(0,0)*\cir<8pt>{};
\endxy
=
\xy
(3,0)*{\circ};(-3,0)*{\bullet};**\crv{(0,-5)} ?(.6)*\dir{>};
(-3,0)*{};(3,0)*{};**\crv{(0,5)} ?(.6)*\dir{>};
\endxy
$$
So let us characterise $1\Cob$ as a category:
\begin{itemize}
\item objects are $0$-dimensional compact oriented manifolds, i.e.,
finite disjoint unions of oriented points. $\N[\bullet,\circ]$.
\item morphisms are $1$-dimensional oriented manifolds, i.e., disjoint
unions of
$$
\xy
(-15,5)*{\circ};(-15,-5)*{\circ};**\dir{-} ?(.6)*\dir{>};
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
(-3,5)*{\circ};(3,5)*{\bullet};**\crv{(-0,0)} ?(.6)*\dir{>};
(9,-5)*{\bullet};(15,-5)*{\circ};**\crv{(12,0)} ?(.6)*\dir{>};
\endxy
\qquad{\rm and}\quad
\xy
(21,0)*\cir<8pt>{};
\endxy
=
\xy
(3,0)*{\circ};(-3,0)*{\bullet};**\crv{(0,-5)} ?(.6)*\dir{>};
(-3,0)*{};(3,0)*{};**\crv{(0,5)} ?(.6)*\dir{>};
\endxy
$$
Two $0$-dimensional oriented manifolds $n\bullet+m\circ$ and
$n'\bullet+m'\circ$ are cobordant---i.e., there is a
$1$-dimensional oriented cobordism between them---if, and only if,
$n-m=n'-m'$.
\item $\otimes$ is the disjoint union $\amalg$ of both $0$-dimensional
and $1$-dimensional manifolds. This product is braided and
symmetric because the manifolds are not assumed to be imbedded in
any dimension, but are considered as abstract manifolds, so the
disjoint union is diffeomorphic to the union of any permutation of
the spaces.
\item duals are as follows:
$$
\bullet^*=\circ\qquad
\circ^*=\bullet\qquad
e_\bullet=\xy
(-3,1)*{\circ};(3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy\qquad
e_\circ=\xy
(3,1)*{\circ};(-3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy\qquad
i_\bullet=\xy
(9,-1)*{\bullet};(15,-1)*{\circ};**\crv{(12,4)}
?(.6)*\dir{>};
\endxy\qquad
i_\circ=\xy
(15,-1)*{\bullet};(9,-1)*{\circ};**\crv{(12,4)}
?(.6)*\dir{>};
\endxy\qquad
$$
and
\item adjoints are
$$
\xy
(-15,5)*{\circ};(-15,-5)*{\circ};**\dir{-} ?(.6)*\dir{>};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(-15,5)*{\circ};(-15,-5)*{\circ};**\dir{-} ?(.6)*\dir{>};
\endxy\qquad
\xy
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-} ?(.6)*\dir{>};
\endxy\qquad
\xy
(-3,1)*{\circ};(3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(15,-1)*{\bullet};(9,-1)*{\circ};**\crv{(12,4)} ?(.6)*\dir{>};
\endxy\qquad
\xy
(9,-1)*{\bullet};(15,-1)*{\circ};**\crv{(12,4)} ?(.6)*\dir{>};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(3,1)*{\circ};(-3,1)*{\bullet};**\crv{(-0,-4)} ?(.6)*\dir{>};
\endxy
$$
\end{itemize}
We now notice that as soon as, say, $\bullet$ is mapped to some object
in a different category, the images of all the other objects and
morphisms are fixed. Therefore, a $1$-dimensional topological quantum
field theory is determined by a single object $Z(\bullet)$.
If we consider unoriented cobordisms, the category simplifies even
more. Indeed, an unoriented $1$-cobordism is just a disjoint union of
circles and segments, and the boundary is a union of (unoriented)
points. As a category, unoriented $1\Cob$ is described as follows:
\begin{itemize}
\item objects are $0$-dimensional compact manifolds, i.e.,
finite disjoint unions of oriented points. $\N[\bullet]$.
\item morphisms are $1$-dimensional unoriented manifolds, i.e., disjoint
unions of
$$
\xy
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-};
(-3,5)*{\bullet};(3,5)*{\bullet};**\crv{(-0,0)};
(9,-5)*{\bullet};(15,-5)*{\bullet};**\crv{(12,0)};
\endxy
\qquad{\rm and}\quad
\xy
(21,0)*\cir<8pt>{};
\endxy
=
\xy
(3,0)*{\bullet};(-3,0)*{\bullet};**\crv{(0,-5)};
(-3,0)*{};(3,0)*{};**\crv{(0,5)};
\endxy
$$
Two $0$-dimensional unoriented manifolds $n\bullet$ and $m\bullet$
are cobordant---i.e., there is a $1$-dimensional unoriented
cobordism between them---if, and only if, $n=m\quad(\mod 2)$.
