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\begin{flushright}
UWThPh-1991-63 \\
\end{flushright}
\vspace{2cm}
\begin{center}
{\Large \bf Covariant QED without Indefinite Metric}\\[40pt]
W. Thirring and H. Narnhofer \\
Institut f\"ur Theoretische Physik \\
Universit\"at Wien
\vfill
{\large \bf Birthday Present for Rudolf Haag}
\vfill
{\bf Abstract} \\
\end{center}
We construct for the linearized Higgs model a representation of the
field operators in a Hilbert space $\Ha$ with the following features:
\begin{enumerate}
\item $\Ha$ has a positive definite metric but is nonseparable.
\item The vacuum is gauge invariant.
\item The gauge variant operators exist only in their exponentiated
form as unitaries.
\item There is a subspace of $\Ha$ where $A_\mu{}^{,\mu}$ is represented
by 0.
\end{enumerate}
\vfill
\end{titlepage}
\section{Introduction}
Quantum electrodynamics is the experimentally most precisely tested
quantum theory and yet in its mathematical formulation there remain some
disturbing features. Even if one disregards the infinite renormalization
constants and the still open question of the convergence of the perturbation
series we are left with some problems typical of gauge theories.
\begin{enumerate}
\item[(i)] The canonical formulation of the theory uses the gauge dependent
potentials $A$ which transform under gauge transformations as
$A \ra A + d\Lambda$. The vacuum should be an invariant state and not
prejudize a particular gauge. Yet a gauge invariant state $|0\rangle$
seems impossible since for a $c$-number $\Lambda$
$$
\langle 0|A|0\rangle \ra \langle 0|A|0 \rangle + d\Lambda. \eqno(1.1)
$$
\item[(ii)] The theory contains the generator of gauge transformations
$\chi = \delta A$ which must satisfy with $A$ commutation rules like $x$
and $p$ since $\chi$ shifts $A$
$$
[A,\chi] = i. \eqno(1.2)
$$
The equivalence of the field equations with Maxwell's equations requires
$\chi = 0$ but as operator equation this contradicts (1.2). So one poses this
constraint as a supplementary condition on the good vectors $|g\rangle$ in
Hilbert space
$$
\chi |g\rangle = 0. \eqno(1.3)
$$
But there is still trouble ahead since v. Neumann's uniqueness theorem
tells us that for the Weyl form of (1.2) there is essentially only the
usual representation where the operators have a purely continuous spectrum.
Thus vectors of the type $|g\rangle$ cannot exist because they would be
eigenvectors of $\chi$. The situation is similar to the ``improper''
eigenfunctions of the operator $x$ which are something like $\delta$-functions.
They are not normalizable, $\int \delta^2 = \infty$, and so one introduces an
indefinite metric $\eta$ in Hilbert space such that
$\langle g|\eta|g\rangle < \infty$ [1,2,3]. This creates more problems in the
physical interpretation (``negative probabilities'') and in formulating
things in a mathematically well defined way (even unitaries $U^* \eta U = \eta$
are no longer bounded operators and the road is paved with domain questions).
\end{enumerate}
Even though people live happily with this construct we want to point out that
there is no need to abandon the standard $C^*$-algebraic formulation of
quantum mechanics which excludes negative probabilities. In fact, there
are general theorems which tell us that the difficulties mentioned before cannot really
exist but must be artefacts of an inappropriate mathematics. The point is that
the gauge transformations generated by $\chi$ are an amenable group and there
is a general theorem if an automorphism group of a $C^*$-algebra is
amenable then there are invariant states $\omega$ [4,5]. Furthermore, since the gauge
transformations are inner automorphism the vector corresponding to the
invariant state must be an eigenvector of the corresponding unitaries.
Thus there must be representations where $\chi$ has a point spectrum and the
goal of this paper is to study them. In doing so we follow a program
pioneered by A. Hurst et al. who have for a long time advocated this
point of view [6,7,8,9]. The novel feature which turns up is that gauge dependent
quantities $S$ do not exist as operators in Hilbert space but only the
associated unitaries $e^{i\alpha S}$. A similar construction of a state
can be found in [10]. Since for the gauge invariant
quantities the representation is the standard one we consider this price
not too high for getting rid of negative probabilities.
\section{The Weyl Algebra}
First we want to recall that von Neumann's uniqueness theorem [11] can be
circumvented and it is possible to find proper eigenvalues to, let's say
$x$, if one is ready to pay a price.
The Weyl algebra $\cal W$ is generated by the unitaries
$W(r,s) = e^{i(pr + xs)}$, $(r,s) \in {\bf R}^2$ which satisfy the relations
$$
W(r,s) W(r',s') = e^{i\sigma((r,s),(r',s'))} W(r+r',s+s') \eqno(2.1)
$$
with the symplectic form
$$
\sigma((r,s),(r',s')) = \frac{1}{2} (s' r - r' s). \eqno(2.2)
$$
We have
$$
W(r,s)^* = W(-r,-s) = W^{-1}(r,s) \eqno(2.3)
$$
and the $C^*$-algebra $\cal W$ consists of the elements $\sum_i c_i W(r_i,s_i)$,
$c_i \in {\bf C}$, $\sum_i |c_i| < \infty$.
Their norm satisfies
$$
\left \| \sum_i c_i W(r_i,s_i) \right\| \leq \sum_i |c_i|, \qquad
\left \| \sum_{i=1}^2 c_i W(r_i,s_i)\right\| = \sum_{i=1}^2 |c_i| . \eqno(2.4)
$$
Note that the map $(r,s) \ra W(r,s)$ is not continuous from ${\bf R}^2$
into $\cal W$ (in the norm topology).
