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\begin{titlepage}
\begin{flushright}
UWThPh-1992-57\\
\end{flushright}
\vspace{4cm}
\begin{center}
{\Large \bf KMS States for the Weyl Algebra}\\[50pt]
Heide Narnhofer and W. Thirring \\
Institut f\"ur Theoretische Physik \\
Universit\"at Wien
\vfill
{\bf Abstract} \\
\end{center}
We consider the possible automorphism groups for the Weyl algebra
over {\bf R} resp. $T$ and classify those for which KMS states,
unique or not unique, exist.
\vfill
\end{titlepage}
\section{Introduction}
In elementary quantum mechanics the canonical equilibrium state is
defined by assigning to an observable $a$ the expectation value
$$
\omega_\beta(a) = \mbox{Tr } e^{-\beta H}a/\mbox{Tr } e^{-\beta H}
\eqno(1.1)
$$
where $H$ is the Hamiltonian which determines the time development by
$$
a_t = e^{itH} a e^{-itH}. \eqno(1.2)
$$
For (1.1) to make sense it is necessary that $H$ has a purely discrete
spectrum bounded from below (for $\beta > 0$). Nevertheless (1.1) has
been generalized to quantum field theory where $H$ has a continuous
spectrum [HHW]. This generalization works via the following abstraction.
$a \ra a_t$ is a one parameter group of automorphisms of the algebra
$\A$ of observables and an equilibrium state $\omega_\beta$ is
characterized by the KMS condition where the time evolution is
analytically continued to imaginary times
$$
\omega_\beta(ab) = \omega_\beta(b a_{i\beta}). \eqno(1.3)
$$
Within this framework one can, for example, also define the equilibrium
state for an infinite system of free fermions or bosons by the
quasifree state over the smeared creation and destruction operators
$a^*_f$, $a_g$
$$
\omega_\beta(a^*_f a_g) = \left\langle g \left| \frac{1}{e^{\beta h} \pm 1}
\right| f \right\rangle. \eqno(1.4)
$$
This defines a state even if the one particle Hamiltonian $h$ has a
continuous spectrum. In the case of fermions $h$ need not even be bounded
from below [RST].
In this note we propose to investigate to what extent the generalization
(1.3) opens new possibilities even in elementary wave mechanics. To get
a simple picture we take a restricted algebra $\A$ so that Aut~$\A$,
the groups of its automorphisms can easily be classified. For $\A$ we take
the algebra generated by the exponentials of the position operators $x$
and momentum operators $p$
$$
W(q,k) = e^{i(kx + qp)}, \qquad (q,k) \in {\bf R}^2.
\eqno(1.5)
$$
The noncommutativity of $x$ and $p$ is encoded in the Weyl relations (2.1)
between these operators and if $\A = W_R$ consists just of the polynomials in
$W(q,k)$ then Aut~$W_R$ are just the inhomogeneous linear symplectic
transformations of $(x,p)$. They are generated by Hamiltonians which are
polynomials in $(x,p)$ up to second order.
Instead of infinite space we can also consider particles on a circle.
The corresponding algebra is $W_T$ with (1.5) valid for $(q,k) \in {\bf R}
\times {\bf Z}$. Here not all polynomials lead to a time automorphism
group.
Our goal is to find out for which automorphisms there are KMS states.
We summarize the result in the following Table.
$$
\begin{tabular}{lccc}
& Automorphisms & KMS state & KMS state \\
& for $W_T$ & for $W_R$ & for $W_T$ \\
1. $H = p$ & yes & no & no \\
2. $H = x$ & yes & no & no \\
3. $H = \frac{1}{2}(p^2 + \nu^2 x^2)$ & yes for $\nu = 0$ & yes & yes, nonunique\\[5pt]
4. $H = \frac{1}{2}(p^2 - x^2)$ & no & no & no \\[5pt]
5. $H = \frac{1}{2}(xp + px)$ & no & no & no \\[5pt]
6. $H = \frac{1}{2} p^2 + gx$, $g \neq 0$ & yes & no & no
\end{tabular}
$$
We can understand the result by the following considerations:
1, 2 give shifts, 3, 4 is a harmonic force (resp. the free motion for
$\nu = 0$), 5 is the dilation and 6 the free fall. According to
arguments given at the beginning only 3 with $\nu \neq 0$ should be
expected to lead to KMS states because for $\nu = 0$ it has a
continuous spectrum in $[0,\infty]$ and the other Hamiltonians even from
$-\infty$ to $\infty$.
