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absolutely continuous spectrum,  
Hamiltonian system, dissipative operators
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\title{Absence of the absolutely continuous spectrum of a first-order
non-selfadjoint Dirac-like system for slowly decaying 
perturbations} 
\author{ 
Marco Marletta 
\\ School of Mathematics, 
\\ Cardiff University, 
\\ PO Box 926, Cardiff CF24 4YH
\\ e-mail: Marco.Marletta@cs.cardiff.ac.uk
\and Roman Romanov\thanks{
On leave from Laboratory of Quantum 
Networks, Institute for Physics, Saint Petersburg State 
University. The work of R. R. has been partially supported by 
EPSRC Grant GR/ R20885 and by RFBR Grant 03-01-00090} 
\\ Department of Computer Science, 
\\ Cardiff University, 
\\ PO Box 916, Cardiff CF24 3XF
\\ e-mail: r.v.romanov@cs.cf.ac.uk} 

\begin{document}
\maketitle

\begin{abstract}
We prove that the absolutely continuous subspace of the completely 
non-selfadjoint part of a first-order dissipative Dirac-like 
system is trivial when the imaginary part of the potential is 
non-integrable. 
\end{abstract}

%\input{amssym}

\noindent {\sc AMS(MOS) Subject classifications:} 34L40, 47B44.

\noindent {\bf Keywords:} absolutely continuous spectrum, 
Hamiltonian system, dissipative operators.

\section*{Introduction}\label{section:1} 
In this article we analyze the structure of the essential spectrum 
of dissipative operator realizations of the ordinary differential 
expression \be \ell_Q := J\frac{d}{dx} + Q(x), \;\;\; x \in 
[0,\infty), \label{eq:1.1} \ee where \be J = 
\left(\begin{array}{cc} 0 & i \\ i & 0 
\end{array}\right), \;\;\; Q(x) = \left(\begin{array}{cc} q_{1}(x) 
& 0 \\ 0 & 
 q_{2}(x) \end{array}\right), \label{eq:1.2} 
\ee in which $ Q ( x ) $ is a bounded function such that $\Im 
q_1(x)$ and $\Im q_{2}(x)$ are non-negative a. e. We shall prove 
the following result. 

\bigskip
\noindent {\bf Theorem 1.} {\it Let $L$ be the operator in 
$L^2(\Rr^+,\Cc^2)$ given by the expression (\ref{eq:1.1}) and with 
domain determined by a self-adjoint boundary condition at $0$. If 
the absolutely continuous subspace of the completely 
non-selfadjoint part of $L$ is nontrivial, then $\Im q_{1}$ and 
$\Im q_{2}$ both lie in $L^1(\Rr^+)$.} 
\bigskip 

This assertion is an analogue of the following theorem in the 
paper of Romanov \cite{Rom} for Schr\"{o}dinger operators on the 
half-line: if a bounded potential $ q $ on the half-line $ x \ge 0 
$ such that $ \Im q ( x ) \ge 0 $, $ \Im q ( x ) \to 0 $ as $ x 
\to \infty $, satisfies $ \Im q \notin L^1 $, then the dissipative 
operator $ l $, $ l u = - u'' + q u $, corresponding to a 
self-adjoint boundary condition at zero has trivial absolutely 
continuous subspace. 

Together with the latter theorem, the result of the present paper 
suggests that the existence of an absolutely continuous spectrum 
for dissipative ordinary differential operators is equivalent to 
having a scattering theory. Let us mention here that the abstract 
local nuclear scattering theory developed in \cite{N,Nab} shows 
that for any pair of operators $ B, B_0 $ such that $ B_0 $ is 
selfadjoint and $ B - B_0 $ is of (relative) trace class the 
absolutely continuous parts of $ B $ and $ B_0 $ are quasi-similar 
by means of wave operators and their spectra coincide. In the 
situation under consideration this implies that the absolutely 
continuous spectrum of $ L $ coincides with that of the 
selfadjoint Dirac operator $ \Re L $ when $ \Im Q \in L^1 $. This 
is to be compared with the situation of the self-adjoint theory, 
where the existence of modified wave operators for the 
Schr\"odinger operator in dimension $1$ has only been established 
for specific classes of slowly decaying potentials $ q $ (see, e. 
g. \cite{Kiselev}), while the absolutely continuous spectrum is 
known to coincide with the positive real line for any real $ q \in 
L^2 $ \cite{DK}. 

The strategy of the proof of the main result follows that in the 
paper \cite{Rom} with one notable exception. We start with an 
abstract re-formulation of triviality of the absolutely continuous 
subspace of a dissipative operator, given in Proposition 
\ref{criterion}, and then obtain in Corollary \ref{corollary:1} a 
convenient sufficient condition in the situation under 
consideration which is expressed in terms of the asymptotics of 
solutions of the equation $ \ell_{ Q^* } y = k y $ with real $ k 
$. Specifically, it says that the absolutely continuous subspace 
of the completely non-selfadjoint part of $L$ is trivial if $ 
\int^\infty \( \Im q_1 | y_1 |^2 + \Im q_2 | y_2 |^2 \) dx $ is 
infinite for a.e. real $ k $. Here $ y = \( y_1 , y_2 \)^T $ is 
the solution of the Cauchy problem for $ \ell_{ Q^* } y = k y $ 
with the initial data $ y(0) = \left( 1 , 0 \right)^T $ (or any 
other $ k $-independent non-zero initial data satisfying a 
selfadjoint boundary condition). In the case of a Schr\"odinger 
operator \cite{Rom}, the next step in the argument used spectral 
averaging \cite{SimonLast} for the selfadjoint problem with the 
potential $ \Re q $. The spectral averaging method, as developed 
by Last-Simon, seems to be only applicable to operators 
semi-bounded below, while a selfadjoint operator corresponding to 
a differential expression of the form (\ref{eq:1.1}) is never 
semi-bounded. To circumvent this difficulty, we first establish 
the result for potentials with "spread" $ \Im Q $ (see Proposition 
\ref{theorem:weak1} and a remark after it), and then show that an 
arbitrary potential can be represented as a sum of a potential 
with spread $ \Im Q $ and an $ L^1 $-potential. An appropriate 
variant of the trace class method, which we develop in the 
abstract part of the paper in Lemma \ref{trace}, allows us to 
handle the $ L^1 $ - perturbation. This argument is similar to 
that used in \cite{Rom} for proving a discrete variant of the main 
result. 

Throughout the paper $ {\bf S}^1 $ stands for the trace class of 
operators. Given an operator $ B $ we write $ B^\prime $ for the 
completely non-selfadjoint part of $ B $. For any vector $ v $ let 
$ v^T $ stand for its transpose. 