In $\cal W$ the inner automorphism group of $\cal W$
isomorphic to ${\bf R}^2$ contains the shifts which are
$$
S_{(\alpha,\beta)} W(r,s) = W(\alpha,\beta) W(r,s) W^{-1}(\alpha,\beta) =
e^{i \alpha s - i \beta r} W(r,s). \eqno(2.5)
$$
Since ${\bf R}^n$ is amenable general theorems tell us that there are
$S$-invariant states. In fact, the only state invariant under $S$ is given by
$$
\omega_{00}(W(r,s)) = \delta_{r,0} \delta_{s,0} \eqno(2.6)
$$
where $\delta_{r,s} = 1$ for $r=s$, 0 otherwise. The state
$$
\omega_0(W(r,s)) = \delta_{r,0} \eqno(2.7)
$$
is invariant only under the subgroup $S_{(0,{\bf R})}$. The tracial state
$\omega_{00}$ is faithful:
$$
\omega_{00}(\sum_i c^*_i w^*(r_i,s_i) \sum_j c_j W(r_j,s_j)) =
\sum_i |c_i|^2 > 0
$$
unless all $c_i$ are zero. $\omega_0$ is not since
$$
\omega_0((1 - W^*(0,s))(1 - W(0,s))) = 0 \quad \forall \, s \in {\bf R}.
\eqno(2.8)
$$
However, $\cal W$ is simple so that each state gives a faithful GNS-representation.
Denoting the cyclic vector in the representation $\pi_{00}$ (resp. $\pi_0$)
given by $\omega_{00}$ (resp. $\omega_0$) by $|0,0\rangle_0$ (resp.
$|0\rangle$) and defining $|r,s\rangle_0 = \pi_{00} W(r,s) |0,0\rangle_0$
we see
$$
{}_0\langle r',s'|r,s\rangle_0 = \delta_{r,r'} \delta_{s,s'}.
$$
Since $(r,s) \in {\bf R}^2$ we have an uncountable set of orthogonal
vectors and thus a representation in a nonseparable
Hilbert space. Its elements are of the form $\sum_{i,j} c_{ij} |r_i,s_j\rangle_0$
with the norm $\sum_{i,j} |c_{ij}|^2 < \infty$ and $\pi_{00}$ is given by
$$
\pi_{00}(W(r,s)) |r',s'\rangle_0 = e^{i\sigma((r,s),(r',s'))}
|r + r',s + s'\rangle_0 . \eqno(2.9)
$$
Note that $(r,s) \ra \pi_{00}(W(r,s))$ is not strongly continuous and thus
one cannot define the infinitesimal generators $p$ and $x$, not even as
unbounded operators. This phenomenon can only occur in the non-separable
Hilbert space. Since $(r,s) \ra \pi_{00}(W(r,s))$ is weakly measurable
in a separable Hilbert space this would be enough to guarantee the
existence of $p$ and $x$.
If we again define $|r,s\rangle = \pi_0 (W(r,s)) |0\rangle$ in the representation
$\pi_0$, the analog of (2.9) is still valid but the scalar product is changed to
$$
\langle r',s'|r,s\rangle = \delta_{r,r'} e^{ir(s'-s)/2}.
$$
Thus $|0,s\rangle = |0\rangle$ and $|r,s\rangle = e^{-irs/2} |r\rangle$
with $\langle r'|r\rangle = \delta_{r,r'}$ and the Hilbert space is still
nonseparable.
In contrast to $\omega_{00}$ now $\omega_0$ corresponds to an
irreducible representation $\pi_0$: Assume $A \pi_0(W(r,s)) = \pi_0
(W(r,s))A$ for all $r,s \in {\bf R}^2$. Then
\begin{eqnarray*}
\lefteqn{A \pi_0 (W(r,s)) | \bar r,0\rangle = \pi_0(W(r,s)) A|\bar r,0
\rangle = } \\
& = & A \pi_0(W(r,s) \cdot W(\bar r,\bar s)) e^{i\bar s \bar r/2}|0,0
\rangle = \pi_0(W(r,s) A \pi_0(W(\bar r,\wh{\bar s})) e^{-i \wh{\bar s}
\bar r/2} |0,0\rangle \\
& = & \pi_0(W(r + \bar r,s + \bar s)) e^{i(\bar s \bar r + r \bar s -
s \bar r)/2} A|0,0\rangle = \pi_0(W(r + \bar r,s + \wh{\bar s}))
e^{i(\wh{\bar s} \bar r + r \wh{\bar s} - s \bar r)/2} A|0,0\rangle.
\end{eqnarray*}
This implies
$$
\pi_0 (W(0,\bar s - \wh{\bar s})) A|0,0\rangle = A|0,0\rangle.
$$
But the only eigenstate to $W(0,s)$ with eigenvalue 1 is $|0,0\rangle$
which is separating for the commutant. Therefore $A = \lambda 1$ and
the representation is irreducible.
Though the representation $\pi_0$ of $W(r,s)$ is not strongly
continuous in both variables it is in the second.
This means that in this representation the generator $p$ does not exist
but $x$ does and is a selfadjoint operator. This corresponds to the fact
that the cyclic vector $|0\rangle$ is the limit $b \ra 0$ of a wave
function $\sim \frac{1}{\sqrt{b}} e^{-(x/b)^2}$ in the standard Schr\"odinger
representation. By compactness there must be weak accumulation points
and in Hilbert space this vector converges weakly to zero. However,
the weak* limit of the corresponding state is $\omega_0$. Note that
now $x$ does not have a continuous spectrum but ${\bf R}$ as a point
spectrum, $|r\rangle$ being the eigenvectors. There is no contradiction
to the Heisenberg commutation relations since $p$ does not exist as
operator.
\section{The Two-Dimensional Oscillator}
As next warming up exercise we ask for which algebraic structures the motion
$$
\vec a(t) = \vec a(0) \cos t + \dot{\vec a}(0) \sin t, \qquad
\vec a \in {\bf R}^2, \eqno(3.1)
$$
is a one parameter group of automorphisms: Classically the equations
$$
a_1 = - \dot a_0, \qquad \qquad a_0 = \dot a_1 \eqno(3.2)
$$
define a time invariant submanifold of phase space but with the usual
quantization one cannot put $a_0 - p_1 = a_1 + p_0 = 0$, this would
contradict the commutation relations $[a_1 + p_0,a_0 - p_1] = -2i$.
