% LATEX FILE
\documentstyle[12pt]{article}
\if@twoside m
\oddsidemargin 14truemm\evensidemargin 0mm
\marginparwidth 85pt
\else
\oddsidemargin 7truemm\evensidemargin 7truemm
\marginparwidth 68pt
\fi
\topmargin 5mm
\headheight 0mm
\headsep 0mm
\textheight 225truemm
\textwidth 150truemm
\parindent=7mm
\renewcommand{\baselinestretch}{1.}
% FONTS
\font\sr = msbm10 scaled\magstep 1
\font\sc = eusm10 scaled\magstep 1
\font\gt = eufm10 scaled\magstep 1
% MACROS
\newcommand{\C}{C\!\!\!\rule[.5pt]{.7pt}{6.5pt}\:\:}
\newcommand{\Cind}{C\!\!\!\rule[.5pt]{.5pt}{4pt}\:\:}
\newcommand{\R}{I\!\!R}
\newcommand{\N}{I\!\!N}
\newcommand{\Q}{Q\hspace{-.6em}\rule[.5pt]{.7pt}{6.5pt}\:\:}
\newcommand{\Z}{Z\!\!Z}
\newcommand{\QED}{\mbox{\rule[-1.5pt]{6pt}{10pt}}}
\newcommand{\Ham}{$\hat H\ $}
\newcommand{\Ha}{$\hat H_}
\newcommand{\LR}{$L^2$({\sr R}}
\newcommand{\INT}{\int_{\R}}
\newcommand{\INTT}{\int_{\R^2}}
\newcommand{\INTp}{\int_{\R_+}}
\newcommand{\LMO}{\Lambda +\Omega^{-1}\Lambda\Omega}
\newcommand{\bI}{{\bf I}}
\newcommand{\tR}{\tilde R}
\newcommand{\ver}{\varepsilon\varrho}
\newcommand{\as}{{\rm argsh}}
\newcommand{\sech}{{\rm sech}}
\newcommand{\sgn}{{\rm sgn}}
\newcommand{\Rj}{R^{(j)}}
\newcommand{\epiaj}{e^{-2\pi i\alpha_j}}
\newcommand{\epiak}{e^{-2\pi i\alpha_k}}
\newcommand{\epia}{e^{-2\pi i\alpha}}
\newcommand{\cD}{{\cal D}}
\newcommand{\cJ}{{\cal J}}
\newcommand{\cV}{{\cal V}}
\newcommand{\cE}{{\cal E}}
\newcommand{\cH}{{\cal H}}
\newcommand{\bef}{{\bf f}}
\newcommand{\tom}{\tilde\omega}
\newcommand{\tl}{\~{\sc L}}
\newcommand{\sC}{{\sr C}}
\newcommand{\sR}{{\sr R}}
\newcommand{\sN}{{\sr N}}
\newcommand{\sQ}{{\sr Q}}
\newcommand{\sZ}{{\sr Z}}
\newcommand{\hscS}{\hbox{\sc S}}
\newcommand{\hsR}{\hbox{\sr R}}
%END OF THE DEFINITION
\begin{document}
\title{\bf Scattering on a finite chain of vortices
\footnote{to appear in Duke Math. J.}}
\date{}
\author{PAVEL \v S\v TOV\'I\v CEK}
\maketitle
\begin{abstract}
\noindent
{\small This problem is related to the Aharonov--Bohm effect. The Hamiltonian
\Ham in \LR$^2)$ is defined as a self--adjoint extension of the symmetric
operator $X=-\Delta$, with $D(X)=C_0^\infty($\sR$_+^2)\oplus
C_0^\infty($\sR$_-^2)$, determined by boundary conditions on the first
coordinate axis. There is proven a perturbative formula for the inverted
operator $[(-\partial_u^2+1)^{1/2}+\exp(-2\pi i\nu(u))\,
(-\partial_u^2+1)^{1/2}\exp(2\pi i\nu(u))]^{-1}$ in $L^2(${\sr R}$,du)$, with
$\nu(u)$ being a piecewise constant function related to the boundary
conditions. This result jointly with the Krein's formula enables one to
construct generalized eigen--functions of $\hat H$ starting from the explicit
form of the unitary mapping between the deficiency subspaces of $X$ defining
the self--adjoint extension. These formulae also make it possible to prove
existence and completeness of the wave operators using the trace class
methods.}
\end{abstract}
\vskip 240pt
\pagebreak
\section{Introduction}
The vortices mentioned in the title should be interpreted as intersections
of the plane with magnetic fluxes concentrated in parallel lines. So the
problem is related to the Aharonov--Bohm effect \cite{AB}. While the
configuration with one vortex enjoys the rotational symmetry and enables
an explicit solution by separation of variables (a very brief summery is
given in the beginning of Sec. 5.1) the case of two and more vortices is
considerably more complicated. Another feature worth of mentioning is that
this problem differs somewhat from the usual potential scattering as the
Hamiltonian involves only gauge field and no potential. As explained in
\cite{PLA89} instead of the formulation employing a vector potential the
Hamiltonian can be defined by boundary conditions on a cut in the plane.
Here, the first coordinate axis serves as the cut since it contains, by our
choice, all the vortices.
The aim of this paper is to study this problem in the framework of the
mathematical scattering theory. This means to compare quantum time evolution
governed by the given Hamiltonian with that governed by the free Hamiltonian.
The two Hamiltonians don't differ by a potential but, on the other hand, they
are self--adjoint extensions of the same symmetric operator. It is clear that
the methods known for potential scattering should be modified or supplied by
additional approaches. In view of the relation between the Hamiltonians, the
Krein's formula suggests itself as a powerful tool, particularly in
connection with the Kato--Birman theory. To be able to apply this approach
effectively one has to determine the deficiency subspaces ${\cal N}(z)$ and to
transcribe the boundary conditions into the unitary mapping $V(z):{\cal N}(z)
\to {\cal N}(\bar z)$ defining the self--adjoint extension.
Evaluating $V(z)$ one is confronted with the following problem. Denote
temporarily by $\hat X$ and $\hat P$ the coordinate and momentum operators
in \LR), $[\hat X,\hat P]=i\bI$. Let $\nu:\,$\sR$\to\,$\sR\ be a measurable
function. The operator
$$
\Xi^{-1}=[(\hat P^2+\bI)^{1/2}+\exp(-2\pi i\,\nu(\hat X))\,
(\hat P^2+\bI)^{1/2}\exp(2\pi i\,\nu(\hat X))]^{-1}
$$
is manifestly positive and bounded. If $\nu\equiv0$ then $\Xi_0^{-1}=(1/2)\,
(\hat P^2+\bI)^{-1/2}$. It is desirable to have some information about
$\Xi^{-1}-\Xi_0^{-1}$. Perhaps the simplest and most straightforward way
is to assume that $\nu\approx0$ and hence $\exp(2\pi i\,\nu(\hat X))$ is
close to $\bI$ and to attempt a perturbative expansion. Most probably, this
naive approach would be possible provided $\nu$ is, say, smooth and compactly
supported. Unfortunately, this is not the case we are interested in. Our
$\nu$ is related to the boundary conditions and is piecewise constant with
finitely many discontinuities. Nevertheless, it turns out that a perturbative
formula exists even in this case though in a much less trivial manner. One
has to tensor the Hilbert space \LR) with the finite--dimensional space
\sC$^N\otimes\,$\sC$^2$, with $N$ being the number of discontinuities,
and to introduce some auxiliary operators $\Rj$ in \sC$^N\otimes\,$\sC$^2
\otimes\,$\LR), one for each discontinuity.
Concerning the history, the author derived an expression
for $\Xi^{-1}$ in \cite{JMP}, but in a bit heuristic manner and in a
form of an infinite series. The approach was based on some
geometric ideas going back to Schulman \cite{Schulman} and making use of the
existence of the universal covering space for \sR$^2\,\backslash\,\{a_1,
\dots,a_N\}$. Afterwards he learned that the same formula can be extracted
with some effort also from the results due to Sato, Miwa and Jimbo and thus
obtained by a
completely different method. These authors developed a powerful machinery
applicable to various problems involving some holonomy condition. After
having studied rotations in finite--dimensional Clifford algebras \cite{SMJI}
they transferred the results formally also to infinite--dimensional spaces.
In connection to our problem, a formula is of interest expressing the Green
function of the operator $-\Delta+w^2,\ w>0$, \cite{SMJIII}. The Laplacian
is defined on \sR$^2\,\backslash\,\{a_1,\dots,a_N\}$ and its definition
involves some holonomy condition. By comparing with the Krein's formula one
can extract from this result the expression for $\Xi^{-1}$, this time in a
closed form.
In the current paper, there is given a direct verification of
this perturbative formula. The advantage is that this way one
avoids completely the heuristic ideas present in any of the two approaches
mentioned above though, on the other hand, the origin of the formula
remains somewhat dimmed. Furthermore, the analysis given in \cite{SMJIII} is not
fully appropriate in the situation considered here and is done from the very
beginning and based on different principles.
Let us point out another difference if compared with the potential scattering.
Instead of using the Lippman-Schwinger equation, one can gain generalized
eigen--functions of the Hamiltonian \Ham as follows. Suppose that one can
choose $\varphi(z)\in{\cal N}(z)$ depending on the spectral parameter
$z\in\,$\sC$\,\backslash\,[0,+\infty)$ in such a way that the function
$\psi_+(z)=\varphi(z)+V(z)\varphi(z)$ admits the limit $z=k^2-i\sigma,\
\sigma\downarrow0$. Since $\hat H(\varphi(z)+V(z)\varphi(z))=
z\,\varphi(z)+\bar z\,V(z)\varphi(z)$ the limiting function $\psi_+(k^2-i0)$
is expected to be a generalized eigen--function. The perturbative formula
plays a key role in this construction of generalized
eigen--functions, announced in a less formal manner in \cite{PLA91}.
Besides, this formula also enables one to apply the trace class methods to
verify existence and completeness of the wave operators.
Let us summarize that the main goals of the paper are a precise verification
of the perturbative formula and a presentation of a construction of
generalized eigen--functions of $\hat H$ giving a stationary picture of
the scattering.
The paper is organized as follows. In Section 2, the Hamiltonian \Ham acting
in \LR$^2)$ is defined and its point spectrum is shown to be empty. Section 3
is devoted to the transcription of the boundary conditions into the unitary
mapping $V(z):{\cal N}(z)\to {\cal N}(\bar z)$ between the deficiency
subspaces. Section 4 is concerned with the
perturbative formula for $\Xi^{-1}$. It is divided into two subsections. In
the first one, some auxiliary operators are introduced and studied, in the
second one, the formula is stated and proven. An application of the
perturbative formula jointly with the Krein's formula to the investigation
of the scattering problem itself is
presented in Section 5, also divided into two subsections. Existence and
completeness of the wave operators for the considered scattering problem are
proven in the framework of the Kato--Birman theory in Subsection 5.1.
Subsection 5.2 presents a construction of the generalized eigen--functions
$\psi_+(k^2-i0)$. It is also shown that the wave
operator $W_-$ sends the plane wave $\psi_0$ to $\psi_+(k^2-i0)$.
\section{The Hamiltonian}
Let us begin with a short comment about notation and conventions. The scalar
product $\langle\cdot,\cdot\rangle$ in a complex Hilbert space is linear in
the second variable. Throughout the paper we exploit frequently the fact that,
by the Schwartz's kernel theorem, at least all bounded operators in $L^2(I,dx)$,
with $I\subset\;$\sR\ being an interval, are determined by their generalized
kernels. In accordance with the physical convention we denote by $\langle x|
A|y\rangle$ the kernel corresponding to an operator $A$. The Fourier transform
is normalized to a unitary operator. We set, as usual,
$\hat\psi(x)=\,${\sc F}$\psi(x)=(2\pi)^{-1/2}\INT dy\;e^{-ixy}\,\psi(y)$
and $\check\psi=\,${\sc F}$^{-1}\psi$. The symbol $\vartheta$ stands
for the Heaviside step function.
Now let us introduce the Hamiltonian \Ham acting in \LR$^2$) . Suppose we
are given a measurable real function $\nu (u)$ defined on \sR. Denote by
\sR$^2_\pm$ the upper (resp. lower) open halfplane in \sR$^2$. Let
${\cal Q}_\nu$ be the closed subspace in the orthogonal sum of Sobolev spaces
${\cal H}^1($\sR$^2_+)\oplus{\cal H}^1($\sR$^2_-)\subset$\LR$^2)$
determined by the boundary condition
$$
\psi (u,0_-)=e^{2\pi i\,\nu (u)}\psi (u,0_+)\, .
\eqno(2.1a) $$
The restriction to ${\cal Q}_\nu$ of the scalar product in the sum of Sobolev
spaces yields a closed form {\gt q}$_\nu$ in \LR$^2$) fulfilling
{\gt q}$_\nu (\psi ,\psi ) \geq \|\psi\|^2$. Hence by a standard construction,
one can relate to {\gt q}$_\nu$ a positive self--adjoint operator \Ha\nu$;
$\hbox{\gt q}_\nu(\psi,\psi)=\langle\psi,\hat H_\nu\psi\rangle+
\langle\psi,\psi\rangle$, for $\psi\in D(\hat H_\nu)$.