\refstepcounter{section}
\subsection*{\S 1. Preliminaries: Absolutely Continuous Subspace}

Let $ H $ be a Hilbert space and $ L $ be a maximal dissipative 
operator in $ H $ of the form $ L = A + i V $, $ A = A^* $, $ V 
\ge 0 $ is bounded. We shall assume that\footnote{By $\sigma_{ess}(L)$
we mean the union of the set of all non-isolated points of $\sigma(L)$
and the set of all isolated points of $\sigma(L)$ of infinite 
multiplicity.} $ \sigma_{ ess } ( L ) 
\subset \Rr $. Define $ H_0 $ to be the maximal 
reducing subspace of $ L $ on which it induces a selfadjoint 
operator, which will be denoted by $ L_0 $. Let $ L^\prime $ be 
the completely non-selfadjoint part of $ L $, $ L^\prime = \left. 
L \right|_{ H^\prime } $, $ H^\prime = H \ominus H_0 $ \cite{N}. 
Define the Hardy classes of vector functions $ \H^2_\pm $ to be 
the collections of $H$-valued analytic functions $f$ on $ \Cc_\pm 
= \{ z \in \Cc \, | \, \pm \Im z > 0 \} $, which satisfy $\sup_{ 
\von 
> 0 } \int_\R \len f ( k \pm i\von ) \rin^2 dk < \infty $, respectively.

\begin{definition}
{\it The absolutely continuous (a. c.) subspace}, $ H_{ac} ( 
L^\prime ) $, of the completely non-selfadjoint operator $ 
L^\prime $ is defined as follows \cite{N,Sachn} \be\label{Hac} 
H_{ac} ( L^\prime ) = {\rm{clos}} \left\{ u \in H: 
\begin{array}{l} 1) \( L - z \)^{ -1 } u \mbox{ is analytic in } \C_+ 
 \\ 2) \, \, V^{1/2} \( L - z \)^{ -1 } u \in \H^2_+  
\end{array} \right\} . \ee

By the a. c. subspace of the operator $ L $ we call the subspace 
$$ H_{ ac } ( L ) = H_{ac} ( L^\prime ) \oplus H_{ac} ( L_0 ) $$ 
where $ H_{ac} ( L_0 ) $ is the a. c. subspace of the self-adjoint 
operator $ L_0 $ defined in the standard way. \end{definition} 

The a. c. subspace of $ L $ is known \cite{N} to be regular 
invariant, that is, $ \overline{( L - z )^{-1} H_{ac} } = H_{ac} $ 
for all $ z \in \rho (L) $. 

Various motivations of this definition and its relation to 
scattering theory are given in \cite{N,Nab,Sachn,Pavli,Pav1}. We 
only mention here\footnote{Theorem \ref{equivl} is not used in the 
paper.} the following "weak" characterization of the subspace $ 
H_{ac} ( L ) $. 

\begin{theorem} \label{equivl}
\cite{equival} $$ H_{ac} ( L ) = {\rm{clos }} \left\{ u \in H: \; 
\begin{array}{l} 1) \( L - z \)^{ -1 } u \mbox{ is analytic in } \C_+ \\ 
2) \left. \left\langle \( L - z \)^{ -1 } u , v \right\rangle 
\right|_{ \C_\pm } \in \H^2_\pm \mbox{ for all } v \in H \cr 
\end{array} \right\}. $$
\end{theorem} 

This theorem shows, in particular, that (\ref{Hac}) can be 
considered as a generalization of the definition of the a. c. 
subspace in the self-adjoint theory. 

The triviality of the subspace $ H_{ac} ( L^\prime ) $ in the 
example studied in this paper will be established on the basis of 
the following criterion which is implicit in \cite{Pav1}. 

\begin{proposition} \label{criterion}
$ H_{ac} ( L^\prime ) = \{ 0 \} $ iff for a. e. $ k \in \R $ we 
have ($ z = k + i \von $) \be\label{crit} D ( z ) \equiv \sqrt 
\von \( L^* - z \)^{ -1 } \sqrt V \stackrel{s} \longrightarrow 0 
\ee as $ \von \searrow 0 $. 
\end{proposition}

Different elementary proofs of the "if" part of the criterion can 
be found in \cite{Rom,equival}. The "only if" part is not used in 
this paper. 

\medskip

\noindent {\bf Remark} {\it \cite{Pav1} The function $ D ( z ) $ 
satisfies $ \| D ( z ) \| \le 1/2 $ for all $ z \in \C_+ $.} 

\medskip

The proof of this fact is by direct calculation. A similar 
calculation is contained in (\ref{contr}) in Lemma \ref{trace} 
below. 

As is well known, if two self-adjoint operators $ {\cal L}_{ 1,2 } 
$ satisfy $ {\cal L}_1 - {\cal L}_2 \in \bS^1 $, then their a. c. 
spectra coincide. In the non-selfadjoint case, the question 
whether $ L_1 - L_2 \in \bS^1 $ for a pair of dissipative 
operators $ L_1 $, $ L_2 $ implies $ \sigma \( L_1 |_{ H_{ac} ( 
L_1 ) } \) = \sigma \( L_2 |_{ H_{ac} ( L_2 ) } \) $ appears to be 
open if $ \Im L_{ 1,2 } $ are not of $ \bS^1 $ separately. We 
shall need a result of this type in a special situation. 

\begin{lemma} \label{trace} Let $ L $ and $ \tL $ be two dissipative
operators such that \be\label{trres} \( L - z \)^{ -1 } - \( \tL - 
z \)^{ -1 } \in \bS^1 , \; \; z \in\rho ( L ) \cap \rho ( \tL ) . 
\ee Write $\tL - L = i \Gamma $, and assume that $ \Gamma \ge 0 $ 
and there exists a bounded operator $ \Pi \ge 0 $ such that $ 
\Gamma = \tV \Pi $, $ \tV = \Im \tL $. If $ H_{ac} ( \tL^\prime ) 
= \{ 0 \} $, then $ H_{ac} ( L^\prime ) = \{ 0 \} $. 
\end{lemma}

This lemma is a variant of Lemma 2.3 from \cite{Rom}, where $ 
\Gamma $ was assumed to be trace class. 