We shall now show how these constraints can be realized by some
representation $\pi_0$.
First we note that (3.1) can be derived from the Hamiltonian
$$
2H = p_1^2 + a_1^2 - p_0^2 - a_0^2 \eqno(3.3)
$$
for which $\dot a_1 = p_1$, $\dot a_0 = - p_0$ and this change in sign
removes the contradiction with the commutation relations (which exist
equally in their better defined Weyl form). The commutation relations
$[a_j,p_n] = i \delta_{jn}$ suggest that the variables
$$
\chi = a_1 - p_0, \qquad S = (a_1 + p_0)/2, \qquad
\Pi_\chi = (a_0 + p_1)/2, \qquad \Pi_S = - a_0 + p_1 \eqno(3.4)
$$
are conjugate
$$
[\chi,S] = [\Pi_\chi,\Pi_S] = [\Pi_\chi,S] = [\Pi_S,\chi] = 0, \qquad
[S,\Pi_S] = [\chi,\Pi_\chi] = i. \eqno(3.5)
$$
In terms of these variables we have
$$
H = S \chi + \Pi_\chi \Pi_S \eqno(3.6)
$$
and the time evolution is
$$
\dot S = \Pi_\chi, \qquad \dot \chi = \Pi_S, \qquad \dot \Pi_\chi = - S,
\qquad \dot \Pi_S = - \chi . \eqno(3.7)
$$
With the commutation relations (3.5) we see that $S(t)$ and $\chi(t)$ generate
time invariant abelian subalgebras but the two do not commute with each
other. To give a meaning to the submanifold (3.2) on which $\chi(t)$ is
zero we replace the ill-defined relations (3.5) by the ones for the Weyl
operators
$$
W(\alpha,\beta,\gamma,\delta) = \exp[i(\alpha S + \beta \Pi_S +
\gamma \chi + \delta \Pi_\chi)]. \eqno(3.8)
$$
In the Hilbert space corresponding to the state
$$
\omega_0(W(\alpha,\beta,\gamma,\delta)) = \delta_{\alpha 0}
\delta_{\delta 0}
$$
we have as in Sect. 1
$$
W(0,\beta,\gamma,0) |0\rangle = |0\rangle \eqno(3.9)
$$
or $\chi$ and $\Pi_S = \dot \chi$ exist as operators and
$$
\chi |0\rangle = \dot \chi |0\rangle = 0. \eqno(3.10)
$$
The state $\omega_0$ is invariant under the automorphism group (3.1) and
it is thus implemented by a one parameter group of unitaries. Formally
in the notation implied by (3.1)
$$
\vec a(t) = U_t^{-1} \vec a(0) U_t . \eqno(3.11)
$$
But since $S$ and $\dot S$ do not exist in this representation one has to
write it
$$
W(\alpha,\beta,\gamma,\delta)(t) = U_t^{-1} W(\alpha,\beta,\gamma,\delta)
U_t \eqno(3.12)
$$
with
$$
W(\alpha,\beta,\gamma,\delta)(t) = W(\alpha \cos t - \delta \sin t,
\beta \cos t + \gamma \sin t, \gamma \cos t - \beta \sin t,
\delta \cos t + \alpha \sin t). \eqno(3.13)
$$
Since the $W$'s are not strongly continuous in $\alpha$ and $\delta$ the
$U$'s cannot be in $t$ either. Thus Stone's theorem does not apply and
$U_t$ cannot be written $e^{-iHt}$. The formal expression (3.6) is not
an operator in this representation only $H|0\rangle = 0$ makes sense.
\paragraph{Remarks (3.14):} In QED [12] instead of $\chi(t) |0\rangle = 0$ the
weaker condition $\chi^-(t)|0\rangle = 0$ is postulated. $\chi^-$ is the
negative frequency part of $\chi(t) = (e^{-i\omega t}/2) (a_1 + i \dot a_1
- i a_0 + \dot a_0) + h.c. \equiv \chi^- + \chi^+$, so in fact $\chi^-$
is the sum of the usual annihilation operators. Therefore with the normal
Hamiltonian $2H = p_1^2 + a_1^2 + p_0^2 + a_0^2$ there is such a vector in the
usual representation $p_k = \frac{1}{i} \frac{\partial}{\partial a_k}$,
namely corresponding to the wave function $|0\rangle \sim
e^{-(a_0^2 + a_1^2)/2}$.
With this Hamiltonian $[\chi(t),S(t')] = 0$ such that all vectors
$|f\rangle \equiv \exp[i \int dt f(t) S(t)] |0\rangle$ are annihilated by
$\chi^-$.
Thus in this representation there is no vector with $\chi(t)|0\rangle = 0$ but
an infinite dimensional subspace spanned by the $|f\rangle$'s with
$\langle f'|\chi(t)|f\rangle = 0$.
\section{Quantumelectrodynamics}
The next model we will study is the electromagnetic field coupled to a
scalar field $\phi$ [13,14]. First we consider the classical equation.