It is easy to see that the intersection of $D($\Ha\nu)$ with the subspace
${\cal H}^2($\sR$^2_+)\oplus{\cal H}^2($\sR$^2_-)\subset$\LR$^2)$
is formed by those functions $\psi$ which fulfill in addition to (2.1a) also
$$
\partial_2\psi (u,0_-)=e^{2\pi i\,\nu (u)}\partial_2\psi (u,0_+)\, .
\eqno(2.1b) $$
In this case, \Ha\nu\psi =-\Delta\psi$ where the generalized derivative on
the RHS is taken in \sR$^2_+\cup$\sR$^2_-$. In this sense, \Ha\nu$ is
determined by the boundary conditions on the first coordinate axis (2.1a,b).
It is also clear that for $\nu (u)\equiv 0$ one obtains the free
Hamiltonian \Ha 0=-\Delta$ in \LR$^2)$.
\proclaim Proposition 2.1.
The point spectrum of \Ha\nu$ is empty.
\noindent{\em Proof.}
Suppose that there exists $\psi\in D($\Ha\nu)$ such that \Ha\nu\psi =
\lambda\psi$. From the equality
$$
\hbox{\gt q}_\nu (\psi ,\psi )-\langle\psi ,\psi\rangle
=\langle\psi,\hat H_\nu\psi\rangle
=\lambda\langle\psi ,\psi\rangle
\eqno(2.2)$$
it follows immediately that, provided $\psi\not= 0$, $\lambda$ is
non--negative. But the value $\lambda=0$ is excluded, too, since
in that case
$$
\int_{\R_-^2\cup\R_+^2}d^2x\ (|\partial_1\psi|^2+|\partial_2\psi|^2)
=\hbox{\gt q}_\nu(\psi,\psi )-\langle\psi,\psi\rangle=0\, .
$$
So $\psi$ would be constant almost everywhere in \sR$_\pm^2$ and,
at the same time, $L^2$--integrable. Consequently $\psi=0$, a
contradiction.
Set $\lambda = a^2,\ a>0$. Let us consider $\psi$ as a tempered
distribution from {\sc S}'(\sR$^2$) and denote by $(\cdot ,\cdot )$ the
pairing between {\sc S}'(\sR$^2$) and {\sc S}(\sR$^2$). Any $\varphi$ from
{\sc S}(\sR$^2$) fulfilling $\varphi (u,0)\equiv 0,\
\partial_2\varphi (u,0)\equiv 0$, belongs to $D($\Ha\nu)$ and it holds
\begin{eqnarray*}
& - & \Big(\,(\partial_1^{\, 2}+\partial_2^{\, 2}+a^2)\psi,
\varphi\Big)=
\int_{\R^2} d^2x\ \psi\,(-\partial_1^2-\partial_2^2-a^2)\varphi \\
& = &
\langle\hat H_\nu\bar\varphi ,\psi\rangle -a^2\langle\bar\varphi ,\psi\rangle
=\langle\bar\varphi ,\hat H_\nu\psi\rangle -a^2\langle\bar\varphi ,\psi\rangle
=0\, .
\end{eqnarray*}
Fix $\eta\in\hscS(\hsR)$ such that $\eta\equiv 1$ in some
neighbourhood of $0$. Any $\varphi\in\hscS(\hsR^2)$ can be written
as
\begin{eqnarray*}
\varphi(x_1,x_2) & = & \varphi(x_1,0)+\partial_2\varphi(x_1,0)\,
x_2+\tilde\varphi(x_1,x_2)\, ,\quad\hbox{where}\\
\tilde\varphi(x_1,x_2) & = & \left(\int_0^1 ds\ (1-s)\,
{\partial^2\over\partial x_2^{\, 2}}\varphi(x_1,x_2s)\right)\,
x_2^{\, 2}\, .
\end{eqnarray*}
Since $\varphi(x_1,x_2)-\varphi(x_1,x_2)\eta(x_2)\equiv 0$ in
some neighbourhood of the first coordinate axis and
$\tilde\varphi(x_1,0)=\partial_2\tilde\varphi(x_1,0)=0$ we can
replace, by the above observation, $\varphi(x_1,x_2)$ first by
$\varphi(x_1,x_2)\eta(x_2)$ and afterwards by
$\varphi(x_1,0)\eta(x_2)+\partial_2\varphi(x_1,0)\, x_2\eta(x_2)$
to get
\begin{eqnarray*}
& & \Big((\partial_1^2+\partial_2^2+a^2)\psi(x_1,x_2),
\varphi(x_1,x_2)\Big)\\
& & = \Big((\partial_1^2+\partial_2^2+a^2)\psi(x_1,x_2),
\varphi(x_1,0)\eta(x_2)+\partial_2\varphi(x_1,0)\, x_2\eta(x_2)
\Big)\, .
\end{eqnarray*}
Define $f_0, f_1\in\,${\sc S}'(\sR ) by ($\xi\in\hscS(\hsR)$),
\begin{eqnarray*}
\Big(f_0(u),\xi(u)\Big) & := &
-\Big((\partial_1^2+\partial_2^2+a^2)\psi(x_1,x_2),
\xi(x_1)\, x_2\eta(x_2)\Big)\, ,\\
\Big(f_1(u),\xi(u)\Big) & := &
-\Big((\partial_1^2+\partial_2^2+a^2)\psi(x_1,x_2),
\xi(x_1)\eta(x_2)\Big)\, .
\end{eqnarray*}
Hence it holds
$$
-\Big((\partial_1^2+\partial_2^2+a^2)\psi,\varphi\Big)=
\Big(f_1(u),\varphi(u,0)\Big)+
\Big(f_0(u),\partial_2\varphi(u,0)\Big)
$$
or, in other words,
$$
-(\partial_1^{\, 2}+\partial_2^{\, 2}+a^2)\psi =f_0\otimes\delta' +f_1\otimes\delta\, .
$$
Using the Fourier transform we get for $|k|\not=a$,
$$
(2\pi )^{1/2}\hat\psi (k)=(|k|^2-a^2)^{-1}(\hat f_1(k_1)+ik_2\hat f_0(k_1))\, .
$$
Since $\hat\psi (\cdot ,k_2)\in\,$\LR ) for almost all $k_2$, $f_0,\, f_1$
must be represented by classical functions. Conversely,
$\hat\psi (k_1,\cdot )\in\,$\LR ) for almost all $k_1$. But if $|k_1|0\, , & \nonumber\cr \cr
& = & (z-\bar z)\,\exp(\sqrt{\kappa^2-\bar z}\, u)\,\Omega\,\Xi (\bar z)^{-1}
(2\,{\rm Re}\,\Lambda(z))^{-1/2}\varphi_+(\kappa)\, , & \nonumber\cr \cr
& & \hspace{95mm} {\rm for}\ u<0\, . & \nonumber\cr
& & & (3.6)\nonumber\cr
\end{eqnarray}
Observe that the operators $\Lambda^{1/2}\Xi^{-1}$ and
$\Omega^{-1}\Lambda^{1/2}\Omega\,\Xi^{-1}$ are everywhere defined
and closed and hence bounded. It follows that
$\kappa\, f(\kappa,u)$ and $\partial_2f(\kappa,u)$ and
consequently both $\partial_1g$ and $\partial_2g$
(the generalized derivatives are taken in \sR$^2_-\cup$\sR$^2_+$)
belong to \LR$^2)$. Recalling (3.2) and letting $u\downarrow 0$ resp.
$u\uparrow 0$ in (3.6) one can also check that $g$ satisfies the boundary
condition (2.1a). Hence $g\in{\cal Q}$({\gt q}).
It remains to show that
$$
\INTT d^2x\,(\partial_1\bar\psi\,
\partial_1g + \partial_2\bar\psi\,\partial_2g)=
\int_{\R_-^2\cup\R_+^2} d^2x\, \bar\psi\, (-\Delta g)
\eqno(3.7)$$
is fulfilled for all $\psi\in{\cal Q}(${\gt q}). As a matter of fact,
it suffices to verify (3.7) only for $\psi$'s from some subspace in
${\cal Q}(${\gt q}) which is dense with respect to the norm induced by
{\gt q}$(\cdot ,\cdot )$. Since the restriction morphisms
${\cal H}^s($\sR$^2_\pm )\to$\LR ) are retractions (they have right
bounded inverses) and since $C_0^\infty($\sR$^2_\pm )$ is dense in
${\cal H}_0^s($\sR$^2_\pm )$ \cite{Triebel} , functions compactly
supported in \sR$^2\,\backslash \{a_1,\dots ,a_N\}$ and smooth on the
closure of both the upper and the lower halfplane are dense in
${\cal Q}(${\gt q}). But if $\psi$ possesses these properties and $g$
is given as above then the equality (3.7) is valid provided $g$
satisfies the second boundary condition (2.1b) in the sense of
distributions. This is actually the case as one can easily see, again
with the help of Lemma 3.1. \QED
\section{A Perturbative Formula for $\Xi (z)^{-1}$}
\subsection{Auxiliary results}
In this subsection we introduce some operators occurring in the formula
for $\Xi (z)^{-1}$ and investigate their main properties. Let $\tR =
(\tR_{\ver}),\ \varepsilon ,\varrho =\pm$,
be a $2\times 2$ matrix of operators in \LR )
depending on two parameters, $\alpha\in (0,1)$ and $w\in\,$\sC, Re$\, w >0$,
and determined by the kernels
\begin{eqnarray}
& & \langle\kappa |\tR_{\ver}(\alpha ;w)|\lambda\rangle & (4.1)\nonumber\cr
& & :={1\over\pi}\, (\kappa^2+w^2)^{-1/4}\,
\frac{\exp[-\alpha (\omega (\kappa )-\omega (\lambda ))-i{\pi\over 2}
\alpha (\varepsilon -\varrho )]}{1+
\exp[-\omega (\kappa )+\omega (\lambda )-i({\pi\over 2}-0)
(\varepsilon -\varrho )]}\,
(\lambda^2+w^2)^{-1/4}\, , & \nonumber\cr
\end{eqnarray}
where
$$
\omega (\kappa )\equiv\omega (w;\kappa ):=\as(w^{-1}\kappa )\, .
$$
Dependence on $\alpha$ will be mostly suppressed in notation.
The kernel (4.1) can be interpreted without difficulties only for
$w>0$ or $(\ver )=(++), (--)$. In the former case, after
the substitution $\kappa =w\,\sinh x$ we get an involution operator and
hence diagonalizable by the Fourier transform. The appearance of
$({\pi\over 2}-0)$ in the denominator indicates a regularization. More
precisely, it holds
\begin{flushleft}\hspace{94pt}
$\langle x|${\sc F}$L_w\,\tR_{\ver}L_w^{\, -1}${\sc F}$^{-1}|y\rangle =
r_{\ver}(x)\,\delta(x-y)\, ,$
\end{flushleft}
$$
r_{\ver}(x)=(\sin \pi(\alpha +ix))^{-1}\exp(-{\pi\over 2}(\varepsilon -
\varrho )x)\, ,
\eqno(4.2)$$
with $L_w$ being the unitary operator in \LR ) defined by
$$
L_w\psi (x):=(w\cosh x)^{1/2}\psi (w\sinh x),\quad w>0\, .
\eqno(4.3)$$
Notice also the scaling property of $\tR$. Let $T(c),\ c\in\,$\sR$_+$, be
the one--parameter unitary group of scaling operators, $T(c)\psi (\kappa):=
c^{1/2}\psi (c\kappa )$. Then $T(c)^{-1}\tR (w)\, T(c)=\tR (cw)$. This is
true verbatimly for $w>0$ and, by now, only symbolically for $w$ non--real.
But with the below given definition of $\tR_{\ver}(w)$ as a bounded operator
it turns out that whatever $w$ is, the norm $\|\tR_{\ver}(w)\|$ depends only
on $\arg w$. For $w>0$,
$$
\tR (w)=T(w)^{-1}\tR (1)\, T(w)\, .
\eqno(4.4)$$
The definition of $\tR_{\ver}(w)$ for $w$ non--real imitates this property. The
operator $T(w),\linebreak 0<\arg w <\pi /2$, is well defined on functions $\psi$
which are sectorial analytic, i.e., analytic on the union of two sectors
$0<\arg z <\arg w\,$ and $\,\pi<\arg z <\pi +\arg w$, continuous on the
closure and, moreover, fulfilling $\psi(e^{i\phi}\kappa )\in\,$\LR ) for
all $\phi\in [0,\arg w]$. A similar situation takes place also for
$-\pi /2<\arg w <0$. Let us first construct a sufficiently rich
family of functions possessing this property. Set ($\omega'=
\partial_{\kappa}\omega$)
$$
\Phi_\zeta [h](\kappa ):=(2\pi )^{-1/2}\,\omega '(\zeta ;\kappa )^{1/2}\INT dx\;
\exp(ix\,\omega(\zeta ;\kappa ))\, h(x)\, .