\noindent {\bf Proof}\, We shall actually prove that if $ 
\widetilde D ( k + i \von ) \equiv \sqrt \von \( \tL^* - z \)^{ -1 
} \sqrt \tV \stackrel{s} \longrightarrow 0 $ as $ \von \to 0 $ for 
a.e. $ k \in \R $, then $ D ( k + i \von ) \stackrel{s} 
\longrightarrow 0 $ for a. e. $ k \in \R $ as well. Taking into 
account the assumed factorization of $ \Gamma $ we find that $$ 
\sqrt \von \( \tL^* - z \)^{ -1 } \sqrt \Gamma \stackrel{s} 
\longrightarrow 0 $$ for a. e. $ k \in \R $. Let us show that $ 
\sqrt \von \( L^* - z \)^{ -1 } \sqrt \Gamma \stackrel{s} 
\longrightarrow 0 $ for a. e. $ k $. We have, $$ \sqrt \von \( L^* 
- z \)^{ -1 } \sqrt \Gamma \cdot G ( z ) = \sqrt \von \( \tL^* - z 
\)^{ -1 } \sqrt \Gamma , $$ where $$ G ( z ) = I + i \sqrt \Gamma 
\( \tL^* - z \)^{ -1 } \sqrt \Gamma . $$ Let $ V = \Im L $. Then 
the following calculation shows that $ G ( z ) $ is a contraction 
for all $ z \in \C_+ $, 
\begin{eqnarray} \label{contr} I - G^* ( z ) G ( z ) = I - \( I - i
\sqrt \Gamma \( \tL - \overline z \)^{ -1 } \sqrt \Gamma \) \( I + 
i \sqrt \Gamma \( \tL^* - z \)^{ -1 } \sqrt \Gamma \) = \nonumber 
\\ i \sqrt \Gamma \( \tL - \overline z \)^{ -1 } \( \( \tL^* - z \) -
\( \tL - \overline z \) + i\Gamma \) \( \tL^* - z \)^{ -1 } \sqrt 
\Gamma = \nonumber 
\\ \sqrt \Gamma \( \tL - \overline z \)^{ -1 } ( 2 \von + 2 V + \Gamma ) \(
\tL^* - z \)^{ -1 } \sqrt \Gamma . 
\end{eqnarray}
Since the right hand side is obviously non-negative, we conclude 
that $ G ( z ) $ is a contractive analytic operator-function in $ 
\C_+ $. We shall now show that this function admits a scalar 
multiple, that is, there exists a scalar contractive analytic 
function, $ g \not\equiv 0 $, such that $ G ( z ) \Omega ( z ) = g 
( z ) \cdot I $ for a certain bounded analytic function $ \Omega 
$. Let $ \left\{ X_N \right\}_{ N=1 }^\infty $ be an arbitrary 
family of finite-dimensional subspaces in $ H $ such that $ 
X_{N-1} \subset X_N $ and $ \bigvee_N X_N = H $. We shall use the 
subscript $ N $ with operators to denote their truncation to $ X_N 
$, $ B_N \colon = \left. P_N B \right|_{ X_N } $ where $ P_N $ is 
the orthogonal projection on $ X_N $. Define $ g_N ( z ) = \det 
G_N ( z ) $. Our aim is to show that there exists a subsequence of 
$ N $ such that $ g_N ( z ) $ converges when $ N \to \infty $ 
along the subsequence for all $ z \in \C_+ $ and $ g( z ) = \lim 
g_N ( z ) $ is the scalar multiple. Notice that the operator $ G 
(z) $ is boundedly invertible for $ \Im z $ large enough, for $ 
G(z) \to I $ in the operator norm when $ \Im z \to + \infty $ 
since $ \tL $ has a bounded imaginary part. Fix an arbitrary $ z_0 
\in \C_+ $ such that $ \len I - G ( z_0 ) \rin \le 1/2 $, so that 
$ T = G ( z_0 ) $ is boundedly invertible. Obviously, $ | \det T_N 
| \le 1 $ since $ T $ is a contraction, and $ T_N $ is boundedly 
invertible for all $ N $. Fix arbitrarily a subsequence of $ N $ 
such that $ \det T_N $ converges. To keep notation to a minimum, 
we do not introduce the corresponding subscript and will always 
assume that $ N \to \infty $ along this subsequence. Then, $ g ( z 
) = \lim g_N ( z ) $ is the scalar multiple. Indeed, the limit 
exists for all $ z \in \C_+ $: we have $$ g_N = \det G_N = \det 
T_N \cdot \det \( I_N+ T_N^{ -1 } \( G_N - T_N \) \) $$; then, 
\begin{eqnarray*} G ( z ) - T = i \sqrt \Gamma \( \( \tL^* - z 
\)^{ -1 } - \( \tL^* - z_0 \)^{ -1 } \) \sqrt \Gamma = & & \\ & 
\hspace{-3cm} i ( z - z_0 ) \sqrt \Gamma \( \tL^* - z \)^{ -1 } 
\cdot \( \tL^* - z_0 \)^{ -1 } \sqrt \Gamma \in \bS^1 & 
\end{eqnarray*} 
since $ \sqrt \Gamma \( \tL^* - z \)^{ -1 } $ is Hilbert-Schmidt, 
which is easily seen to be equivalent to the assumption 
(\ref{trres}); hence, $ G_N - T_N \rightarrow G - T $ in the $ 
\bS^1 $-norm, and therefore $$ \det \( I_N+ T_N^{ -1 } \( G_N - 
T_N \) \) \longrightarrow \det \( I+ T^{ -1 } \( G - T \) \) . $$ 
Since $ \det T_N $ converges by construction, we obtain that the 
limit $ g( z ) $ exists. Obviously, the limit is a bounded 
function of $ z $ in $ \C_+ $ because $ | g_N ( z ) | \le 1 $ for 
all $ N $. An application of the Montel theorem shows that $ g ( z 
) $ is an analytic function. 

Let us check that $ g \not\equiv 0 $. If $ \det \( I+ T^{ -1 } \( 
G ( z ) - T \) \) = 0 $, then $ \ker G ( z ) $ is nontrivial which 
is only possible for a discrete set of values of $ z \in \C_+ $ 
since $ G ( z ) $ is a compact operator. It remains to show that $ 
\lim \det T_N \ne 0 $. Indeed, consider $ \left| \det T_N 
\right|^2 = \det \( T_N^* T_N \) $. It follows from (\ref{contr}) 
that $ I - T^* T \in \bS^1 $, and a calculation similar to 
(\ref{contr}) gives 
$$ I_N - T_N^* T_N = \left. P_N \sqrt \Gamma 
\( \tL - \overline z_0 \)^{ -1 } \( 2 \von + 2 \tV - \sqrt \Gamma 
P_N \sqrt \Gamma \) \( \tL^* - z_0 \)^{ -1 } \sqrt \Gamma 
\right|_{ X_N }. $$ 
This shows that the right hand side (extended to 
$ X_N^\perp $ by zero) converges in the trace norm to $ I - T^* T 
$. Therefore, $ \lim \det \( T_N^* T_N \) = \det T^* T \ne 0 $ 
since $ \ker T $ is trivial by the choice of $ z_0 $. 

Now, $ \Omega ( z ) = g ( z ) G^{ -1 } ( z ) $ satisfies $ \len 
\Omega ( z ) \rin \le 1 $ since $ | g_N ( z ) | \len G^{ -1 }_N ( 
z ) \rin \le 1 $ for each $ N $ and all $ z $ such that $ g_N ( z 
) \ne 0 $, and thus $ g (z) $ is a scalar multiple. 