As in the
Higgs model we assume the absolute value of $\phi$ to be fixed by some
mechanism so that we set $\phi(x) = e^{iS(x)}$. In this case the
Lagrangean of the scalar field becomes ($g_{00} = + 1$)
$$
{\cal L}_s =
\frac{1}{2} (\phi_{,\mu} + ie A_\mu \phi)(\phi^*_{,\mu} - i e A_\mu \phi^*)
= \frac{1}{2} (S_{,\mu} + e A_\mu)(S^{,\mu} + e A^\mu) \eqno(4.1)
$$
which leads to a current
$$
J_\mu = - e(S_{,\mu} + e A_\mu). \eqno(4.2)
$$
Thus Maxwell's equations
$$
F_{\mu\nu}^{,\nu} = J_\mu \eqno(4.3)
$$
are complemented by
$$
F_{\mu\nu} \equiv A_{\mu,\nu} - A_{\nu,\mu} = e^{-2}(J_{\nu,\mu} - J_{\mu,\nu}),
\qquad J_\mu{}^{,\mu} = 0. \eqno(4.4)
$$
>From this we infer from (4.3) and (4.4)
$$
e^2 F_{\mu\nu}{}^{,\nu} = - J_{\mu,\nu}{}^\nu = e^2 J_\mu \eqno(4.5)
$$
and thus again from (4.4)
$$
F_{\mu\nu,\sigma}{}^\sigma = - e^2 F_{\mu\nu} \eqno(4.6)
$$
For a canonical formalism one uses frequently for the electromagnetic
Lagrangean
$$
{\cal L}_e = - \frac{1}{2} A_{\mu,\nu} A^{\mu,\nu}. \eqno(4.7)
$$
The Euler equations of the total Lagrangean ${\cal L} = {\cal L}_e +
{\cal L}_s$ are
$$
A_{\mu,\nu}{}^\nu = J_\mu = - \frac{\partial \cL}{\partial A_\mu},
\qquad J_\mu{}^{,\mu} = 0. \eqno(4.8)
$$
They are equivalent to (4.3) iff
$$
\chi := A_\mu{}^{,\mu} = const. \eqno(4.9)
$$
Keeping $\chi$ as dynamical variable we derive from the Euler equation
(4.8) $J_\mu{}^{,\mu} = 0$ in addition to (4.6) the relations
$$
\chi_{,\sigma}{}^\sigma = 0 , \qquad S_{,\sigma}{}^\sigma = - e \chi. \eqno(4.9)
$$
\paragraph{Remarks:}
\begin{enumerate}
\item Whereas the original system of equations is invariant under the gauge
transformations $A_\mu \ra A_\mu + \Lambda_{,\mu}$, $S \ra S - e \Lambda$
(4.8) allows only for gauge transformations with $\Lambda_{,\mu}{}^\mu = 0$.
$\chi$ is then gauge invariant, $S$ is not.
\item The equations (4.9) were studied in [15] as the prototype of a dipole
ghost. There it was shown that for its quantization there is no need for a
Hilbert space with indefinite metric. There we had no constraint
$\chi = 0$ and we could use a standard representation in terms of creation
and annihilation operators. This corresponds in the toy model in Section 3
where both $x$ and $p$ exist as operators. A discussion in the framework
of indefinite metric can be found in [16,17].
\item It has to be remarked that the particles of the $F$-field acquire a
mass $|e|$ purely by the equations of motion (4.5). There is no
breaking of the gauge group $A_\mu \ra A_\mu + \Lambda_{,\mu}$,
$S \ra S - e \Lambda$ and in (4.20) we shall specify a gauge invariant
vacuum.
\end{enumerate}
For the quantization we shall restrict ourselves for simplicity to one
space dimension since the transversal degrees of freedom do not add
anything to our argument. In two dimensions the dual of $F_{\mu\nu}$ is
a scalar which we denote by
$$
F = F_{10} = \dot A_1 - A'_0 . \eqno(4.10)
$$
Since the equations for $F$ and for $\chi$ and $S$ decouple we shall
introduce these variables instead of $A_0$, $A$ and $S$. From the
Lagrangean (4.8) we have the canonical variables
$$
\Pi_0 = - \dot A_0, \qquad \Pi_1 = \dot A_1 \quad \mbox{and} \quad
\Pi_S = \dot S + e A_0 \eqno(4.11)
$$
and using the field equations we find that formally
$$
S,\qquad \chi = - \Pi_0 - A'_1, \qquad F = \Pi_1 - A'_0 \eqno(4.12)
$$
obey with
$$
\dot S = \Pi_S - eA_0, \qquad \dot \chi = A''_0 - e\Pi_S - \Pi'_1,
\qquad \dot F = A''_1 - e S' - e^2 A_1 + \Pi'_0
$$
at equal times the commutation relations
$$
i \delta(x - x') = [S(x),\dot S(x')] = e^{-1}[\dot S(x),\chi(x')] =
e^{-1}[\dot \chi(x),S(x')] = e^{-2}[F(x),\dot F(x')], \eqno(4.13)
$$
all the other commutators being zero. Because of the field
equations
$$
F_{,\nu}{}^\nu = - e^2 F, \qquad \chi_{,\nu}{}^\nu = 0, \qquad
S_{,\nu}{}^\nu = - e \chi \eqno(4.14)
$$
we can express the fields arbitrary times by their initial values with
the usual invariant functions $\Delta$, $D$, $\cal D$ which satisfy
$$
\begin{array}{lll}
\Delta_{,\nu}{}^\nu + e^2 \Delta = 0, &
D_{,\nu}{}^\nu = 0, & {\cal D}_{,\nu}{}^\nu = D, \\[12pt]
\Delta(-x) = - \Delta(x),\qquad & D(-x) = - D(x),\qquad &
{\cal D}(-x) = - {\cal D}(x), \qquad x \in {\bf R}^2,
\end{array}
$$
$$
\delta(x) = \dot \Delta(0,x) = \dot D(0,x). \eqno(4.15)
$$
and therefore
$$
D(t,x) = \frac{1}{2}[\Theta(t-|x|) - \Theta(|x|-t)], \qquad
{\cal D}(t,x) = \frac{1}{4}(t^2 - x^2) D(t, x).
$$
Thus for arbitrary times (4.13) generalizes to
$$
\begin{array}{lcl}
[F(x),F(x')] &=& i \Delta(x-x')e^2 \\[12pt]
[\chi(x),\chi(x')] &=& 0 \\[12pt]
[\chi(x),S(x')] &=& i e D(x-x') \\[12pt]
[S(x),S(x')] &=& iD(x-x') + e^2 i{\cal D}(x-x').
\end{array} \eqno(4.16)
$$
To give these relations a meaning for elements of a $C^*$-algebra we
define the Weyl algebra $W$ by
$$
W(f,c,s) = \exp[ i\int d^2x(f(x)F(x) + c(x)\chi(x) + s(x)S(x))], \qquad
f,c,s \in C^\infty_0({\bf R}^2),
$$
$$
\|W(f,c,s) - W(f',c',s')\| = \begin{array}{ll}
0 & \mbox{if } f - f' = (\Box + e^2)g_1, c-c' = \Box g_2,
s-s' = - e\Box(c-c'),\, g_i \in {\cal S} \\
2 & \mbox{ otherwise.} \end{array} \eqno(4.17)
$$
($\Box = \partial^2/\partial x^\mu \partial x_\mu$.)