\eqno(4.5)$$
\proclaim Lemma 4.1.
Suppose that $e^{\theta |x|}h\in\,$\LR ), for some $\theta\in
(0,{\pi\over 2}]$, and $\zeta\in\,$\sC$\,\backslash\{0\},\
\linebreak |\arg\zeta | <\theta$.
Then the function $\Phi_\zeta [h]$ is $L^2$--integrable and
it holds
$$
\| \Phi_\zeta [h]\|\leq C\,\| e^{\theta |x|}h\|\, ,
\eqno(4.6)$$
where the bound $C$ can be chosen uniformly provided $\zeta$ runs over some
compact subset in the sector. For a fixed $\zeta$, functions
$\Phi_\zeta [h]$, with $h\in C_0^\infty($\sR ), are dense in \LR ).
\noindent{\em Proof.}
$L^2$--integrability follows from the asymptotic behaviour
$$
\as(\pm z)=\pm (\ln 2z \, +\gamma (z)),\hbox{ with }|\gamma(z)|\leq C'|z|^{-2}\, .
\eqno(4.7)$$
valid in the sector $|\arg z|\leq\theta_0 <\pi /2$.
To show the density observe first that for any $\zeta,\ {\rm Re}\,\zeta >0$,
the func\-tions $\kappa^n\,\sech(\zeta\kappa ),\ n\in\,$\sN$_0$, form a
total subset in \LR ,$d\kappa )$. Actually, suppose that $f\in\,$\LR ) is
orthogonal to all these functions. But the Fourier image of
$\sech(\zeta\kappa )\bar f(\kappa )$ has an analytic extension to the strip
$|{\rm Im}\, z|<{\rm Re}\,\zeta$ and the orthogonality means that all
its derivatives vanish at the point $z=0$. Whence $f=0$ in \LR ).
>From the analycity and decay properties of the function $(\cosh x)^{1/2}
(\sinh x)^n\,\linebreak
\sech(\sinh x)$ one can conclude that its Fourier image $h_n$
fulfills $e^{\theta |x|}h_n\in\,$\LR ) for any $\theta\in [0,{\pi\over 2})$.
Since $\Phi_\zeta [h_n]=\zeta^{-n-1/2} \kappa^n\,\sech(\zeta^{-1}\kappa )$
and owing to the above observation, functions $\Phi_\zeta [h]$, with
$e^{\theta |x|}h\in\,$\LR ) for some $\theta >|\arg \zeta|$, are dense
in \LR ). The inequality (4.6) then implies that the functions
$h\in C_0^\infty ($\sR ) will do as well. \QED
Assuming again $e^{\theta |x|}h\in\,$\LR ) and $|\arg \zeta|<\theta$, the
function $\Phi_\zeta [h]$ has an analytic extension to the sector
$|\arg z -\arg \zeta |<\theta$. Particularly, if
$|\arg w -\arg \zeta |<\theta$ then
$$
T(w)\,\Phi_\zeta [h] =\Phi_{w^{-1}\zeta}[h] \, .
\eqno(4.8)$$
Respecting the relation (4.4), we define
$$
\tR_{\ver}\,\Phi_w[h] :=\Phi_w[r_{\ver}h]\, .
\eqno(4.9)$$
By Lemma 4.1, the operators $\tR_{\ver}$ are densely defined. We have to
show that they are bounded. Moreover, it will turn out that they depend
on $w$ analytically.
Let us consider shortly a general operator $Q$ of this kind, that means
defined by
$$
Q\,\Phi_w[h] :=\Phi_w[qh]\, ,
$$
with $q$ being a bounded measurable function. Let us also suppose that the
function $\check q$ can be extended analytically to the strip
$|{\rm Im}\, z|<2|\arg w|$. Notice that (4.5) means $\Phi [h](\kappa )=
\omega '(\kappa )^{1/2}\check h(\omega (\kappa ))$. Expressing
{\sc F}$^{-1}(qh)$ as a convolution and assuming that the integration curve
can be deformed in the complex plane, from the real axis to the curve
$z=\omega(\lambda ),\ \lambda\in\,$\sR, one gets the generalized kernel
of $Q$ written as
$$
\langle\kappa |Q|\lambda\rangle =(2\pi )^{-1/2}\,\omega '(\kappa )^{1/2}\,
\check q(\omega (\kappa )-\omega (\lambda ))\,\omega '(\lambda )^{1/2}\, .
\eqno(4.10)$$
It is easy to see that this procedure works provided $|q(x)|\leq C\,
e^{-\pi |x|}$. Another important example allowing this treatment
is $q(x)=\vartheta (\pm x)$. Then
$\check q(z)=\mp (\sqrt{2\pi}\, iz)^{-1}$, with Im$\, z>0$ resp. Im$\, z<0$.
Though $\check q(z)$ has a pole at $z=0$, the integration curve can be
deformed provided one obeys the regularization indicated in the following
formula
$$
\langle\kappa |Q|\lambda\rangle =\mp{1\over 2\pi i}\,
\omega '(\kappa )^{1/2}\,\frac{1}{\omega (\kappa )-\omega (\lambda )\pm i0}
\,\omega '(\lambda )^{1/2}\, .
$$
Returning to the operators $\tR_{\ver}$, one finds immediately that the
kernel of $\tR_{++}=\tR_{--}$ is given by the formula (4.1). Furthermore,
one can write
\begin{eqnarray}
& & r_{\pm\mp}(x)=p_\pm(x)\pm 2i\, e^{\mp i\pi\alpha}\vartheta(\mp x)\, ,
\quad\hbox{where} & (4.11)\nonumber\cr \cr
& & \sin \pi(\alpha +ix)\; p_\pm(x)=e^{\mp\pi x}\vartheta (\pm x)+
e^{\mp2\pi i\alpha}\, e^{\pm\pi x}\vartheta(\mp x)\, , & \nonumber\cr
\end{eqnarray}
and so $e^{2\pi |x|}p_\pm$ is bounded. It follows that
\begin{eqnarray}
& & \langle\kappa |\tR_{\pm\mp}|\lambda\rangle = & (4.12)\nonumber\cr
& & \omega '(\kappa )^{1/2}\,
\left( (2\pi )^{-1/2}\check p_\pm(\omega (\kappa )-\omega (\lambda ))+
e^{\mp i\pi\alpha}\,\frac{1}{\pi (\omega (\kappa )-\omega (\lambda )\mp i0)}
\right)\,\omega '(\lambda )^{1/2}\, .
& \nonumber\cr
\end{eqnarray}
Notice also that for $-2\pi <{\rm Im}\, z<0$ resp. $0<{\rm Im}\, z<2\pi$,
it holds
$$
(\pi /2)^{1/2}\,\check p_\pm (z)+e^{\mp i\pi\alpha}z^{-1}=
e^{-\alpha z\mp i\pi\alpha}(1-e^{-z})^{-1}\, .
\eqno(4.13)$$
\proclaim Lemma 4.2.
Suppose that {\rm supp}$\, q\subset [0,+\infty )$, the restriction of $q$
to the positive half--axis belongs to
$C^2([0,+\infty))$ and $e^{\pi x}q,\ e^{\pi x}q',\ e^{\pi x}q''$ are bounded.
Then the operator $Q$ with the kernel (4.10) is bounded. The bound is
uniform provided $w$ runs over some compact subset in the halfplane
{\rm Re}$\, w>0$.
\noindent{\em Proof.}
Let us consider $L_1Q\, L_1^{\, -1}$ rather than $Q$; $L_1$ is given in
(4.3). Since\linebreak
$|\cosh(x)\,\omega '(\sinh x)|$ is bounded we can restrict
ourselves to the kernel $\check q(\omega (\sinh x)-\omega (\sinh y))$. Fix
some $\theta ,\ |\arg w|<\theta <\pi /2$. By integrating by parts one finds
that
$$
(2\pi )^{1/2}\,\check q(z)=q(0)\,
(2\theta-iz)^{-1}+g(z),\quad\hbox{with}\
|g(z)|\leq C_1\, |2\theta -iz|^{-2}\, ,
$$
provided $|{\rm Im}\, z|\leq 2|\arg w|$. It is elementary to check that there
exists $C_2>0$ such that
$$
|{\rm Re}\;\as(w^{-1}\sinh x)-{\rm Re}\;\as(w^{-1}\sinh y)|\geq C_2|x-y|\, .
\eqno(4.14)$$
Consequently,
$$
|g(\omega (\sinh x)-\omega (\sinh y))|\leq C_3\, (1+(x-y)^2)^{-1}\, .
$$
Owing to the Young inequality, the RHS is a kernel of some bounded
convolution operator. Furthermore, the Fourier transform shows that the
kernel $[2\theta +i\ln w\, (\sgn\, x-\sgn\, y)-i(x-y)]^{-1}$ corresponds to a
bounded operator. As one can verify with the help of (4.7), subtracting this
kernel from $[2\theta -i(\omega (\sinh x)-\omega (\sinh y))]^{-1}$ one gets
a Hilbert--Schmidt kernel. \QED
The following Lemma is an adaptation of Proposition 2.3.8 in \cite{SMJII}.
\proclaim Lemma 4.3.
For any $\sigma\in\,${\sr R}$\,\backslash\,\{ 0\} $, the operator $Q$
corresponding to the kernel
$$
\omega '(\kappa )^{1/2}\,(\omega (\kappa )-\omega (\lambda )+i\sigma )^{-1}\,
\omega '(\lambda )^{1/2}
$$
is bounded in \LR ). There exist strong limits for $\sigma\downarrow 0$ and
$\sigma\uparrow 0$.
\noindent{\em Proof.}
To be specific, we shall consider the case $\sigma >0$. Again rather than on
$Q$ we shall concentrate on $L_1Q\, L_1^{\, -1}$. Let us denote $\eta (x)=
\as (w^{-1}\sinh x)$. Since $|\eta '(x)|$ is bounded from below as well as
from above, we can switch to the kernel $(\eta (x)-\eta (y)+i\sigma )^{-1}
\eta '(y)$. The splitting of the Hilbert space, \LR $)=\;$\LR$_+)\oplus$\LR$_-)
$, induces a splitting of the operator into four blocks. We shall focus just
on the case of operator acting in \LR$_+)$. The remaining cases can be
treated similarly. The idea is to compare the kernel with another one,
namely $(x-y+i\sigma )^{-1}$. The latter one corresponds to a bounded
operator and the strong limit $\sigma\downarrow 0$ exists, as manifested by
the Fourier transform. The asymptotic behaviour (4.7) means that $\eta (x)=
-\ln w +x+\gamma_1(x)$, $e^{2x}\gamma_1(x)$ is bounded, and the same is true
for all derivatives of $\gamma_1$. One can write
\begin{eqnarray}
& & \frac{1}{x-y+i\sigma}-\frac{1}{\eta (x)-\eta (y)+i\sigma}\,\eta '(y)
=\frac{x-y+i\sigma}{\eta (x)-\eta (y)+i\sigma} & (4.15)\nonumber\cr
\cr \cr
& & \times\ \left\{ \frac{(x-y)^2}{(x-y+i\sigma )^2}\,
\frac{\eta (x)-\eta (y)-\eta '(y)\, (x-y)}{(x-y)^2}
+\frac{i\sigma}{(x-y+i\sigma)^2}\,(1-\eta '(y))\right\}\, . & \nonumber\cr
\end{eqnarray}
Next we are going to estimate the absolute value of the two summands
on the RHS of (4.15)
by kernels corresponding to manifestly bounded operators. One can easily
derive from (4.14) that the factor standing in front of the composite
bracket is bounded independently of $\sigma$. To treat the first summand,
let us rewrite
$$
\frac{\eta (x)-\eta (y)-\eta '(y)\, (x-y)}{(x-y)^2}=\int_0^1 ds\,
(1-s)\;\eta ''(sx+(1-s)y)\, .
$$
>From the asymptotic behaviour of $\eta$ one can conclude that
$\eta ''(sx+(1-s)y)$ is a Hilbert--Schmidt kernel and the norm is bounded
by $C\,s^{-1/2}(1-s)^{-1/2}$. Whence the first summand corresponds to a
Hilbert--Schmidt operator. Since the prefactor depends on $\sigma$
continuously and is bounded uniformly, the limit $\sigma\downarrow 0$
exists even in the Hilbert--Schmidt norm.
Let us turn to the second summand. By the Young inequality, the kernel
$\sigma |x-y+i\sigma |^{-2}$ corresponds to a bounded operator. Moreover,
the bound can be chosen independently of $\sigma$ since the $L^1$--norm
of the function $\sigma |x+i\sigma |^{-2}$ does not depend on $\sigma$. To
get the strong limit $\sigma\downarrow 0$ for the kernel
$\sigma (x-y+i\sigma )^{-2}$, it is enough to apply the Fourier transform
sending a convolution operator to an operator acting by multiplication. It
turns out that the limit is the zero operator. \QED
\proclaim Proposition 4.4.
The operators $\tR_{\ver}(w)$ are bounded and so, after having been closed,
defined everywhere on \LR ). The bound can be chosen uniformly with respect
to $w$ running over some compact subset in the halfplane Re$\; w>0$.
