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Trace formulae, spectral asymptotics, Stark operator, spectral shift function
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% TRACE FORMULAE AND HIGH ENERGY ASYMPTOTICS
% FOR THE PERTURBED THREE-DIMENSIONAL STARK OPERATOR
%
%E. KOROTYAEV and A. PUSHNITSKI
%
% Version 8 
% 9 January 2002
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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\begin{document}



\title{Trace formulae and high energy asymptotics
for the perturbed three-dimensional Stark operator}




\date{January 2002}



\author{E.~Korotyaev\thanks{
Institut f\"ur  Mathematik,  Humboldt Universit\"at zu Berlin,
Rudower Chaussee 25, 12489, Berlin, Germany.
e-mail: ek@mathematik.hu-berlin.de}\mbox{ }
and A.~Pushnitski\thanks{
Department of Mathematical Sciences, Loughborough University,  
Loughborough, LE11 3TU, U.K.
e-mail: a.b.pushnitski@lboro.ac.uk}}



\maketitle



\begin{abstract}
\noindent In $L^2(\R^3)$, we consider the unperturbed Stark operator $H_0$
(i.e., the Schr\"odinger operator with a linear potential)
and its perturbation $H=H_0+V$ by an infinitely smooth
compactly supported potential $V$.
The large energy asymptotic expansion for the modified
perturbation determinant for the pair $(H_0,H)$ is obtained
and explicit formulae for the coefficients in this expansion are given.
By a standard procedure, this expansion yields trace formulae
of the Buslaev--Faddeev type.
\end{abstract}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction and Main Results}
\label{sec.a}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\textbf{1. The Stark operator.}
In $L^2(\R^3,d\x)$ we consider the unperturbed and perturbed Stark operators
\begin{equation}
H_0=-\laplace+\inprod{\bb}{\x}
\text{ and }
H=H_0+V
\text{ with }
V=\overline{V}\in C_0^\infty(\R^3),
%..............................................................................................................................(a.1)
\label{a.1}
\end{equation}
where $\bb\in\R^3$, $\bb\not=0$ is a constant vector (the strength of the
external homogeneous electic field) and $\inprod{\cdot}{\cdot}$ 
is the inner product in $\R^3$.
The operators $H_0$ and $H$ are  self-adjoint on the same domain
and $C_0^\infty(\R^3)$ is a core for both $H_0$ and $H$.
The spectrum of both $H_0$ and $H$ is purely absolutely continuous
and coincides with $(-\infty,\infty)$ (see \cite{AH,H}).

\noindent \textbf{2. Notation.}
We denote $\C_+=\{z\in\C\mid \Im z>0\}$.
The variable $\l$ has the meaning of a spectral parameter
and varies over $\R$, whereas the variable $z$ usually varies over $\C_+$.
By $\SS_1$ and $\SS_2$ we denote respectively the trace class and the
Hilbert-Schmidt class of compact operators.
We use the standard notation for the Heaviside function:
$\theta(t)=(\abs{t}+t)/2$.
By $C$ we denote various constants whose value is not important for
our construction.
%Finally, $\Z_+=\{0,1,2\dots\}$.

\noindent \textbf{3. The modified perturbation determinant.}
The main objects of the present paper are the modified perturbation
determinant $\D_{H/H_0}(z)$, the spectral shift function $\xi(\l)$
for the pair \eqref{a.1}, and the harmonic conjugate $h(\l)$ of
$\xi(\l)$.
We start by defining the modified perturbation determinant:
\begin{equation}
\D_{H/H_0}(z)=\det\left(\frac{H-z}{H+i}\frac{H_0+i}{H_0-z}\right),
\quad z\in\C_+.
%..............................................................................................................................(a.2)
\label{a.2}
\end{equation}
\begin{lemma}
\label{l.a.1}
(i) For any  $z\in\C_+$,
\begin{equation}
(H-z)(H+i)^{-1}(H_0+i)(H_0-z)^{-1}-I=
(z+i)(H+i)^{-1}V(H_0-z)^{-1}\in\SS_1.
%..............................................................................................................................(a.3)
\label{a.3}
\end{equation}
Thus, the determinant \eqref{a.2} is well defined.

\noindent (ii)
The finction $\D_{H/H_0}(z)$ is analytic in $z\in\C_+$
and the limit $\D_{H/H_0}(\l+i0)$ exists for all $\l\in\R$.

\noindent (iii) The function $\D_{H/H_0}(\l+i0)$ is locally H\"older continuous with
any exponent $\g\in(0,1)$. I.e., for any $\g\in(0,1)$
and any $\l\in\R$,
there exists a constant $C>0$ such that for all $\mu\in(\l-1,\l+1)$,
$$
\abs{\D_{H/H_0}(\l+i0)-\D_{H/H_0}(\mu+i0)}\leq C\abs{\l-\mu}^\g.
$$

\noindent (iv) One has
\begin{equation}
\D_{H/H_0}(z)\not=0\quad \forall z\in\C_+
\quad\text{ and }\quad
\D_{H/H_0}(\l+i0)\not=0, \quad \forall \l\in\R.
%..............................................................................................................................(a.4)
\label{a.4}
\end{equation}
\end{lemma}
The proof is a straightforward application of
standard techniques of scattering theory.
For completeness, we give the proof in the Appendix.

By \eqref{a.4}, the analytic function
$\log\D_{H/H_0}(z)$
(for some choice of the branch of the logarithm) is well defined.
Changing the choice of the branch results
in adding a constant to $\log\D_{H/H_0}(z)$.
By \lma{a.1}(ii)-(iv),
the limit $\log\D_{H/H_0}(\l+i0)$ exists for all $\l\in\R$ and is a
H\"older continuous function of $\l$.
The derivative of $\log\D_{H/H_0}(z)$
admits a simple representation
\begin{equation}
\frac{d}{dz}\log\D_{H/H_0}(z)=\Tr((H_0-z)^{-1}-(H-z)^{-1}),\quad
z\in\C_+;
%..............................................................................................................................(a.4a)
\label{a.4a}
\end{equation}
this formula is valid for any pair of self-adjoint operators $H_0$, $H$
such that the inclusion \eqref{a.3} holds true.

\noindent \textbf{4. The case $\bb=0$.}
First we recall the large energy asymptotic
expansion for $\log\D_{H/H_0}(z)$ in the case $\bb=0$.
Consider the operators $T_0=-\Delta$ and $T=-\Delta+V$ in
$L^2(\R^3)$ (the potential $V$ is the same as in \eqref{a.1})
and denote
$$
\D_{T/T_0}(z)=
\det\left(\frac{T-z}{T+i}\frac{T_0+i}{T_0-z}\right).
$$
Similarly to \lma{a.1}, it is well known that 
the determinant $\D_{T/T_0}(z)$ is well defined,
$\D_{T/T_0}(z)\not=0$ for $z\in\C_+$
and $\D_{T/T_0}(z)$ is an analytic function of $z\in\C_+$.
Let us define the function $\log\D_{T/T_0}(z)$
for some choice of the branch of the logarithm.
We fix the branch of the square root of $z\in\C$, $\Im z\geq0$ by
\begin{equation}
\sqrt{z}=\sqrt{\abs{z}}e^{i\arg(z)/2},\quad
\arg(z)\in[0,\pi].
%..............................................................................................................................(a.5)
\label{a.5}
\end{equation}
It is well known that the following asymptotic expansion holds
true for $\log\D_{T/T_0}(z)$:
\begin{equation}
\log\D_{T/T_0}(z)\sim
iS_{-1}\sqrt{z}+
C+\sum_{k=0}^\infty\frac{iS_k}{z^k\sqrt{z}},
\quad \abs{z}\to\infty,\quad z\in\C_+.
%..............................................................................................................................(a.6a)
\label{a.6a}
\end{equation}
Here $C\in\C$ is a constant which depends on the choice of the branch
of the logarithm in the definition of $\log\D_{T/T_0}(z)$.
The coefficients $S_k$ are real numbers given by
\begin{equation}
S_k=\int_{\R^3}F_k[V(x)]dx,
%..............................................................................................................................(a.6b)
\label{a.6b}
\end{equation}
and $F_k[V]$ are polynomials (with real coefficients)
of $V$ and its derivatives.
Explicit formulae for $F_k$ are known; in particular,
\begin{equation}
F_{-1}[V]=\frac{1}{4\pi}V,\quad
F_0[V]=-\frac{1}{16\pi}V^2,\quad
F_1[V]=-\frac{1}{3\cdot 2^5\pi}V^3
-\frac{1}{3\cdot 2^6\pi}\abs{\nabla V}^2.
%..............................................................................................................................(a.6c)
\label{a.6c}
\end{equation}
The expansion \eqref{a.6a} has been proven in \cite{Buslaev1,Buslaev2}
(see also \cite{ColindV} and \cite{Guillope,Popov,Robert}).
The expansion \eqref{a.6a} can be considered as a consequence
of the asymptotic expansion of the integral kernel of $(T-z)^{-1}$
on the diagonal (see \eqref{a.4a}).
A system of recurrence relations for computing the polynomials
$F_k$ was given in \cite{Buslaev2}; explicit formulae
can be found in the recent paper \cite{HitrikPolterovich}.