Their algebraic properties are determined by
$$
\begin{array}{rcl}
W(f,c,s) W(\wh f,\wh c,\wh s) &=& W(f+\wh f,c+\wh c,s+\wh s) \cdot \\[12pt]
&& \cdot \exp\{i \int d^2x[f(x)e^2 \Delta(x-x') \wh f(x') +
e(c(x) D(\wh x - x') \wh s(x') \\[12pt]
&& \mbox{} - s(x) D(x-x')\wh c(x')) +
s(x)(D(x - x') + e^2 {\cal D}(x - x'))\wh s(x)] \}.
\end{array} \eqno(4.18)
$$
In generalization of our previous state (2.7) the vacuum state is given by
$$
\omega_0 (W(f,c,s)) = \begin{array}{ll}
e^{-e^2 f \Delta^+ f} & s \equiv 0 \\[7pt]
0 & s \not\equiv 0.
\end{array} \eqno(4.19)
$$
\paragraph{Remarks (4.20)}
\begin{enumerate}
\item The usual [3] argument for the necessity of an indefinite metric runs
as follows. From (4.16) one infers
$$
\langle 0| S(x) S(x')|0\rangle = D^+ (x-x') + e^2 {\cal D}^+(x - x')
$$
where
$$
{\cal D}^+ (x) = \int \frac{dp}{2\pi} \Theta(p_0) e^{ipx} \delta'(p^2).
$$
Since $\delta'$ does not have a definite sign this formula is not possible
in Hilbert space. In our formulation $S$ is not an operator but only
$\exp [i\int dx S(x) s(x)]$ exists and its expectation value is not
$\exp [- \frac{1}{2} \int s(x) {\cal D}^+ (x-x') s(x')]$ but zero if
$s \neq 0$.
\item Naively one might think that the Higgs model contradicts Goldstone's
theorem [18--20] because one has a broken symmetry and the physical particles have mass
$> 0$. There is no formal contradiction in our approach since our state is gauge
invariant.
\item Note that for $e = 0$ the transformation from $A_i$, $\Pi_i$ to
$(F,\chi)$, $(\dot F, \dot \chi)$ becomes singular $(F' \equiv \dot \chi)$
and the latter algebra abelian. The $A$'s are expressible as $A_\mu = e^{-1} S_{,\mu} +
\ve_{\mu\nu} F^{,\nu}$ but for $e \ra 0$ this does not tend to a limit.
\end{enumerate}
We define a Poincar\'e transformation
$$
\sigma_P W(f,c,s) = W(f_P,c_P,s_P) \eqno(4.21)
$$
with
$$
\begin{array}{lll}
f_P(x) & = & f(\Lambda^{-1} x - \lambda) \\
c_P(x) & = & c(\Lambda^{-1} x - \lambda) \\
s_P(x) & = & s(\Lambda^{-1} x - \lambda).
\end{array} \eqno(4.22)
$$
Our state is $\sigma_P$ invariant and therefore $\sigma_P$ is unitarily
implementable. However it is not strongly continuous because
$$
\| (W(0,0,t) - W(0,0,s)) |\Omega\rangle \| = 2 \quad \mbox{for }
t \neq 0, \; s \equiv 0. \eqno(4.23)
$$
Therefore there is no generator for the Poincar\'e group. This
corresponds to the fact that the Hamiltonian should contain
$\int (\nabla S)^2$ which in our representation is not defined.
However, if we restrict our interest on $W(f,c,0)$ in the corresponding
subspace $W(f,c,0) |\Omega\rangle$ the Poincar\'e transformation acts
strongly continuously, such that the restricted generators are given
by $H = \frac{1}{2} \int ((\nabla F)^2 + e^2 F^2)dx$. Furthermore,
$\omega_0$ is gauge invariant and this group is strongly continuous
in the $\omega_0$-representation. It is implemented by
$W(0,c,0)$.
$\omega_0$ corresponds to an irreducible representation for
$\M = \{ W(f,c,s)\}$, though in a nonseparable Hilbert space.
It corresponds to an irreducible though not faithful representation
of $\M_{\rm phys} = \{W(f,c,0)\}$ in the same way as $\omega$ did
for $\{ e^{isx}\}$. And again it can be obtained as the limit of
pure regular states:
We follow the calculations in [15]. There we considered the dipole
ghost, though without the operator $F$, that anyhow commutes with $\chi$
and $S$ and therefore lives in a separate factor of the tensor product.
If we define $B = S + \chi/2$ (where for simplicity we now put $e=1$),
then $[\chi,\dot \chi] = [B,\dot B] = 0$, whereas
$[\chi,\dot B] = [B,\dot \chi] = i \delta(x-x')$.
We can use the result of [15] to write
\begin{eqnarray*}
\wt \chi(k) &=& \{ \alpha_1(k) + \alpha_1^\dg(k) -
i(\alpha_2(k) - \alpha_2^\dg (k))\}/2\sqrt{k} \\
\dot{\wt \chi}(k) &=& -\{ \alpha_1(k) + \alpha_1^\dg(k) +
i(\alpha_2(k) - \alpha_2^\dg (k))\}\sqrt{k}/2 \\
\wt B(k) &=& \{ \alpha_2(k) + \alpha_2^\dg(k) -
i(\alpha_1(k) - \alpha_1^\dg (k))\}/2\sqrt{k} \\
\dot{\wt B}(k) &=&- \{ \alpha_2(k) + \alpha_2^\dg(k) +
i(\alpha_1(k) - \alpha_1^\dg (k))\}\sqrt{k}/2
\end{eqnarray*}
with $\alpha_1(k)$, $\alpha_2(k)$ bosonic annihilation operators.