Moreover, they depend analytically on $w$ in this halfplane.
\noindent{\em Proof.}
The boundedness follows immediately from (4.12), Lemma 4.2 and Lemma 4.3.
Let us turn to the analycity. We claim that to any $\theta\in (0,\pi /2)$
one can find $\sigma >0$ such that the equation
$$
\Phi_\zeta [g]=\Phi_1[h]
$$
has a unique solution $g$, $e^{\theta|x|}g\in\;$\LR ), whenever $|\zeta -1|
<\sigma$ and $h\in C^\infty_0($\sR ). Actually, using the analycity of both
functions and some elementary estimates one can check that, provided $\zeta$
is close enough to $1$, the equation allows a substitution $\kappa =\zeta\,
\kappa '$. Hence it can be rewritten as
$$
(1+\kappa ^2)^{-1/4}\,\check g(\as\,\kappa)\equiv\Phi_1[g](\kappa )=
\Phi_{\zeta^{-1}}[h](\kappa )\, .
$$
Observe also that provided $\psi_1,\,\psi_2$ belong to the domain of the
scaling operator $T(w^{-1})$, as specified after the formula (4.4), then
$$
\langle\psi_1,T(w^{-1})\psi_2\rangle =\langle T(\bar w)\psi_1,\psi_2\rangle\, .
$$
To get this equality one has to rotate the integration path in the complex
plane.
Suppose that we are given some $w\in\,$\sC, Re$\; w>0$. By the above
observations and owing to (4.8),
there exists $\sigma >0$ such that for any couple $h_1, h_2\in
C^\infty_0($\sR ), it holds
$$
\langle\Phi_{\bar w}[h_1],\tR_{\ver}(\zeta )\,\Phi_w[h_2]\rangle =
\langle\Phi_{\bar\zeta^{-1}\bar w}[h_1],\tR_{\ver}(1)\,
\Phi_{\zeta^{-1}w}[h_2]\rangle
$$
and this expression
depends analytically on $\zeta$ in the neighbourhood $|\zeta^{-1}w-1|<\sigma$.
Taking into account the uniform boundedness of $\tR_{\ver}(\zeta )$ and the
density of functions $\Phi_w[h],\ h\in C^\infty_0$, one obtains the weak
analycity. But the weak analycity is equivalent to the strong analycity
\cite{RSI}. \QED
\subsection{Statement and proof of the formula}
Let $\Pi$ be the parity operator in \LR ), $\Pi\psi (\kappa):=\psi (-\kappa)$,
and $Y_c,\ c\in\,$\sR , a unitary operator acting by
$$
Y_c\psi (\kappa ):=e^{-ic\kappa}\psi (\kappa )\, .
$$
For $z\in\;$\sC$\,\backslash\,[0,+\infty )$ and $j=1,2,\dots ,N$, set
$$
R^{(j)}(z):=-\sin(\pi\alpha_j)\, Y_{a_j}\,{\rm diag}(\Pi ,\bI )\,
\tR(\alpha_j ;\sqrt{-z})\,{\rm diag}(\bI ,\Pi )\, Y_{-a_j}\, .
\eqno(4.16)$$
Here we identify $Y$ with $\bI_2\otimes Y$. So $\Rj (z)=
(\Rj_{\ver})_{\varepsilon ,\varrho =\pm}$
is an operator in \sC$^2\otimes$\LR ) related to
the j$^{\rm th}$ vortex. Where not necessary, the indication of dependence
on $z$ will be suppressed. Symbolically,
\begin{eqnarray}
\langle\kappa |\Rj_{\ver}|\lambda\rangle
& = & -{\sin \pi\alpha_j\over\pi}\, (\kappa^2+w^2)^{-1/4}\,\exp(-ia_j\kappa)
\nonumber\cr \cr & \times &
\frac{\exp[\alpha_j (\varepsilon\,\omega (\kappa )+\varrho\,\omega (\lambda ))
-i{\pi\over 2}\alpha_j (\varepsilon -\varrho )]}
{1+\exp[\varepsilon\,\omega (\kappa )+\varrho\,\omega (\lambda )
-i({\pi\over 2}-0)(\varepsilon -\varrho )]}\,
\exp(ia_j\lambda)\,(\lambda^2+w^2)^{-1/4}\, , \nonumber\cr
\end{eqnarray}
with $w=\sqrt{-z}$. Clearly, $\Rj (z)^\ast =\Rj (\bar z)$. Furthermore,
let $E_a$ be the Fourier transform of the orthogonal projector onto
$L^2((-\infty ,a))$, i.e.,
$$
E_a\hat\psi :=(\vartheta (a-x)\,\psi )^\wedge\, .
\eqno(4.17)
$$
We set also $E^c_a:=\bI-E_a$ and write $E_j$ instead of $E_{a_j}$. As an
intermediate step leading to the formula, we shall prove some algebraic identities
relating $\Rj,\ E_k$ and $\Lambda$. Verifications can be simplified provided
the following relations and scaling properties, valid for $w>0$, are used
\begin{eqnarray}
& & E_a=Y_aE_0\,Y_{-a}\, , & (4.18a)\nonumber\cr
& & T(w^{-1})\Lambda (-1)\, T(w)=w^{-1}\Lambda (-w^2)\, , & (4.18b)\nonumber\cr
& & T(w)\, E_0\, T(w^{-1})=E_0\, . & (4.18c)\nonumber\cr
\end{eqnarray}
\proclaim Lemma 4.5.
Let $F\in\;${\sc S}'(\sR) be the Fourier image of $\exp(i\tau\,
\as (\kappa))$, with $\tau$ being a real parameter, in the space of
tempered distributions. Then the pairing between $F$ and some $\varphi\in\;$
{\sc S}(\sR ) reads
\begin{eqnarray}
(F,\varphi ) & = & \left(\sqrt{2\pi}-\sqrt{{2\over\pi}}\,
\sinh({\pi\tau\over 2}) \INT ds\; e^{-\cosh s}\,\tanh s\;\sin{\tau s}\right)\,
\varphi (0)\nonumber\\
& + & \sqrt{{2\over\pi}} \INT dx\,{\tau\over x}\,K_{i\tau}(|x|)\,
\exp({1\over 2}\,\pi\tau\;\sgn\, x)\, (\varphi (x)-\vartheta (1-|x|)\,
\varphi (0))\, ,\nonumber
\end{eqnarray}
where $K_{i\tau}(z)$ is the Macdonald function. Particularly, restricting
$F$ to \sR$\,\backslash\,\{ 0\}$ one obtains
$$
F(x)=\sqrt{{2\over\pi}}\,{\tau\over x}\, K_{i\tau}(|x|)\,
\exp({1\over 2}\,\pi\tau\;\sgn\, x)\, .
\eqno(4.19)$$
\noindent{\em Proof.}
One starts from the equality
\begin{flushright}
{\sc F}$((\kappa^2+1)^{-1/2}\exp(i\tau\,\as\,\kappa ))(x)=
\sqrt{2/\pi}\, K_{i\tau}(|x|)\,\exp({1\over 2}\,\pi\tau\;\sgn\, x)\, .$
\hspace{9pt} (4.20)
\end{flushright}
Rewriting $1$ as $(\kappa^2+1)(\kappa^2+1)^{-1/2}\cdot (\kappa^2+1)^{-1/2}$,
one can express $F$ as $(-d^2/dx^2+1)$ applied to a convolution integral
equaling, up to a constant factor,
\begin{eqnarray}
& & {1\over\pi}\INT d\xi\; K_0(|x-\xi|)\, K_{i\tau}(|\xi |)\,
e^{\pi\tau\,\vartheta (\xi )} \nonumber\\
& & ={\pi\over 2}\, e^{-|x|+\pi\tau\,\vartheta(x)}-{1\over 2}\;\sgn\, x\;
e^{\pi\tau\,\vartheta (x)}
\INT du\; e^{-|x|\,\cosh u}\,\frac{\sin {\tau u}}{\sinh u}\, .\nonumber
\end{eqnarray}
Now it is enough to make use of the simple fact that, provided supp$\, f
\subset\,$\sR$_+$, $f,f'\in L^1_{\rm loc}($\sR$_+)$ and $\lim_{x\downarrow 0}
x f(x)=0$, it holds
$$
({d\over dx}f,\varphi )=f(1)\,\varphi (0)+\int_{\R_+} dx\, f'(x)\,
(\varphi (x)-\vartheta (1-x)\,\varphi (0))\, .
$$
We omit the details. \QED
\proclaim Proposition 4.6.
For any $j=1,2,\dots ,N$, and $z\in\,$\sC$\,\backslash\, [0,+\infty )$, it
holds
\begin{eqnarray}
& \Rj_{\varepsilon ,-}\,\Lambda^{-1/2}(E_j+\epiaj E_j^c)-
\epiaj \Rj_{\varepsilon ,+}\,\Lambda^{-1/2}=(\epiaj -1)\,\Lambda^{-1/2}E_j\, ,
& (4.21) \nonumber\cr\cr
& \Rj_{\varepsilon ,-}\,\Lambda^{1/2}(E_j+\epiaj E_j^c)+
\epiaj \Rj_{\varepsilon ,+}\,\Lambda^{1/2}=\varepsilon\,
(\epiaj -1)\,\Lambda^{1/2}E_j\, ,& (4.22) \nonumber\cr
\end{eqnarray}
with $\varepsilon =\pm$, and (4.22) should be considered on the functions
from {\sc F}$\big(C^\infty_0((-\infty ,a_j)) \linebreak \oplus
C^\infty_0((a_j,+\infty ))\big)$.
\noindent{\em Proof.}
Since both sides of (4.21) as well as of (4.22) (after having been applied
to some function from the above indicated domain) depend on $z$ analytically,
we can restrict ourselves to the case $z<0$. But owing to the properties
(4.18), it is sufficient to consider the case $a_j=0$ and $z=-1$. Further
in this proof, the index $j$ is suppressed. The composition from the right
with $E$ and $E^c$ results in a couple of equations equivalent to the
original equality, namely
\begin{eqnarray}
& & (R_{\varepsilon ,-}-\epia R_{\varepsilon ,+}-(\epia -1)\,\bI )\,
\Lambda^{-1/2}E=0\, , & (4.23a) \nonumber\cr \cr
& & (R_{\varepsilon ,-}-R_{\varepsilon ,+})\,\Lambda^{-1/2}E^c=0\, , &
(4.23b) \nonumber\cr
\end{eqnarray}
for (4.21), and
\begin{eqnarray}
& & (R_{\varepsilon ,-}+\epia R_{\varepsilon ,+}-\varepsilon\,
(\epia -1)\,\bI )\,\Lambda^{1/2}E=0\, , & (4.24a) \nonumber\cr \cr
& & (R_{\varepsilon ,-}+R_{\varepsilon ,+})\,\Lambda^{1/2}E^c=0\, , &
(4.24b) \nonumber\cr
\end{eqnarray}
for (4.22). Verification is done by explicit calculations. In the case of
(4.23a), expressing the kernel $R_{\ver}$ with the help of the integral
identity
$$
\frac{e^{-\alpha u}}{1+e^{-u}}={1\over 2}\INT d\tau\;
\frac{e^{iu\tau}}{\sin \pi (\alpha +i\tau )},\quad |{\rm Im}\, u|<\pi\, ,
\eqno(4.25)$$
and applying (4.20), one finds that for $x<0$ it holds
\begin{flushleft}
\hspace{20pt}
$\sqrt{2\pi}\;\langle\kappa |(R_{\varepsilon ,-}-\epia R_{\varepsilon ,+}-
(\epia -1)\,\bI )\,\Lambda^{-1/2}${\sc F}$|x\rangle$\\
\hspace{20pt}
$$=(1-\epia )\, (\kappa^2+1)^{-1/4}\left(e^{-i\kappa x}-{1\over\pi}\INT
d\tau\;e^{-i\varepsilon\,\omega (\kappa)\,\tau}\,e^{-\varepsilon\pi\tau /2}\,
K_{i\tau}(|x|)\right)\, ,$$
\end{flushleft}
where $\omega (\kappa )=\as\,\kappa$. But since $K_{i\tau}(|x|)$ is even in
$\tau$ and
$$
{1\over\pi}\INT d\tau\; e^{-iu\tau}\, K_{i\tau}(|x|)=e^{-|x|\,\cosh u},\quad
|{\rm Im}\, u|<{\pi\over 2}\, ,
\eqno(4.26)$$
the last expression is zero (in the sense of distributions).
Now it suffices to recall (4.17). Similarly one
can proceed also in the remaining three cases. Verifying (4.24a,b), one has
to use (4.19) rather than (4.20). We omit the details. \QED
To get the formula for $\Xi^{-1}$ one has to tensor \LR ) with the
finite--dimensional space \sC$^N\otimes$\sC$^2$. We denote, as before, by
$\{ e_+,e_-\}$ the standard basis in \sC$^2$ and $\{\epsilon_1,\dots ,
\epsilon_N\}$ stands for the standard basis in \sC$^N$. To simplify the
notation we identify, where appropriate, an operator $Z$ in \LR ) with
$\bI_N\otimes\bI_2\otimes Z$ and similarly for other factors. Equipped with
the algebraic relations (4.21), (4.22) one can prove
\proclaim Theorem 4.7.
