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% Geometric Approach to Inverse Scattering for the Schroedinger Equation
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% Silke Arians
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\begin{document}

\parindent=0cm
\baselineskip=14pt
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\title{Geometric Approach to Inverse Scattering for
the Schr\"odinger Equation with Magnetic and Electric Potentials}
\author{Silke Arians\thanks{This research was supported by a grant from
Deutsche Forschungsgemeinschaft, hereby gratefully acknowledged.}\\
Institut f\"ur Reine und Angewandte Mathematik\\
RWTH Aachen\\
D-52056 Aachen, Germany\\
e-mail: silke@iram.rwth-aachen.de}
\date{}
\maketitle

\begin{abstract}
\noindent
We consider the Hamiltonian $H=(p-A(x))^2/(2m)+V(x)$ of a quantum particle
in a magnetic field $B=rot A$ and a potential $V$ in space dimensions
$\nu\ge 2$. If $V$ is of short range then the high velocity limit of the
scattering operator uniquely determines the magnetic field $B$ and the
potential $V$. If, in addition, long--range potentials $V^l$ are present,
some knowledge of (the far out tail of) $V^l$ is needed to define a modified
Dollard wave operator and a scattering operator $S^D$. Again its high velocity
limit uniquely determines $B$ and $V=V^s+V^l$. Moreover, we give explicit
error bounds which are inverse proportional to the velocity.
\end{abstract}


\section{Notations and Main Results}
\setcounter{equation}{0}


We study a 2--body quantum mechanical system in $\nu\ge 2$ dimensions. Here the
interactions are given by a vector potential $\A$ and a scalar potential
$V=V^s+V^l$.\\
$V^s$ is assumed to be short--range, i.e. $V^s\in{\cal{V}}_{SR}$ with
\begin{eqnarray*}
{\cal{V}}_{SR}& :=& \left\{V^s\,|\,V^s\mbox{ is Kato--small with relative
bound }<1,\right.\\
& & \left.\int_0^\infty dR\,\|V^s(\y)\,F(|\y|\ge R)\,(-\Delta+\E)^{-1}\|
<\infty\right\}.
\end{eqnarray*}
$V^l$ is assumed to be
long--range (i.e., $V^l\in C^1(\RR^\nu)$, $V^l$ tends to zero asymptotically
and $|\nabla V^l(\y)|\le \const\,(1+|\y|)^{-\gamma}$ with $\gamma>3/2$).
Each component of
the field strength tensor $F=d\A$ is Kato--small with relative bound $<1$, and
the decay for each component is assumed as
\beq\label{decayfij}
\|F_{ij}(\y)\,F(|\y|\ge R)\,(-\Delta+\E)^{-1}\|\le\const\,(1+R)^{-(2+\delta)},
\,\,\delta>0.
\eeq
In addition we assume
\beq\label{decaydifij}
\|\partial_i F_{ij}(\y)\,F(|\y|\ge R)\,(-\Delta+\E)^{-1}\|\le\const\,
(1+R)^{-(2+\tilde{\delta})},\,\, \tilde{\delta}>0.
\eeq
In $\nu=3$ dimensions the
vector $(\sum_{i=1}^{3}\partial_i F_{ij})_{j=1,2,3}=-\rot\B$ and up
to a constant equal to the current density. Each component of the vector
potential is supposed to be a continuous, short--range $L^2$--function with
short--range $\div\A$. For $\A_i$ and $F_{ij}$
a regularization with $(-\Delta+\E)^{-1/2}$ is sufficient.
We remark that for the proof of the reconstruction formulas the decay
assumptions on the vector potential can be weakened (see Generalizations).
By $H_0=\frac{1}{2m}\p^2$ we denote the free Hamiltonian and by
\begin{eqnarray*}
H & = & \frac{1}{2m}(\p-\A(\x))^2+V(\x) \\
& := & H_0-\frac{1}{m}\A(\x)\p+\frac{i}{2m}\div\A+\frac{1}{2m}|\A|^2(\x)+V(\x)
\end{eqnarray*}
the interacting Hamiltonian with reduced mass $m$ and momentum
$\p=-i\nabla_\x$. Then the interacting Hamiltonian is self--adjoint on
the domain ${\cal{D}}(H)={\cal{D}}(H_0)=W^{2,2}(\RR^\nu)$ in the Hilbert space
${\cal{H}}=L^2(\RR^\nu)$.\\
In the special case, when $V$ is of short range the wave operators
\beq \Omega_\pm:=\slim e^{iHt}e^{-iH_0 t} \eeq
exist and are complete (see, e.g., \cite[Theorem 4.1]{E 92} and
\cite[Theorem 3.7]{AD}). The scattering operator $S$ is
\beq S:=(\Omega_+)^*\Omega_-. \eeq
Let $\Phi_0\in{\cal{H}}$ be a normalized asymptotic configuration with compact
momentum support in the ball of radius $m\eta,\,\eta>0$, and momentum space
wave function $\hat{\phi}_0\in C_0^\infty(B_{m\eta}(0))$. We can find a
function $f\in C_0^\infty(\RR^{\nu})$ such that $\Phi_0=f(\p)\Phi_0$.
By $\Phi_\v$ we denote the boosted configuration translated
by $m\v$ in momentum space, where $\v$ is the velocity:
\beq\label{phiv}
\Phi_\v=e^{im\v\x}\Phi_0 \,\Leftrightarrow \,\hat{\phi}_\v(\p)=
\hat{\phi}_0(\p-m\v),
\eeq
which has compact velocity support in $B_\eta(\v)$. Let $v:=|\v|$ and
without loss of generality let $v\ge 16\eta$. We will
obtain asymptotics of $S$ for the high velocity limit in an arbitrary
direction $\hatv:=\v/v$, $v\rightarrow\infty$.