We consider quasifree states with
$$
\langle \alpha_1(k) \alpha_1^\dg(k')\rangle =
\langle \alpha_2(k) \alpha_2^\dg(k')\rangle =
\rho(k) \delta(k-k')
$$
$$
\langle \alpha_1(k) \alpha_1(k')\rangle =
- \langle \alpha_2(k) \alpha_2(k')\rangle =
\sigma(k) \delta(k-k')
$$
$$
\langle \alpha_1^{(\dg)}(k) \alpha_2^{(\dg)}(k')\rangle = 0.
$$
Calculating
$$
\langle \chi^2\rangle = (4\rho - 2 -2\sigma -2\bar \sigma)/4k, \qquad
\langle \dot \chi^2\rangle = (4\rho - 2 -2\sigma - 2\bar\sigma)k/4
$$
and
$$
\langle B^2\rangle = (4\rho - 2 + 2\sigma + 2\bar \sigma)/4k,
$$
we can choose $\sigma = \sqrt{\rho(\rho-1)}$ such that $\omega$ is pure
and take the limit $\rho \ra \infty$. Then $\lim \langle \chi^2\rangle_\rho
= 0$ whereas $\lim \langle B\rangle_\rho$ diverges, in fact, we should
calculate $\langle e^{iB\gamma}\rangle$ which oscillates faster and
faster.
In [15] it was mentioned that there does not exist a pure state that is
space- and also time-invariant. But silently is was assumed that in the
representation the Weyl operators should be strongly continuous (regular).
A typical term for the time dependence is
$$
\langle \alpha_1^\dg(k,t) \alpha_1(k',t)\rangle =
\delta(k-k') [\rho_1 \cos^2 kt + \rho_2 \sin^2 kt -1
+ (2\rho_1 - 1 - 2\sigma_1) \frac{t^2}{k^2} \cos^2 kt +
$$
$$
+ (2 \rho_2 - 1 - 2\sigma_2)
\left( \frac{t}{k} + \frac{1}{k^2}\right)^2 \sin^2 k t].
$$
If $\rho_1 = \rho_2 \ra \sigma_1 + 1/2 = \sigma_2 + 1/2$ this becomes time independent
and $\omega$ invariant.
This should be compared with the result of [17]: In the representation
with indefinite metric the time evolution is implementable but in terms
of creation and annihilation operators with pseudocanonical commutation
relations.
We should also compare this result with the model discussed in [21]:
Again there is a fourth order differential equation. This makes the
Hamiltonian unbounded from below. But the singularity is not drastic
enough to forbid time independent regular representations. Only the
positivity condition is violated. Again this can be cured by
escaping to indefinite metric, and again this indefinite metric
corresponds to an irregular representation that is regular only in
the physical subspace on the physical observables.
\section{External Perturbations}
One might be afraid that representations like $\pi_0$ where the Weyl
operators are not strongly continuous in all variables will lead to
difficulties in perturbation theory when one wants to expand in powers
of the parameter of a perturbation. We shall investigate how serious
these troubles might become in the simplest typical cases. For this
purpose we shall couple the oscillators to an external ($= c$-number)
force and the Higgs model to an external current. In both cases we
shall see that for good couplings the perturbation expansion exists
already for the time evolution of the algebra independent of a
particular state and is not related to the vacuum satisfying a constraint.
In the case of oscillators we generalize the Hamiltonian (3.6) to
$$
H = S \chi + \Pi_\chi \Pi_S - j \chi \eqno(5.1)
$$
where $j(t)$ is some $c$-number $L^1({\bf R})$ function. This changes
the canonical equations to
$$
\dot S = \Pi_\chi, \qquad \dot \Pi_\chi = j - S, \qquad
\dot \chi = \Pi_S, \qquad \dot \Pi_S = - \chi. \eqno(5.2)
$$
They are solved by
$$
\begin{array}{l}
S(t) \equiv S \cos t + \Pi_\chi \sin t + \int_0^t dt' j(t') \sin(t-t') \\[12pt]
\chi(t) = \chi \cos t + \Pi_S \sin t \\[12pt]
\Pi_S(t) = - \chi \sin t + \Pi_S \cos t \\[12pt]
\Pi_\chi(t) = - S \sin t + \Pi_\chi \cos t + \int_0^t dt' j(t')
\cos (t-t').
\end{array} \eqno(5.3)
$$
In quantum theory these classical solutions induce an automorphism
$\Theta_j(t)$ of the Weyl algebra ($\Theta_j(t) e^{i\alpha S} :=
e^{i\alpha S(t)}$, etc.) and keeping the notation (3.13), we see
$W(\alpha,\beta,\gamma,\delta)(t) = \Theta_0(t) W(\alpha,\beta,\gamma,
\delta)$ and
$$
\Theta_j(t)(W(\alpha,\beta,\gamma,\delta)) = W(\alpha,\beta,\gamma,\delta)(t)
\exp[i \int_0^t dt' j(t')(\alpha \sin(t-t') + \delta \cos (t-t'))].
\eqno(5.4)
$$
Since even $\Theta_0(t)$ was not norm continuous in $t$ we see that
$\Theta_j(t)$ cannot be either. On the other hand, the $j$-dependence is
contained in a finite phase factor such that the map
$L^1({\bf R}) \times {\cal W} \ra {\cal W} : j,W \ra \Theta_j(t) W$
is norm continuous. The $j$ part is even continuous in $t$ such that
$\Theta_0(-t) \Theta_j(t)$ as map $\W \ra \W$ is norm continuous in $t$.