It follows from the existence of a scalar multiple and the Fatou 
theorem that the strong boundary values, $ G^{ -1 } ( k ) = 
\mathop{\mbox{s-lim}}_{ \von \downarrow 0 } G^{ -1 } ( k + i \von 
) $, of the function $ G^{ -1 } $ exist for a. e. $ k \in \R $. We 
infer that for a. e. $ k \in \R $ $$ \sqrt \von \( L^* - z \)^{ -1 
} \sqrt \Gamma = \sqrt \von \( \tL^* - z \)^{ -1 } \sqrt \Gamma 
\cdot G^{ -1 }( z ) \mathop{\stackrel{s} \longrightarrow}_{ \von 
\downarrow 0 } 0 . $$ 

One can now verify the condition (\ref{crit}) for the operator $ L 
$. We have, \be\label{lv} \sqrt \von \( L^* - z \)^{ -1 } \sqrt V 
= \sqrt \von \( \tL^* - z \)^{ -1 } \sqrt V - i \sqrt \von \( L^* 
- z \)^{ -1 } \sqrt \Gamma \cdot Q ( z ) , \ee where $ Q ( z ) = 
\sqrt \Gamma \( \tL^* - z \)^{ -1 } \sqrt V $ is a bounded 
analytic function in $ \C_+ $. The latter follows from a 
calculation similar to (\ref{contr}) giving that $ I - G ( z ) G^* 
( z ) $ equals the right hand side of (\ref{contr}) with $ \( 
\tL - \overline z \)^{ -1 } $ and $ \( \tL^* - z \)^{ -1 } $ 
swapped, and therefore $ Q ( z ) Q^* ( z ) \le I - G ( z ) G^* ( z 
) \le I $. The first term on the right hand side of (\ref{lv}) 
converges strongly to zero by the assumption, and an application 
of the Fatou theorem to the function $ Q (z) $ shows that the 
second term converges strongly to zero as well. \hfill $\Box$ 

In the next section we establish triviality of the absolutely 
continuous subspace for an operator $ L $ by verifying the 
condition (\ref{crit}) for an operator $ \tL $ obeying the 
assumptions of this lemma. Since the {\it proof} of the lemma 
consists of demonstrating that the condition (\ref{crit}) for the 
operator $ L $ is satisfied if it is satisfied for the operator $ 
\tL $, the final result then does not depend on the "only if" part 
of the criterion of Proposition \ref{criterion}. 

\refstepcounter{section}
\subsection*{\S 2. The operator corresponding to $ \ell_Q $}

Let $\TTh(\cdot,z)$ and $\TPh(\cdot,z)$ be the solutions of the 
initial value problems 
\[ \ell_{Q^*}\TTh = z \TTh, \;\;\; \TTh(0,z) = \left(\begin{array}{c}
 1 \\ 0 \end{array}\right), \]
\[ \ell_{Q^*}\TPh = z\TPh, \;\;\; \TPh(0,z) = \left(\begin{array}{c}
 0 \\ 1 \end{array}\right). \]

It is standard that the functions $ \TTh ( x , z ) $ and $ \TPh ( 
x , z ) $ are entire functions of $ z $ for all $ x $ and are 
continuous in $ z $ uniformly in $ x \in I $ for any compact 
interval $ I $. 

\begin{lemma} \label{lemma1}For each $z$ in the upper half plane 
$\Cc_{+} $, there exists a unique (up to scalar multiples) 
solution of the differential equation $\ell_{Q^*}\TPs = z \TPs$ 
which lies in $L^2(\Rr^+,\Cc^2)$, and which can be chosen to be of 
the form \be \TPs(x,z) = \TTh(x,z) + m (z)\TPh(x,z) . 
\label{eq:2.1} \ee 
\end{lemma}
\noindent {\bf Proof}\, Multiplying the equation $\ell_{Q^*} y = 
zy$ by $J$ we obtain an equation of the form $y' = Ay$ in which 
$A$ has zero diagonal. The fundamental matrix of this equation 
therefore has a constant determinant; it follows that any two 
linearly independent solutions, say $u=\left(\begin{array}{c} u_1 
\\ u_2 
\end{array}\right)$ and $v=\left(\begin{array}{c} v_1 \\ v_2 \end{array}\right)
$ satisfy $W[u,v] \equiv u_1(x)v_2(x)-u_2(x)v_1(x) = const. \neq 0$.
Integrating this identity in $x$ and applying the Schwarz inequality
shows that at most one of the solutions lies in $L^2(\Rr^+,\Cc^2)$.
This is a version of a standard argument: see, e.g., Levitan and
Sargsjan \cite{kn:LevitanSargsjan}.

Let $L_{min}$ be the closure of the operator defined by the 
differential expression $\ell_{Q}$ on $ C_0^\infty( \Rr^+,\Cc^2 ) 
$ in the space $ L^2(\Rr^+ , \Cc^2 ) $. An integration by parts 
shows that $ L_{min} $ is a dissipative operator. Since $ L_{min} 
$ is obviously not maximal dissipative, $\dim\ker(L_{min}^*-zI)$ 
is either 1 or 2 for all $z\in \Cc_+$. By the argument of the 
previous paragraph, dimension 2 is impossible. Thus there exists 
exactly one solution $\TPs$ of the equation $\ell_{Q^*}y = zy$ in 
$L^2$. 

To show that $\TPs$ can be chosen of the form (\ref{eq:2.1}) it 
remains to note that $\TPh(\cdot,z)\not\in L^2(\Rr^+,\Cc^2)$ for 
all $z\in \Cc_{+}$: if it were, for some $z$, then $ -z $ would be 
an eigenvalue in the lower half plane of a dissipative operator 
corresponding to the differential expression $ -\ell_{ Q^* } $. 
\hfill $\Box$ 

\begin{definition}
Let $h\in i\Rr$ be fixed. By $L$ we denote the operator in the 
Hilbert space $ H = L^2(\Rr^+ , \Cc^2 ) $ given by the 
differential expression $ \ell_Q $ with domain 
\[ D(L) = \left\{ y = \left(\begin{array}{c} y_1 \\ y_2 \end{array}\right) 
 \in H \, | \, y\in AC_{loc}(\Rr^+), \;\;
 \ell_Qy \in  H, \;\; y_2(0) = h y_1(0) \right\} . \]
\end{definition}

Using Lemma \ref{lemma1} it is easy to see that $ L $ is a maximal 
dissipative operator, and so $ \sigma ( L ) \subset \overline 
{\C_+ } $. As the following example shows, this operator is not 
completely non-selfadjoint in general. 

\medskip
\noindent{\bf Example.} We construct a potential $Q$ with $\Im Q 
\neq 0$, such that $L$ has a real eigenvalue. The boundary 
condition will be $y_1(0)=0$, corresponding to $h= \infty $. 

Let $X>0$ be fixed and let $ y $ be given on $[0,X]$ by 
\[ y(x) = \left(\begin{array}{c} 0 \\ 1 \end{array}\right), 
\;\;\; 0 \leq x \leq X. \] 
Fix a real non-zero $k$ and choose $q_2 = k$ on $[0,X]$; let 
$q_1$ satisfy $\Im(q_1)\not\equiv 0$ and be otherwise arbitrary on 
$[0,X]$. Clearly $y$ satisfies the differential equation 
$\ell_{Q}y = k y$ on $[0,X]$ together with the boundary condition. 