\begin{theorem}\label{th1}
Suppose that $V=V^s$.
Then the following formula for the asymptotics of $S$ holds for all $\Phi_\v,
\,\Psi_\v$ as in Eq. (\ref{phiv}):
\beq
\label{ffA} \lim_{v\to\infty}(S\Phi_\v,\Psi_\v)=
\left(e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}\Phi_0,\Psi_0\right)
.
\eeq
In particular, the scattering operator $S$ uniquely determines the magnetic
field $\B$.
\end{theorem}

Regarding $S$ as a mapping from the set of short--range potentials
${\cal{V}}_{SR}$ into the set of bounded operators $L({\cal{H}})$ we have
the following

\begin{theorem}\label{th2}
Suppose that $V=V^s$.
Then the following reconstruction formula holds for all
$\Phi_\v,\,\Psi_\v$ as in Eq. (\ref{phiv}):
\begin{eqnarray}
&& \lim_{v\to\infty}iv\left(\left(e^{-i\int_{-\infty}^\infty
\hatv\A(\x+\xi\hatv)\,d\xi}S-\E\right)\Phi_\v,\Psi_\v\right)
-(F(\B)\Phi_0,\Psi_0) \nn \\
& = & \label{ffVs} \int_{-\infty}^\infty (V^s(\x+\tau\hatv)\Phi_0,\Psi_0)\,
d\tau,
\end{eqnarray}
where $F(\B)$ is an explicit operator which is independent of $V^s$, see
(\ref{defK}) and (\ref{exactformula}).
In particular, for a given magnetic field $\B$ the scattering
map
\[ S(\B,\cdot):{\cal{V}}_{SR}\rightarrow L({\cal{H}}) \]
is injective.
\end{theorem}

For $V=V^s+V^l$, the modified Dollard wave operators
\beq
\Omega^D_\pm:=\slim e^{iHt}U^D(t,0)\,\,\mbox{ with }\,U^D(t,0)=e^{-iH_0 t}
e^{-i\int_0^t V^l(s\p/m)\,ds}
\eeq
exist and are complete (see, e.g., \cite[Theorem 4.1]{E 92} and
\cite[Theorem 3.14]{AD}).
\beq S^D:=(\Omega^D_+)^*\Omega^D_- \eeq
is the scattering operator $S^D$. Since the splitting of $V$ into a short--
and a long--range part is not unique the Dollard wave operators
are not uniquely defined. Nevertheless, the magnetic field and the scalar
potential are uniquely obtained from {\em any} of the scattering operators
$S^D$.

\begin{theorem}\label{th3}
The same formula for the asymptotics of $S^D$ as in the short--range case
holds for all $\Phi_\v,\,\Psi_\v$ as in Eq. (\ref{phiv}):
\beq
\label{ffAD} \lim_{v\to\infty}(S^D\Phi_\v,\Psi_\v)=
\left(e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}\Phi_0,
\Psi_0\right).
\eeq
In particular, the scattering operator $S^D$ uniquely determines the magnetic
field.
\end{theorem}

\begin{theorem}\label{th4}
The following reconstruction formula
holds for all $\Phi_\v,\,\Psi_\v$ as in Eq. (\ref{phiv}):
\begin{eqnarray}
&& \lim_{v\to\infty}iv\left(\left(e^{-i\int_{-\infty}^\infty
\hatv\A(\x+\xi\hatv)\,d\xi}S^D-\E\right)\Phi_\v,\Psi_\v\right)
-(F(\B,V^l)\Phi_0,\Psi_0)\nn\\
& = & \label{ffVsD} \int_{-\infty}^\infty (V^s(\x+\tau\hatv)\Phi_0,\Psi_0)\,
d\tau,
\end{eqnarray}
where $F(\B,V^l)$ is an explicit operator which is independent of $V^s$, see
(\ref{defK}) and (\ref{exactformulaD}).
In particular, for a given magnetic field $\B$ and a long--range potential
$V^l$ the scattering map
\[ S^D(\B,V^l,\cdot):{\cal{V}}_{SR}\rightarrow L({\cal{H}}) \]
is injective.\\
Moreover, the following formula holds for all $\Phi_\v,\,\Psi_\v$ as in Eq.
(\ref{phiv}):
\begin{eqnarray}
&& \lim_{v\to\infty}iv\left(\left[e^{-i\int_{-\infty}^\infty
\hatv\A(\x+\xi\hatv)\,d\xi}S^D,\p_1\right]\Phi_\v,\Psi_\v\right)
-(F(\B)\Phi_0,\Psi_0)\nn\\
& = & \int_{-\infty}^\infty \{(V^s(\x+\tau\hatv)\p_1\Phi_0,\Psi_0)
-(V^s(\x+\tau\hatv)\Phi_0,\p_1\Psi_0) \nn\\
& & \label{ffV} \mbox{}+i((\partial_1 V^l)(\x+\tau\hatv)\Phi_0,\Psi_0)\}\,
d\tau,
\end{eqnarray}
where $F(\B)$ is an explicit operator which is independent of $V$. In
particular, if $\B$ is given any
of the scattering operators $S^D$ uniquely determines the total potential
$V$.
\end{theorem}

All formulas hold in the high velocity limit. For large, but fixed, $v$ we
obtain an error term of order ${\cal{O}}(v^{-1})$, see (\ref{errorbound}) and
(\ref{errorboundD}) for details.

\begin{rem}
As each component of $\A$ is of short range, we can find a function $\lambda
\in C^2(\RR^\nu)$ tending to zero asymptotically with
$\B=\rot\A=\rot(\A+\nabla\lambda)$. Then $S^{(D)}$ and
$\exp(\pm\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi)$ are gauge
invariant under these gauge transformations and all formulas are independent
of the special choice of the gauge.
\end{rem}