\noindent \textbf{5. Asymptotic expansion for $\log\D_{H/H_0}(z)$.}
We return to the case $\bb\not=0$.
Central to our construction is the asymptotic expansion of
$\log\D_{H/H_0}(z)$ for $\abs{z}\to\infty$, $z\in\C_+$.
As above, we fix the branch of the square root by \eqref{a.5}.
\begin{theorem}
\label{t.a.2}
The following asymptotic expansion holds true:
\begin{equation}
\log\D_{H/H_0}(z)\sim
iQ_{-1}\sqrt{z}+
C+\sum_{k=0}^\infty\frac{iQ_k}{z^k\sqrt{z}},
\quad \abs{z}\to\infty,\quad z\in\C_+.
%..............................................................................................................................(a.6)
\label{a.6}
\end{equation}
Here $C\in\C$ is a constant which depends on the
choice of the branch of the logarithm in the definition of $\log\D_{H/H_0}(z)$.
The coefficients $Q_k$, $k=-1,0,1,2,\dots$ are real numbers given by
\begin{equation}
Q_k=\int_{\R^3}(F_k[\inprod{\bb}{\x}+V(\x)]-F_k[\inprod{\bb}{\x}])d\x,
%..............................................................................................................................(a.7)
\label{a.7}
\end{equation}
where $F_k$ are the polynomials from the expansion
\eqref{a.6a}, \eqref{a.6b}.
\end{theorem}
In particular, by \eqref{a.6c}, one has
\begin{equation*}
\begin{split}
Q_{-1}=&\frac{1}{4\pi}\int_{\R^3}V(\x)d\x,\\
Q_0=&-\frac{1}{16\pi}\int_{\R^3}V^2(\x)d\x
-\frac{1}{8\pi}\int_{\R^3}V(\x)\inprod{\bb}{\x}d\x,\\
Q_1=&-\frac{1}{3\cdot 2^5\pi}\int_{\R^3}V^3(\x)d\x-
\frac{1}{3\cdot 2^6\pi}\int_{\R^3}\abs{\nabla V(\x)}^2d\x
-\frac{1}{2^5\pi}\int_{\R^3} V^2(\x)\inprod{\bb}{\x}d\x\\
&-\frac{1}{2^5\pi}\int_{\R^3} V(\x)\inprod{\bb}{\x}^2d\x.
\end{split}
\end{equation*}


The important feature of the expansion \eqref{a.6}
is that it holds uniformly over $\arg(z)\in(0,\pi)$,
and thus also holds for the boundary values
$\log\D_{H/H_0}(\l+i0)$.


The coefficient $C$ in \eqref{a.6} is not important for
our construction;
essentially, we are working with the function
\begin{equation}
F(z):=\log\D_{H/H_0}(z)-C.
%..............................................................................................................................(a.7a)
\label{a.7a}
\end{equation}

We use the following method of proof of the expansion
\eqref{a.6}, \eqref{a.7}.
In Section \ref{sec.e}, using the method similar to
the one of \cite{ColindV}, we prove that the asymptotic expansion
\eqref{a.6} holds true \emph{with some coefficients} $Q_k\in\C$;
the approach of Section \ref{sec.e} does not give simple formulae
\eqref{a.7} for these coefficients.
Next, in Section \ref{sec.f}, using different technique, we prove that
\emph{in any sector} $\{z\in\C\mid \e<\arg z<\pi-\e\}$,
$\e>0$, the expansion \eqref{a.6} holds true with the coefficients
$Q_k$ given by \eqref{a.7}; the approach of Section \ref{sec.f}
does not allow us to prove the expansion \eqref{a.6} uniformly
over $\arg(z)\in(0,\pi)$.
Clearly, the combination of the two  above statements  leads to
\thm{a.2}.


\noindent \textbf{6. The spectral shift function.}
It appears that $\log\D_{H/H_0}(z)$ admits the integral representation
\begin{equation}
\log\D_{H/H_0}(z)=\int_{-\infty}^\infty\xi(\l)
\left(\frac{1}{\l-z}-\frac{1}{\l+i}\right)d\l,
%..............................................................................................................................(a.8)
\label{a.8}
\end{equation}
where $\xi$ is the spectral shift function for the pair \eqref{a.1}.
We refer to the survey paper \cite{BYa} or to the book \cite{Yafaev}
for exposition of the spectral shift function theory.
The spectral shift function is related to the scattering matrix of
the pair \eqref{a.1}; we postpone the discussion of this subject
till Section \ref{sec.d}.

By \eqref{a.8}, the spectral shift function can be determined by the formula
\begin{equation}
\xi(\l)=\frac1\pi\Im\log\D_{H/H_0}(\l+i0)-\frac1{2\pi}\Im\log\D_{H/H_0}(i),
\quad\l\in\R.
%..............................................................................................................................(a.9)
\label{a.9}
\end{equation}
The change of the branch of the logarithm results in
adding an integer constant to $\xi$.
In order to fix this constant, we need the following
\begin{lemma}
\label{l.a.3}
One has
\begin{equation}
\xi(\l)=n+O(\abs{\l}^{-N}),\quad \l\to-\infty,\quad \forall N>0,
%..............................................................................................................................(a.10)
\label{a.10}
\end{equation}
where $n$ is an integer, which depends
on the choice of the branch of the logarithm in \eqref{a.8}.
\end{lemma}
The proof is given in Section \ref{sec.d}.

Clearly, we can choose the branch of the logarithm so that
the constant $n$ in \eqref{a.10} equals zero.
Thus, $\xi$ is normalized by
\begin{equation}
\xi(\l)\to0,\quad \l\to-\infty.
%..............................................................................................................................(a.11)
\label{a.11}
\end{equation}
In terms of the function $F(z)$ (see \eqref{a.7a}) this implies
\begin{equation}
\xi(\l)=\frac1\pi\Im F(\l+i0),\qquad \l\in\R.
%..............................................................................................................................(a.11a)
\label{a.11a}
\end{equation}

The asymptotic expansion of \thm{a.2}
immediately yields the large energy asymptotic expansion for $\xi$:
\begin{align}
\xi(\l)&\sim\frac{1}{\pi}\sum_{k=-1}^\infty
\frac{Q_k}{\l^k\sqrt{\l}},\quad
\l\to+\infty,
%..............................................................................................................................(a.12)
\label{a.12}
\\
\xi(\l)&=O(\abs{\l}^{-N}),\quad \l\to-\infty,\quad \forall N>0.
%..............................................................................................................................(a.13)
\label{a.13}
\end{align}

\noindent \textbf{7. The harmonic conjugate $h$.}
Apart from $\xi(\l)$, the trace formulae for the pair \eqref{a.1}
involve a function $h(\l)$, which is a properly normalized
harmonic conjugate of $\xi(\l)$.
We define $h(\l)$ by  (compare with \eqref{a.11a})
$$
h(\l)=\frac1\pi \Re F(\l+i0),\qquad \l\in\R.
$$
By \lma{a.1}(iii),(iv), the function $h(\l)$ is
H\"older continuous.
The asymptotic expansion of \thm{a.2} yields
(compare with \eqref{a.12},\eqref{a.13})
\begin{align}
h(\l)&=O(\abs{\l}^{-N}),\qquad \lambda\to+\infty, \quad\forall N>0,
%..............................................................................................................................(a.14)
\label{a.14}
\\
h(\l)&\sim\frac{1}{\pi}\sum_{k=-1}^\infty
\frac{Q_k}{\l^k\sqrt{\abs{\l}}},\quad
\l\to-\infty.
%..............................................................................................................................(a.15)
\label{a.15}
\end{align}
\textbf{Remark}
One can define the spectral shift function
by using the Birman--Krein formula
(see the discussion in Section \ref{sec.d})
without any reference
to the perturbation determinant.
Then $h(\l)$ can also be defined without any reference to $\D_{H/H_0}(z)$
by the following two conditions:

(i) $h(\l)+i\xi(\l)$ is a boundary value of an analytic function in $\C_+$;

(ii) the following normalization condition holds true:
$h(\l)\to0$ as $\l\to+\infty$.
%\end{remark}

\noindent \textbf{8. Trace formulae.} Now that $\xi$ and $h$ have been defined,
we are ready to state the result on the trace formulae.
Let $Q_k$ be the coefficients from the expansion \eqref{a.6}.

\begin{theorem}
\label{t.a.4}
The following trace formulae hold true:
\begin{gather}
Q_{-1}=\int_0^\infty\frac{h(\l)-h(0)}{\l\sqrt{\l}}d\l-
\int_{-\infty}^0\frac{\xi(\l)-\xi(0)}{\abs{\l}\sqrt{\abs{\l}}}d\l,
%..............................................................................................................................(a.15a)
\label{a.15a}
\\
Q_n=\int_0^\infty h(\l)\l^{n-\frac12}d\l
+(-1)^n\int_{-\infty}^0\abs{\l}^{n-\frac12}\xi(\l)d\l,
\quad n=0,1,2,\dots
%..............................................................................................................................(a.16)
\label{a.16}
\\
\int_{-\infty}^\infty\biggl(\xi(\l)-\frac{1}{\pi}
\sum_{k=-1}^n\frac{Q_k}{\l^k\sqrt{\l}}\theta(\l)\biggr)\l^nd\l=0,
\quad n=1,2,\dots
%..............................................................................................................................(a.17)
\label{a.17}
\end{gather}
\end{theorem}
Note that by \eqref{a.12},  \eqref{a.13} and \eqref{a.14},
the above integrals converge absolutely.