This turns out to be not quite so, the free Hamiltonian $p^2$ also leads
to KMS states $\Omega_\beta$ but the representation given by $\omega_\beta$
is not the standard representation and acts instead in a nonseparabel Hilbert
space. This is connected with the infinite volume and disappears when
one goes to a finite configuration space. Periodic boundary conditions
correspond to considering the subalgebra generated by
$W(q,k)$, $k \in {\bf Z}$, but now 3 ($\nu \neq 0$) and
4, 5 do not lead to automorphisms. In this case only
$H_3(\nu = 0) = H_6(g=0)$ leads to KMS states but a new phenomenon
appears which one knows only from phase transitions in infinite systems.
For given $\beta$ there is a one parameter family of KMS states for the
same time evolution. This is related to the fact that on the torus the
formal Hamiltonian $\frac{1}{2} p^2$ has a one parameter family of
selfadjoint extensions but they all lead to the time evolution
$x \ra x + pt$, $p \ra p$.
The different selfadjoint extensions of $p$ correspond to different
inequivalent irreducible representations of $W_T$. In them different
values are assigned to the central elements
$e^{2\pi inp} = e^{2\pi in\gamma}$, $\gamma \in [0,1)$
and in each of them we can construct a KMS state. These are all the
extremal KMS states. This phenomenon is not typical of the
thermodynamic limit but appears already for finite matrix algebras.
In contradistinction to phase transitions the value $\gamma$ for the central
elements is independent of $\beta$.
One can even consider a finite phase space by taking
$$
(q,k) \in \left( \frac{1}{g_1}{\bf Z}, \frac{1}{2\pi g_2} {\bf Z}\right),
\qquad g_i \in {\bf Z}.
$$
According to naive arguments such a phase space has only $g_1g_2$ cells
(of volume $h = 1$) and indeed $\A$ has then $g_1 \cdot g_2$ dimensional
irreducible representations. This kind of quantum system has been
studied for comparison with classical transformations on $T^2$ (as
phase space) and it appears in the motion on $T^2$ (as configuration
space) with a perpendicular magnetic field. In this case
the situation is as for the Hamiltonian 3
and we do have KMS states.
However, if there is in addition an electric field we are in the
situation 1 and there are no KMS states.
Of course, there are always invariant states like the tracial state
which corresponds to infinite temperature. In the cases where the KMS
states exist it will turn out to be easier to get their explicit form
by looking at the consequences of (1.3) than to calculate the traces
in (1.1).
Summarizing we can say that the spectral type of the Hamiltonian
(discrete or continuous) is not relevant for the existence of KMS states.
This might be expected since $H$ is a representation dependent object
unless the automorphism is inner. Nevertheless in our examples we
found that there are no KMS states in those cases where $H$ was not
semibounded. So only in case 3, i.e. for harmonic forces or free time
evolution KMS states exist. If we restrict ourselves to subalgebras,
then there can exist several KMS states, but one should not talk about
phase transition, because the parametrization of the KMS states is
independent of the temperature.