For $x>X$, let $y_2(x) = 1/\cosh(x-X)$ and $q_1(x)\equiv 0$.
Define $y_1(x)$ for $x>X$ by
\[ y_1(x) = \frac{iy_2'(x)}{k} = -\frac{i\sinh(x-X)}{k\cosh^2(x-X)}. \]
Obviously $y_1$ and $y_2$ are continuous at $x=X$ and decay exponentially
for large $x$; therefore $y\in D(L)$. The equation $iy_2'=(k-q_1)y_1$ is 
satisfied for all $x$ by construction. If we now choose $q_2$ for
$x>X$ so that the equation $iy_1'= (k-q_2)y_2$ holds, then $ y $ will be
an eigenfunction of $ L $. We thus let
\[ q_2 = k + k^{-1}\frac{y_2''}{y_2}. \]
The function $q_2$ is asymptotically a real constant at infinity and so 
$Q$ satisfies all our hypotheses. \hfill $\Box$

\medskip

We shall require an expression for the action of the resolvent 
$(L^*-z)^{-1} $ on compactly supported vectors. 

\begin{definition} Let $\TPs_h(\cdot,z)$ denote the solution of the differential 
equation $\ell_{Q^*}\TPs_h = z \TPs_h$ with the initial condition 
given by $\TPs_h(0,z) = \left(\begin{array}{c} 1 
\\ h \end{array}\right) $, so that
\[ \TPs_h(\cdot,z) = \TTh(\cdot,z) + h\TPh(\cdot,z) . \] \end{definition}

Notice that $ m (z) \ne h $ for any $ z \in \Cc_+ $, for otherwise 
$ \TPs ( x , z ) $ would be an eigenfunction of $ L^* $, and $ z $ 
would be an eigenvalue. Let $ E = \mbox{diag} ( 1 , -1 ) $. Then a 
straightforward calculation taking into account that $ \det ( 
\TPs_h , \TPs ) = m ( z ) - h $ shows that for any compactly 
supported $ u \in H $ \be\label{resol} \( L^* - z \)^{ -1 } u = 
\frac i{m(z)-h}\left\{ \TPs_h(x,z)\int_x^\infty\TPs^T(s,z)E u(s)ds 
 + \TPs(x,z)\int_0^x\TPs_h^T(s,z)E u(s)ds\right\}. \ee

\begin{lemma}\label{lemma:3}
The function $ m(z)$ defined in lemma \ref{lemma1} is analytic in 
$\Cc_{+}$. For all $ z \in \C_+ $ it satisfies $\Re m (z) < 0$. 
Moreover, for all $ z $ with $ \epsilon = \Im z > 0 $, 
\begin{eqnarray} - \Re m (z) & = &\epsilon \int_0^\infty 
\left( |\TPs_1(x,z)|^2 + |\TPs_2(x,z)|^2 \right) dx + \nonumber \\ 
& & \int_0^\infty \left( \Im q_{1}(x) |\TPs_1(x, z )|^2 + \Im 
q_{2}(x) |\TPs_2(x, z)|^2 \right) dx , \label{eq:2.4} 
\end{eqnarray} 
and therefore 
\[ \limsup_{\epsilon \searrow 
0 }\( \epsilon \len \TPs ( \cdot , k + i \epsilon ) \rin_H^2 \) 
\le - \Re m ( k ) \] whenever the limit $ m ( k ) = \lim_{ 
\epsilon \searrow 0 } m ( k + i \epsilon )$ exists. 
\end{lemma}
 
\noindent {\bf Proof}\, It follows from (\ref{resol}) that for any 
compactly supported $ u,v \in H $ $$ F ( z ) \equiv \llangle \( 
L^* - z \)^{ -1 } u , v \rrangle = \frac {A m ( z ) + B }{m(z)-h} 
$$ where $ A, B $ are complex constants depending on $ u $ and $ v 
$. Then, $ B \ne - h A $ for a suitable choice of $ u $ and $ v $, 
since otherwise the fact that $ F ( z ) \to 0 $ when $ \Im z \to 
\+ \infty $ would imply that $ F ( z ) \equiv 0 $ for any 
compactly supported $ u, v \in H $, and therefore $ \( L^* - z 
\)^{ -1 } = 0 $. We now infer that $ m ( z ) $ is analytic in $ 
\C_+ $ from the analyticity of $ F ( z ) $. Alternatively, the 
analyticity of $m$ can be proved through the nesting circles 
analysis in the same way as in the case of a dissipative second 
order Schr\"{o}dinger operator \cite{Sims}. 

We shall now establish (\ref{eq:2.4}). The differential equations 
satisfied by $\TPs_1(x,k+i\epsilon)$ and $\TPs_2(x,k+i\epsilon)$ 
are 
\[ i \TPs_2' + (\Re q_1(x) - i \Im q_1(x) - k - i\epsilon)\TPs_1 = 0, \]
\[ i \TPs_1' + (\Re q_2(x) - i \Im q_2(x) -k - i\epsilon)\TPs_2 = 0. \]
Multiply the first equation by $\overline{\TPs_1}$ and the complex
conjugate of the second equation by $\TPs_2$, subtract and obtain
\[  - i (\Im q_1+\epsilon)|\TPs_1|^2 - i(\Im 
q_2+\epsilon)|\TPs_2^2|^2 
 + (\Re q_1 -k)|\TPs_1|^2 - (\Re q_2 -k)|\TPs_2|^2
 + i (\overline{\TPs_1}\TPs_2)' = 0. \]
Now on taking imaginary parts and integrating over $[0, X ] $, \[ 
\Re (\overline{\TPs_1 ( X ) }\TPs_2 ( X )) = \int_0^X \left\{ (\Im 
q_1+\epsilon)|\TPs_1|^2 + (\Im q_2+\epsilon)|\TPs_2|^2 \right\} dx 
 + \Re m ( k + i \epsilon ) . \]
This shows that the left hand side has a limit, finite or 
infinite, when $ X \to \infty $. This limit must be zero, because 
$ \TPs \in L^2 $, and therefore, $ \overline{\TPs_1 }\TPs_2 \in 
L^1 $. The result follows. \hfill $\Box$ 

In order to use Proposition \ref{criterion} we shall need an 
asymptotic of $ \( L^* - z \)^{ -1 } u $ when $ \epsilon \searrow 
0 $ for compactly supported $ u $, given by the following lemma. 

\begin{lemma}\label{lemma:6}
For any $ u \in L^2 ( \Rr_+ , \C^2 ) $ having compact support and 
$z=k+i\epsilon$, $ \epsilon > 0 $, \be \begin{array}{c} \( L^* - z 
\)^{ -1 } u = \beta_z [u] \TPs(x,z) + r ( x , z ) , \\ \\ 
{\displaystyle 
 \beta_z [u] = \frac{i}{m (z)-h} \int_0^\infty \TPs_h^T(s,z)Eu(s)ds }
 , \end{array} \label{eq:2.5}
\ee where $ r ( x , z ) $ satisfies $ \limsup_{ \epsilon \searrow 
0 } \len r ( \cdot , z ) \rin_H < \infty $ for all $ k $ such that 
the finite boundary value $ m ( k ) = \lim_{ \epsilon \searrow 0 } 
m ( k + i \epsilon )$ exists and $ m ( k ) \neq h $. 
\end{lemma}