Given the asymptotic expansion \eqref{a.6}, the proof
of the trace formulae \eqref{a.15a}--\eqref{a.17}
becomes a simple exercise in complex analysis
(see e.g. \cite{Buslaev2,ColindV}).
For completeness, we give this proof in Section~\ref{sec.b}.

Trace formulae for the case $\bb=0$  were obtained by
V.~S.~Buslaev and L.~D.~Faddeev in \cite{BuslaevFaddeev}
in the one-dimensional 
situation and by  V.~S.~Buslaev in \cite{Buslaev1,Buslaev2}
in the three-dimensional situation.
In the case $\bb=0$ the integrals over $(-\infty,0)$ can be replaced by the
sums of the appropriate powers of eigenvalues of $-\D+V$.

\noindent \textbf{9. Remark}
Fix any $a\in\R$. The asymptotic expansion \eqref{a.6} is equivalent
to the asymptotic expansion in powers of $(z-a)$:
\begin{equation}
\log\D_{H/H_0}(z)\sim
iQ_{-1}^{(a)}\sqrt{z-a}+
C+\sum_{k=0}^\infty\frac{iQ_k^{(a)}}{(z-a)^k\sqrt{z-a}},
\quad \abs{z}\to\infty,\quad z\in\C_+.
%..............................................................................................................................(a.18)
\label{a.18}
\end{equation}
The coefficients $Q_k^{(a)}$ are real and can be expressed
in terms of the coefficients $Q_k$.
For example,
$$
Q_{-1}^{(a)}=Q_{-1},\quad
Q_{0}^{(a)}=Q_{0}+\frac{a}{2}Q_{-1},\quad
Q_{1}^{(a)}=Q_{1}-\frac{a}{2}Q_{0}-\frac{a^2}{8}Q_{-1}.
$$
In the same way as the expansion \eqref{a.6} yields
the trace formulae \eqref{a.15a}--\eqref{a.17},
the expansion \eqref{a.18} yields
\begin{gather*}
Q_{-1}=\int_a^\infty\frac{h(\l)-h(a)}{(\l-a)\sqrt{\l-a}}d\l-
\int_{-\infty}^a\frac{\xi(\l)-\xi(a)}{\abs{\l-a}\sqrt{\abs{\l-a}}}d\l,
\\
Q_n^{(a)}=\int_a^\infty h(\l)(\l-a)^{n-\frac12}d\l
+(-1)^n\int_{-\infty}^a\abs{\l-a}^{n-\frac12}\xi(\l)d\l,
\quad n=0,1,2,\dots
\\
\int_{-\infty}^\infty\biggl(\xi(\l)-\frac{1}{\pi}
\sum_{k=-1}^n\frac{Q_k^{(a)}}{(\l-a)^k\sqrt{\l-a}}
\theta(\l-a)\biggr)(\l-a)^n
d\l=0,
\quad n=1,2,\dots
\end{gather*}
\textbf{10. The structure of the paper.}
In Section \ref{sec.b} we prove \thm{a.4}, assuming \thm{a.2}
and \lma{a.3}.
In Sections \ref{sec.e} and \ref{sec.f} we prove \thm{a.2}.
In Section \ref{sec.d} we recall the connection between the spectral shift
  function and the scattering matrix, and, using this connection,
prove \lma{a.3}.
In Appendix we prove \lma{a.1}.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of  Theorem \protect\ref{t.a.4}}
\label{sec.b}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\textbf{1. Proof of \eqref{a.15a}.}
Let $F$ be the function
defined by \eqref{a.7a}.
Then
$\frac1\pi F(\l+i0)=h(\l)+i\xi(\l)$ for all $\l\in\R$.
By \lma{a.1}(iii),(iv),  the function $F(\l+i0)$ is H\"older continuous.

Next, consider an auxiliary function
$$
L(z)=i\frac{F(z)-F(0)}{z\sqrt{z}}.
$$
The function $L$ is analytic in $\C_+$;
the asymptotic expansion of \thm{a.2} yields
\begin{equation}
L(z)=-\frac{Q_{-1}}{z}+O(\abs{z}^{-3/2}),\quad \abs{z}\to\infty,
\quad z\in\C_+.
%..............................................................................................................................(b.2)
\label{b.2}
\end{equation}
The boundary values $L(\l+i0)$ belong to $L^p(\R)$ for any
$p\in(1,2)$ (by the H\"older continuity of $F$).
Thus, $L$ belongs to the Hardy class $H^p(\C_+)$, $p\in(1,2)$ and
therefore can be represented as a Cauchy type integral
\begin{equation}
L(z)=\frac1\pi\int_{-\infty}^\infty\frac{\Im L(\l+i0)}{\l-z}d\l.
%..............................................................................................................................(b.3)
\label{b.3}
\end{equation}
 From here we get
\begin{equation}
L(z)=-\frac{1}{\pi z}\int_{-\infty}^\infty\Im L(\l+i0)d\l
+o(\abs{z}^{-1}),\quad \Im z\to+\infty,
\quad z\in\C_+.
%..............................................................................................................................(b.4)
\label{b.4}
\end{equation}
Comparing this to \eqref{b.2}, one obtains
\begin{multline*}
Q_{-1}=\frac1\pi\int_{-\infty}^\infty\Im L(\l+i0)d\l
=\int_{-\infty}^\infty\Im\left(i
\frac{h(\l)+i\xi(\l)-h(0)-i\xi(0)}{\l\sqrt{\l}}\right)d\l\\
=\int_0^\infty\frac{h(\l)-h(0)}{\l\sqrt{\l}}d\l-
\int_{-\infty}^0\frac{\xi(\l)-\xi(0)}{\abs{\l}\sqrt{\abs{\l}}}d\l.
\end{multline*}

\textbf{2. Proof of  \eqref{a.16}.}
Consider an auxiliary function
$$
L(z)=iz^{n-1}\sqrt{z} \left(F(z)-
\sum_{k=-1}^{n-1}\frac{iQ_k}{z^k\sqrt{z}}\right).
$$
The function $L$ is analytic in $\C_+$;
the asymptotic expansion of \thm{a.2} yields
\begin{equation}
L(z)=-\frac{Q_{n}}{z}+O(\abs{z}^{-2}),\quad \abs{z}\to\infty,
\quad z\in\C_+.
%..............................................................................................................................(b.5)
\label{b.5}
\end{equation}
The boundary values $L(\l+i0)$ belong to $L^p(\R)$ for any
$p>1$ if $n>0$ and for any $p\in(1,2)$ if $n=0$.
Thus, $L$ belongs to the Hardy class $H^p(\C_+)$,
the integral representation \eqref{b.3} holds true and we get \eqref{b.4}.
Comparing \eqref{b.4} and \eqref{b.5}, one obtains
$$
Q_n=\frac1\pi\int_{-\infty}^\infty
\Im\left[ iz^{n-1}\sqrt{z} \left(F(z)-
\sum_{k=-1}^{n-1}\frac{iQ_k}{z^k\sqrt{z}}\right)\right]
=\int_0^\infty h(\l)\l^{n-\frac12}d\l
+(-1)^n\int_{-\infty}^0\xi(\l)\abs{\l}^{n-\frac12}d\l.
$$

\noindent \textbf{3. Proof of  \eqref{a.17}.}
Consider an auxiliary function
$$
L(z)=z^n\left(F(z)-\sum_{k=-1}^n
\frac{iQ_k}{z^k\sqrt{z}}\right).
$$
The function $L$ is analytic in $\C_+$;
the asymptotic expansion of \thm{a.2} yields
\begin{equation}
L(z)=O(\abs{z}^{-3/2}),\quad \abs{z}\to\infty,
\quad z\in\C_+.
%..............................................................................................................................(b.6)
\label{b.6}
\end{equation}
The function $L$ belongs to the Hardy class $H^p(\C_+)$
for any $p\in[1,2)$.
As above, the integral representation \eqref{b.3}
and the asymptotic formula \eqref{b.4} hold true;
comparing \eqref{b.4} and \eqref{b.6}, we get
$$
0=\frac1\pi\int_{-\infty}^\infty\Im\left(\l^nF(\l)
-\l^n\sum_{k=-1}^n\frac{iQ_k}{\l^k\sqrt{\l}}\right)d\l
=\int_{-\infty}^\infty\left(\xi(\l)-\frac{1}{\pi}
\sum_{k=-1}^n\frac{Q_k}{\l^k\sqrt{\l}}\theta(\l)\right)\l^nd\l.
$$
This completes the proof. \qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of the expansion \protect\eqref{a.6}}
\label{sec.e}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section we prove that the expansion \eqref{a.6}
holds true with some coefficients $Q_k\in\C$.
Formulae \eqref{a.7} for these coefficients will be proven
in the next section.


\noindent \textbf{1. Notation}
Throughout the rest of the paper we use the following notation.
For the vector $\bb$ from the definition of $H_0$,
we denote $\b=\abs{\bb}$.
For a vector $\x\in\R^3$, we denote by $\xpar\in\R$ and $\xper\in\R^2$
the components of $\x$ parallel and orthogonal to
the direction of $\bb$.

We factorize the potential $V$ as follows:
\begin{equation}
V=W_1W_2,\quad W_1=\sqrt{\abs{V}},
\quad W_2=\sign(V)\sqrt{\abs{V}}.
%..............................................................................................................................(c.1)
\label{c.1}
\end{equation}
Our main objects in this section are the free resolvent
$R_0(z)=(H_0-z)^{-1}$ and the sandwiched resolvent $T(z)=W_1R_0(z)W_2$.