\section{The Weyl Algebra and Its Automorphisms}
If we exponentiate the operators $x$ and $p$ to $W(q,k) = e^{i(kx+qp)}$
these Weyl operators satisfy the multiplication law
$$
W(q,k) W(\bar q,\bar k) = e^{i(q\bar k-k\bar q)/2} W(q + \bar q, k + \bar k),
\qquad (q,k) \in {\bf R}^2. \eqno(2.1)
$$
We shall consider the algebra $W$ generated by these operators (i.e.
their finite linear combinations), the $C^*$-algebra $\bar W$ (the norm
closure of $W$), and the v. Neumann algebras $\pi(W)''$ (the weak closure
in a representation $\pi$). Furthermore we shall consider the
subalgebras $W_T$ (``a particle on a torus'') where $(q,k) \in {\bf R}
\times {\bf Z}$ and $W_C$ (``the rotation algebra'') where
$(q,k) \in \Theta {\bf Z} \times {\bf Z}$. $W$ is simple, $W_T$ has the
center ${\cal Z} = \{W(n,0), n \in 2\pi {\bf Z}\}$ and similarly
$W_C$ if $\Theta$ is rational. If $\Theta$ is irrational, $W_C$ is
again simple. The automorphisms of $W$ correspond to inhomogeneous
linear symplectic transformations of $(x,p)$ and generate $(z = (q,k)$)
$$
\alpha_{L,S} W(z) \ra e^{i \langle L|z\rangle} W(Sz), \eqno(2.2)
$$
with $S \in SL(2,R)$, $L$ linear, real.
Since $\alpha_{L,S} \circ \alpha_{\bar L,\bar S} =
\alpha_{\bar L + L \bar S,S \circ \bar S}$
we see Aut~$W = {\bf R}^2 \times\!\!\!\!\!\!\supset SL(2,R)$.
This group is five-dimensional and there are the following one parameter
subgroups.
\begin{enumerate}
\item Shift in $x$: $L = (0,t)$, $S = {\bf 1}$,
\item Shift in $p$: $L = (t,0)$, $S = {\bf 1}$,
\item Harmonic oscillator: $L = 0$, $S = \left( \ba{cc}
\cos t & \sin t \\ \sin t & \cos t \ea \right)$,
\item Repulsive harmonic force: $L = 0$,
$S = \left( \ba{cc} \cosh t & \sinh t \\ \sinh t & \cosh t \ea \right)$,
\item Dilation: $L = 0$, $S = \left( \ba{cc} e^t & 0 \\ 0 & e^{-t} \ea
\right)$,
\item Free fall: $L = (gt,\dfrac{g}{2} t^2)$, $S = \left( \ba {cc}
1 & t \\ 0 & 1 \ea \right)$.
\end{enumerate}
The groups 1 and 2 (resp. 4 and 5) are conjugate. In $\bar W$ there are
other inner automorphisms, f.i. the ones generated by $e^{it(W+W^*)}$
but they shall not be considered here. 1 and 2 leave $W_T$ and $W_C$
and 6 leaves $W_T$ invariant. Thus they also give automorphisms of these
subalgebras though $x$ is not a globally defined function on the
torus. Since $\|W(z) - W(\bar z)\| = 2$ for $z \neq \bar z$ only
1 and 2 are normcontinuous.
\section{States}
A state $\omega$ is a positive linear functional over $W$ with
$\omega({\bf 1}) = {\bf 1}$. It gives a representation $\pi_\omega$ of
$W$ in a Hilbert space $\Ha_\omega$ the vectors of which we shall
denote by $|q,k\rangle := W(q,k)|0,0\rangle$. $\omega$ is determined
by
$$
f(q,k) := \omega(W(q,k)) = \langle 0,0|W(q,k)|0,0\rangle
$$
and the scalar product of two vectors is
$$
\langle \bar q,\bar k|q,k\rangle =
\langle 0,0|W^*(\bar q,\bar k) W(q,k)|0,0\rangle =
e^{i(q \bar k - k \bar q)} f(q - \bar q, k - \bar k). \eqno(3.1)
$$
Thus $f$ defines a state over $W$ iff $|f(q,k)| \leq 1$ and
$e^{i(q \bar k - k \bar q)/2} f(q-\bar q,k - \bar k)$ is a positive
kernel. If $f$ is continuous then $(q,k) \ra \pi_\omega(W(q,k))$ is
weakly continuous and v. Neumann's uniqueness theorem says that in
this case $\pi_\omega(W)$ is isomorphic to the standard representation.