Our proofs use a geometric, time--dependent method developed by Enss and
Weder.
This method has been used in \cite{EW 95} to prove Theorems \ref{th2} and
\ref{th4} in the case without magnetic field.\\To our knowledge the case with
magnetic fields was previously treated by Novikov and Khenkin in \cite{NK}.
Eskin and Ralston proved in \cite{ER} that the scattering
operator at fixed energy uniquely determines the magnetic field and the
scalar potential by stationary phase methods for exponentially decaying $\A$
and $V^s$. The results of Nicoleau in \cite{N 96} and \cite{N2 96} are
similar to our Theorems 1--4: By stationary methods he proves that the high
energy limit uniquely determines the magnetic field and the scalar potential,
where the potentials have to be $C^\infty$--functions with stronger decay
assumptions on higher derivatives, but not necessarily short--range vector
potential. In these papers singularities of the short--range part are not
included.


\section{Proof in the short--range case $V=V^s$}
\setcounter{equation}{0}


We first compute $\Omega_-\Phi_\v$ using the abbreviation
\begin{eqnarray}
I(\x,\p)& := & H-H_0 \nn \\
& = & V^s(\x)+\frac{i}{2m}\div \A+\frac{1}{2m}|\A|^2(\x)
-\frac{1}{m}\A(\x)\p
\end{eqnarray}
for the short--range interaction by the scalar and the vector potentials.
\begin{eqnarray}
\!\!\!\!\!\!\!\!& & \!\!\Omega_-\Phi_{\v} \nn\\
\!\!\!\!\!\!\!\!& = & \!\!\Phi_{\v}-\int_{-\infty}^0 dt\, \frac{d}{dt}\left(
e^{iHt}e^{-iH_0 t}\right)\Phi_{\v}\nn\\
\!\!\!\!\!\!\!\!& = & \!\!\Phi_{\v}-\int_{-\infty}^0 dt\, e^{iHt}i \,I(\x,\p)
\, e^{-iH_0 t}\Phi_{\v}\nn\\
\!\!\!\!\!\!\!\!& = & \!\!\label{integrand}e^{im\v\x}\Phi_{0}-\int_{-\infty}^0
dt\, e^{im\v\x}e^{i(H+(\p-\A(\x))\v)t}
i \,I(\x,\p+m\v)\,e^{-i(H_0+\p\v)t}\Phi_{0},
\end{eqnarray}
where the relation $e^{-im\v\x}f(\p)\,e^{im\v\x}=f(\p+m\v)$ is used. With the
abbreviations
\beq
\label{defH2barH1} {H}_2 := H/v+(\p-\A(\x))\hatv \,\,\mbox{ and }\,\,
H_1 := H_0/v+\p\hatv
\eeq
and the substitution $\tau=vt$ we obtain:
\[\Omega_-\Phi_{\v}=e^{im\v\x}\Omega_-({H}_2,H_1)\Phi_0.\]
If there is an integrable bound of the integrand in (\ref{integrand})
uniformly in $\v$ we can interchange
the high velocity limit and the integral by the dominated convergence theorem.
Since $\A,\,\div\A,\,|\A|^2$
and $V^s$ are of short range the existence of such an integrable bound can be
shown as in the case without magnetic field (see \cite[Lemma 2.2]{EW 95}).
We conclude
\begin{eqnarray}
\lim_{v\to\infty}e^{-im\v\x}\Omega_-\Phi_{\v} & = &
\Phi_0-\int_{-\infty}^0
d\tau \, e^{i(\p-\A(\x))\hatv\tau}(-i\A(\x)\hatv)e^{-i\p\hatv\tau}\Phi_0 \nn\\
& = & \lim_{\tau\to -\infty}e^{i(\p-\A(\x))\hatv\tau}e^{-i\p\hatv\tau}\Phi_0
\nn\\
& =:& \label{limvOmegam}\Omega_-((\p-\A(\x))\hatv,\p\hatv)\Phi_0,
\end{eqnarray}
where $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i {H}_2\tau)=
\exp(i(\p-\A(\x))\hatv\tau)$
and $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i H_1 \tau)=
\exp(i\p\hatv\tau)$ for fixed $\tau$ is used.
It is obvious that the same computation can be done for $\Omega_+$ (by
replacing $-\infty$ with $+\infty$). Analogously we obtain
\beq
\label{limvOmegap} \Omega_+((\p-\A(\x))\hatv,\p\hatv):=
\mathop{\mbox{\rm s-lim}}_{\tau\to +\infty}e^{i(\p-\A(\x))\hatv\tau}
e^{-i\p\hatv\tau}.
\eeq
We can prove that
\[ e^{i(\p-\A(\x))\hatv\tau} e^{-i\p\hatv\tau}
e^{i\int_{0}^{\tau}\hatv\A(\x+\xi\hatv)d\xi}=\E\]
holds for all $\tau\in\RR$ by showing that the derivatives $d/d\tau$ of both
sides are equal.
Consequently we get the following explicit expressions for (\ref{limvOmegam})
and (\ref{limvOmegap}):
\begin{eqnarray}
\Omega_-((\p-\A(\x))\hatv,\p\hatv)\Phi_0 &=& e^{+i\int_{-\infty}^{0}
\hatv\A(\x+\xi\hatv)d\xi}\Phi_0 \\
\Omega_+((\p-\A(\x))\hatv,\p\hatv)\Phi_0 &=& e^{-i\int_{0}^{\infty}
\hatv\A(\x+\xi\hatv)d\xi}\Phi_0.
\end{eqnarray}
Therefore the scattering operator $S$ is equal to
$\exp(i\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi)$ in the high velocity
limit $v\to\infty$ and we obtain formula (\ref{ffA}):
\[ \lim_{v\to\infty}(S\Phi_\v,\Psi_\v)=
\left(e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}\Phi_0,\Psi_0\right)
.\]
So far we have only used that $\div\A$ and the components of $\A$ are in
${\cal{V}}_{SR}$.\\
As $\A$ is continuous we infer $\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,
d\xi$ from above. This integral transform uniquely determines the magnetic
field $\rot\A=\B$ (see \cite[Lemma 3.4]{JP} for $\A\in (L^2(\RR^\nu))^\nu\cap
(C^0(\RR^\nu))^\nu$ and \cite[Lemma 3.4]{I} for $\A\in (C^2(\RR^\nu))^\nu$ and
more decay assumptions on $\A$). Hence Theorem \ref{th1} is proved. \qed