Furthermore since
$$
\Theta_0(-t) \Theta_j(t) W(\alpha,\beta,\gamma,\delta) =
W(\alpha,\beta,\gamma,\delta)e^{i(\alpha a(t) + \delta d(t))}
\eqno(5.5)
$$
with the $c$-numbers
$$
a(t) = \int_0^t dt' j(t') \sin (t-t'), \qquad
d(t) = \int_0^t dt' j(t') \cos(t-t'),
$$
$\Theta_0(-t) \Theta_j(t)$ is just a shift. This shows that it is an inner
automorphism and thus in any representation unitarily implemented:
$$
\Theta_0(-t) \Theta_j(t) W(\alpha,\beta,\gamma,\delta) =
W(0,-a/2,d/2,0) W(\alpha,\beta,\gamma,\delta)W(0,a/2,-d/2,0).
$$
Therefore for these perturbations a power series expansion in $j$
exists even on the level of the dynamical $C^*$-system ($\W,\Theta$)
without reference to a particular state. The state
$\omega_0(W(\alpha,\beta,\gamma,\delta)) = \delta_{\alpha 0}
\delta_{\delta 0}$ is invariant under $\Theta_j(t)$ and the
corresponding vector $|0\rangle$ (3.9) has eigenvalue 1 for the
generators $W(0,\beta,\gamma,0)$ for $\Theta_0(-t) \Theta_j(t)$. This
means that the constraint is still that the $\Theta_j(t)$ invariant
subalgebra generated by $\chi(t)$ annihilates $|0\rangle$. We could
also consider a representation generated by
$$
\bar \omega_0(W(\alpha,\beta,\gamma,\delta)) = \delta_{\beta 0}
\delta_{\gamma 0}. \eqno(5.6)
$$
This state is not invariant under $\Theta_j(t)$ but
$$
\bar \omega_0(\Theta_j(t) W(\alpha,\beta,\gamma,\delta)) =
e^{i(\alpha a(t)+\delta d(t))} \bar \omega_0(W(\alpha,\beta,\gamma,
\delta)). \eqno(5.7)
$$
For this representation the cyclic vector is no longer an eigenvector
for the generator $W(0,\beta,\gamma,0)$ of $\Theta_0(-t) \Theta_j(t)$.
This could be anticipated from the inhomogeneous time dependence of
$S$ (5.3) and thus for $j \neq 0$ the cyclic vector
$|\bar 0\rangle$ for (5.6) cannot satisfy $S(t) |\bar 0\rangle = 0 \;
\forall \, t$. Since we assumed $j(t) \in L^1({\bf R})$ we note that
$\lim_{t \ra \pm \infty} \Theta_0(-t) \Theta_j(t)$ exists as norm limit
when applied to $W \in \W$ and thus the M\o ller operators and the
$S$ matrix exist in any representation. Turning to the Higgs model we
couple to an external current by adding to (4.1) and (4.7)
$$
\cL_i = - A_\mu j_\mu . \eqno(5.8)
$$
To keep gauge invariance we impose on $j$ the conservation equation
$j^\mu{}_{,\mu} = 0$ in which case the Euler equations from
$\cL = \cL_e + \cL_s + \cL_i$, namely,
$$
A_{\mu,\nu}{}^\nu = - e(S_{,\mu} + e A_\mu) + j_\mu, \qquad
(S_{,\mu} + e A_\mu)^{,\mu} = 0 \eqno(5.9)
$$
imply again for $\chi = A_\mu{}^{,\mu}$, $\Box S \equiv S_{,\mu}{}^\mu$
etc.
$$
\Box \chi = 0, \qquad \Box S = - e \chi. \eqno(5.10)
$$
(5.9) are equivalent to Maxwell's equations
$$
F_{\mu\nu}{}^{,\nu} = J_\mu + j_\mu \eqno(5.11)
$$
whenever the constraint $\chi = const.$ holds. In two dimensions with the
notation (4.10), in addition to (5.10) we find
$$
(\Box + e^2)F = j, \qquad j = j_{1,0} - j_{0,1} .\eqno(5.12)
$$
We have seen the $\W$ tensors into $W(f,0,0)$ and $W(0,c,s)$, though
the field algebra is no longer characterized by (4.17) but by initial
data like (3.8) which anyhow is a one to one correspondence.
Only $W(f,0,0)$ is affected by $j$ whereas the rest is not since (5.10)
is unchanged.
Since again the scattering automorphism corresponds to a phase for the
Weyl operators, the automorphism is inner and the representations
equivalent for the full field algebra. For the gauge invariant observable
algebra this is not necessarily so (compare [7]).
If $e \neq 0$, then $\omega_0$ (4.19) on $W(f,0,0)$ is irreducible and
every shift is inner. Therefore the representations are equivalent.
For $e = 0$ the relations between $F$ and $\chi$, given by the evolution
equation, involve $j$ and the representations of $W(f,c,0)$ for
different $j$ become inequivalent.
\section{Conclusions}
In the usual canonical formulation of QED the field equations are
equivalent to Maxwell's equation provided $\chi = A_\mu{}^{,\mu}$
vanishes. In quantum theory one faces the difficulty that not every
ideal is a twosided ideal hence there is no representation
$\Pi$ with $\Pi_{|\chi} = 0$. There can be just a subspace
$\{ |g\rangle\}$ in Hilbert space (the ``physical space'') such that
$\Pi(\chi) |g\rangle = 0$. Even the existence of such vectors $|g\rangle$
is problematic and usually one tries to reach mathematical consistency
by introducing an indefinite scalar product in Hilbert space to obtain
$\langle g|g\rangle < \infty$. In such a representation one loses
positivity and norm and with them important tools of analysis.
Therefore we propose a conservative procedure where the $C^*$-algebra
of the fields $\A$ obeys the canonical commutation rules in their
Weyl form and is represented in a Hilbert space $\Ha$ by an
isomorphism $\pi: \A \ra \B(\Ha)$.
It turns out that the existence of a physical subspace
$\{ |g\rangle : \chi |g\rangle = 0\}$ requires $\Ha$ to be non-separable
and the Weyl operators are not strongly continuous in some arguments.
As a consequence the gauge variant fields do not exist as selfadjoint
operators but only their associated unitaries. However, there is no real
difficulty since observables as gauge invariant quantities exist and
their representation in the physical space [22] is the usual one. To this
extent the results of this paper only assure that (disregarding
divergences in perturbation theory) the $C^*$-algebraic
formulation is adequate for QED. There are of course other methods of
treating the problem and even in our conservative procedure there is
a one-parameter family of Lagrangeans which lead to different field
equations which are equivalent mod~$\chi = 0$ and are equally good.