\noindent \textbf{2. Strategy}
Our reasoning is similar to that of \cite{ColindV} and is based 
on the following representation:
\begin{equation}
\log\D_{H/H_0}(z)=\log\D_{H/H_0}(i)+
\int_i^z\Tr\br{\frac{d}{dw}T(w)}dw+
\sum_{n=1}^\infty\frac{(-1)^n}{n+1}\Tr T^{n+1}(z)
%..............................................................................................................................(e.11)
\label{e.11}
\end{equation}
(below we will prove that \eqref{e.11} holds true for all 
$z\in\C$ with large enough modulus).
In order to derive \eqref{a.6} from \eqref{e.11},
we will need (i) estimates for the remainder of the series in the r.h.s.
of \eqref{e.11}; (ii) asymptotic expansion for large $\abs{z}$
of every term in the r.h.s. of \eqref{e.11}. 

\noindent \textbf{3. Estimates for $T(z)$}
We will need  estimates of the operator norm $\norm{T(z)}$
and the Hilbert-Schmidt norm $\norm{T(z)}_{\SS_2}$ of $T(z)$.


\begin{lemma}\cite{Korotyaev}
\label{l.d.1}
For any $\e\in(0,1/4)$,
$$
\norm{T(z)}\leq C(1+\abs{z})^{-\frac14+\e},
\qquad z\in\C_+,
$$
where the constant $C$ depends on $V$, $\b$, and $\e$,
but not on $z$.
\end{lemma}

\begin{lemma}
\label{l.e.1}
For any $z\in\C_+$, the sandwiched resolvent $T(z)$ belongs
to the Hilbert-Schmidt class and
\begin{equation}
\norm{T(z)}^2_{\SS_2}\leq
C_1\int_{\R^3}\int_{\R^3}
\frac{\abs{V(\x)V(\y)}}{\abs{\x-\y}^2}d\x d\y
+C_2\b^{2/3}\left(\int_{\R^3}\abs{V(x)}d\x\right)^2,
%..............................................................................................................................(e.1)
\label{e.1}
\end{equation}
where $C_1$ and $C_2$ are absolute constants.
\end{lemma}
(The actual expression in the r.h.s. of \eqref{e.1}
is of no importance to our construction;
it is essential that this expression does not depend on $z$).
\begin{proof}
For the resolvent $R_0(z)$ one has
$$
R_0(z)=i\int_0^\infty e^{izt}e^{-itH_0}dt.
$$
Using the Avron-Herbst formula for $e^{-itH_0}$ (see \cite{AH}),
one finds the integral kernel $R_0(z)[\x,\y]$ of $R_0(z)$:
\begin{equation}
R_0(z)[\x,\y]=\frac{e^{-i\pi/4}}{8\pi\sqrt{\pi}}\int_0^\infty
  t^{-3/2} e^{-i\b^2t^3/12}e^{izt}e^{-i(\xpar+y^\parallel)\b t/2}
e^{i\abs{\x-\y}^2/4t}dt.
%..............................................................................................................................(e.2)
\label{e.2}
\end{equation}
By the same pattern, for any $\eta\in\C_+$ one has
the following representation for the integral kernel of the
resolvent $(-\laplace-\eta)^{-1}$ in $L^2(\R^3)$:
\begin{equation}
\frac{e^{i\sqrt{\eta}\abs{\x-\y}}}{4\pi\abs{\x-\y}}
=\frac{e^{-i\pi/4}}{8\pi\sqrt{\pi}}\int_0^\infty
  t^{-3/2} e^{i\eta t}
e^{i\abs{\x-\y}^2/4t}dt.
%..............................................................................................................................(e.3)
\label{e.3}
\end{equation}
Denoting $\eta:=z-(\xpar+y^\parallel)\b /2$ and subtracting
\eqref{e.3} from \eqref{e.2}, we obtain:
$$
R_0(z)[\x,\y]=
\frac{e^{i\sqrt{\eta}\abs{\x-\y}}}{4\pi\abs{\x-\y}}+
\frac{e^{-i\pi/4}}{8\pi\sqrt{\pi}}\int_0^\infty
  t^{-3/2} (e^{-i\b^2t^3/12}-1)e^{i\eta t}
e^{i\abs{\x-\y}^2/4t}dt.
$$
Thus,
\begin{align*}
\abs{R_0(z)[\x,\y]}&\leq\frac1{4\pi\abs{\x-\y}}+
\frac{1}{8\pi\sqrt{\pi}}
\int_0^\infty
  t^{-3/2} \abs{e^{-i\b^2t^3/12}-1}dt
\\
&=\frac1{4\pi\abs{\x-\y}}+
\b^{1/3}\frac{1}{8\pi\sqrt{\pi}}
\int_0^\infty
  t^{-3/2} \abs{e^{-it^3/12}-1}dt.
\end{align*}
Now a direct estimate of the Hilbert-Schmidt norm of $T(z)$ 
yields \eqref{e.1}.
\end{proof}

\noindent \textbf{4. Asymptotics for $T(z)$.}
Below we use the notation
$T'(z):=\frac{d}{dz}T(z)=W_1R_0^2(z)W_2$.
It is easy to show  (see, e.g., the proof of  Theorem 4.2 in \cite{AH}) that
\begin{equation}
W_1(H_0-z)^{-1}\in\SS_2, \quad \forall z\in\C_+.
%..............................................................................................................................(c.2)
\label{c.2}
\end{equation}
 From here it follows that the operator $T'(z)$ belongs to the trace class.


\begin{lemma}
\label{l.e.2}
The following asymptotic expansion holds true:
\begin{equation}
\Tr T'(z)\sim
\sum_{k=0}^\infty\frac{ia_k}{z^k\sqrt{z}},
\quad \abs{z}\to\infty,\quad z\in\C_+,
%..............................................................................................................................(e.4)
\label{e.4}
\end{equation}
with some  coefficients $a_k\in\C$.
\end{lemma}

\begin{lemma}
\label{l.e.3}
For any $n=2,3,\dots$,
the following asymptotic expansion holds true:
\begin{equation}
\Tr (T(z))^n\sim
\sum_{k=n-2}^\infty\frac{ia^{(n)}_k}{z^k\sqrt{z}},
\quad \abs{z}\to\infty,\quad z\in\C_+,
%..............................................................................................................................(e.6)
\label{e.6}
\end{equation}
with some coefficients $a^{(n)}_k\in\C$.
\end{lemma}

We postpone the proof of Lemmas \ref{l.e.2} and \ref{l.e.3}
till the end of this section.

\noindent \textbf{5. Proof of the expansion \eqref{a.6}}
First note that a simple computation (see formula \eqref{c.7}
in the Appendix) yields
$$
\D_{H/H_0}(z)=\det(I+T(z))(I+T(-i))^{-1}.
$$
Next, by \lma{d.1}, one can choose $R>0$ big enough so that
\begin{equation}
\norm{T(z)}\leq 1/2\quad
\forall z\in\C_+,\quad \abs{z}>R.
%..............................................................................................................................(e.9)
\label{e.9}
\end{equation}
For such $z$, differentiating $\log\D_{H/H_0}(z)$
and expanding the result, we get
\begin{multline}
\frac{d}{dz}\log\D_{H/H_0}(z)=
\frac{d}{dz}\log\det(I+T(z))(I+T(-i))^{-1}
\\
=\Tr\bigl(T'(z)(I+T(z))^{-1}\bigr)
=\sum_{n=0}^\infty (-1)^n\Tr(T'(z)T^n(z)),
%..............................................................................................................................(e.10)
\label{e.10}
\end{multline}
where the series converge absolutely by \eqref{e.9}:
$$
\abs{\Tr T'(z) T^n(z)}\leq
\norm{T'(z)}_{\SS_1}\cdot 2^{-n}.
$$
Integrating \eqref{e.10}, one obtains
\eqref{e.11}.
Note that by \lma{d.1}, \lma{e.1} and \eqref{e.9}, one has
the following bound on the remainder of the series
in the r.h.s. of \eqref{e.11}:
\begin{multline*}
\bigl\lvert\sum_{n=N+1}^\infty
\frac{(-1)^n}{n+1}\Tr T^{n+1}(z)\bigr\rvert\leq
\norm{T(z)}^2_{\SS_2}\sum_{n=N+1}^\infty
\norm{T(z)}^{n-1}\leq 2
\norm{T(z)}^2_{\SS_2}\norm{T(z)}^{N}
\\
=O(\abs{z}^{-(\frac14-\e)N}),
\quad\forall \e\in(0,1/4),\quad N=0,1,2,\dots
\end{multline*}
Due to the last estimate, the required asymptotic expansion
\eqref{a.6} follows from the convergent series expansion \eqref{e.11}
and the asymptotic expansions given by \lma{e.2}
and \lma{e.3}.
\qed


\noindent \textbf{6. Proof of \lma{e.2}.}

\noindent \underline{1. Integral representation for $\Tr T'(z)$}
Differentiating \eqref{e.2} over $z$, we obtain the formula
for the integral kernel of $R_0^2(z)$:
$$
R^2_0(z)[\x,\y]=\frac{e^{i\pi/4}}{8\pi\sqrt{\pi}}\int_0^\infty
  t^{-1/2} e^{-i\b^2t^3/12}e^{izt}e^{-i(\xpar+y^\parallel)\b t/2}
e^{i\abs{\x-\y}^2/4t}dt.
$$
Now $\Tr T'(z)$ can be explicitly computed:
\begin{align}
\Tr T'(z)&=\int_{\R^3} V(\x) R^2_0(z)[\x,\x]d\x
=\frac{e^{i\pi/4}}{8\pi\sqrt{\pi}}\int_{\R^3}d\x\, V(\x)
\int_0^\infty dt\,
  t^{-1/2} e^{-i\b^2t^3/12}e^{izt}e^{-i\xpar\b t}
\notag\\
&=\frac{e^{i\pi/4}}{2\sqrt{2}}
\int_0^\infty\wh V(t\bb)e^{-i\b^2t^3/12}
e^{izt}t^{-1/2}dt,
%..............................................................................................................................(e.12)
\label{e.12}
\end{align}
where $\wh V$ is the (unitary) Fourier transform of $V$.