Now $\B$ is given and after choosing a short--range $\A$ with $\rot\A=\B$
especially $\Omega_\pm((\p-\A(\x))\hatv,\p\hatv)$ is known. Because of
(\ref{limvOmegam}) we are interested in the difference
\beq\label{diffOm}
\Omega_{\pm}({H}_2,H_1)-\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv).
\eeq
We need the following intertwining properties of
$\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv)$ in order to show that the norm of
this difference is of order ${\cal O}(v^{-1})$:
\beqa
& & \left(\p-\A(\x)\right)\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv) \nn\\
& = & \label{intom1}
\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv)\left(\p-\A(\x)-\nabla\int_0^{\pm\infty}
\hatv\A(\x+\xi\hatv)d\xi\right) \\
& = & \label{intom2}
\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv)\left(\p-\int_0^{\pm\infty}
(\hatv\times\B)(\x+\xi\hatv)d\xi\right),
\eeqa
and for finite $t\in\RR$:
\beqa
& & \left(\p-\A(\x)\right)e^{-i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi}
\nn\\
& = & \label{intom1f}
e^{-i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi}
\left(\p-\A(\x)-\nabla\int_0^{t}
\hatv\A(\x+\xi\hatv)d\xi\right) \\
& = & \label{intom2f}
e^{-i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi}\left(\p-\A(\x+t\hatv)-\int_0^{t}
(\hatv\times\B)(\x+\xi\hatv)d\xi\right).
\eeqa
Equalities (\ref{intom1}) and (\ref{intom1f}) can be proven by showing
\beqa
& & \frac{d}{ds}\left(e^{i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi\,s}
(\p-\A(\x))e^{-i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi\,s}\right) \nn\\
& = & \frac{d}{ds}\left(\p-\A(\x)-\nabla\int_0^{t}
\hatv\A(\x+\xi\hatv)d\xi\,s\right).\nn
\eeqa
An easy calculus gives $\nabla\int_0^{\pm\infty}
\hatv\A(\x+\xi\hatv)d\xi= \int_0^{\pm\infty}
(\hatv\times\B)(\x+\xi\hatv)d\xi-\A(\x)$, respectively $\nabla\int_0^{t}
\hatv\A(\x+\xi\hatv)d\xi= \int_0^{t}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)-\A(\x)$
and thus equalities (\ref{intom2}) and (\ref{intom2f}),
where we use the 3--dimensional notation $(\hatv\times\B)$ for all dimensions
with $(\hatv\times\B)$ as a notation for
\beq (\hatv\times\B)_i=\sum_{j=1}^{\nu}F_{ij}\hatv_j. \eeq
In the following we use the abbreviation
\beq\label{fh}
\hat{\F}_{\pm}(\x):=\int_0^{\pm\infty}(\hatv\times\B)(\x+\xi\hatv)d\xi.
\eeq


\begin{la}\label{Omegarest}
For ${H}_2$ and $H_1$ as defined in (\ref{defH2barH1})
\beq
\left\|\left\{\Omega_{\pm}({H}_2,H_1)-\Omega_{\pm}((\p-\A(\x))\hatv ,\p\hatv )
\right\}\Phi_0\right\|={\cal O}(v^{-1}).
\eeq
\end{la}

\bew
We formally compute the difference
\begin{eqnarray}
& & \Omega_{\pm}({H}_2,H_1)- \hat{\Omega}_{\pm}\nn\\
& = & \lim_{t\to\pm\infty} e^{iH_2 t}e^{-iH_1 t}-
e^{-i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi}\nn\\
& = & \lim_{t\to\pm\infty} e^{iH_2 t}\left\{\E-e^{-iH_2 t}
e^{-i\int_{0}^{t}\hatv\A(\x+\xi\hatv)d\xi}e^{iH_1 t}\right\}e^{-iH_1 t}\nn\\
& = & -\lim_{t\to\pm\infty} e^{iH_2 t}\int_0^t d\tau\,\frac{d}{d\tau}
\left\{e^{-iH_2 \tau}e^{-i\int_{0}^{\tau}\hatv\A(\x+\xi\hatv)d\xi}
e^{iH_1 \tau}\right\}e^{-iH_1 t}\nn\\
& = & \lim_{t\to\pm\infty} e^{iH_2 t}\int_0^t d\tau\,
e^{-iH_2 \tau}
i\left\{H_2 e^{-i\int_{0}^{\tau}\hatv\A(\x+\xi\hatv)d\xi}\right.\nn\\
& & \left.\mbox{}-
e^{-i\int_{0}^{\tau}\hatv\A(\x+\xi\hatv)d\xi}[H_1-\hatv\A(\x+\tau\hatv)]
\right\}e^{iH_1 \tau}e^{-iH_1 t}\nn\\
& = & \lim_{t\to\pm\infty} \int_0^t d\tau\,
e^{iH_2 \tau}
i\left\{H_2 e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)d\xi}\right.\nn\\
& & \left.\label{diffn}\mbox{}-
e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)d\xi}[H_1-\hatv\A(\x+(t-\tau)\hatv)]
\right\}e^{-iH_1 \tau}
\end{eqnarray}
Now we calculate the following difference with the help of (\ref{intom2f}):
\pagebreak