For any $z\in\;$\sC$\,\backslash\, [0,+\infty )$, it holds
$$
\Xi (z)^{-1}={1\over 2}\,\Lambda (z)^{-1}+{1\over 2}\,\Lambda (z)^{-1/2}\,
(\bef^t\otimes e_-^{\,t}\,(\bI -\cD\cJ )^{-1}\cD\;\bef\otimes e_-)\,
\Lambda (z)^{-1/2}\, ,
\eqno(4.27)$$
where
$$
\cD ={\rm diag}(R^{(1)}(z),\dots ,R^{(N)}(z))
$$
is an operator in \sC$^N\otimes$\sC$^2\otimes$\LR ),
$$
\cJ=\pmatrix{0 & K & \dots & K \cr
K^t & 0 & \dots & K \cr
\dots & \dots & \dots & \dots \cr
\dots & \dots & \dots & \dots \cr
K^t & K^t & \dots & 0 \cr},\qquad with\quad
K=\pmatrix{0 & 1 \cr 0 & 0 \cr}\, ,
\eqno(4.28)$$
is a $2N\times 2N$ matrix, and $\bef =\epsilon_1+\epsilon_2+\dots +
\epsilon_N$.
In the next Lemma as well as several times in the rest of the paper, we apply
the following well--known criterion to show that an operator in \LR )
belongs to the trace class \cite{RSIII}. Suppose that $(1+|x|)^{\delta /2}f,\,
(1+|x|)^{\delta /2}g\in\;$\LR ) for some $\delta >1$, then the kernel
$f(x)\,\hat g(x-y)$ corresponds to a trace class operator.
\proclaim Lemma 4.8.
For $z<0$, the operator $(\cD\cJ )^{2N}$ belongs to the trace class and its
norm can be estimated by
$$
\| (\cD\cJ )^{2N}\|\leq C\; (\max_j\;\sin(\pi\alpha_j))\; K_0(wd)\, ,
\eqno(4.29)$$
where $w=\sqrt{-z}>0$ and $d=\min_{j\not =k}\, |a_j-a_k|$.
\noindent{\em Proof.}
First observe that
$$
\epsilon^t_j\otimes e^t_\varepsilon\,(\cD\cJ )^{M-1}\cD\,
\epsilon_k\otimes e_\varrho
=\sum_{(j_1,\dots ,j_M)} R^{(j_1)}_{\varepsilon ,\mu_2}\,
R^{(j_2)}_{-\mu_2,\mu_3}\,\dots\,
R^{(j_M)}_{-\mu_M,\varrho}\, ,
\eqno(4.30) $$
where we have put $\mu_m=\sgn (j_m-j_{m-1})$ and the sum runs over all finite
sequences $(j_1,\dots ,j_M ),\ j_i\in\{1,\dots ,N\}$, $j_i\not =j_{i+1}$,
and $j_1=j,\ j_M=k$.
If $M=2N$, one can find in any sequence $(j_1,\dots ,j_{2N})$ three subsequent
indices $j_m,j_{m+1},j_{m+2},$ such that $j_mj_{m+2}$. Thus any
summand in (4.30) contains a neighbouring couple $R^{(m)}_{\mu ,+}
R^{(n)}_{-,-}$. Set $d_{mn}=|a_m-a_n|$. For $w>0$, we have (c.f. (4.2))
$$
\| \Rj_{\ver}\| =|\sin\pi\alpha_j|\,\| \tR_{\ver}(\alpha_j;w)\|=
|\sin\pi\alpha_j|\,\|r_{\ver}\|_\infty\, ,
$$
and so $\|\Rj_{\ver}\|\leq 1$, if $\varepsilon =\varrho$, and $\leq 2$, if
$\varepsilon\not =\varrho$. From the form of (4.16) it is clear that it
suffices to show that the kernel
\begin{center}
$\langle x|${\sc F}$\, L_w\tR_{\mu ,+}(\alpha_m;w)\,Y_{d_{mn}}
\tR_{-,-}(\alpha_n;w)\,L_w^{\, -1}${\sc F}$^{-1}| y\rangle$
\end{center}
corresponds to a trace class operator and to get the sought estimate for it.
But this kernel can be expressed as
$$
{1\over\pi}\,\sin\pi\alpha_m\;\sin\pi\alpha_n\;
\frac{e^{-\mu\pi x/2}}{\sin\pi (\alpha_m+ix)}\;K_{i(x-y)}(w\,d_{mn})\;
\frac{e^{\pi y/2}}{\sin\pi (\alpha_n+iy)}\, .
$$
By the above criterion, $e^{-\tau |x|}K_{i(x-y)}(wd)$ corresponds to a trace
class operator for any $\tau >0$ (c.f. (4.26)). Besides, one can estimate
the Hilbert--Schmidt norm using $|K_{i\tau}(wd)|\leq K_0(wd)$ and
$$
\INT dx\; |\sin\pi (\alpha +ix)|^{-2}\,e^{\pm\pi x}=(\sin\pi\alpha )^{-1}\, .
\qquad \QED$$
\noindent{\em Remark}
It is easy to see that if $A(z)$ is a family of bounded operators in some
Hilbert space, analytic in $z$ on the disk $|z|<1$ and such that $A(z)$ is
compact for all $z\in (-1,1)$ then $A(z)$ is compact for all $z$ from the
disk. Actually, since the norm limit of compact operators is a compact
operator, all the derivatives of $A(z)$ at the point $z=0$ are compact. The
assertion then follows from the Taylor expansion. Using repeatedly this
argument one concludes that $(\cD\cJ )^{2N}$ is compact in the region
\sC$\,\backslash\, [0,+\infty )$. By the analytic Fredholm theorem \cite{RSI},
$$
(\bI -\cD\cJ )^{-1}=\left(\bI +\sum_{k=1}^{2N-1} (\cD\cJ )^k\right)\,
(\bI -(\cD\cJ)^{2N})^{-1}
$$
is a meromorphic function in this region. For $z$ negative, the estimate
(4.29) tells us that a perturbative expansion into a geometric series is
possible if the numbers characterizing the strength of the vortices,
$\sin\pi\alpha_j$, are small enough or, on the other hand, if $|z|$ is large
enough.
\noindent{\em Proof of Theorem 4.7.}
The following operator identities are thought as being applied to a vector
from {\sc F}$\big(C^\infty_0(I_0)\oplus\dots\oplus C^\infty_0(I_N)\big)$.
First note that combining (4.21) and (4.22) one obtains
$$
\Rj_{\varepsilon ,-}\Lambda^{-1/2}[(E_j+\epiaj E_j^c)\,\Lambda +\Lambda
(E_j+\epiaj E_j^c)]=(\epiaj -1)\,\Lambda^{-1/2}(E_j\Lambda +\varepsilon\,
\Lambda\, E_j)\, .
\eqno(4.31)$$
Suppose that $k>j$ and set $\mu_{kj}=\sgn(k-j)=+$. As $E_j^cE_k^c=E_k^c$, one
can derive from (4.23b), (4.24b) that
$$
\Rj_{\varepsilon ,-}\Lambda^{-1/2}(E_k^c\,\Lambda +\Lambda E_k^c)=
-\Rj_{\varepsilon ,+}\Lambda^{-1/2}(E_k\,\Lambda -\Lambda E_k)\, .
$$
Combining this equality with (4.31) one finds that
\begin{eqnarray}
& & (e^{-2\pi i\alpha_k}-1)\,
\Rj_{\varepsilon ,-}\Lambda^{-1/2}(E_k^c\,\Lambda +\Lambda E_k^c)
& (4.32)\nonumber\cr \cr
& & = -\Rj_{\varepsilon ,\mu_{kj}} R^{(k)}_{-\mu_{kj},-}\Lambda^{-1/2}
[(E_k+\epiak E_k^c)\,\Lambda +\Lambda\, (E_k+\epiak E_k^c)]\, . &
\nonumber\cr
\end{eqnarray}
Suppose now, on the contrary, that $kj}(e^{-2\pi i\,\nu_k}-e^{-2\pi i\,\nu_{k-1}})\,E_k^c\, . &
\nonumber\cr
\end{eqnarray}
Using the notation
\begin{center}
$\cV={\rm diag}(e^{-2\pi i\,\nu_1},e^{-2\pi i\,\nu_2},\dots ,
e^{-2\pi i\,\nu_N}),\quad \cV'={\rm diag}(1,e^{-2\pi i\,\nu_1},
\dots ,e^{-2\pi i\,\nu_{N-1}})\, ,$\\ \vspace{12pt}
$\cE={\rm diag}(E_1,E_2,\dots ,E_N),\quad \cE^c=
{\rm diag}(E_1^c,E_2^c,\dots ,E_N^c)\, ,$
\end{center}
one can derive from the expression for $\Omega^{-1}$ and from (4.32), (4.33)
that
\begin{eqnarray}
& & \cD\,\Lambda^{-1/2}(\Omega^{-1}\Lambda +\Lambda\,\Omega^{-1})\;
\bef\otimes e_-\nonumber\cr\cr
& & =(\bI -\cD\cJ )\cD\,\Lambda^{-1/2}[(\cV'\cE+\cV\cE^c)\,\Lambda+\Lambda\,
(\cV'\cE+\cV\cE^c)]\;\bef\otimes e_-\, .\nonumber\cr
\end{eqnarray}
As remarked above, $(\bI -\cD\cJ )^{-1}$ exists, possibly up to some isolated
values of the spectral parameter $z$. Applying once more (4.31) one obtains
\begin{eqnarray}
& & (\bef^t\otimes e_-^t\,(\bI -\cD\cJ )^{-1}\cD\;\bef\otimes e_-)\,
\Lambda (z)^{-1/2}\, (\Omega^{-1}\Lambda +\Lambda\,\Omega^{-1})\nonumber\cr\cr
& & =\bef^t\otimes e_-^t\,
\cD\,\Lambda^{-1/2}[(\cV'\cE+\cV\cE^c)\,\Lambda+\Lambda\,
(\cV'\cE+\cV\cE^c)]\;\bef\otimes e_-\nonumber\cr\cr
& & =\Lambda^{-1/2}(\bef^t\;(\cV-\cV')\,(\cE\,\Lambda -\Lambda\,\cE)\;\bef )
\, .\nonumber\cr
\end{eqnarray}
Since
$$
\bef^t\;(\cV-\cV')\,\cE\;\bef =e^{-2\pi i\,\nu_N}-\Omega^{-1}\, ,
$$
one arrives, after some simple manipulations, at
$$
{1\over 2}\,[\Lambda (z)^{-1}+\Lambda (z)^{-1/2}\,
(\bef^t\otimes e_-^t\,(\bI -\cD\cJ )^{-1}\cD\;\bef\otimes e_-)\,
\Lambda (z)^{-1/2}]\, (\Lambda +\Omega^{-1}\Lambda\Omega )=\bI\, .
$$
Now it is enough to reexamine the proof of Lemma 3.1 to observe that
{\sc F}$\big(C^\infty_0(I_0)\oplus\dots\oplus C^\infty_0(I_N)\big)\subset
D(\Lambda +\Omega^{-1}\Lambda\Omega )$ is a core for $\Xi$\ (as
$C_0^\infty (I)$ is dense in $\cH^{1/2}_0(I)$). \QED
\section{Wave Operators}
\subsection{Existence and completeness}
Let us begin with recalling some basic known facts about the one--vortex
scattering. This problem was solved in the stationary picture already
in the original paper \cite{AB} and later reconsidered several times from
different points of view, e.g., \cite{Rui,Hagen,Jackiw}. One can assume that the
vortex is sitting in the origin. Respecting the rotational symmetry,
it is natural to pass to the polar coordinates, i.e., to identify
\LR$^2,d^2x)\simeq\;$\LR$_+,r\,dr)\otimes L^2((0,2\pi ),d\phi )$. The
boundary conditions (2.1a,b), determining the domain of the Hamiltonian
$\hat H=-(\partial_r^{\, 2}+r^{-1}\partial_r+r^{-2}\partial_\phi^{\, 2})$,
now read
$$
\psi (r,2\pi )=e^{2\pi i\alpha}\psi (r,0),\quad
\partial_\phi\,\psi (r,2\pi )=e^{2\pi i\alpha}\partial_\phi\,\psi (r,0)\, .
$$
Though in this case much more usual choice of the Hamiltonian is
$\hat H'=-(\partial_r^{\, 2}+r^{-1}\partial_r+r^{-2}
(\partial_\phi +i\alpha )^2)$ with the $2\pi$--periodic boundary conditions.
The both Hamiltonians are unitarily equivalent, $\hat H'=G_\alpha^{\, -1}
\hat H\, G_\alpha$, with $G_\alpha\psi (r,\phi )=e^{i\alpha\phi}
\psi (r,\phi )$.