\noindent \underline{2. Stationary phase expansion}
Denote $g(t)=\wh V(t\bb)e^{-i\b^2t^3/12}$.
Due to our assumption  $V\in C_0^\infty(\R^3)$,
the function $g$ belongs to the Schwartz class on $\R$.
The standard stationary phase technique gives
the expansion
\begin{equation}
\int_0^\infty g(t) e^{izt}t^{-1/2}dt\sim
e^{i\pi/4}\sum_{k=0}^\infty
\frac{i^k\Gamma(k+\frac12)}{k!z^k\sqrt{z}}g^{(k)}(0),
\quad \abs{z}\to\infty,\quad z\in\C_+.
%..............................................................................................................................(e.13)
\label{e.13}
\end{equation}
By \eqref{e.12}, this yields the required expansion \eqref{e.4}.
\qed


\noindent \textbf{7. Proof of \lma{e.3}}

\noindent \underline{1. Strategy}
The strategy of the proof is the same as that of the proof of \lma{e.2},
but the technical details are more complicated.
We are going to prove the following integral representation:
\begin{equation}
\Tr T^n(z)=\int_0^\infty t^{-1/2}e^{izt}g(t)dt,
\quad
g(t)=t^{n-2}f(t),
%..............................................................................................................................(e.14)
\label{e.14}
\end{equation}
where $f$ is a Schwartz class function on $\R$.
The stationary phase method gives the expansion \eqref{e.13}.
As $g^{(k)}(0)=0$ for
$k=0,1,\dots,n-3$, the expansion actually starts from
the term with $k=n-2$, so we get \eqref{e.6}.

\noindent \underline{2. The kernel of $\Phi R_0(z) V\Phi^*$}
First note that by \eqref{c.2},
$(R_0(z)V)^n$ is a trace class operator for $n\geq2$, and so
$\Tr T^n(z)=\Tr(R_0(z)V)^n$.
Next,  $\Tr(R_0(z)V)^n=\Tr(\Phi R_0(z) V\Phi^*)^n$,
where $\Phi$ is the (unitary) Fourier transform in $L^2(\R^3)$.
Let us compute the integral kernel of the operator
$\Phi R_0(z)V\Phi^*$.
For the resolvent $R_0(z)$ one has (see \cite{AH}):

$$
R_0(z)=\Phi^* e^{i(\xpar)^3/3\b}
(i\b\frac{\partial}{\partial\xpar}+\abs{\xper}^2-z)^{-1}
e^{-i(\xpar)^3/3\b}\Phi.
$$
The integral kernel of the resolvent of
$i\b\frac{\partial}{\partial\xpar}$
can be easily computed.
This gives the following expression for
the integral kernel of $\Phi R_0(z)V\Phi^*$:
\begin{gather}
(\Phi R_0(z)V\Phi^*)[\x,\y]=i(2\pi)^{-3/2}
\int_0^\infty\wh V(\x-\y+\b t) e^{izt}e^{-i\phi_0},
%..............................................................................................................................(e.15)
\label{e.15}
\\
\phi_0=\frac{1}{3\b}(\xpar-\b t)^3-\frac{1}{3\b}(\xpar)^3
+t\abs{\xper}^2
=t\abs{\x}^2+\b t^2 \xpar+\b^2 t^3/3.
%..............................................................................................................................(e.16)
\label{e.16}
\end{gather}
Iterating the above kernel and computing trace, one obtains:
\begin{gather}
\begin{split}
\Tr(\Phi R_0(z)V\Phi^*)^n=&
i^n(2\pi)^{-3n/2}\ithree d\x_1
\biggl[\int_0^\infty dt_1\dots\int_0^\infty dt_n
\ithree d\x_2\dots \ithree d\x_n
\\
&\times e^{iz(t_1+\dots +t_n)}
\wh V(\x_1-\x_2+t_1\bb)\dots\wh V(\x_n-\x_1+t_n\bb)
e^{-i\phi}
\biggr],
\end{split}
%..............................................................................................................................(e.17)
\label{e.17}
\\
\phi=\sum_{j=1}^n t_j\abs{\x_j}^2+
\b\sum_{j=1}^n t_j^2\xpar_j+
(\b^2/3)\sum_{j=1}^n t_j^3.
%..............................................................................................................................(e.18)
\label{e.18}
\end{gather}
It is easy to see that the integral in the square brackets in
\eqref{e.17} converges absolutely.
The integral over $\x_1$ also converges absolutely
(as $(\Phi R_0(z)V\Phi^*)^n$ belongs to the trace class),
but we may not be able to legitimately interchange
the order of integration.
As we still would like to do this, we use the following
regularization:
$$
\Tr(\Phi R_0(z)V\Phi^*)^n=i^n(2\pi)^{-3n/2}
\lim_{R\to\infty}\int_{[-R,R]^3} d\x_1
\bigl[\dots\bigr],
$$
where $[\dots]$ stands for the integral
in the square brackets in \eqref{e.17}.

\noindent \underline{3. Change of variables}
Let us perform the following changes of variables
in the integral \eqref{e.17}.
First, instead of $t_1,\dots,t_n$,
we introduce the variables
$t_0,\o_1,\o_2,\dots,\o_{n-1}$,
%\begin{align*}
%t_0&=t_1+\dots +t_n,
%\\
%\o_j&=t_j/t_0,\quad j=1,2,\dots,n-1.
%\end{align*}
$$
t_0=t_1+\dots +t_n,
\quad
\o_j=t_j/t_0,\quad j=1,2,\dots,n-1.
$$
We also use the notation
$\o_n:=1-\o_1-\dots-\o_{n-1}$,
but $\o_n$ is treated as a function of the variables
$\o_1,\o_2,\dots,\o_{n-1}$, rather than as an independent variable.
We denote by $\o$ the vector
$\o=(\o_1,\dots,\o_{n-1})\in\R^{n-1}$.
The integral over $dt_1\dots dt_n$ then becomes
$$
\int_0^\infty dt_1\dots\int_0^\infty dt_n
=
\int_0^\infty dt_0\, t_0^{n-1}
\int_0^1d\o_1\int_0^{1-\o_1}d\o_2
\int_0^{1-\o_1-\o_2}d\o_3\dots
\int_0^{1-\o_1-\dots-\o_{n-2}}d\o_{n-1};
$$
we will use the shorthand notation
$\int_\O d\o$
for the above integral over $d\o_1\dots d\o_{n-1}$.

Next, instead of $\x_1,\dots,\x_n$, we introduce the variables
%\begin{align*}
%\x_0&=\x_1+\dots+\x_n,
%\\
%\y_j&=\x_j-\x_{j+1}+t_0\o_j\bb,\quad
%j=1,\dots,n-1.
%\end{align*}
$$
\x_0=\x_1+\dots+\x_n,
\qquad
\y_j=\x_j-\x_{j+1}+t_0\o_j\bb,\quad
j=1,\dots,n-1.
$$
We also use the notation
$y^\parallel=(y_1^\parallel,
y_2^\parallel,\dots,y_{n-1}^\parallel)
\in\R^{n-1}$,
$y^\perp=(y_1^\perp,
y_2^\perp,\dots,y_{n-1}^\perp)
\in\R^{2(n-1)}$,
$y=(\y_1,\y_2\dots,\y_{n-1})\in\R^{3(n-1)}$.