\beqa
\!\!\!\!\!\!& & \!\!\!\!\left\{H_2
e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)d\xi}-
e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)d\xi}[H_1-
\hatv\A(\x+(t-\tau)\hatv)]\right\}\nn\\
\!\!\!\!\!\!& = & \!\!\!\!v^{-1}\,e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)
d\xi}\left\{-\frac{1}{2m}\p\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+(t-\tau)\hatv)\right)\right. \nn\\
\!\!\!\!\!\!& & \!\!\!\!\mbox{}-\frac{1}{2m}\left(
\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+(t-\tau)\hatv)\right)\p \nn\\
\!\!\!\!\!\!& & \!\!\!\!\left.\mbox{}+\frac{1}{2m}\left(
\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+(t-\tau)\hatv)\right)^2+V^s(\x)\right\}
\nn\\
\!\!\!\!\!\!& \stackrel{t\to\pm\infty}{\approx} & \!\!\!\!v^{-1}\,
e^{-i\int_{0}^{\pm\infty}\hatv\A(\x+\xi\hatv)d\xi}\left(-\frac{1}{2m}\p
\hat{\F}_{\pm}(\x)-\frac{1}{2m}\hat{\F}_{\pm}(\x)\p+
\frac{1}{2m}\hat{\F}_{\pm}^2(\x)+V^s(\x)\right),\nn
\eeqa
where we used $(\hatv\times\B)\hatv=0$ which comes from the anti--symmetry of
the field strength tensor $F$.
Since $\|(1+\x^2)^{3/2}\Phi_0\|\le\const<\infty$ the norm of (\ref{diffn})
is bounded by
\beqa
& & v^{-1}\int_0^{t}d\tau \left\|\left\{-\frac{1}{2m}(\p-m\v)
\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)\right)\right.\right.\nn\\
& & \mbox{}-\frac{1}{2m}\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)\right)(\p-m\v)\nn\\
& & \left.\mbox{}+\frac{1}{2m}\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)\right)^2 +V^s(\x)\right\}\nn\\
& & \label{termFVs}\left.\mbox{ }e^{-iH_0\tau/v}f(\p-m\v)
(1+\x^2)^{-3/2}\right\|\\
& \stackrel{!}{\le}&  v^{-1}\int_0^{t}d\tau \,h(|\tau|).\nn
\eeqa
If there is such an integrable majorant $h\in L^1([0,\pm\infty),d\tau)$ for
the norm (\ref{termFVs}) uniformly in $\v$ we finished the proof.\\
The integrability of the summand with $V^s$ is proven in
\cite[Lemma 2.2]{EW 95}. We know that for a function
$\bar{f}\in C_0^\infty(\RR^\nu)$ with $\bar{f}\equiv 1$ on the support of $f$
$\|\int_0^{t-\tau}d\xi\,(\hatv\times\B)_i(\x+\xi\hatv)\,F(|\x-\tau\hatv|\le
|\tau|/4)\bar
{f}(\p)\|\le \const\, (1+|\tau|)^{-(1+\delta)}$, $\delta>0$, because of
\beqa
& & \int_0^{t-\tau}d\xi \left\|(\hatv\times\B)_i(\x+\xi\hatv)\,
F(|\x-\tau\hatv|\le |\tau|/4)\bar{f}(\p)\right\| \nn\\
& \le & \int_\tau^{t}d\xi \left\|(\hatv\times\B)_i(\x+\xi\hatv)\,
F(|\x|\le|\tau|/4)\bar{f}(\p)\right\| \nn\\
& \le & \int_\tau^{t}d\xi \left\|(\hatv\times\B)_i(\x)\,
F(|\x-\xi\hatv|\le |\xi|/4)\bar{f}(\p)\right\| \nn\\
& \le & \sum_{j=1}^{\nu}\int_\tau^{t}d\xi \left\|\hatv_j F_{ij}(\x)\,
F(|\x|\ge 3|\xi|/4)\bar{f}(\p)\right\| \nn\\
& \le & \label{srvb}\const \sum_{j=1}^{\nu}\int_\tau^{t}d\xi\,
(1+|\xi|/2)^{-(2+\delta)} \le \const\, (1+|\tau|)^{-(1+\delta)},
\eeqa
where we used the decay assumption on each component $F_{ij}$. Besides
\beqa
& & \left\|\A_i(\x+(t-\tau)\hatv)\,
F(|\x-\tau\hatv|\le |\tau|/4)\bar{f}(\p)\right\| \nn\\
& = & \left\|\A_i(\x)\,F(|\x-t\hatv|\le |\tau|/4)\bar{f}(\p)\right\| \nn\\
& \le & \label{sra} \left\|\A_i(\x)\,F(|\x|\ge 3|\tau|/4)\bar{f}(\p)\right\|
\le h_1(|\tau|),
\eeqa
where we used that each component of $\A$ is of short range. For the terms in
(\ref{termFVs}) with $\p-m\v$ on the left side we have to calculate the
commutator with $\p$ and infer $\sum_{i,j=1}^\nu \partial_i F_{ij}\hatv_j$ and
$\div\A$. In the same way as above we can show that
\beqa
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!& &  \label{srgvb}\left\|\int_0^{t-\tau}d\xi
\sum_{i,j=1}^\nu \partial_i F_{ij}\hatv_j(\x+\xi\hatv)\,
F(|\x-\tau\hatv|\le |\tau|/4)\bar{f}(\p)\right\|\le h_2(|\tau|)\mbox{ and}\\
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!& &
\label{srga}\left\|\div\A(\x+(t-\tau)\hatv)\,
F(|\x-\tau\hatv|\le |\tau|/4)\bar{f}(\p)\right\| \le h_3(|\tau|)
\eeqa
by using the decay assumptions on $\partial_i F_{ij}$ and $\div\A$. We
remark that in $\nu=3$ dimensions $\sum_{i,j=1}^3 \partial_i
F_{ij}\hatv_j=-\rot\rot\A\hatv=-\rot\B\hatv=-\mu_0\, \j\hatv$ with
permeability constant $\mu_0$ and current density $\j$.
Then the integrable bound can be shown as in
\cite[Proof of Lemma 2.2]{EW 95} by using
\beqa
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!& & \left\|F(|\x|\ge |\tau|/8)
(1+\x^2)^{-3/2}\right\|\le \const\, (1+|\tau|)^{-2}\mbox{ and}\\
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!& & \label{ppH0}\left\|F(|\x-\tau\hatv|
\ge |\tau|/4)e^{-iH_0 \tau/v}f(\p-m\v)
F(|\x|\le |\tau|/8)\right\|\le C_l\,(1+|\tau|)^{-l}
\eeqa
uniformly in $\v$ for $l=1,2,\dots$, which is a propagation property
expressing the fact that the solutions of the free Schr\"odinger equation
have rapid decay away from the classically forbidden region (see
\cite[Proposition 2.10]{E 83} or \cite[Proposition 2.1]{EW 95}). \qed