$\hat H'$ can be diagonalized by the Henkel transformation ($\tau >0$),
\begin{center}
{\sc H}$_\tau :\,$\LR$_+,k\, dk)\to\,$\LR$_+,r\,dr)\quad$, {\sc H}$_\tau f(r)=
\int_{\R_+} dk\; k\; J_\tau(rk)\, f(k)\, .$
\end{center}
That means ($n\in\,$\sZ )
\begin{center}
$e^{-it\hat H'}\;${\sc H}$_{|n+\alpha|}(f)\otimes e^{in\phi}=\;$
{\sc H}$_{|n+\alpha|}(e^{-it\,k^2}\,f)\otimes e^{in\phi}\, .$
\end{center}
The stationary phase method then yields explicitly both the wave operators
\linebreak $W_\pm (\hat H',\hat H_0)$ and $W_\pm (\hat H_0,\hat H')$,
\begin{eqnarray}
& W_\pm (\hat H',\hat H_0)=s-\lim_{t\to\pm\infty}e^{it\,\hat H'}
e^{-it\,\hat H_0}\, , & \nonumber\cr \cr
& W_\pm (\hat H',\hat H_0)\; f\otimes e^{in\phi}=
\exp(\pm\, i{\pi\over 2}(|n+\alpha|-|n|))\, f\otimes e^{in\phi}\, . &
\nonumber\cr
\end{eqnarray}
It is worth of emphasizing that the existence of the wave operators
$W_\pm (\hat H,\hat H_0)$ does not follow immediately. The point is that
while $\hat H'$ is unitarily transformed \Ha 0$ is kept fixed. Writing
$$
e^{it\,\hat H}e^{-it\,\hat H_0}=G_\alpha \left( e^{it\,\hat H'}
e^{-it\,\hat H_0}\right) \left( e^{it\,\hat H_0}G_\alpha^{-1}
e^{-it\,\hat H_0}\right)\, ,
$$
one is lead to considering the limit
$$
s-\lim_{t\to\pm\infty} e^{it\,\hat H_0}G_\alpha^{-1}
e^{-it\,\hat H_0} \, .
$$
But its existence can be verified by an explicit calculation, again with the
help of the stationary phase method \cite{PLA91}. So let us close this summary
by pointing out that, in the case $N=1$, the wave operators
$W_\pm (\hat H,\hat H_0)$ exist and are complete.
\proclaim Theorem 5.1.
For arbitrary number of vortices $N$, the wave operators\linebreak
$W_\pm (\hat H,\hat H_0)$ exist and are complete.
\noindent{\em Proof.}
We shall show the existence and completeness of $W_\pm (\hat H,\hat H_1)$,
with \Ha 1$ being a properly chosen one--solenoid Hamiltonian. Provided this
is true, the theorem follows from the just recalled result and the chain
rule. By the Kato--Birman theory \cite{RSIII}, it suffices to show that
$(\hat H+\bI )^{-1}-(\hat H_1+\bI)^{-1}$ is a trace class operator. Using
the Krein's formula (3.1) and Proposition 3.2, one finds ($z=-1$, Re$\,
\Lambda =\Lambda$ and $U$ was introduced in (3.4))
\begin{eqnarray}
& & (\hat H+\bI )^{-1}-(\hat H_1+\bI)^{-1} \nonumber\cr
& & ={1\over 2}\, P^\ast U\,\Lambda^{-1/2}\,
\pmatrix{\Xi^{-1}-\Xi_1^{-1} & \Xi^{-1}\Omega^{-1}-\Xi_1^{-1}\Omega_1^{-1}\cr
\Omega\,\Xi^{-1}-\Omega_1\Xi_1^{-1} &
\Omega\,\Xi^{-1}\Omega^{-1}-\Omega_1\Xi_1^{-1}\Omega_1^{-1} \cr}\,
\Lambda^{-1/2}U^{-1}P\, .
\end{eqnarray}
We claim that $\Xi^{-1}-{1\over 2}\,\Lambda^{-1}$ is a trace class operator.
In view of Theorem~4.7 and Lemma 4.8, the statement will hold provided the
operators $\Lambda^{-1/2}(\bef^t\otimes e_-^t\,(\cD\cJ)^{M-1}\cD\,
\newline
\bef\otimes e_-)\,\Lambda^{-1/2}$, $M=1,2,\dots$, are shown to belong to the
trace class. To this end, let us check the summands on the RHS of (4.30)
with $\varepsilon =\varrho =-$. One of the following two possibilities occurs:
(i) $j_1>j_2$ or $j_{M-1}j_M$. Considering
the case (i), let us assume that, say, $j_1>j_2$. In view of (4.16), it is
enough to show that $\Lambda^{-1/2}\tR_{--}$ is a trace class operator. But
this can be done with the help of the criterion preceding Lemma 4.8 since
(c.f. (4.3))
$$
\langle x|L_1\Lambda^{-1/2}\tR_{--}\, L_1^{\, -1}|y\rangle =(\cosh x)^{-1/2}
\,\frac{e^{-\alpha (x-y)}}{1+e^{-x+y}}\, .
$$
In the case (ii), there exist three subsequent indices $j_m,\,
j_{m+1},\, j_{m+2}$, such that $j_mj_{m+2}$. But this situation
was considered already in the course of the proof of Lemma 4.8.
So one can replace, where appropriate, $\Xi^{-1}$ as well as
$\Xi_1^{-1}$ by $(1/2)\Lambda^{-1}$ modulo trace class operators.
Consequently, $\Xi^{-1}-\Xi_1^{-1}$ is a trace class operator.
But the same is true also for
$\Omega\,\Xi^{-1}\Omega^{-1}-\Omega_1\Xi_1^{-1}\Omega_1^{-1}$
as can be seen when replacing
$\Omega$ by $\Omega^{-1}$ and $\Omega_1$ by $\Omega_1^{-1}$. The proof
will be finished provided one can choose \Ha 1$ and hence $\Omega_1$ in
such a way that $(\Omega -\Omega_1)\,\Lambda^{-3/2}$ and $\Lambda^{-3/2}
(\Omega^{-1} -\Omega_1^{-1})$ are trace class operators. But applying
once more the same criterion as above and recalling (3.2), it is easy to see
that this is the case provided $\exp(2\pi i\nu)-\exp(2\pi i\,\nu_1)$ has a
bounded support. \QED
\subsection{Generalized eigen--functions}
As explained already in Introduction, the main idea consists in considering
the functions $\psi_+(z)=\varphi(z)+V(z)\varphi(z)$, with $\varphi(z)\in
{\cal N}(z)$ being properly chosen so that the limit
$z=k^2-i\sigma ,\ \sigma\downarrow0$, is possible. To be able to apply
this approach one needs some information about the operators $R(z)$ in this
limit. We assume that $k>0$. Let us first
consider the limiting operators $\tR_{\ver}(w)$, with $w=\sqrt{-z}=ik+0$.
As $\as (ix)=i\arcsin x$, for $|x|\leq 1$, and $\as (ix\pm 0)=i(\pi /2)\mp\,
\as\, (x^2-1)^{1/2}$, for $x\geq 1$, there exists the limiting curve
$\lambda\mapsto\omega (\lambda )\equiv\omega (w;\lambda )$, with $w=ik+0$.
It is continuous though it is not smooth as it has two wedges in the points
$z=\pm\, i\pi /2$. The reparameterization operator $L_1$ given in (4.3) is
useless in this situation and must be redefined. Thus we put $L\psi :=\,$
{\sc L}'$(x)^{1/2}\psi(${\sc L}$(x))$, where
\begin{flushleft}
\hspace{55pt} {\sc L}$(x)=\mp\, k\,\cosh(x\pm{\pi\over 2})\, ,
\quad\hbox{for}\quad x\leq
-{\pi\over 2}\quad\hbox{resp.}\quad x\geq {\pi\over 2}\,$,
\hspace{41pt}(5.1)\\ \vspace{5pt}
$\hspace{87pt} =k\,\sin x\, ,\hspace{58pt}\hbox{for}\ -{\pi\over 2}\leq x\leq
{\pi\over 2}\, .$
\end{flushleft}
Clearly, {\sc L}$\,\in C^1($\sR ). Setting $\eta =\omega\circ${\sc L}, we
have
\begin{flushleft}
\hspace{55pt} $\eta (x)=x\pm{\pi\over 2}\pm i{\pi\over 2}\, ,
\quad\hbox{for}\quad x\leq
-{\pi\over 2}\quad resp.\quad x\geq {\pi\over 2}\,$,\\ \vspace{5pt}
$\hspace{85pt} =-ix\, ,\hspace{45pt}\hbox{for}\ -{\pi\over 2}\leq x\leq
{\pi\over 2}\, .$
\end{flushleft}
The functions $\Phi [h]\equiv\Phi_w[h]$ (c.f. (4.5)), with $w=ik+0$, are well
defined and belong to \LR ) provided there exists $\theta >\pi /2$ such that
$e^{\theta |x|}h\in\,$\LR ). The estimate (4.6) is again valid. This is easy
to see from the equality $L\,\Phi [h](x)=\eta '(x)^{1/2}\,\check h(\eta (x))$.
Hence then definition (4.9) of $\tR_{\ver}(w)$ makes sense also in this
limiting case.
\proclaim Lemma 5.2.
The limiting operators $\tR_{\ver}(w)$, with $w=ik+0$, are densely defined,
bounded and hence, after having been closed, defined everywhere. Moreover,
it holds
$$
s-\lim_{\sigma\downarrow 0} \tR_{\ver}(ik+\sigma )=\tR_{\ver}(ik+0)\, .
$$
\noindent{\em Proof.}
Concerning the density, the argument given in Lemma 4.1 is not applicable
here. Suppose that $f\in\,$\LR ) is orthogonal to all functions $L\,\Phi [h]$,
with $h$ fulfilling $e^{\theta |x|}h\in\,$\LR ) for some $\theta >\pi /2$.
Given any $\tau >0$, we shall show that $f(x)=0$ for almost all $x,\ x<-
{\pi\over 2}-\tau$. Set $\phi =\arctan(\pi\,\tau^{-1})\in (0,\pi /2)$. To any
$c$ positive relate the function $F_c(x)=M_c\,\exp(-e^{i\phi}\,c^{-2}x^2)$.
The normalization constant $M_c=c^{-1}M_1$ is chosen so that $\INT dx\;
F_c(x)=1$. $F_c$ extends as an entire function and $F_c(\cdot +iy)\in\,$\LR )
for all $y\in\,$\sR. It follows that to any $b<-{\pi\over 2}-\tau\ $,
there exists a unique function $h_{bc}$ fulfilling
$e^{\theta |x|}h_{bc}\in\,$\LR ), for all $\theta >0$, and
$L\,\Phi [h_{bc}](x)=F_c(x-b)$, for all $x<-\pi /2$. Moreover, since
$\arg(\eta (x)-{\pi\over 2}-i{\pi\over 2}-b)\in [-\phi ,0]$, for all
$x\in\,$\sR , it holds
$$
|L\,\Phi [h_{bc}](x)|\leq M_c\,\exp\left( -\cos(\phi)\, c^{-2}\, |\eta (x)
-{\pi\over 2}-i{\pi\over 2}-b|^2\right)\, .
$$
Now it is easy to find that $\int dx\,\bar f(x)\,\varphi (x)=0$, for each
$\varphi\in C_0^\infty ((-\infty ,-{\pi\over 2}-\tau ))$. This is a
consequence of the fact that $F_c(x-b)$ converges to $\delta (x-b)$, as
$c\downarrow 0$, in the space of distributions and $f$ is orthogonal to
$L\,\Phi [h_{bc}]$ and hence also to
$\int db\ F_c(x-b)\,\varphi(b)$.
Thus the above assertion is verified. But since $\tau >0$
was arbitrary $f(x)=0$ for a. a. $x<-\pi /2$. Similarly, $f(x)=0$ for a. a.
$x>\pi /2$. To show that $f(x)$ vanishes also on the interval $|x|<\pi /2$,
it is enough to consider the entire functions $F_n(z)=z^n\,\exp(-z^2)$,
$n=0,1,2,\dots$, to relate to them functions $h_n$ so that $\check h_n=F_n$
and to apply the orthogonality argument.
Concerning the fact that $\tR_{\ver}(w)$ are bounded, note that the formula
(4.12) is still true. The only slight change required is to adjust the
regularization since the complex number $i$ is tangent to the central part of
the curve $\lambda\mapsto\omega(\lambda )$. To be precise one has to write,
say, $(\omega (\kappa )-\omega (\lambda )\mp e^{i\pi /4}\epsilon ),\
\epsilon\downarrow 0$.
Considering the operators $L\,\tR_{\ver}(w)\, L^{-1}$ rather than
$\tR_{\ver}(w)$ and splitting the Hilbert space as \LR )$\, =L^2((-\infty ,
-\pi /2))\oplus L^2((-\pi /2,\pi /2))\oplus L^2((\pi /2,+\infty ))$, one
can proceed exactly the same way as in the proof of Lemmas 4.2 and 4.3. We
omit the details.