Expressing $\x_1,\dots,\x_n$ in terms of the new variables,
we get
\begin{align*}
\xpar_j&=\frac1n\xpar_0+l_j^\parallel(y^\parallel-\b t_0\o),
\quad j=1,\dots,n,
\\
\xper_j&=\frac1n\xper_0+l_j^\perp(y^\perp),
\quad j=1,\dots,n,
\end{align*}
where $l^\parallel_j:\R^{n-1}\to\R$ and
$l_j^\perp:\R^{2(n-1)}\to\R^2$ are some linear functions.
The integral over $d\x_1\dots d\x_n$ can be transformed as
$$
\int_{[-R,R]^3} d\x_1
\ithree d\x_2\dots \ithree d\x_n=
\frac1{n^3}\ithree d\y_1\dots\ithree d\y_{n-1}\int_{Q_R}d\x_0,
$$
where
$Q_R=[-nR,nR]^3-(nl_1^\parallel(y^\parallel-\b t_0\o),
nl_1^\perp(y^\perp))$.
Thus, we get
\begin{align}
\Tr&(\Phi R_0(z)V\Phi^*)^n=
i^n(2\pi)^{-3n/2}n^{-3}
\lim_{R\to\infty}\int_0^\infty dt_0\,
t_0^{n-1} e^{izt_0}\int_\O d\o
\notag\\
&\times
\ithree d\y_1\dots\ithree d\y_{n-1}
\wh V(\y_1)\dots\wh V(\y_{n-1})
\wh V(t_0\bb-\y_1-\dots-\y_{n-1})
\int_{Q_R}e^{-i\phi}d\x_0.
%..............................................................................................................................(e.19)
\label{e.19}
\end{align}

\noindent \underline{4. Computing $\phi$}
Expressing the phase function $\phi$
in terms of the new variables gives:
\begin{align*}
\phi=&t_0\sum_{j=1}^n \o_j(\tfrac1n\xpar_0
+l_j^\parallel(y^\parallel-\b t_0\o))^2
+t_0\sum_{j=1}^n \o_j\abs{\tfrac1n\xper_0+l_j^\perp(y^\perp)}^2
\\
&+\b t_0^2\sum_{j=1}^n \o_j^2(\tfrac1n \xpar_0
+l_j^\parallel(y^\parallel-\b t_0\o))
+\frac{\b^2}{3}t_0^3\sum_{j=1}^n \o_j^3.
\end{align*}
Expanding and completing squares, we get
\begin{equation}
\phi=\frac{t_0}{n^2}\abs{\x_0+
P(\b t_0,\o,y)}^2+\psi(t_0,\o,y),
%..............................................................................................................................(e.20)
\label{e.20}
\end{equation}
where $P$ and $\psi$ are polynomials.


\noindent \underline{5. Integrating over $\x_0$}
Using the formula
$$
\lim_{R\to\infty}\int_{-R}^R e^{-is^2}ds
=e^{-i\pi/4}\sqrt{\pi}
$$
and the expression \eqref{e.20} for
the phase function $\phi$, we obtain
$$
\lim_{R\to\infty}\int_{Q_R}
e^{-i\phi}d\x_0=
e^{-i3\pi/4}\left(\frac{n\sqrt{\pi}}{\sqrt{t_0}}\right)^3e^{-i\psi(t_0,\o,y)}.
$$
Due to the fast decay of the integrand in \eqref{e.19}
in the variables $t_0,\y_1,\dots,\y_{n-1}$,
one can pass to the limit as $R\to\infty$,
which gives the required integral representation \eqref{e.14}
with
\begin{equation}
\begin{split}
f(t)&=i^n\pi^{-3(n-1)/2}2^{-3n/2}e^{-i3\pi/4}f_0(t)
\\
f_0(t)&=\int_\O d\o
\ithree d\y_1\dots\ithree d\y_{n-1}
\wh V(\y_1)\dots\wh V(\y_{n-1})
\wh V(t_0\bb-\y_1-\dots-\y_{n-1})
e^{-i\psi(t,\o,y)}.
\end{split}
%..............................................................................................................................(e.21)
\label{e.21}
\end{equation}
As $\wh V$ belongs to the Schwartz class on $\R^3$,
the function $f_0$ belongs to the Schwartz class on $\R$.
\qed


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Formulae \protect\eqref{a.7} for the coefficients $Q_k$}
\label{sec.f}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
First we introduce the following notation.
For $\e>0$, denote $\C_\e=\{z\in\C\mid \e<\arg z<\pi-\e\}$.
Choose $N>0$ big enough so that the ball
$\{\x\in\R^3\mid \abs{\x}<N\}$ contains the support of $V$.
Let $G_0\in C_0^\infty(\R^3)$ be a function such that
$G_0(\x)=\inprod{\bb}{\x}$ for all $\x$, $\abs{\x}\leq N$.
Denote $G(\x)=\inprod{\bb}{\x}-G_0(\x)$.
Note that by definition, $Vg\equiv0$,
where $g$ is $G$ or any of the derivatives of $G$.

Next, consider the operators
$$
\wt H_0=-\D+G_0,\quad \wt H=-\D+G_0+V.
$$
Similarly to \eqref{a.2}, let us define the perturbation determinants
$\D_{\wt H_0/T_0}(z)$ and $\D_{\wt H/T_0}(z)$
(recall that we have denoted $T_0=-\D$).
The logarithms of these determinants admit asymptotic expansions
of the type \eqref{a.6a}, \eqref{a.6b} (with $G_0$ and $G_0+V$ respectively
instead of $V$).
Subtracting these expansions one from another, we get
\begin{equation}
\log\D_{\wt H/T_0}(z)-\log\D_{\wt H_0/T_0}(z)
\sim
iQ_{-1}\sqrt{z}+
C+\sum_{k=0}^\infty\frac{iQ_k}{z^k\sqrt{z}},
\quad \abs{z}\to\infty,\quad z\in\C_+,
%..............................................................................................................................(f.1)
\label{f.1}
\end{equation}
where the coefficients $Q_k$ are the same as in \eqref{a.7}:
$$
Q_k=\int_{\R^3}(F_k[G_0(\x)+V(\x)]-F_k[G_0(\x)])d\x=
\int_{\R^3}(F_k[\inprod{\bb}{\x}+V(\x)]-F_k[\inprod{\bb}{\x}])d\x.
$$

In this section, our aim is to prove that for any $\e>0$, any $n>0$ and
some constant $C\in\C$,
\begin{equation}
\log\D_{H/H_0}(z)-
(\log\D_{\wt H/T_0}(z)-\log\D_{\wt H_0/T_0}(z))
=C+O(\abs{z}^{-n}),
\quad \abs{z}\to\infty,\quad z\in\C_\e.
%..............................................................................................................................(f.2)
\label{f.2}
\end{equation}
 From \eqref{f.1} and \eqref{f.2} we get the expansion \eqref{a.6} with the
coefficients \eqref{a.7} in the sector $\C_\e$. As the coefficients of the
expansions of $\log\D_{H/H_0}(z)$ in $\C_+$ and $\C_\e$ coincide, we 
get formulae
\eqref{a.7} for the coefficients $Q_k$ of the expansion \eqref{a.6}.

Using \eqref{a.4a}, we see that the derivative of the l.h.s. of \eqref{f.2}
with respect to $z$ equals to
$$
\Tr\bigl[\bigl((H-z)^{-1}-(H_0-z)^{-1}\bigr)
-\bigl((\wt H-z)^{-1}-(\wt H_0-z)^{-1}\bigr)\bigr].
$$
Instead of proving \eqref{f.2}, we shall prove a stronger statement:
\begin{lemma}
\label{l.f.1}
For any $\e>0$ and any $n>0$,
\begin{equation}
\norm{\bigl((H-z)^{-1}-(H_0-z)^{-1}\bigr)
-\bigl((\wt H-z)^{-1}-(\wt H_0-z)^{-1}\bigr)}_{\SS_1}=
O(\abs{z}^{-n}),\quad
\abs{z}\to\infty,\quad z\in\C_\e.
%..............................................................................................................................(f.3)
\label{f.3}
\end{equation}
\end{lemma}

Before proving \lma{f.1}, let us prove an auxiliary statement.
Denote
$$
\wt R_0(z)=(\wt H_0-z)^{-1}.
$$
\begin{lemma}
\label{lma.f.2}
Let $\f_1,\f_2\in C_0^\infty(\R^3)$ and suppose that $\f_1\f_2\equiv0$.
Then for any $\e>0$ and any $n>0$,
\begin{equation}
\norm{\f_1\wt R_0(z)\f_2}_{\SS_2}=
O(\abs{z}^{-n}),\quad
\abs{z}\to\infty,\quad z\in\C_\e.
%..............................................................................................................................(f.4)
\label{f.4}
\end{equation}
  \end{lemma}
\begin{proof}
Throughout the proof, $\e>0$ is an arbitrary fixed number
and $z\in\C_\e$.
It is straightforward to see that $\f_1(T_0-z)^{-1}\f_2\in\SS_2$
and therefore
$\f_1\wt R_0(z)\f_2\in\SS_2$.
 From here one easily obtains that
$$
\norm{\f_1\wt R_0(z)\f_2}_{\SS_2}=
O(1),\quad
\abs{z}\to\infty.
$$
Next, we will prove \eqref{f.4} by induction in $n$
(more precisely, by induction in $2n$).
Assume that \eqref{f.4} holds true for some value of $n$.
One has:
\begin{multline}
\f_1\wt R_0(z)\f_2=\f_1(\wt R_0(z)\f_2-\f_2\wt R_0(z))
=\f_1\wt R_0(z)(\f_2\wt H_0-\wt H_0\f_2)\wt R_0(z)
\\
=\f_1\wt R_0(z)(\D \f_2)\wt R_0(z)
+\sum_{j=1}^3\f_1\wt R_0(z)
\left(\tfrac{\partial\f_2}{\partial x_j}\right)D_j\wt R_0(z),
%..............................................................................................................................(f.5)
\label{f.5}
\end{multline}
where $D_j=\tfrac{\partial}{\partial x_j}$.
It is easy to see that
\begin{equation}
\norm{D_j\wt R_0(z)}=O(\abs{z}^{-1/2}),\quad
\norm{\wt R_0(z)}=O(\abs{z}^{-1}),\quad
\text{ as }\abs{z}\to\infty
%..............................................................................................................................(f.5a)
\label{f.5a}
\end{equation}
and so \eqref{f.5} gives:
\begin{multline*}
\norm{\f_1\wt R_0(z)\f_2}_{\SS_2}\leq
\norm{\f_1\wt R_0(z)(\D\f_2)}_{\SS_2}O(\abs{z}^{-1})
\\
+\norm{\f_1\wt R_0(z)\left(\tfrac{\partial\f_2}{\partial x_j}\right)}_{\SS_2}
O(\abs{z}^{-1/2})=
O(\abs{z}^{-n-\frac12}),
\quad \abs{z}\to\infty.
\end{multline*}
This completes the proof.
\end{proof}

\noindent \emph{Proof of Lemma \ref{l.f.1}.}
Throughout the proof, $\e>0$ is an arbitrary fixed number
and $z\in\C_\e$.
As above, we use the notation $\wt R_0(z)=(\wt H_0-z)^{-1}$.