$W$ has also a tracial state $\omega_0$ with $f_0(q,k) = \delta_q
\delta_k := 1$ for $q = k = 0$ and 0 otherwise. $\pi_{\omega_0}$ is not
weakly continuous and it acts in a nonseparable Hilbert space since
$\langle \bar q,\bar k|q,k\rangle = \delta_{q-\bar q} \delta_{k -
\bar k}$ (compare also [NT], [AMS].
Thus the $|q,k\rangle$ are a nondenumerable orthogonal basis.
Since $W_T$ has a nontrivial center the uniqueness theorem does no longer hold
and there is a one parameter family of irreducible representations
$\pi_\gamma$. Physically $\gamma$ is the Bohm-Aharonov phase and
mathematically it gives the $c$-number assigned to the central elements
$\pi_\gamma(W(2\pi n,0)) = e^{2\pi in\gamma}$. Note that $\pi_\gamma$
is generated by $\omega_\gamma$,
$$
\omega_\gamma(W(q,k)) = e^{i\gamma q} \delta_k \eqno(3.2)
$$
which gives
$$
\langle \bar q,\bar k|q,k\rangle = \delta_{k \bar k}
e^{i(q - \bar q)(\gamma + k/2)}
$$
and thus
$$
|q,k\rangle = e^{iq(\gamma + k/2)} |0,k\rangle.
$$
Therefore
$$
\pi_\gamma(W(2\pi n,0))|q,k\rangle = e^{2\pi in\gamma} |q,k\rangle .
$$
We shall now turn to KMS states. Since the automorphisms are of the form
$W(z) \ra e^{i \langle z|L_t\rangle} W(S_t z)$ the KMS condition is
$(\sigma(z,\bar z) = \frac{1}{2}(q \bar k - k \bar q))$
$$
\omega(W(z) W(\bar z)) = \omega(W(\bar z) e^{i\langle z|L_{i\beta}\rangle}
W(S_{i\beta} z)) \eqno(3.3)
$$
which means
$$
e^{i\sigma(z,\bar z)} f(z + \bar z) =
e^{i\sigma(\bar z,S_{i\beta}z) +i \langle z|L_{i\beta}\rangle}
f(S_{i\beta} z + \bar z) \quad \forall\, z,\bar z \in {\bf R}^2. \eqno(3.4)
$$
For $\bar z = - z$ we conclude
$$
1 = e^{-i\sigma(z,S_{i\beta}z) +i \langle z|L_{i\beta}\rangle}
f((S_{i\beta} - 1)z), \qquad \forall \, z \in {\bf R}^2. \eqno(3.5)
$$
For the subgroups 1 and 2 where $S_t = 1$, $L \neq 0$ we see that there
are no KMS states.
Case 3:
We have $L = 0$ and
$$
S_{i\beta} = \left( \ba{rr} \cosh \beta & i \sinh \beta \\
-i \sinh \beta & \cosh \beta \ea \right)
$$
and thus
$$
f(z) = \exp\left[ i\sigma (z,\frac{1}{S_{i\beta} -1} z)\right]
$$
or in components
$$
f(q,k) = \exp\left[-\frac{1}{2} (q^2+k^2)\coth \frac{\beta}{2}\right].
\eqno(3.6)
$$
This satisfies the requirements (3.4), for $\beta > 0$ in addition that
$0 \leq f \leq 1$, for $\beta \ra 0$
($T \ra \infty$) it gives the tracial state and for $\beta \ra \infty$
($T \ra 0$) the standard harmonic oscillator ground state representation.
Since conjugate automorphisms have conjugate KMS states it remains to
consider 4 or 5.