Let us now turn to the strong limit. Here we focus only on the most
troublesome case of the kernel (4.10), with $\omega (\kappa )=
\as (w^{-1}\kappa )$ and
$(2\pi )^{-1/2}\, \check q(z)=(z-e^{i\pi /4}\epsilon ),\
\epsilon\downarrow 0$. Owing
to the scaling properties we can assume that $k=1$. So we are going to
consider the limit $w^{-1}=(-1-i\sigma )^{1/2},\ \sigma\downarrow 0$. The
difficulties are caused by the appearance of the two wedge points of the
curve $\lambda\mapsto\omega (\lambda )$ in the limiting case. After cutting
off some $\delta$--neighbourhoods of the points $\kappa =\pm 1$ in the real
line, one can proceed the same way as in Lemma 4.3 not to show only the
existence of the strong limit $\epsilon\downarrow 0$ when regularizing the kernel
but also to verify even the norm continuity in the parameter $\sigma ,\
\sigma\downarrow 0$. Since $C_0^\infty ($\sR$\,\backslash\,\{ -1,1\} )$ is
dense in \LR ) it is enough to show that the operator $Q$ is bounded uniformly
in the parameter $\sigma$. Furthermore, we can concentrate only on some
vicinity of the wedge points. To be specific, let us restrict $Q$ (by means
of sandwiching it by orthogonal projectors) to the subspace $L^2((1-\delta ,
1+\delta ))$. The idea is to approximate the curves $\lambda\mapsto
\omega (\lambda )$, as $\sigma$ tends to zero, by hyperbolas for they are
comparatively easy to handle.
Set $\omega_A(\kappa )=-(-1-i\sigma )^{1/2}(\kappa^{-2}-1-i\sigma )^{1/2}$.
Clearly, $\omega_A\,'(\kappa )=\kappa^{-2}\omega '(\kappa )$. Let us rotate
the curves by $\pi /4$,
$$
\tom (\kappa ):=e^{i\pi /4}(1+i\sigma )^{-1/2}[\omega (\kappa )-\omega (1)],
\quad \tom_A(\kappa ):=e^{i\pi /4}(1+i\sigma )^{-1/2}\omega_A(\kappa )\, .
$$
The kernel can be now written as $\langle\kappa |Q|\lambda\rangle =
\tom '(\kappa )^{1/2}\, (\tom (\kappa )-\tom(\lambda )-i0)^{-1}\,
\tom'(\lambda )^{1/2}$. Let \tl$(x)$ be the inverse to the function $\kappa
\mapsto\;$Re$\,\tom (\kappa )$ and, similarly, \tl$_A(x)$ is the inverted
function to $\kappa\mapsto\;$Re$\,\tom_A(\kappa )$. Set $\beta =
\tom\circ$\tl, $\beta_A=\tom_A\circ$\tl$_A$ and $\varphi =\,$
\tl$_A^{-1}\circ$\tl . It holds $\varphi (0)=0,\ \varphi '(x)=\,$\tl$(x)^{-2}$
and so $|\varphi (x)|\geq C_1\, |x|$, with $C_1 >0$. Furthermore, Re$\,
\beta (x)=\,$Re$\,\beta_A(x)=x,\ \beta '(x)=\beta_A\,'(\varphi (x))$ and one
can check that $\beta_A(x)=x+i\,(\sigma +x^2)^{1/2}$. Reparameterizing and
taking onto account that $1\leq |\beta '(x)|\leq\sqrt{2}$, one can switch to
the kernel $\langle x|\tilde Q|y\rangle =(\beta (x)-\beta(y)-i0)^{-1}
\beta '(y)$, with $\tilde Q$ acting in $L^2((-\delta_1,\delta_2)),\
(-\delta_1,\delta_2)=\,$\tl$^{-1}((1-\delta ,1+\delta ))$. Now the method of
the proof of Lemma 4.3 is again applicable since $\sqrt2\, |x-y|\geq
|\beta (x)-\beta (y)|\geq|{\rm Re}\,\beta (x)-{\rm Re}\,\beta (y)|=|x-y|$,
$|1-\beta '(x)|$ is bounded independently of $\sigma$ and $|\beta ''(x)|^2
\leq (1-\delta )^{-2}|\beta_A''(\varphi (x))|^2\leq C_2\,\sigma^2(\sigma +
x^2)^{-3}$. \QED
We need also some estimates to show that $(\bI -\cD\cJ)^{-1}$ exists also
in the limiting case.
\proclaim Lemma 5.3.
Let $d_0,d_1,\dots,d_n$ and $k_0$ be some positive constants, $u,v\in\,$\sR\
and $k\geq k_0$ be a parameter. Then the absolute value of the integral
$$
\int_{\R^n}d^n\tau\; K_{i(u-\tau_1)}(ik\,d_0)\;\sgn\,\tau_1\;
K_{i(\tau_1-\tau_2)}(ik\,d_1)\dots\sgn\,\tau_n\; K_{i(\tau_n-v)}(ik\,d_n)
\eqno(5.2)$$
can be estimated from above by the expression
$$
C\,k^{-1/2}\,(1+k^{-1/2}|v|)^n\,\exp\left({\pi\over 2}\,(|u|+|v|)
\right)\, ,
$$
where the constant $C>0$ depends only on $d_0,\dots,d_n$ and $k_0$. The
estimate is valid also for $n=0$ provided the integral is replaced by
$K_{i(u-v)}(ik\,d_0)$.
\noindent{\em Proof.}
Recalling (4.26) and the fact that the Fourier image of $\,\sgn\,\tau\,$ equals
\linebreak $-i(2/\pi)^{1/2}\,P(1/x)$, with $P$ indicating the regularization
in the sense of Cauchy principal value, one can rewrite (5.2) with the help
of the Fourier transform as
$$
{1\over 2}\,(-i)^n\int_{\R^{n+1}}d^{n+1}x\;\exp(iu\,x_0-iv\,x_n)\,
\exp\left(-ik\sum_{j=0}^n d_j\cosh x_j\right)\,\prod_{j=1}^n P{1\over x_{j-1}
-x_j}\, .
$$
The substitution $x_j=2\,\as(k^{-1/2}s_j)$ yields
\begin{eqnarray}
& & k^{-1/2}\,(-i)^n\,\exp\left(-ik\sum_{j=0}^n d_j\right)\int_{\R^{n+1}}
d^{n+1}s\;\nonumber\cr
& & \times\exp(2iu\,\as(k^{-1/2}s_0)-2iv\,\as(k^{-1/2}s_n))\,
\prod_{j=0}^n (1+k^{-1}\,s_j^{\,2})^{-1/2} \nonumber\cr
& & \times\prod_{j=1}^n\left\{k^{-1/2}(s_{j-1}-s_j)\,[\as(k^{-1/2}s_{j-1})-
\as(k^{-1/2}s_j)]^{-1}\right\}\,\nonumber\cr
& & \times\prod_{j=1}^n P{1\over s_{j-1}-s_j}\;\,
\exp\left(-2i\sum_{j=0}^n d_j\,s_j^{\,2}\right)\, .\nonumber\cr
\end{eqnarray}
Now it is possible to rotate the variables $s_j$, each in its own complex
plane, by the angle $-\pi/4$. As a result, the Gaussian term $\exp(-2\sum
d_j\,s_j^{\,2})$ appears in the integrand. To get rid of the singular terms,
perform another substitution $s_m=\sum_{j\leq m}t_j$ and recall that, in
the sense of distributions, $P(1/t)=d/dt\,\ln |t|$. Integration by parts
produces a polynomial of degree $n$ in the variable $k^{-1/2}v$. Since
$|{\rm Im}\;\as(e^{-i\pi/4}s)|<\pi/4$, for $s\in\,$\sR , it is not difficult
to see that the integrand can be estimated by $(1+k^{-1/2}|v|)^n\exp(\pi(
|u|+|v|)/2)\,F(t)$, with $F\in L^1($\sR$^{n+1})$ being independent of
$u,\,v$ and $k$. \QED
\proclaim Lemma 5.4.
In the limiting case $z=k^2-i0$, $\cD(\cJ\cD)^{4N-3}$ is a Hilbert--Schmidt
operator and its norm decays as $k^{-1/2}$ for $k\to+\infty$.
\noindent{\em Proof.}
First note that when evaluating the kernel of the composition $Q_1Q_2$, with
$Q_1$ and $Q_2$ being of type (4.10), one has to integrate along the curve
$z=\omega(\lambda),\ \lambda\in\,$\sR, in the complex plane. This concerns
also the operators $\tR_{\ver}(w),\ w=\sqrt{-z}=ik+0$. The corresponding
kernels are given by (4.1), for $\varepsilon=\varrho$, and by (4.12), for
$\varepsilon\not=\varrho$. From the form of (4.16) it is clear that when
evaluating the entries of the matrix of operators $\cD\cJ\cD$ one is
confronted with the expressions of the type
$$
\tR_{\varepsilon\mu}(\alpha_i;w)\,Y_d\,\tR_{-\mu,\varrho}(\alpha_j;w)\, ,
$$
where $d=|a_i-a_j|$. The operator $Y_d$ contributes to the integration by the
weight $\exp(kd\,\sinh z)$. Now it is easy to see that the integration
curve $z=\omega(\lambda),\ \lambda\in\,$\sR, can be deformed to $z=-i
{\pi\over2}+s,\ s\in\,$\sR. It may happen that while being deformed the curve
passes through a pole of the integrand and thus, by the Cauchy theorem,
additional summands appear in the final expression. This case occurs when
$\varepsilon=+,\, \mu=-$ and $\varrho=+$ but it is avoided when $\varepsilon$
is arbitrary, $\mu=+,\, \varrho=-$, and also when $\varepsilon=\mu=-$ and
$\varrho$ is arbitrary. Nevertheless, whatever case happens it is important
to observe that the kernels resulting in the process of deformation, namely
($\kappa,x\in\,$\sR\ are the variables)
\begin{eqnarray}
& \omega'(\kappa)^{1/2}\,\exp(-\alpha(\omega(\kappa)-x-i\varepsilon{\pi\over
2}))\,[1+\exp(-\omega(\kappa)+x+i\varepsilon({\pi\over2}-0))]^{-1}\, , &
\nonumber\cr \cr &
\omega'(\kappa)^{1/2}\,\check p_+(\omega(\kappa)-x+i{\pi\over2}),
\quad\hbox{and}\quad
\omega'(\kappa)^{1/2}\,(\omega(\kappa)-x+i{\pi\over2}+i0)^{-1}\, , &
\nonumber\cr
\end{eqnarray}
correspond to bounded operators with norms independent of $k$.
Next let us examine the equality (4.30) with $M=4N-2$. It is not difficult
to see that in a
finite sequence of couples of signs, $(\varepsilon,\mu_2)(-\mu_2,\mu_3)
\dots(-\mu_M,\varrho)$, there exists a subsequence of the type $(--)\,(+-)^m
\,(++)\,(-+)^n\,(--)(+\varrho')$, with $m,n\in\,$\sN$_0$. Consequently, each
summand on the RHS of (4.30) contains a subchain corresponding to this
subsequence and when evaluating the kernel of this composite operator one
can deform
integration curves in the complex plane as described above. The most decisive
part of the subsequence is $(++)\,(-+)^n\,(--)$. The role of the terms
standing to the left and to the right is to enable additional deformation of
the integration curves. Bearing in mind that the norm of $\Rj(z)$ does not
depend on $k$, we conclude that is is enough to show that the kernel
($x,y$ are the variables)
\begin{eqnarray}
& & \int_{\R^{n+1}}d^{n+1}s\;\frac{\exp(-\beta_0(x-s_1))}{1+\exp(-x+s_1)}\,
\nonumber\cr
& & \times\left(\prod_{j=1}^n\exp(-ik\,d_j\,\cosh s_j)\,
\frac{\exp(-\beta_j(s_j-s_{j+1}-i\pi))}{1+\exp(-s_j+s_{j+1}+i(\pi-0))}\,
\right) \nonumber\cr
& & \times\exp(-ik\,d_{n+1}\,\cosh s_{n+1})\,
\frac{\exp(-\beta_{n+1}(s_{n+1}-y))}{1+\exp(-s_{n+1}+y)}\, ,\nonumber\cr
\end{eqnarray}
where the $\beta$'s belong to $(0,1)$ and the $d$'s are positive, corresponds
to a Hilbert--Schmidt operator with a norm decaying as $k^{-1/2}$ for $k\to
+\infty$.
Applying the Fourier transformation and the relations (4.25) and (4.26), the
latter one with $u$ real but with $|x|$ replaced by $ik$, one obtains another
kernel in the variables $u,v\in\,$\sR,
\begin{eqnarray}
& & \pi\,\frac{1}{\sin\pi(\beta_0+iu)}\int_{\R^n}d^n\tau\;
K_{i(u-\tau_1)}(ik\,d_1)\,\frac{\exp(\pi\tau_1)}{\sin\pi(\beta_1+i\,\tau_1)}
\nonumber\cr
& & \times\left(\prod_{j=2}^n K_{i(\tau_{j-1}-\tau_j)}(ik\,d_j)\,
\frac{\exp(\pi\tau_j)}{\sin\pi(\beta_j+i\,\tau_j)}\,\right)\,
K_{i(\tau_n-v)}(ik\,d_{n+1})\,
\frac{1}{\sin\pi(\beta_{n+1}+iv)}\, .\nonumber\cr
\end{eqnarray}
Next decompose each multiplier $e^{\pi\tau}/\sin\pi(\beta+i\tau)\,$ as it was
done in (4.11) and write $1+\sgn\,\tau$ instead of $2\,\vartheta(\tau)$.