\item $\otimes$ is the disjoint union $\amalg$ of both $0$-dimensional
and $1$-dimensional manifolds. For the same reasons as in the case
of $1$-dimensional oriented cobordisms, this product is braided and
symmetric.
\item duals are as follows:
$$
\bullet^*=\bullet\qquad
e_\bullet=\xy
(-3,1)*{\bullet};(3,1)*{\bullet};**\crv{(-0,-4)};
\endxy\qquad
i_\bullet=\xy
(9,-1)*{\bullet};(15,-1)*{\bullet};**\crv{(12,4)};
\endxy
$$
and
\item adjoints are
$$
\xy
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(-9,-5)*{\bullet};(-9,5)*{\bullet};**\dir{-};
\endxy\qquad
\xy
(-3,1)*{\bullet};(3,1)*{\bullet};**\crv{(-0,-4)};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(15,-1)*{\bullet};(9,-1)*{\bullet};**\crv{(12,4)};
\endxy\qquad
\xy
(9,-1)*{\bullet};(15,-1)*{\bullet};**\crv{(12,4)};
\endxy
\buildrel \dagger\over\rightarrow
\xy
(3,1)*{\bullet};(-3,1)*{\bullet};**\crv{(-0,-4)};
\endxy
$$
Note, however, that the adjoint of a morphism is the same as its
dual, so this will be a special case in which a topological
quantum field theory will automatically be unitary.
\end{itemize}
Oriented $1\Cob$ goes to $\C\Hilb$ and $\C\Vect$
Unoriented $1\Cob$ goes to $\R\Vect=\R\Hilb$ (see the remark about
adjoints being equal to duals).
A $1$-cobordism is a space (vector or Hilbert).
[does this mean that a $2$-cobordism should map to a subcategory of
$\Vect$ or $\Hilb$?]
\chapter{Quantum Mechanics from a Category Theoretic Viewpoint (III)}
\section{Fields on $(1+1)$-dimensional spacetime (I)}
Last time we quantised a free particle on a line. Now let's quantise
the vacuum Maxwell equations on a $(1+1)$-dimensional spacetime with
the topology of a cylinder. We will find that this is isomorphic to a
particle on a line!
Our spacetime is $\R\times S^1(l)$ with metric $ds^2=-dt^2+l^2dx^2$,
where $l$ is the ``perimeter of the Universe'' and $x\in\R/\Z$. The
volume form is $d{\rm Vol}=l\,dt\wedge dx$ and the induced metric on
$1$-forms is $\langle dt,dt\rangle=-1$ and $\langle
dx,dx\rangle=1/l^2$. The Hodge dual, satisfying
$\omega\wedge *\omega=\langle\omega,\omega\rangle d{\rm Vol}$, is
therefore
$$
\cases{*1=l\,dt\wedge dx,\cr
*dt=-l\,dx,\cr
*dx=-{1\over l}dt,\cr
*dt\wedge dx=-{1\over l}.\cr}
$$
Note that $*^2=-1$.
Maxwell's theory is Yang--Mills theory with gauge group $\R^+$ or
$U(1)$. These two cannot be distinguished at the level of local
degrees of freedom since their Lie algebras are isomorphic. However,
as we shall see they have different global properties that affect the
quantum theory.
We first tackle the case of $\R^+$ gauge group, which is easier
because the topology of the group itself is trivial. The exterior
derivative is $d=d\theta{\partial\over\partial
x}+dt{\partial\over\partial t}$, and the exterior covariant derivative
is
$$
\nabla=d+A\wedge\qquad{\rm with}\quad A=a_tdt+a_x dx.
$$
The curvature of the connection is
\begin{eqnarray*}
F&=&\nabla^2=dA=(\partial_ta_x-\partial_x a_t)dt\wedge dx,\qquad{\rm
and}\\
*F&=&{1\over l}(\partial_x a_t-\partial_t a_x).\\
\end{eqnarray*}
We see that in $2D$ there is a one-component electric field and no
magnetic field. Now, $*F$ is a {\it function\/} on space-time, and
Maxwell's equation $d{*F}=0$ simply implies that ${1\over
l}(\partial_ta_x-\partial_x a_t)$ is a constant, say $e$. On $\R^2$
this would be the end of the story, since $\partial_ta_x-\partial_x
a_t=el$ uniquely determines $A$ up to gauge transformations. On
$\R\times S^1$, however, we have nontrivial solutions even if $e=0$,
as in the Aharonov-Bohm effect.