1. First recall  that (cf. \eqref{c.2})
\begin{equation}
\f R_0(z)\in\SS_2,\quad
\f \wt R_0(z)\in\SS_2,\quad \forall \f\in C_0^\infty(\R^3).
%..............................................................................................................................(f.5aa)
\label{f.5aa}
\end{equation}
By a standard argument, it follows that the Hilbert-Schmidt norms
of the above operators tend to zero as $\abs{z}\to\infty$.
Thus, one can choose $M>0$ big enough so that
\begin{equation}
\norm{VR_0(z)}_{\SS_2}\leq1/2,\quad
\norm{V\wt R_0(z)}_{\SS_2}\leq1/2,\quad
\forall z\in\C_\e,\quad \abs{z}\geq M.
%..............................................................................................................................(f.5b)
\label{f.5b}
\end{equation}

2.   By \eqref{f.5b}, for $\abs{z}\geq M$
the operator on l.h.s. of \eqref{f.3} can be presented as the
sum of convergent series:
\begin{multline}
\bigl((H-z)^{-1}-(H_0-z)^{-1}\bigr)
-\bigl((\wt H-z)^{-1}-(\wt H_0-z)^{-1}\bigr)\\
=\sum_{k=1}^\infty(-1)^k
R_0(z)(VR_0(z))^k-(-1)^k\wt R_0(z)(V\wt R_0(z))^k.
%..............................................................................................................................(f.6)
\label{f.6}
\end{multline}
First let us estimate the norm of the remainder term in \eqref{f.6}:
$$
\sum_{k=n+1}^\infty
\norm{R_0(z)(VR_0(z))^k}_{\SS_1}\leq
\norm{R_0(z)}\norm{VR_0(z)}_{\SS_2}^n=O(\abs{z}^{-n}),
\quad \abs{z}\to\infty.
$$
In the same way,
$$
\sum_{k=n+1}^\infty
\norm{\wt R_0(z)(V\wt R_0(z))^k}_{\SS_1}\leq
\norm{\wt R_0(z)}\norm{V\wt R_0(z)}_{\SS_2}^n=O(\abs{z}^{-n}),
\quad \abs{z}\to\infty.
$$
Thus, it suffices to prove that for any $k\in\N$ and
any $n>0$,
\begin{equation}
\norm{
R_0(z)(VR_0(z))^k-\wt R_0(z)(V\wt R_0(z))^k}_{\SS_1}=
O(\abs{z}^{-n}),\quad
\abs{z}\to\infty.
%..............................................................................................................................(f.6a)
\label{f.6a}
\end{equation}

3. One has
\begin{multline*}
R_0(z)(VR_0(z))^k-\wt R_0(z)(V\wt R_0(z))^k
=(R_0(z)-\wt R_0(z))(VR_0(z))^k
\\
+\wt R_0(z)V(R_0(z)-\wt R_0(z))(VR_0(z)))^{k-1}
+\dots+\wt R_0(z)(V\wt R_0(z))^{k-1}V(R_0(z)-\wt R_0(z)).
\end{multline*}
Thus, for $k\geq 2$ the relation \eqref{f.6a}
will follow if we prove that
\begin{equation}
\norm{V(R_0(z)-\wt R_0(z))}_{\SS_2}=O(\abs{z}^{-n}),\quad
\abs{z}\to\infty,\quad \forall n>0.
%..............................................................................................................................(f.7)
\label{f.7}
\end{equation}
For $k=1$, we use the following simple trick: find
$\f\in C_0^\infty(\R^3)$ such that $\f V=V$.
Then
\begin{multline*}
R_0(z)VR_0(z)-\wt R_0(z)V\wt R_0(z)
=(R_0(z)-\wt R_0(z))V\f R_0(z)
\\
+\wt R_0(z)\f V(R_0(z)-\wt R_0(z))
+\wt R_0(z)\f V(R_0(z)-\wt R_0(z)).
\end{multline*}
By \eqref{f.5aa}, we see that for $k=1$ the relation \eqref{f.6a}
will also follow from \eqref{f.7}.

4. Let us prove \eqref{f.7}.
One has
$$
V(R_0(z)-\wt R_0(z))=-V\wt R_0(z)GR_0(z).
$$
Performing the computation analogous to \eqref{f.5} twice, one obtains:
\begin{multline*}
V\wt R_0(z) G=V\wt R_0(z)(\D G)\wt R_0(z)+
\sum_{j=1}^3 V\wt R_0(z)\left(\tfrac{\partial}{\partial x_j}\D G\right)
D_j\wt R_0(z)
\\
+\sum_{j,l=1}^3 V\wt R_0(z)\left(\tfrac{\partial^2}{\partial 
x_j\partial x_l} G\right)
D_j\wt R_0(z)D_l\wt R_0(z).
\end{multline*}
Note that the functions $\D G$, $\tfrac{\partial}{\partial x_j}\D G$, and
  $\tfrac{\partial^2}{\partial x_j\partial x_l} G$
belong to the class $C_0^\infty(\R^3)$ and the support of either of them
is disjoint with the support of $V$. Recalling \eqref{f.4} and \eqref{f.5a},
we obtain that for any $n>0$,
$\norm{V\wt R_0(z) G}_{\SS_2}=O(\abs{z}^{-n})$.
This proves \eqref{f.7}.
\qed



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The scattering matrix}
\label{sec.d}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section we recall the connection between
the spectral shift function and the scattering matrix
and, using this connection, prove \lma{a.3}.


\textbf{1. The spectral representation for $H_0$.}
We will need some facts concering the spectral decomposition
of $H_0$, which we denote by $E_0(\l)\equiv\theta(\l-H_0)$.
First we recall formulae for $E_0(\l)$ due to \cite{AH}.
Let $f\mapsto \wt f$ be
the unitary transformation in $L^2(\R^3)$, which can be defined
on $L^1(\R^3)\cap L^2(\R^3)$ by the explicit formula
$$
\wt f(\p)=(2\pi)^{-1}\int_{\R^3}e^{-i\inprod{p^\perp}{\xper}}
\b^{1/3} \Ai(\b^{1/3}(\xpar-p^\parallel))f(\x)d\x,
$$
where $\Ai$ is the Airy function.
This unitary transformation carries over $H_0$ into multiplication by
$\b p^\parallel+\abs{p^\perp}^2$ in $L^2(\R^3,d\p)$:
\begin{equation}
(H_0f)\wt{\mbox{ }}(\p)=(\b p^\parallel+\abs{p^\perp}^2)\wt f(\p).
%..............................................................................................................................(c.3)
\label{c.3}
\end{equation}
Thus, for any $f\in L^2(\R^3)$, the quadratic form of $E_0(\l)$
can be presented as
\begin{equation}
(E_0(\l)f,f)=
\int_{\b p^\parallel+\abs{p^\perp}^2<\l}
\abs{\wt f(\p)}^2d\p
=\int_{\R^2}dp^\perp\int_{-\infty}^{(\l-\abs{p^\perp}^2)/\b}
\abs{\wt f(\p)}^2dp^\parallel.
%..............................................................................................................................(c.4)
\label{c.4}
\end{equation}
Differentiation with respect to $\l$ gives
$$
\frac{d}{d\l}(E_0(\l)f,f)=(J(\l)f,J(\l)f),
\quad f\in L^1(\R^3)\cap L^2(\R^3),
$$
where $J(\l):L^1(\R^3)\cap L^2(\R^3)\to  L^2(\R^2)$
is given by
\begin{equation}
(J(\l)f)(p^\perp)=
(2\pi)^{-1}\int_{\R^3}e^{-i\inprod{p^\perp}{\xper}}
\b^{-2/3} \Ai(\b^{1/3}\xpar+\b^{-2/3}\abs{p^\perp}^2
-\b^{-2/3}\l) f(\x)d\x.
%..............................................................................................................................(c.5)
\label{c.5}
\end{equation}
Recall (see \eqref{c.1}) that $W_1=\sqrt{\abs{V}}$.
\begin{lemma}
\label{l.c.1}
For any $\l\in\R$, the operator
$$
J(\l)W_1: L^2(\R^3)\to L^2(\R^2)
$$
belongs to the Hilbert-Schmidt class $\SS_2$ and is continuous
and differentiable in the $\SS_2$ norm.
One has
\begin{equation}
\norm{J(\l)W_1}_{\SS_2}=O(\abs{\l}^{-N}),
\quad \l\to-\infty,
\quad \forall N>0.
%..............................................................................................................................(c.6)
\label{c.6}
\end{equation}
\end{lemma}
\begin{proof}
The proof is a direct estimate of the Hilbert-Schmidt norm:
\begin{align*}
\norm{J(\l)W_1}_{\SS_2}^2=&
(2\pi)^{-2}\int_{\R^3}d\x\int_{\R^2}dp^\perp
\b^{-4/3}\abs{\Ai(\b^{1/3}\xpar+\b^{-2/3}
\abs{p^\perp}^2-\b^{-2/3}\l)}^2
\abs{V(\x)};
\\
\norm{\frac{d}{d\l}J(\l)W_1}_{\SS_2}^2=&
(2\pi)^{-2}\int_{\R^3}d\x\int_{\R^2}dp^\perp
\b^{-8/3}\abs{\Ai'(\b^{1/3}\xpar+\b^{-2/3}
\abs{p^\perp}^2-\b^{-2/3}\l)}^2
\abs{V(\x)}.
\end{align*}
Due to the superexponential decay of $\Ai(t)$ and $\Ai'(t)$
as $t\to+\infty$, the integrals converge and the estimate
\eqref{c.6} holds true.
\end{proof}