Case 5:
We have $L = 0$,
$$
S_{i\beta} = \left( \ba{cc} e^{i\beta} & 0 \\ 0 & e^{-i\beta} \ea \right)
$$
and (3.4) gives the condition
$$
f(z) = \exp\left[i \sigma(z, \frac{1}{S_{i\beta} - 1} z) \right],
$$
that is
$$
f(q,k) = \exp\left[ \frac{i}{2} qk \left( \frac{e^{-i\beta}}{e^{-i\beta} -1}
- \frac{e^{i\beta}}{e^{i\beta} - 1}\right) \right] =
\exp\left[-qk \frac{1}{2} \cot \frac{\beta}{2} \right].
$$
Since this does not satisfy $|f(q,k)| \leq 1$ there are no KMS states
for 4 and 5.
Case 6:
$$
S_{i\beta} - 1 = \left( \ba{cc} 0 & i\beta \\ 0 & 0 \ea \right)
$$
and
$$
L_{i\beta} = \left( i\beta g, - \frac{g}{2} \beta^2\right)
$$
$S_{i\beta} -1$ is not invertible but (3.5) can be used and reads
$$
1 = \exp\left[- \beta \frac{k^2}{2} - \frac{g}{2} \beta^2 k - i\beta gq
\right] f(i\beta k,0) \qquad \forall\, (q,k).
$$
Thus for $g \neq 0$ there are no KMS states.
Case 6, $g=0$:
We see that
$f(q,0) = \exp[-q^2/2\beta]$ which requires again $\beta > 0$. For
$k \neq 0$ we have to go back to (3.4),
$$
\exp\left[ \frac{i}{2}(q \bar k - k \bar q) \right]
f(q + \bar q,k + \bar k) =
\exp\left[ \frac{i}{2}(\bar q k - \bar k(q + i\beta k)) \right]
f(q + i \beta k + \bar q,k + \bar k)
$$
or for $\bar q = k = 0$
$$
\exp\left[ \frac{i}{2} \bar k q\right] f(q,\bar k) =
\exp\left[-\frac{i}{2} \bar k q \right] f(q,\bar k)
$$
and thus $f(q,k) = 0$ for $k \neq 0$. The KMS state
$$
f_\beta(q,k) = \delta_k \exp \left[ \frac{-q^2}{2\beta}\right]
\eqno(3.7)
$$
gives again a representation of $W$ which is not strongly continuous in
$k$ and thus $x$ does not exist as operator, only $e^{ikx}$.
However, the time evolution $(q,k) \ra (q + kt,k)$ is represented
strongly continuously and thus possesses a generator $H$. On the
contrary for $W_T$ this representation is strongly continuous ({\bf Z}
has discrete topology). In this case the central elements
$W(2\pi n,0)$ are not represented by multiples of unity since the
vectors $|q,k\rangle$ for different $q$ point in different directions
and
$$
\pi_\beta (W(2\pi n,0))|\bar q,\bar k\rangle =
e^{i\pi (n\bar k)} | \bar q + 2\pi n,\bar k\rangle.
$$
Thus (3.7) does not give an extremal KMS state but can be decomposed as an
integral over extremal KMS states over $W_T$.
The latter are superpositions of the states (3.2)
$$
f_{g,\gamma}(q) = e^{i(g + \gamma)q}, \qquad g \in {\bf Z}, \;
\gamma \in [0,1),
$$
$$
f_{\beta,\gamma}(q) = \left.\sum_{g \in {\bf Z}} f_{g,\gamma}(q)
e^{- \beta(g + \gamma)^2/2} \right/ \sum_{g \in {\bf Z}}
e^{- \beta(g + \gamma)^2/2}, \eqno(3.8)
$$
$$
f_\beta (q) = \int_0^1 d\gamma \: f_{\beta,\gamma}(q).