To get shorter formulas we define
\begin{eqnarray}
&&  \label{defK}K_\pm(\p,\x):=-\frac{1}{2m}\p\hat{\F}_{\pm}(\x)-
\frac{1}{2m}\hat{\F}_{\pm}(\x)\p+\frac{1}{2m}\hat{\F}_{\pm}^2(\x)
\end{eqnarray}
with $\hat{\F}_\pm$ as defined in (\ref{fh}). In particular the proof of Lemma
\ref{Omegarest} shows
\begin{eqnarray}
\!\!\!\!\!\!& & v\left\{\Omega_{\pm}({H}_2,H_1)-\Omega_{\pm}((\p-\A(\x))\hatv,
\p\hatv)\right\}\Phi_0\nn\\
\!\!\!\!\!\!& \stackrel{v\to\infty}{\approx} &\label{diffofomega}
i\int_0^{\pm\infty}
d\tau\,e^{-i\int_{0}^{\pm\infty}\hatv\A(\x+\xi\hatv)d\xi}\left(
V^s(\x+\tau\hatv)+K_\pm(\p,\x+\tau\hatv)\right)\Phi_0,
\end{eqnarray}
where $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i {H}_2\tau)=
\exp(i(\p-\A(\x))\hatv\tau)$
and $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i H_1\tau)=
\exp(i\p\hatv\tau)$ for fixed $\tau$ is used.
Hence we already know that
\begin{eqnarray*}
\Omega_{\pm}({H}_2,H_1)-\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv)=:R_{\pm}
={\cal{O}}(v^{-1})
\end{eqnarray*}
and $\mathop{\mbox{\rm s-lim}}_{v\to\infty}v R_{\pm}$ is explicitly known. Now
we complete the proof of the reconstruction formula for $V^{s}$ and calculate
the following limit:
%\pagebreak

\begin{eqnarray}
& & \lim_{v\to\infty}iv\left(\left(S-e^{i\int_{-\infty}^{\infty}
\hatv\A(\x+\xi\hatv)d\xi}\right)
\Phi_{\v},\Psi_{\v}\right) \nn\\
& = & \lim_{v\to\infty}\left\{iv\left(S\Phi_{\v},\Psi_{\v}\right)-iv\left(
e^{i\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi}\Phi_{\v},\Psi_{\v}\right)
\right\}\nn\\
& = & \lim_{v\to\infty}\left\{iv(R_-\Phi_0,\Omega_{+}((\p-\A(\x))\hatv,\p\hatv)
\Psi_0)\right.\nn\\
& & \mbox{}+iv(\Omega_{-}((\p-\A(\x))\hatv,\p\hatv)\Phi_0,R_+\Psi_0)\nn\\
& & \label{error}\mbox{}+iv(R_{-}\Phi_0,R_+\Psi_0)\\
& & \mbox{}+iv\underbrace{(\Omega_{-}((\p-\A(\x))\hatv,\p\hatv)\Phi_0,
\Omega_{+}((\p-\A(\x))\hatv,\p\hatv)\Psi_0)}
_{=\,\left(e^{i\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi}\Phi_{0},
\Psi_{0}\right)}\nn\\
& & \left.\mbox{}-iv\left(e^{i\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi}
\Phi_{0},\Psi_{0}\right)\right\}\nn\\
& = & \lim_{v\to\infty}\left\{iv\left(R_-\Phi_0,e^{-i\int_{0}^{\infty}
\hatv\A(\x+\xi\hatv)d\xi}\Psi_{0}\right)
+ iv\left(e^{i\int_{-\infty}^{0}\hatv\A(\x+\xi\hatv)d\xi}\Phi_{0},R_+\Psi_0
\right)\right\}.\nn
\end{eqnarray}
We subtract the terms of $R_\pm$ which only depend on $\B$, multiply in the
scalar product with $\exp(-i\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi)$
and obtain
\begin{eqnarray}
\!\!\!&& \!\!\!\!\lim_{v\to\infty}iv\left(e^{-i\int_{-\infty}^\infty
\hatv\A(\x+\xi\hatv)\,
d\xi}\left(S-e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}\right)
\Phi_\v,\Psi_\v\right) \nn\\
\!\!\!&& \!\!\!\!\mbox{}-\int_{0}^\infty
\left(e^{-i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}
K_+(\p,\x+\tau\hatv)e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}
\Phi_0,\Psi_0\right)\,d\tau\nn\\
\!\!\!&& \!\!\!\!\label{exactformula} \mbox{}-\int_{-\infty}^0
(K_-(\p,\x+\tau\hatv)\Phi_0,\Psi_0)\,d\tau \\
\!\!\!& = & \!\!\!\!\label{formelVs}\int_{-\infty}^\infty (V^s(\x+\tau\hatv)
\Phi_0,\Psi_0)\,d\tau,
\end{eqnarray}
which is the desired reconstruction formula in Theorem \ref{th2}. The error
term can be calculated from (\ref{error}) and is bounded by
\beq \label{errorbound}
v^{-1} \|v R_-\Phi_0\|\,\|v R_+\Psi_0\|.
\eeq
Therefore the error bounds are of order ${\cal{O}}(v^{-1})$.\\
$S$ is known and $\B$ is known. Then the terms only
depending on $\B$ can be computed and (\ref{formelVs}) is known for all
$\Phi_0$, $\Psi_0$ in a dense set.
The injectivity results from the fact that the reconstruction of the potential
$V^s$ can be reduced to the inversion of the Radon transform of a bounded,
continuous, square integrable function on $\RR^2$ (see
\cite[Proof of Theorem 1.1]{EW 95}) which uniquely determines the function
(see \cite[Chapter I, Theorem 2.17]{SH 84}).\\
Hence Theorem \ref{th2} is proved. \qed