\noindent {\bf Proof}\, Let $ X < \infty $ be such that $ u $ is 
supported on $ [ 0 , X ] $, and consider the expression 
(\ref{resol}) for $ \( L^* - z \)^{ -1 } u $. Notice that the 
first term in the curly brackets vanishes for $ x > X $, and 
therefore $ r = \( L^* - z \)^{ -1 } u -\beta_z \TPs $ is 
supported on $ [0,X] $ for any $ z \in \Cc_+ $. Then, the standard 
perturbation theory for initial value problems implies that $ 
\TPs_h ( x, z ) $ and $ \TPs ( x, z ) $ converge to $ \TPs_h ( x, 
k ) $ and $ \TPs ( x, k ) $ when $ \epsilon \searrow 0 $ uniformly 
in $ x < X $ for all $ k $ such that the finite boundary value $ m 
( k ) $ exists. Combining these facts, we obtain that $ r ( x , z 
) $ converges when $ \von \searrow 0 $ uniformly in $ x < X $, 
provided that the common factor in the right hand side of 
(\ref{resol}) converges to a finite limit. The result follows. 
\hfill $ \Box $ 

\refstepcounter{section}
\subsection*{\S 3. Proof of the main result} 

Define the non-negative functions $r_1$ and $r_2$ by 
\[ r_1(x)^2 := \Im q_1(x), \;\;  r_2(x)^2 := \Im q_2(x) . \]
Then the operator $ \( \Im L \)^{ 1/2 } $ is given by 
mulptiplication by the matrix \be R(x) = \left(\begin{array}{cc} 
r_1(x) & 0 
\\                                        0 & r_2(x) 
                                  \end{array}\right). 
\label{eq:Rdef} \ee 

\begin{lemma} \label{theorem:2}
Let $k\in\Rr$ be such that $ m (k)$ exists finitely and is not 
equal to $h$. If $R \TPs_h(\cdot,k) \not\in L^2(\Rr^+,\Cc^2)$, 
then $ 
  D(k+i\epsilon) \stackrel{s}{\rightarrow} 0 $ as $\epsilon \searrow 0 $.
\end{lemma}
\noindent {\bf Proof}\, Observe that the linear set \[ {\cal D} = 
\left\{ u \in L^2(\Rr^+,\Cc^2) \, | \, u \mbox{ is compactly 
supported and } \int_0^\infty \TPs_h^T (s,k) E R(s)u(s)ds = 0 
\right\} 
\] is dense in $ L^2(\Rr^+,\Cc^2)$ if $ 
R(\cdot)\TPs_h(\cdot,k)\not\in L^2(\Rr^+,\Cc^2)$. Since the 
function $ D ( z ) $ is bounded in $ \C_+ $ by a remark after 
Proposition \ref{criterion}, it suffices to prove \be\label{Dlim} 
\lim_{\epsilon\rightarrow 0}D(k+i\epsilon)u = 0 \ee for $u\in 
{\cal D}$. From Lemma \ref{lemma:6}, we have for any $u \in {\cal 
D}$, with $ z = k + i\epsilon $, $$ D(k+i\epsilon)u = \epsilon^{ 
1/2 } \beta_z [ R u ] \TPs( \cdot ,z) + o ( 1 ) , \; \; \epsilon 
\searrow 0 , $$ where the $ o $-symbol refers to the $ L^2 $-norm. 
Now, Lemma \ref{lemma:3} shows that $ \epsilon^{ 1/2 } \len \TPs ( 
k + i \epsilon ) \rin_H $ is bounded above when $ \epsilon 
\searrow 0 $. Since $ m ( k ) \neq h $, one can pass to the limit 
$ \epsilon \searrow 0 $ in the expression (\ref{eq:2.5}) for $ 
\beta_z $ to find that $ \beta_{ k + i \epsilon } $ converges to a 
multiple of $ \int_0^\infty \TPs_h^T(s,k)E R ( s ) u(s)ds $ which 
is zero because $ u \in {\cal D} $. Combining these, we obtain 
(\ref{Dlim}). \hfill $ \Box $ 

Note that the complement of the set of all $k\in \Rr$ for which $ 
m (k)$ exists and is not equal to $h$ is a set of (Lebesgue) 
measure 0, by the uniqueness theorem for Nevanlinna functions. 
Taking into account the criterion of Proposition \ref{criterion}, 
we arrive at the following. 

\begin{corollary}\label{corollary:1}
If $ R\TPs_h(\cdot,k)\not\in L^2(\Rr^+,\Cc^2)$ for Lebesgue almost 
all $k\in\Rr$, then\footnote{Recall that $ L^\prime $ denotes the 
completely non-selfadjoint part of the operator $ L $.} $ H_{ac} ( 
L^\prime ) = \{ 0 \}$. 
\end{corollary}

Thus, our main result will be established if we prove that 
$R(\cdot)\TPs_h(\cdot,k)\not\in L^2(\Rr^+,\Cc^2)$ for a. a. 
$k\in\Rr$ if $ \Im Q\notin L^1 . $ Before doing this, we would 
like to indicate a simple argument showing that $ H_{ ac } $ is 
trivial when $ \Im Q \notin L^1 $ for a class of potentials $ Q $. 
Notice first that if $ m ( k ) $ exists finitely, then the 
solution $ \TPs ( x , k ) = \lim_{ \epsilon \searrow 0 } \TPs (x,k 
+ i \epsilon ) $ to $ \ell_{ Q^* } \Psi = k \Psi $ exists. As is 
shown in the asymptotic theory of linear differential systems 
\cite{kn:Eastham}, if the potential $ Q $ decays at infinity in a 
sufficiently regular way the asymptotics of solutions of the 
equation $ \ell_{ Q^* } \Psi = k \Psi $ can be calculated 
explicitly. In particular, an application of the general theory in 
the situation under consideration gives the following. 

\begin{proposition} \label{wkbcond} \cite[Theorem 
1.8.3]{kn:Eastham} Let $ \Re Q = 0 $, and suppose that $ Q \to 0 
$, $ Q^\prime \in L^1 $ and $ Q \in L^2 $. Then for any $ k \in \R 
$ there exist two solutions, $ \Psi_\pm $, of the equation $ 
\ell_{ Q^* } \Psi = k \Psi $ with the asymptotics of the form (the 
WKB asymptotic) \be\label{wkb} \Psi_\pm \sim \exp \left[ \pm i \( 
k x + \frac 12 \int^x \trace Q ( \xi ) d \xi \) \right] 
\left(\begin{array}{c} 1 
\\ \mp 1 \end{array}\right), \; \; x 
\to \infty . \ee \end{proposition} 