$$
The states defined by $f_{\beta,\gamma}$ are already KMS states since
(3.4) requires
$$
f(q) = e^{iqk - \beta k^2/2} f(q + i\beta k)
$$
which follows from (3.8) since $k \in {\bf Z}$. Furthermore $f_{\beta,\gamma}$ satisfies
$f(q + 2\pi n) = e^{2 \pi in\gamma} f(q)$. Since
$$
\langle q + 2\pi n,\bar k|q,k\rangle_\gamma = \delta_{\bar k,k}
e^{in\pi(k- 2\gamma)} \langle q,k|q,k\rangle_\gamma
$$
we see that $\forall\,k$
$$
|q + 2\pi n,k\rangle = e^{i\pi nk + 2\pi in\gamma}|q,k\rangle
$$
and
$$
\pi_\gamma(W(2\pi n,0))|q,k\rangle = e^{2\pi in\gamma}|q,k\rangle,
$$
that is the central elements are represented in $\pi_\gamma$ by
$c$-numbers. The same still holds in $\pi_{\beta,\gamma}$ but not in
$\pi_\beta$. In $\Ha_{\beta,\gamma}$ we have
$$
\langle q + 2\pi n,\bar k|q,k\rangle_{\beta,\gamma} =
\delta_{\bar k,k} \left. \sum_g e^{-in\pi(k+2\gamma+2g)}
e^{-\beta(\gamma + g)^2/2} \right/ \sum_g e^{-\beta(\gamma + g)^2/2}
= \delta_{\bar k k} e^{-in\pi(k+2\gamma)}.
$$
Thus again
$$
\pi_{\beta,\gamma}(W(2\pi n,0))|q,k\rangle = e^{2\pi in\gamma} |q,k\rangle
$$
and $\pi_{\beta,\gamma}(W(2\pi,0)) = e^{2\pi i\gamma}$. Integrating over
$\gamma$ makes $|\langle q + 2\pi n,k|q,k\rangle| < 1$ and thus
$\pi_\beta(W(2\pi n,0))$ is no longer a $c$-number.
We can conversely show that in this way we have obtained all possible
KMS states for the shift on $W_T$. As already stated the KMS condition
(3.4) implies ($\beta = 1$)
$$
\omega(W(q,k)) = \exp\left[-i(k-n)q + ikb -
\frac{(k-n)}{2}n \right] \omega(W(q + in,k))
$$
$\forall\,b,n$ which demands
$$
\omega(W(q,k)) = \delta_{k} f(q)
$$
and
$$
f(q) = e^{inq - n^2/2} f(q + in).
$$
We have to assume that $f$ is analytic and it decreases so that we
can pass to the Fourier transform
$$
\wt f(p) = \int f(q) e^{iq p} dq =
\int_{-\infty}^\infty e^{in(q + in) + n^2/2} f(q + in)
e^{i(q + in)p + np} dq
$$
or
$$
\wt f(p) = \exp\left[ \frac{n^2}{2} + np \right] \wt f(p + n)
$$
Thus we can choose $\wt f(p)$ arbitrary in the interval [0,1) and this
determines then $\wt f(p)$ resp. $f(q)$ completely. The choice
$\wt f(p) = \delta(p - \gamma)$ gives the extremal KMS state
$\omega_\gamma$, and their convex combinations give all the other
choices.
\section{Application to the Quantum Hall Current}
The Weyl algebra is the appropriate description for the quantum Hall
effect, i.e. a particle on a two-dimensional torus in a magnetic field
orthogonal to the plane [B,GNT,T]. Then
$$
\exp[i \alpha v_x] \exp[i \gamma v_y] = \exp [i\gamma \alpha B]
\exp [i \gamma v_y] \exp[i \alpha v_x]
$$
and
$$
\exp[i\alpha v_x] \exp[inx] = \exp[i\alpha n] \exp[inx]
\exp[i \alpha v_x],
$$
$$
\exp[i \beta v_y] \exp[imy] = \exp[i\beta m] \exp[imy]
\exp[i \beta v_y],
$$
$$
\alpha,\gamma \in {\bf R}^2, \qquad (mn,n) \in {\bf Z}^2.
$$
Introducing new variables
$$
\exp[in \bar x] = \exp[inx] \exp\left[-i \frac{n}{B} v_y\right]
$$
$$
\exp[im \bar y] = \exp[imy] \exp\left[ i \frac{m}{B} v_x\right]
$$
the new variables $\bar x,\bar y$ describe the centers of the Larmor
precession. Now
$$
\left[ e^{i(n \bar x + m \bar y)}, e^{i(\alpha v_x + \beta v_y)}
\right] = 0
$$
whereas
$$
\exp[in \bar x] \exp[i m \bar y] = \exp\left[ i \frac{nm}{B}\right]
\exp[im\bar y] \exp[in \bar x].