\section{Proof in the long--range case $V=V^s+V^l$}
\setcounter{equation}{0}


Now $H=(\p-\A(\x))/(2m)+V(\x)$ with $V=V^s+V^l$. We use the abbreviation
\beq I^l(\x,\p,t) :=  V^l(\x)-V^l(t\p/m) \eeq
for the long--range interaction and compute $\Omega^D_-\Phi_\v$:
\begin{eqnarray}
& & \Omega^D_-\Phi_{\v} \nn\\
& = & \Phi_{\v}-\int_{-\infty}^0 dt\, \frac{d}{dt}\left(e^{iHt}U^D(t)\right)
\Phi_{\v}\nn\\
& = & e^{im\v\x}\Phi_{0}-\int_{-\infty}^0 dt\, e^{im\v\x}
e^{i(H+(\p-\A(\x))\v)t}
i \,\left\{I(\x,\p+m\v)+I^l(\x,\p+m\v,t)\right\} \nn\\
& & \label{intD}\mbox{ }e^{-i\left(H_0 t+\p\v t+\int_0^t V^l(s(\p+m\v)/m)ds
\right)}\Phi_{0}.
\end{eqnarray}
With the abbreviations
\beq
\label{defH2barH1D} {H}_2 := H/v+(\p-\A(\x))\hatv \,\,\mbox{ and }\,\,
H^D_1(\tau) := H_0/v+\p\hatv +v^{-1}V^l\left(\frac{p+m\v}{m}\frac{\tau}
{v}\right),
\eeq
where $H^D_1(\tau)$ generates
\beq\label{defbU}
\bar{U}^D(\tau/v,0):=\exp\left\{-i\left(H_0 \tau/v+\p\hatv\tau
+v^{-1}\int_0^\tau V^l\left(\frac{\p+m\v}{m}\frac{s}{v}\right)ds\right)\right\}
,\eeq
we obtain the following relation:
\[\Omega^D_-\Phi_{\v}=e^{im\v\x}\Omega_-({H}_2,H^D_1)\Phi_0.\]
If there is an integrable bound of the integrand in (\ref{intD})
uniformly in $\v$ we can interchange
the high velocity limit and the integral. Since $\A,\,\div\A,\,|\A|^2$ and
$V^s$ are of short range the existence of such an integrable bound
for $I$ can be shown as in the case without magnetic field (see
\cite[Lemma 3.2]{EW 95}). The integrable bound for $I^l$ is given in
\cite[Lemma 3.3]{EW 95}. Similar to the short--range case we conclude
\begin{eqnarray}
\lim_{v\to\infty}e^{-im\v\x}\Omega^D_-\Phi_{\v} & = &
\Omega_-((\p-\A(\x))\hatv,\p\hatv)\Phi_0,
\end{eqnarray}
where $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i {H}_2\tau)=
\exp(i(\p-\A(\x))\hatv\tau)$
and $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i H^D_1(\tau))=
\exp(i\p\hatv\tau)$ for fixed $\tau$ is used.
It is obvious that the same computation can be done for $\Omega^D_+$ (by
replacing $-\infty$ with $+\infty$).\\
Then the scattering operator $S^D$ yields the same high velocity limit as in
the short--range case, that is $S^D\stackrel{v\to\infty}{\approx}
\exp(i\int_{-\infty}^{\infty}\hatv\A(\x+\xi\hatv)d\xi)$, and we obtain formula
(\ref{ffAD}):
\[ \lim_{v\to\infty}(S^D\Phi_\v,\Psi_\v)=
\left(e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}\Phi_0,\Psi_0\right)
.\]
So far we have only used that $\div\A$ and the components of $\A$ are in
${\cal{V}}_{SR}$.\\
For $\A\in (L^2(\RR^\nu))^\nu\cap (C^0(\RR^\nu))^\nu$
formula (\ref{ffAD}) uniquely determines the magnetic field $\B$, which has
been noted above. Hence Theorem \ref{th3} is proved. \qed

Now $\B$ is given and after choosing a short--range $\A$ with $\rot\A=\B$
especially $\Omega_\pm((\p-\A(\x))\hatv,\p\hatv)$ is known.
We are again interested in the difference
\beq
\label{diffOmD}\Omega_{\pm}({H}_2,H^D_1)-\Omega_{\pm}((\p-\A(\x))
\hatv,\p\hatv)
\eeq
and show

\begin{la}\label{OmegarestD}
For ${H}_2$ and $H^D_1$ as defined in (\ref{defH2barH1D})
\beq
\left\|\left\{\Omega_{\pm}({H}_2,H^D_1)-\Omega_{\pm}((\p-\A(\x))\hatv,
\p\hatv )\right\}\Phi_0\right\|={\cal O}(v^{-1}).
\eeq
\end{la}