Assume that the asymptotics (\ref{wkb}) holds for a. e. $ k \in \R 
$, and that $ \Im Q \notin L^1 $. Then, obviously, the solution $ 
\Psi_- $ is growing, $ | \Psi_- | \sim \exp \( \frac 12 \int^x 
\trace ( \Im Q ) \) $, while $ \Psi_+ $ decays. Notice that if the 
asymptotics holds for a given $ k $, and a solution $ v $ to $ 
\ell_{ Q^* } \Psi = k \Psi $ satisfies $ R v \in L^2 $, then $ v $ 
must be a multiple of $ \Psi_+ $, for 
\begin{eqnarray*} \int^\infty \left\| R \Psi_- \right\|^2 dx \sim 
\int^\infty \trace \( \Im Q ( x )\) \exp \( C \int^x \trace \( \Im 
Q ( \xi ) \) \, d \xi \) dx = \\ \exp \( C \int^\infty \trace \( 
\Im Q ( x ) \) \, dx \) = \infty . \end{eqnarray*} It follows that 
if (\ref{wkb}) holds for a $ k \in \R $ such that $ m ( k ) $ 
exists and $ R \TPs_h ( \cdot , k ) \in L^2 $, then the solutions 
$ \TPs_h ( \cdot , k ) $ and $ \TPs ( \cdot , k ) $ must coincide. 
This is, however, only possible on the set of $ k $ such that $ m 
( k ) = h $, which has zero measure. Therefore $ R \TPs_h ( \cdot 
, k ) \notin L^2 $ for a. e. $ k $. Applying corollary 
\ref{corollary:1}, we thus obtain the main result for potentials $ 
Q $ such that the asymptotics (\ref{wkb}) holds for a. e. $ k \in 
\R $. Notice that although the conditions of Proposition 
\ref{wkbcond} can be sharpened, the WKB asymptotics cannot be 
justified for generic potentials, and the argument above has 
nothing to do with the actual proof of Theorem 1. Moreover, even 
in the case of Schr\"odinger operators it is well-known 
\cite{Pearson} that the WKB asymptotic may fail for a. e. positive 
value of the spectral parameter. 

We shall establish Theorem 1 by successively reducing it to a 
partial case which we now proceed to prove. 
 
\begin{proposition}\label{theorem:weak1}
If $r_1r_2\notin L^1(\Rr^+)$ then $ R \TPs_h ( \cdot , k ) \notin 
L^2 $ for a. e. $ k \in \R $. 
\end{proposition}
\noindent{\bf Proof} \, We prove the result in the contrapositive 
form. Suppose that $R\TPs_h (\cdot,k)\in L^2 $ for a certain real 
$ k $ such that $ m ( k ) \neq h $. Multiply the Wronskian 
identity $\det(\TPs_h(x,k),\TPs(x,k)) = m (k)-h \ne 0 $ by 
$r_1r_2$ and integrate in $ x $ to obtain $$ \(\mbox{const.}\) 
\int^N \hspace{-2mm} r_1(x)r_2(x) dx \leq 
 \int^N |r_1 \TPs_{h,1}|\cdot |r_2 \TPs_2| dx 
+ \int^N |r_2\TPs_{h,2}|\cdot |r_1\TPs_1| dx . $$ 
Now $r_1 
\TPs_{h,1} \in L^2(\Rr^+)$ and $r_2\TPs_{h,2}\in L^2(\Rr^+)$ by 
choice of $k$, while $r_1\TPs_{1}\in L^2(\Rr^+)$ and 
$r_2\TPs_{2}\in L^2(\Rr^+)$ by Lemma \ref{lemma:3}. By the Schwarz 
inequality it follows that $r_1r_2\in L^1(\Rr^+)$. \hfill $\Box$ 
\vspace{2mm} 

In particular, this proposition shows that if $r_1r_2\notin 
L^1(\Rr^+)$, then $H_{ac}(L^\prime) = \{ 0 \}$ in view of 
Corollary \ref{corollary:1}. 

We would now like to reduce the problem under consideration to a 
problem where only one of the entries of $ \Im Q $ is non-zero. We 
start with a Gronwall-type lemma comparing the solutions $ \TPs $ 
and $ \TPs_h $ corresponding to different $ \Im Q $ at large $ x 
$. 

\begin{lemma} \label{lemma:5.2}
Let $k\in\Rr$ be such that $ m ( k ) $ exists finitely, and $ m ( 
k ) \neq h $. Let the function $$ \hQ (x) = 
\left(\begin{array}{cc} \hat{ q_1 } (x) & 0 
\\ 0 & \hat{q_2} (x) 
                                  \end{array}\right) $$ satisfy $ 0 
\le \Im \hQ ( x ) \le \Im Q ( x ) $ and $ \Re \hQ ( x ) = \Re Q ( 
x ) $ for a.e. $ x $. If $R(\cdot)\TPs_h(\cdot,k)\in 
L^2(\Rr^+,\Cc^2)$, then the equation $\ell_{\hQ^* }y = k y$ has 
solutions $\hat{\Psi}^{(1)}$ and $\hat{\Psi}^{(2)}$ such that 
\be
\left. 
\begin{array}{rcl}
\hat{\Psi}^{(1)}(x,k) & = & \TPs_h(x,k) + o \( \left| \TPs_h(x,k) 
\right| + \left| \TPs(x,k) \right|\) \\ \hat{\Psi}^{(2)}(x,k) & = 
& \TPs(x,k) + o \( \left| \TPs_h(x,k) \right| + \left| \TPs(x,k) 
\right|\) 
\end{array}\right\} \;\; (x \rightarrow \infty).
\label{eq:5.5} \ee 
\end{lemma}
\noindent{\bf Proof}\, Let the function $ B ( x ) \ge 0 $ be 
defined by $ B^2 = \Im ( Q - \hQ ) $, and $ Y $ denote the $ 2 
\times 2 $-matrix $(\TPs_h,\TPs)$. Write 
$(\hat{\Psi}^{(1)},\hat{\Psi}^{(2)}) = Y Z $. Elementary 
manipulations show that the matrix $Z$ must satisfy the 
differential equation \be JYZ' = -i B^2 YZ. \label{eq:5.6} \ee 

As the argument from the beginning of the proof of Lemma 
\ref{lemma1} shows\footnote{This is the first time that we use in 
an essential way the fact that we deal with diagonal $ \Im Q $.}, 
the determinant of the matrix $ Y $ is constant in $ x $, $ \det Y 
= m ( k ) - h \ne 0 $. Now, let $ J_0 = \left(\begin{array}{cc} 0 
& -1 
\\ 1 & 0 \end{array}\right) $. A straightforward calculation 
shows that 
\[ Y^T J_0 Y = ( \det Y ) J_0  . \] 
Thus, $(JY)^{-1} = J_0^{-1}Y^TJ_0J^{-1}/\det Y$ and hence from 
(\ref{eq:5.6}), together with the fact that 
$J_0J^{-1}=iE=i\,\mbox{diag}(1,-1)$,\be Z' = - \frac{i}{\det Y} 
J_0^{-1}Y^TJ_0J^{-1}B^2 YZ = - \frac{1}{\det Y}J_0(BY)^TE(BY)Z. 
\label{eq:5.7} \ee 
 