$$
Therefore
$$
\M = W({\bf R}^2,B) \otimes W\left({\bf Z}^2,\frac{1}{B}\right).
$$
The time evolution of particles in a magnetic field only affects
$W({\bf R}^2,B)$ and here corresponds to case 3 with
$H = \frac{1}{2}(v^2_x + v_y^2)$ so that we already found the possible
KMS states
$$
\omega_\beta(W(\alpha,\beta,n,m) = f(\alpha,\beta) \bar f(n,m)
$$
with $f(\alpha)$ from (3.6) and $\bar f$ can be chosen arbitrarily.
If we add an electric field in the $y$-direction this corresponds to
the Hamiltonian
$$
H = \frac{1}{2}(v_x^2 + v_y^2) - \frac{E}{B} v_x + E \bar y =
\frac{1}{2} \left( \left( v_x - \frac{E}{B}\right)^2 + v_y^2\right)
- \frac{E^2}{2B^2} + E \bar y
$$
which means
$$
\tau_t \exp[im \bar x + i n \bar y] =
\exp\left[i \frac{E}{B} mt + im \bar x + i n \bar y \right].
$$
Therefore in the velocity space the Hamiltonian is only affected by a
shift and we can get the same class of KMS states as for case 3 but
in the $(\bar x,\bar y)$ space we are in case 1 and know that there
does not exist a KMS state. This result should be compared with the
observation [GNT] that in the presence on an electric field time
evolution rotates between inequivalent irreducible representations
(for $B$ rational they exist). Only for $\beta = 0$ exists the
invariant tracial state which leads to a type II$_1$-representation.
The reason why there are no KMS states for $E \neq 0$ is the same why
there are no KMS states for the shift on the torus. There the
Hamiltonian $H = p$ has eigenvalues $g \in {\bf Z}$. Thus the
occupancy of the levels has the Boltzmann factor $e^{-\beta g}$ and
this does not lead to a (normalized) probability distribution. As
mentioned at the beginning in the second quantized theory there is no
difficulty for fermions. There the Fermi distribution $\sim 1/(1 + e^{\beta g})$
gives a KMS state. However, it is not locally normal (relative to the
vacuum) which means that there is an infinite particle density. This
is clear since all levels for $g \ll - \beta^{-1}$ are practically
filled. In reality an electric field has only a finite range and then
stability of matter guarantees that $H - \mu N$ is positive and this
obstacle to KMS states does not exist. These facts are not reproduced
by the motion on the torus.
\bibliographystyle{abbrv}
\begin{thebibliography}{WWW}
\bibitem[AMS]{} F. Acerbi, G. Morchio, F. Strocchi, SISSA preprint 39/92
\bibitem[B]{} J. Bellisard, in Statistical Mechanics and Field Theory,
ed. T.C. Dorlas, N.M. Hugenholtz, M. Winnink (Springer Verlag, Berlin
1986), p. 99--157
\bibitem[GNT]{} H.R. Gr\"umm, H. Narnhofer, W. Thirring, Acta Phys.
Austr. {\em 57}, 175 (1985)
\bibitem[HHW]{} R. Haag, N.M. Hugenholtz, M. Winnink, Commun. Math.
Phys. {\em 5}, 215--236 (1967)
\bibitem[NT]{} H. Narnhofer, W. Thirring, Vienna preprint UWThPh-1991-63,
to be publ. in Rev. Math. Phys. (1992), in honour of R. Haag
\bibitem[RST]{} F. Rocca, M. Sirugue, D. Testard, Commun. Math. Phys.
{\em 13}, 317--334 (1969); {\em 19}, 119--141 (1970)
\bibitem[T]{} W. Thirring, A Course in Mathematical Physics III
(Springer, New York 1981)
\end{thebibliography}
\end{document}