\textbf{2. The scattering matrix.}
The wave operators $W_\pm=W_\pm(H,H_0)$ for the pair
\eqref{a.1} exist and are complete (see \cite{AH,H})
and thus the scattering operator
$S=W_+^*W_-$ is well defined and unitary.
The operators $H_0$ and $S$ commute and thus
are simultaneously diagonalizable:
$$
L^2(\R^3)=\int_\R^\oplus\H(\l) d\l,\quad
H_0=\int_\R^\oplus\l I_\l d\l,\quad
S=\int_\R^\oplus S(\l)d\l;\quad
$$
here $I_\l$ is the identity in the fiber space $\H(\l)$ and
$S(\l)$ is the scattering matrix for the pair \eqref{a.1}.
According to the diagonalization formulae
\eqref{c.3}, \eqref{c.4}, all the
fiber spaces $\H(\l)$ coincide with  $L^2(\R^2)$.
The stationary representation for the scattering matrix
reads (see e.g. \cite{Yajima}):
\begin{equation}
S(\l)=I-2\pi i (J(\l)W_2)(I+T(\l+i0))^{-1}(J(\l)W_1)^*,
\quad \l\in\R;
%..............................................................................................................................(d.1)
\label{d.1}
\end{equation}
note that the operator $I+T(\l+i0)$ is invertible for all $\l\in\R$
(see the proof of \lma{a.1}(iv)).

By a standard argument (cf. the proof of  \lma{a.1}(iii)
in the Appendix), \lma{c.1} implies that $T(\l+i0)$
is continuous in $\l\in\R$ in the operator norm.
Therefore, we obtain that $S(\l)-I$ is
continuous in $\l$ in the trace norm.
Thus, the function $\det S(\l)$ is continuous in $\l$.

\textbf{3. The Birman-Krein formula.}
According to the Birman-Krein formula \cite{BK},
\begin{equation}
\det S(\l)=e^{-2\pi i\xi(\l)},
\quad \l\in\R.
%..............................................................................................................................(d.2)
\label{d.2}
\end{equation}
As both $\xi(\l)$ and $\det S(\l)$ are continuous in $\l$,
formula \eqref{d.2} determines $\xi(\l)$ up to a constant integer term.

 From \lma{d.1}, \lma{c.1} and the stationary representation
\eqref{d.1} we immediately obtain the estimate
$$
\norm{S(\l)-I_\l}_{\SS_1}=O(\abs{\l}^{-N}),
\quad \l\to-\infty,
\quad \forall N>0.
$$
Thus, by \eqref{d.2}, the spectral shift function $\xi(\l)$
satisfies \eqref{a.10}.
This proves \lma{a.3}.







%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Appendix: Proof of  Lemma \protect\ref{l.a.1}}
\label{sec.c}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The proof is a combination of standard results and techniques
of scattering theory for trace class and smooth perturbations.

1. Let us prove the statement (i).
The equality in \eqref{a.3} is a matter of direct calculation.
In order to  check that the operator in \eqref{a.3} belongs to the trace class,
first recall the inclusion \eqref{c.2}.
Applying the resolvent identity yields
$$
W_1(H-z)^{-1}\in\SS_2, \quad \forall z\in\C_+,
$$
which gives the required result.

2. Let us prove the statements (ii), (iii).
For $z\in\C_+$ one has
$$
T(z)-T(-i)=(z+i)W_1R_0(z)
R_0(-i)W_2\in\SS_1
$$
by \eqref{c.2}.
Denote for brevity $R(z)=(H-z)^{-1}$.
A straightforward computation yields
\begin{align}
\D_{H/H_0}(z)&=\det(I-(z+i)R(-i))(I+(z+i)R_0(z))\notag\\
&=\det[I+(z+i)(R_0(z)-R(-i)-(z+i)R(-i)R_0(z))]\notag\\
&=\det[I+(z+i)R(-i)VR_0(z)]
\quad\text{(using $\det(I+AB)=\det(I+BA)$)}\notag\\
&=\det[I+(z+i)W_1R_0(z)R(-i)W_2]
=\det[I+W_1(R_0(z)-R(-i)-R_0(z)VR(-i))W_2]\notag\\
&=\det(I+W_1R_0(z)W_2)(I-W_1R(-i)W_2)
=\det(I+T(z))(I+T(-i))^{-1}\notag\\
&=\det[I+(T(z)-T(-i))(I+T(-i))^{-1}].
%..............................................................................................................................(c.7)
\label{c.7}
\end{align}
The operator $T(z)-T(-i)$ is analytic in $z\in\C_+$,
which proves that $\D_{H/H_0}(z)$ is analytic in $\C_+$.
In order to prove the rest of the statements, below we to check that
for any finite open interval $\d\subset \R$ and any $\l\in\d$,
the limit $T(\l+i0)-T(-i)$ exists in the trace norm and
is locally H\"older continuous in $\l\in\d$ in the trace norm.

Let $E_0(\R\setminus\d)$ be the spectral projection of $H_0$,
associated to the complement of $\d$ in $\R$.
One has
\begin{equation}
T(z)-T(-i)=(z+i)\int_\d
\frac{(J(\l)W_1)^*(J(\l)W_2)}{(t+i)(t-z)}dt
+(z+i)W_1R_0(z)E_0(\R\setminus\d)R_0(-i)W_2,
%..............................................................................................................................(c.8)
\label{c.8}
\end{equation}
where $J(\l)$ is defined by \eqref{c.5}.
By \lma{c.1}, the integral in \eqref{c.8} has a limit as $z\to\l+i0$,
$\l\in\d$, and this limit is H\"older continuous (with any
H\"older exponent $\g\in(0,1)$).
The second term in the r.h.s. of \eqref{c.8} is smooth in $z\in\d$.


3. Let us prove the statement (iv).
Recall \eqref{c.7}:
$$\D_{H/H_0}(z)=\det(I+T(z))(I+T(-i))^{-1}.
$$
As $(I+T(-i))$ is bounded and invertible,
it is sufficient to check that $(-1)$ is not an eigenvalue of
$T(z)$ for $z\in\C_+$ or for $z=\l+i0$, $\l\in\R$.
If $T(z)\psi=-\psi$ for $\psi\in L^2(\R^3)$ and $z\in\C_+$,
then
$H\phi=z\phi$, where $\phi=R_0(z)W_2\psi\in L^2(\R^3)$.
This contradicts the self-adjointness of $H$.

4. Finally, let us check that $(-1)$ is not an eigenvalue of
$T(\l+i0)$ for  $\l\in\R$.
  Suppose that
\begin{equation}
T(\l+i0)\psi=-\psi\quad
\text{for } \psi\in L^2(\R^3)\quad
\text{and }\l\in\R.
%..............................................................................................................................(c.9)
\label{c.9}
\end{equation}
Let us use a standard argument (due to S.~Agmon) to show that
$\l$ is an eigenvalue of $H$.
It is sufficient to prove that
\begin{equation}
R_0(\l+i0)W_2\psi\in L^2(\R^3),
%..............................................................................................................................(c.10)
\label{c.10}
\end{equation}
because \eqref{c.9} and  \eqref{c.10} yield
$H(R_0(\l+i0)W_2\psi)=\l(R_0(\l+i0)W_2\psi)$.

In order to prove \eqref{c.10}, first take the inner product of
\eqref{c.9} with $W_2R_0(\l+i0)W_2\psi$
and then take the imaginary part of the resulting equation.
This gives $J(\l)W_2\psi=0$
(where $J(\l)$ is defined by \eqref{c.5}).
Next, let $\d=(\l-1,\l+1)$.
The norm of  $R_0(\l+i0)W_2\psi$
can be computed as follows:
$$
\norm{R_0(\l+i0)W_2\psi}^2=
\norm{E_0(\R\setminus\d)R_0(\l+i0)W_2\psi}^2
+\int_\d\frac{\norm{J(t)W_2\psi}^2}{\abs{t-\l}^2}dt.
$$
The first term in the r.h.s. is finite.
The second term can be estimated by using \lma{c.1}:
$$
\norm{J(t)W_2\psi}^2=\norm{(J(t)-J(\l))W_2\psi}^2
\leq C(\l)\abs{t-\l}^2\norm{\psi}^2,
$$
so the integral converges, the norm is finite,
and we get \eqref{c.10}.

Thus, $\l$ is an eigenvalue of $H$.
But, by a result of \cite{AH}, the
operator $H$ has no eigenvalues. Therefore, \eqref{c.9}
is impossible.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Acknowledgments}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The research was supported by the London Mathematical Society grant
under Scheme 4 ref. 4535 and by the SFB grant no. 288.
The second author is grateful to R.~Halburd for useful discussions.

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