The ``physical field'' is $A$ modulo gauge transformations, which are
of the form $A\rightarrow A'=A-df$. By taking
$f(t,x)=\int_0^tds\,a_t(s,x)$ we get $A'=a(t,x)dx$. This is called
``temporal gauge''. Since $\partial_ta(t,x)=el$ is constant, we
actually have $A'=\bigl(a(x)+elt\bigr)dx$. However there is still some
gauge freedom left, because by taking a gauge transformation generated
by a suitable $f(x)$ we can cancel the variation in $a(x)$. To be
precise,
$$
f(x)=\int_0^x dy\,a(y)-x\oint_0^1 dy\,a(y)
$$
generates the gauge transformation
$$
a(x)\rightarrow a'(x)=a(x)-{\partial\over\partial x}f(x)=\oint_0^1
dy\,a(y),
$$
which is a gauge-invariant constant. Therefore, a solution up to gauge
transformations of the vacuum Maxwell equations on $\R\times S^1$ is
$A=(a+let)dx$. This, of course, looks like the trajectory of a free
particle with initial position $a$ and velocity $le$. The analogy is
supported by the observation that the equation $d(*F)=0$, which
implies that $e$ is constant, is like Newton's law stating that
momentum is conserved for a particle of mass $1/l$.
The above is the way a gauge is traditional chosen in
physics. However, this is just an explicit example of computing the
first de~Rham cohomology of a spacetime. Now, we can obtain the same
parametrization of the space of connections modulo gauge
transformations in a slicker way as follows. We observe that the first
cohomology class of $\R\times S^1$ is generated by $dx$, therefore
$A=a(t)dx$ up to gauge transformations. The curvature of the
connection (which is gauge-invariant and therefore observable) is
$F=\dot a(t)dt\wedge dx$, so $*F=-{1\over l}\dot a(t)=-e(t)$ and the constant
part of the connection can be observed by sending a test particle
around a loop in the $x$ direction and comparing it with a particle at
rest. A more invariant experiment can be performed by sending two test
particles at the speed of light in opposite directions and having them
interact when they meet. Mathematically, these are much more
complicated than the equivalent $a(t)=\oint_{t=\rm const}A$. So we
have two functions $a(t)$ and $e(t)$ related by the constraint
$e(t)=\dot a(t)/l$. The analogy with classical mechanics is clear, as in
that case $q(t)$ and $p(t)$ are related by the constraint $p=m\dot q$.
The Maxwell Lagrangian density for $A=a(t)dx$ is
$$
{\cal L}={-1\over 2}F\wedge *F={1\over 2l}\dot a^2 dt\wedge
dx={1\over 2l^2}\dot a^2 d{\rm Vol},
$$
so integrating over $dx$ we obtain the Lagrangian
$$
L={1\over 2l}\dot a^2dt.
$$
The momentum conjugate to $a(t)$ is $e(t)={\dot a}/l$, so
define $E=-{*F}$. The Hamiltonian
$$
H={e\dot a}dt-L
$$
can be obtained by integrating the following Hamiltonian density over
$dx$:
$$
{\cal H}=E\wedge dA-{\cal L}.
$$
Now, if we express the Hamiltonian and Hamiltonian density in terms of
$(a,e)$ and $(A,E)$ respectively, we get
$$
H={l\over 2}{e^2}
$$
and
$$
{\cal H}={1\over 2}{E\wedge *E}.
$$
Now, in terms of independent fields $(A,E)$, the Lagrangian density is
$$
{\cal L}=E\wedge dA-{1\over 2}E\wedge *E=-{1\over
2}F\wedge{*F}
$$
This is $EF$ theory with an extra term involving the Hodge~$*$. From
this discussion it is clear that $\R^+$ Maxwell's theory on the
cylinder of radius $l$ is analogous to a free particle of mass $1/l$
on a line, where the $1$-form $A=a(t)dx$ plays the role of position,
and the $0$-form (function) $E$ plays the role of momentum conjugate
to $A$. If we quantise the Hamiltonian $H={l\over 2}{e^2}$ we obtain
the following Schr\"odinger equation:
$$
i{\partial\over\partial_t}\psi(t,a)=-{l\over
2}{\partial^2\over\partial a^2}\psi(t,a).
$$