However they lead to different commutation relations for the $A$
and it would be
unpleasant if the procedure described here were only to work for some of
them. As will be examined in the appendix this is not so but as long
as one of the canonical conjugate variables does not vanish identically,
our method works equally well.
\section*{Appendix A}
Here we shall verify that the commutation relations from a particular
class of Lagrangeans
$$
\cL = \frac{1}{2} \{ -A_{\mu,\nu} A^{\mu,\nu} + \alpha A_{\mu,\nu}
A^{\nu,\mu} + (S_{,\mu} + e A_\mu)^2 \} \eqno(A.1)
$$
with the field equations
$$
A_{\mu,\nu}{}^\nu - \alpha A^\nu{}_{,\mu\nu} = - e S_{,\mu} - e^2 A_\mu
\eqno(A.2)
$$
leads $\forall \, \alpha \neq 1$ to commutative relations which are
consistent with the constraint $A^\mu{}_{,\mu} |0\rangle = \chi |0\rangle
= 0$ $\forall \, (x,t)$. (For simplicity we shall use the Heisenberg
version.) If we start with 2 dimensions the 3 conjugate fields are
$$
\Pi_0 = - (1-\alpha) \dot A_0, \qquad
\Pi_1 = \dot A_1 - \alpha A'_0, \qquad
\Pi_S = \dot S + e A_0.
$$
This gives for the constraint variable $\chi = \dot A_0 - A'_1$
\begin{eqnarray*}
[\chi(x,t),\dot \chi(x',t)] &=& \left[ - \frac{1}{1-\alpha} \Pi_0 - A'_1,
\frac{1-\alpha}{1-\alpha} A''_0 - \frac{1}{1-\alpha} \Pi'_1 -
\frac{e}{1-\alpha} \Pi_S \right] \\
&=& - \frac{1}{1-\alpha} [\Pi_0(x,t),A''_0(x',t)] +
\frac{1}{1-\alpha} [A'_1(x,t),\Pi'_1(x',t)] \\
&=& i \delta''(x-x') \left( \frac{1}{1-\alpha} - \frac{1}{1-\alpha}\right)
= 0
\end{eqnarray*}
\begin{eqnarray*}
[F(x,t),\dot F(x',t)] &=& [\Pi_1 - (1-\alpha) A'_0, (1-\alpha) A''_1 +
\Pi'_0 - e S' - e^2 A_1] \\
&=& [\Pi_1,(1-\alpha) A''_1 - e^2 A_1] - (1-\alpha)[A'_0,\Pi'_0] \\
&=& i e^2 \delta(x-x')
\end{eqnarray*}
$$
[\chi(x,t),F(x',t)] = - \left[ \frac{1}{1-\alpha} \Pi_0 + A'_1,
\Pi_1 - (1-\alpha) A'_0\right] = 0
$$
\begin{eqnarray*}
[\dot F(x,t),\dot \chi(x',t)] &=& \left[(1-\alpha) A''_1 + \Pi'_0 - e S' - e^2A_1,
A''_0 - \frac{1}{1-\alpha} \Pi'_1 - \frac{e}{1-\alpha} \Pi_S\right] \\
&=& \frac{1}{1-\alpha} [(1-\alpha) A''_1 - e^2 A_1,\Pi'_1] +
[\Pi'_0,A''_0] + \frac{e^2}{1-\alpha} [S',\Pi_S] = 0.
\end{eqnarray*}
The two remaining commutators $[\dot F,\chi]$ and $[F,\dot \chi]$ vanish
obviously since no conjugate quantities occur. Thus $\forall \, \alpha \neq 1$,
$\chi$ with $\dot \chi$ generate an abelian subalgebra and $F$ and $\dot F$
generate a massive bosonic field which becomes abelian for $e^2 \ra 0$.
For $e = 0$, in addition, the fields $\chi$, $\dot \chi$ and $F$ and $\dot F$
become linearly dependent. So the constraint $\chi \equiv 0$ implies $F \equiv 0$.
By a somewhat more involved calculation one verifies that also
$3+1$ dimensions $\chi = A_k{}^{,k}$ and $\dot \chi$ commutes for all
$\alpha$ such that $\chi(t) |0\rangle = 0$ can be imposed consistently.
\section*{Appendix B}
We want to discuss on the example of the Weyl algebra how a state that
satisfies the constraint can be constructed.
We consider the automorphism group
$$
\alpha_t W(r,s) = e^{irt} W(r,s).
$$
Starting with some state $\omega(W(r,s))$ we can take some invariant
mean (which exists) that will have the form
$$
\bar \omega(W(r,s)) = \delta_{r0} f(s).
$$
For this state we can construct its GNS representation and find in its
commutant operators $M(0,s)$ with
$$
\langle \Omega|\pi (W^*(r,t)) M(0,s) \pi (W(\bar r,\bar t))|\Omega\rangle =
\delta_{r \bar r} f(s + \bar t - t) e^{ir(t-\bar t)/2}
$$
or
$$
M(0,s) \pi(W(\bar r,\bar t))|\Omega\rangle = \pi(W(\bar r,\bar t)
\cdot W(0,s))|\Omega\rangle
$$
as well as operators $M(p,0)$ with
$$
M(p,0) \pi (W(\bar r,\bar t))|\Omega\rangle =
\pi(W(\bar r,\bar t) \cdot W(p,0))|\Omega\rangle.
$$
$\{ M(0,s)\}''$ is invariant under
$\alpha_t$ and maximal abelian in the commutant of $\pi(\W)$.
Therefore it implies a unique decomposition of $\bar \omega$ into extremal
$\alpha_t$ invariant states.
Since $\{M(0,s)\}''$ is maximal abelian in $\{M(r,s)\}''$ they are at
the same time irreducible for
$\{ W(r,s)\}''$. Furthermore they correspond to vectors with pure
point spectrum of $W(0,s)$.
\newpage
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