The identity
$$
\INT d\tau\;K_{i(u-\tau)}(ic_1)\,K_{i(\tau-v)}(ic_2)=\pi\,
K_{i(u-v)}(i(c_1+c_2))
$$
then implies that the resulting expression is a finite sum of kernels and
each of them, by the estimate given in Lemma 5.3, has a finite
Hilbert--Schmidt norm decaying as $k^{-1/2}$ for $k\to+\infty$. \QED
As a final auxiliary result we shall need another algebraic identity.
\proclaim Lemma 5.5.
For any $z\in\;$\sC$\,\backslash\, [0,+\infty )$, it holds
$$
\Omega\,\Xi^{-1}={1\over 2}\,e^{2\pi i\,\nu_N}\,[\Lambda^{-1}+\Lambda^{-1/2}\,
(\bef^t\otimes e_+^{\,t}\,\cD(\bI -\cD\cJ )^{-1}\;\bef\otimes e_-)\,
\Lambda^{-1/2}]\, ,
\eqno(5.3)$$
\noindent{\em Proof.}
The proof follows the same lines as the proof of Theorem 4.7 and so we omit
it. Note only that instead from (4.23a,b) one has to start from the adjoint
relations, namely
\begin{eqnarray}
& & E\,\Lambda^{-1/2}(R_{-,\varepsilon}-e^{2\pi i\alpha} R_{+,\varepsilon}
-(e^{2\pi i\alpha}-1)\,\bI )\,=0\, , \nonumber\cr \cr
& & E^c\Lambda^{-1/2}(R_{-,\varepsilon}-R_{+,\varepsilon})=0\, .
\qquad\qquad\QED \nonumber\cr
\end{eqnarray}
Let us now turn to the construction of generalized eigen--functions. We
assume that $z=k^2-i\sigma,\ \sigma>0$, and $k>k_0$ is large enough so that,
by Lemma 5.4, $(\bI-\cJ\cD)^{-1}$ exists also in the limit $\sigma\downarrow
0$. In the relation (3.6) perform the substitution $\varphi_+=(2\,{\rm Re}
\,\Lambda)^{-1/2}h$, with $h\in C_0^{\infty}((-k,k))$, to get
\begin{eqnarray}
f(z;\kappa ,u) & = &
(\exp(-\sqrt{\kappa^2-z}\,u)-\exp(-\sqrt{\kappa^2-\bar z}\,u))\,h(\kappa)
& (5.4) \nonumber\cr \cr
& + & \exp(-\sqrt{\kappa^2-\bar z}\,u)\,\Xi(\bar z)^{-1}
(\Lambda(\bar z)-\Lambda(z))\,h(\kappa)\, ,
\qquad\hbox{for}\ u>0\, ,& \nonumber\cr \cr
& = & \exp(\sqrt{\kappa^2-\bar z}\, u)\,\Omega\,\Xi (\bar z)^{-1}
(\Lambda(\bar z)-\Lambda(z))\,h(\kappa)\, ,
\qquad\hbox{for}\ u<0\, .& \nonumber\cr \cr
\end{eqnarray}
Redenote
\begin{flushright}
$\psi_+(z;x):=\,${\sc F}$^{-1}_{\kappa\to x_1}f(z;\kappa,x_2)\,$.
\hspace{120pt}(5.5)
\end{flushright}
Hence $\psi_+(z;x)=U(z)\varphi +U(\bar z)\tilde V(z)\varphi$ with
$\varphi=(\varphi_+,0)\in\,$\sC$^2\otimes$\LR ). Let us perform the
substitution $\kappa=k\,${\sc L}$(s)$ (c.f. (5.1)) in the Fourier integral
(5.5) and observe that the operator $A$ with the kernel $\langle s|A|\lambda
\rangle=k\,${\sc L}$'(s)\,\langle k\,${\sc L}$(s)|\Lambda^{-1/2}\Rj_{\ver}|
\lambda\rangle$ is bounded in \LR) even in the limiting case $\sigma=0$. On
the basis of identities (4.27), for $\Xi(\bar z)^{-1}$, and (5.3), for
$\Omega\,\Xi(\bar z)^{-1}$, one can conclude that the function
{\sc L}$'(\cdot)\,f(z;k\,${\sc L}$(\cdot),u)$ belongs to \LR), for all $u\in
\,$\sR. Owing to the rapid decay in the infinity, this function is
$L^1$--integrable, too. This is true also in the limit $\sigma\downarrow0$
and so $\psi_+(z;x)$, with $z=k^2-i0$, makes sense as well. To any density
$g\in C_0^\infty((k_0,+\infty))$ relate the integral
$$
F_\sigma(x):=\INTp dk\;g(k)\,\psi_+(k^2-i\sigma;x)\, .
\eqno(5.6)$$
Decomposing $\psi_+=\psi_1+\psi_2$, with
$$
\psi_2(x)={1\over\sqrt{2\pi}}\int_{|\kappa|>k} d\kappa\;e^{i\kappa\,x_1}\,
f(z;\kappa,x_2)\, ,
$$
and using the substitution $\kappa=k\,\sin\theta$, one can rewrite
\begin{eqnarray}
\INTp dk\;g(k)\,\psi_1(k^2-i\sigma;x) & = & {1\over\sqrt{2\pi}}
\int_{-\pi/2}^{\pi/2} d\theta\INTp dk
\sum_{\varepsilon=\pm}e^{ik(x_1\sin\theta+\varepsilon\,x_2
\cos\theta)} & \nonumber\cr
& \times & k\,\cos\theta\;g(k)\,\tilde f_\varepsilon(k^2-i\sigma;\theta)\, .
& (5.7) \nonumber\cr
\end{eqnarray}
The functions $\tilde f_\varepsilon(k^2-i\sigma;\theta)$ can be easily
extracted from (5.4). As $\psi_2\in\,$\LR$^2)$ and the RHS of (5.7) is a
Fourier--Plancherel integral rewritten in the polar coordinates $(k,\theta)$,
one finds that $F_\sigma$ is $L^2$--integrable. Moreover, $F_\sigma$
converges to $F_0$, as $\sigma\downarrow0$, in the Hilbert space \LR$^2)$.
Because
$$
\hat H(U(z)\varphi+U(\bar z)\,\tilde V(z)\varphi)=
z\,U(z)\varphi+\bar z\,U(\bar z)\,\tilde V(z)\varphi
$$
and \Ham is closed one obtains in the limit
$$
\hat H\,F_0(x)=\INTp dk\;k^2\,g(k)\,\psi_+(k^2-i0;x)\, .
\eqno(5.8)$$
Notice that, because of the definition of $\Lambda(z)$,
$f(k^2-i0;\kappa,u)$ is supported in $\{|\kappa|\leq k\}$. Set
$$
\psi_0(k;x):={1\over\sqrt{2\pi}}\int_{|\kappa|0$ and
$\theta\in(\pi,2\pi)$ for $x_2<0$, one obtains the asymptotics as $|x_2|\to
+\infty$,
\begin{eqnarray}
\psi_+(k^2-i0;x) & \simeq & \psi_0(k;x)\,\vartheta(x_2)
+ \int_0^\pi d\theta\;
\exp [ik(x_1\cos\theta +x_2\sin\theta)]\,S(\theta)\,
\vartheta(x_2) & \nonumber\cr
& + & \int_\pi^{2\pi} d\theta\;
\exp [ik(x_1\cos\theta +x_2\sin\theta)]\,S(\theta)\,
\vartheta(-x_2)\, . & (5.10)\nonumber\cr
\end{eqnarray}
$S(\theta)$ is some function belonging to $L^2((0,2\pi))$.
\proclaim Theorem 5.6.
The functions $\psi_+(k^2-i0;x)$ given in (5.5), with $k>k_0$ large enough,
are generalized eigen--functions of the Hamiltonian \Ham corresponding to the
value $k^2$. Moreover,
it holds $W_-\psi_0(k;x)=\psi_+(k^2-i0;x)$ in the
usual sense, i.e., for an arbitrary density $g\in C_0^\infty((k_0,+\infty))$
it is true
$$
W_-\,\left(\INTp dk\;g(k)\,\psi_0(k;x)\,\right)=
\INTp dk\;g(k)\,\psi_+(k^2-i0;x)\, .
\eqno(5.11)$$
\noindent{\em Proof.}
It remains to show that the limit $t\to-\infty$ of the expression
$$
\INTT d^2x\;\left|\INTp dk\;e^{-it\,k^2}\,g(k)\,(\psi_+(k^2-i0;x)-
\psi_0(k;x))\right|^2
$$
is zero. Compare (5.4), (5.5) with (5.9) to find
that one can approximate this integral with an error,
arbitrarily small and independent of $t$, by replacing $\psi_+(k^2-i0;x)-
\psi_0(k;x)$ by $\xi(z;x)=\,${\sc F}$^{-1}_{\kappa\to x_1}
\rho(z;\kappa,x_2)$, where
%\pagebreak
\begin{eqnarray}
\rho(z;\kappa,u) & = &
-\exp(-\sqrt{\kappa^2-\bar z}\,u)\,h(\kappa)
+ 2\,\exp(-\sqrt{\kappa^2-\bar z}\,u)\,\Xi(\bar z)^{-1}
\Lambda(\bar z)\,h(\kappa)\, ,\nonumber\cr \cr
& & \hspace{90mm}\qquad\hbox{for}\ u>0\, , \nonumber\cr \cr
& = & -\exp(\sqrt{\kappa^2-\bar z}\,u)\,h(\kappa)
+ 2\,\exp(\sqrt{\kappa^2-\bar z}\, u)\,\Omega\,\Xi (\bar z)^{-1}
\Lambda(\bar z)\,h(\kappa)\, , \nonumber\cr \cr
& & \hspace{90mm}\qquad\hbox{for}\ u<0\, , \nonumber\cr
\end{eqnarray}
$z=k^2-i\sigma$ and $\sigma>0$ is sufficiently small. Consequently, it is
enough to verify that the limit is zero after this replacement. By the
Lebesgue criterion, one can interchange the limit and the integration in the
variable $x_1$. Concerning the variable $x_2$, the stationary phase method is
applicable since $\xi(k^2-i\sigma;x)$ depends on $k$ smoothly (even
analytically). \QED
\begin{thebibliography}{article}
\bibitem[1]{AB} Aharonov A., Bohm D.: Phys. Rev. {\bf115}, 485 (1959)
\bibitem[2]{PLA89} \v S\v tov\'\i\v cek P.: Phys. Lett. A~{\bf142},
5--10 (1989)
\bibitem[3]{JMP} \v S\v tov\'\i\v cek P.: J. Math. Phys. {\bf32}, 2114--2122
(1991)
\bibitem[4]{Schulman} Schulman L. S.: Phys. Rev. {\bf176}, 1558 (1968)
\bibitem[5]{SMJI} Sato M., Miwa T., Jimbo M.: Holonomic quantum fields I,
Publ. RIMS, Kyoto Univ. {\bf14}, 223--267 (1978)
\bibitem[6]{SMJIII} Sato M., Miwa T., Jimbo M.: Holonomic quantum fields III,
Publ. RIMS, Kyoto Univ. {\bf15}, 577--629 (1979)
\bibitem[7]{PLA91} \v S\v tov\'\i\v cek P.: Phys. Lett. A~{\bf161},
13--20 (1991)
\bibitem[8]{Kato} Kato T.: Perturbation theory for linear operators.
Heidelberg: Springer 1966
\bibitem[9]{Triebel} Triebel H.: Interpolation theory, function spaces,
differential operators. Amsterdam, New York: North-Holland 1978
\bibitem[10]{SMJII} Sato M., Miwa T., Jimbo M.: Holonomic quantum fields II,
Publ. RIMS, Kyoto Univ. {\bf15}, 201--278 (1979)
\bibitem[11]{RSI} Reed M., Simon B.: Methods of modern mathematical physics I:
Functional analysis. New York: Academic Press 1972
\bibitem[12]{RSIII} Reed M., Simon B.: Methods of modern mathematical physics III:
Scattering theory. New York: Academic Press 1979
\bibitem[13]{Rui} Ruijsenaars S. N. M.: Ann. Phys. (NY) {\bf146}, 1 (1983)
\bibitem[14]{Hagen} Hagen C. R.: Phys. Rev. D {\bf41}, 2015 (1990)
\bibitem[15]{Jackiw} Jackiw R.: Ann. Phys. {\bf201}, 83--116 (1990)
\end{thebibliography}
\vskip 24pt
DEPARTMENT OF MATHEMATICS AND DOPPLER INSTITUTE,\\
FACULTY OF NUCLEAR SCIENCE, CTU,\\
TROJANOVA~13, 120~00~PRAGUE, CZECH REPUBLIC
\end{document}