It follows from Lemma \ref{lemma:3} that \[ \limsup_{ \epsilon 
\searrow 0 } \int_0^\infty \len R ( x ) \TPs (x, z ) \rin^2 dx \] 
is finite when $ m ( k ) $ exists finitely. Hence, by the Fatou 
lemma $R\TPs(\cdot,k) \in L^2(\Rr^+,\Cc^2)$. Together with the 
hypothesis of the current lemma this implies that $ R Y\in L^2 $, 
and, moreover, $ B Y \in L^2 $, since $ B ( x ) \le R ( x ) $ a.e. 
Thus, $(BY)^T E (BY) $ is summable, and by the Levinson theorem 
(see, e.g., \cite[Theorem 1.3]{kn:Eastham}) it follows that 
(\ref{eq:5.7}) possesses a solution $Z$ satisfying \be Z(x) = I + 
o(1), \;\;\; x\rightarrow\infty \label{eq:5.8} , \ee and 
(\ref{eq:5.5}) is proved. \hfill $\Box$ 

Let $ \hat{R} = \( \Im \hat{Q} \)^{ 1/2 } $. The asymptotic 
(\ref{eq:5.5}) shows, in particular, that in the situation of the 
lemma, $ \hat{R} \hat{\Psi} \in L^2 $ for any solution $ \hat \Psi 
$ to $\ell_{\hQ^* }y = k y$. Applying Corollary \ref{corollary:1}, 
we now obtain the following assertion. 

\begin{corollary}\label{lemma:5.7} Let $L_j$ denote the operators defined by the same boundary 
conditions as $L$ but given by 
\[ L_j y = \ell_{Q_j}y, \]
where 
\[ Q_1(x) = \left(\begin{array}{cc} q_{1}(x) & 0 \\
0 & \Re q_{2}(x) \end{array}\right), \;\; Q_2(x) = 
\left(\begin{array}{cc} \Re q_{1}(x) & 0 \\ 0 & q_{2}(x) 
\end{array}\right) , \] and let $ R_j = \( \Im Q_j \)^{ 1/2 } $, 
and $ \TPs_{h,j} ( x , k ) $ be the solutions to $\ell_{Q_j^*}y = 
k y $ with the Cauchy data $ y ( 0 ) = \( 1 , h \)^T $. If at 
least one of the operators $ L_j $ satisfies the condition $ R_j 
\TPs_{h,j} ( \cdot , k ) \notin L^2 $ for a. e. $ k \in \R $ (and 
so the corresponding subspace $ H_{ac} (L_j^\prime) $ is trivial), 
then $ H_{ ac } ( L^\prime ) $ is trivial. 
\end{corollary}

\noindent {\bf Proof of Theorem 1}\, In view of Corollary 
\ref{lemma:5.7} it is sufficient to show that $ R \TPs_h ( \cdot , 
k ) \notin L^2 $ for a. e. $ k \in \R $ for the operator $ L $ 
corresponding to a potential $ Q $ such that $\Im(q_1)\equiv 0$ 
and $\Im(q_2)\notin L^1$. Since $\Im(q_2)\notin L^1$ we have $r_2 
\notin L^2$ and so by the Uniform Boundedness Principle there 
exists a function $d\in L^2$ such that $ r_2 d \notin L^1 $. 
Obviously we can choose $d$ to be non-negative. Let $L_d $ denote 
the operator with the potential 
\[ Q_d = \left(\begin{array}{cc} \Re q_{1} + id^2 & 0 \\ 
0 & q_{2}(x) \end{array}\right) , \] and let $ R_d = \( \Im Q_d 
\)^{ 1/2 } $, and $ \TPs_h^d ( x , k ) $ be the solution of the 
differential equation $\ell_{Q^*_d}\TPs_h = z \TPs_h$ with the 
initial condition $\TPs_h^d (0,z) = \left(\begin{array}{c} 1 
\\ h \end{array}\right)$. Since 
$r_2 d \notin L^1$, we have $ R_d \TPs_h^d ( \cdot , k ) \notin 
L^2 $ for a. e. $ k \in \R $ by Proposition \ref{theorem:weak1}, 
and therefore $ H_{ ac } ( L_d^\prime ) = \{ 0 \} $. We now wish 
to apply Lemma \ref{trace} with $\tL = L_d $ so that $ \Gamma $ is 
given by multiplication by $ \mbox{diag} \( d^2 , 0 \) $. We only 
have to check that $ \( L - z \)^{ -1 } - \( L_d - z \)^{ -1 } \in 
\bS^1 $, since all the other assumptions of the Lemma are 
satisfied trivially. Let $ A_0 $ be the operator corresponding to 
the potential $ Q \equiv 0 $. The fact that $ d\in L^2$ and the 
assumption that $ \Re Q $ is bounded imply that, in the notation 
of Lemma \ref{trace}, the operator $ \sqrt \Gamma \( A_0 - z \)^{ 
-1 } $ is Hilbert-Schmidt for any non-real $ z $ since the 
integral matrix kernel $K(x,s)$ of this operator satisfies $$ \len 
K(x,s) \rin_{ \mbox{Mat}_2 ( \C ) } \leq C 
d(x)\exp(-\epsilon|x-s|), \; \; \; \epsilon = | \Im z | 
> 0, 
$$ for all $x$ and $s$, where $C$ is a constant depending on $ z $ 
only. Applying the resolvent identity repeatedly, we now have (for 
any non-real $ z \in \rho ( L_d ) \cap \rho ( L ) $) $$ \( L - z 
\)^{ -1 } - \( L_d - z \)^{ -1 } = \Xi_1 ( z ) \( A_0 - z \)^{ -1 
} \sqrt \Gamma \cdot \sqrt \Gamma \( A_0 - z \)^{ -1 } \Xi_2 ( z ) 
$$ where $ \Xi_{ 1,2 } ( z ) $ are bounded operators. This formula 
shows that its left hand side is of the trace class. The result 
follows. \hfill $\Box$ 

Apparently, Theorem 1 cannot be obtained in this way without a 
condition on the behavior of $ \Re Q $ at infinity, which 
guarantees that $ \( \Re L - z \)^{ -1 } - \( A_0 - z \)^{ -1 } $ 
is a bounded operator and makes it possible to apply the resolvent 
identity to show that the difference of the resolvents of $ L $ 
and $ L_d $ is trace class. 

\section*{Concluding Remarks}

Notice that the general matrix differential operator $ L $ 
corresponding to the differential expression $ \ell_V := 
J\frac{d}{dx} + V(x) $ with the same boundary conditions as above 
and arbitrary bounded matrix potential $ V = \left( 
\begin{array}{cc} v_{11} & v_{12} \\ v_{21} & 
 v_{22} \end{array} \right) $ is unitarily 
equivalent to an operator with a diagonal potential $ Q $ if $ v_{ 
12 } = v_{ 21 } $ and is real. The equivalence is given by a 
unitary gauge transformation $ U $ (see \cite{kn:LevitanSargsjan}) defined by 
\[ U y  = \exp\( i\int_0^x v_{12}(t)dt \) y . \]
Then $ U^* L U = L_Q $ where $ L_Q $ is operator corresponding to 
the differential expression $ \ell_Q $ with 
\[ Q(x) = \left(\begin{array}{cc} v_{11}(x) & 0 \\ 0 & 
 v_{22}(x) \end{array}\right). 
\]
Since the operator $ U $ transforms the absolutely continuous 
subspaces of $ L_Q $ into that of $ L $, 
$H_{ ac } ( L ) =  U H_{ac} ( L_Q ) $. 


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