\bew
We compute the difference and obtain similar to (\ref{diffn})
\begin{eqnarray}
\!\!\!\!& & \Omega_{\pm}({H}_2,H^D_1)-\Omega_{\pm}((\p-\A(\x))\hatv
,\p\hatv ) \nn\\
\!\!\!\!& = & -\lim_{t\to\pm\infty} e^{iH_2 t}\int_0^t
d\tau\,\frac{d}{d\tau}\left\{e^{-iH_2\tau}
e^{-i\int_{0}^{\tau}\hatv\A(\x+\xi\hatv)d\xi}\bar{U}^D((t-\tau)/v,t/v)\right\}
\bar{U}^D(t/v,0)\nn \\
\!\!\!\!& = & \lim_{t\to\pm\infty} \int_0^t d\tau\, e^{iH_2 \tau}
i\left\{H_2 e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)d\xi}
-e^{-i\int_{0}^{t-\tau}\hatv\A(\x+\xi\hatv)d\xi}\right.\nn\\
\!\!\!\!& & \left.\label{diffnshortD}\mbox{}[H_1^D(t-\tau)-\hatv\A(\x+(t-\tau)
\hatv)]\right\} \bar{U}^D(\tau/v,0)
\end{eqnarray}
with $\bar{U}^D(\tau/v,0)$ as defined in (\ref{defbU}).
Then (\ref{diffnshortD}) is bounded by
\beqa
& & v^{-1} \lim_{t\to\pm\infty} \int_0^t d\tau
\left\|\left\{-\frac{1}{2m}(\p-m\v)\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)\right)\right.\right.\nn\\
& & \mbox{}-\frac{1}{2m}\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)\right)(\p-m\v)\nn\\
& & \left.\mbox{}+\frac{1}{2m}\left(\int_0^{t-\tau}
(\hatv\times\B)(\x+\xi\hatv)d\xi+\A(\x+t\hatv)\right)^2 +V^s(\x)\right.\nn\\
& & \label{termFVsD}\left.\left.\mbox{}+V^l(\x)
-V^l\left(\frac{\p}{m}\frac{\tau}{v}\right)
\right\} U^D(\tau/v,0) f(\p-m\v)(1+\x^2)^{-2}\right\| \\
& \stackrel{!}{\le}&  v^{-1}\int_0^{\pm\infty}d\tau \,h(|\tau|).\nn
\eeqa
If we find an integrable majorant $h\in L^1([0,\pm\infty),d\tau)$ for the norm
(\ref{termFVsD}) uniformly in $\v$ we finished the proof.\\
The integrability of the summand with $V^s$ is proven in
\cite[Lemma 3.2]{EW 95} and of the summand with
$V^l(\x)-V^l(\frac{\p}{m}\frac{\tau}{v})$
in \cite[Lemma 3.3]{EW 95}. With (\ref{srvb})--(\ref{srga}) the
integrability of the remaining terms can be shown as in
\cite[Proof of Lemma 3.2]{EW 95} with the help of (\ref{ppH0}) and
\beq \label{ppU}\!
\left\|F(|\x|\ge |\tau|/8)e^{-i\int_0^{\tau/v} V^l(s\p/m)ds}f(\p-m\v)
(1+\x^2)^{-2}\right\|\le C\,(1+|\tau|)^{-(2+\varepsilon)}
\eeq
with $\varepsilon>0$. For the propagation property of the Dollard
correction (\ref{ppU}) see \cite[Proposition 3.1]{EW 95}. \qed

Similar to (\ref{diffofomega}) the proof of Lemma \ref{OmegarestD} shows
with $K_\pm$ as defined in (\ref{defK})
\begin{eqnarray*}
& & v\left\{\Omega_{\pm}({H}_2,H^D_1)-\Omega_{\pm}((\p-\A(\x))\hatv ,
\p\hatv )\right\}\Phi_0\\
& \stackrel{v\to\infty}{\approx} & i\int_0^{\pm\infty} d\tau\,
e^{-i\int_{0}^{\pm\infty}\hatv\A(\x+\xi\hatv)d\xi}
\left\{V^s(\x+\tau\hatv)+V^l(\x+\tau\hatv)-V^l(\tau\hatv)\right\}\Phi_0 \\
&& \mbox{}+i\int_{0}^{\pm \infty}d\tau\,
e^{-i\int_{0}^{\pm\infty} \hatv\A(\x+\xi\hatv)\,d\xi}
K_\pm(\p,\x+\tau\hatv)\Phi_0,
\end{eqnarray*}
where $\mathop{\mbox{\rm s-lim}}_{v\to\infty}\exp(i {H}_2\tau)=
\exp(i(\p-\A(\x))\hatv\tau)$ and $\mathop{\mbox{\rm s-lim}}_{v\to\infty}
\exp(i H_1^D(\tau))=\exp(i\p\hatv\tau)$ for fixed $\tau$ is used. Hence we
already know that
\begin{eqnarray*}
\Omega_{\pm}({H}_2,H^D_1)-\Omega_{\pm}((\p-\A(\x))\hatv,\p\hatv)
=:R^D_{\pm}={\cal{O}}(v^{-1})
\end{eqnarray*}
and $\mathop{\mbox{\rm s-lim}}_{v\to\infty}v R^D_{\pm}$ is explicitly known.
As in the short--range case we complete the proof of the reconstruction
formula for $V^{s}$ and obtain
\begin{eqnarray}
&& \!\!\!\!\!\!\lim_{v\to\infty}iv\left(e^{-i\int_{-\infty}^\infty
\hatv\A(\x+\xi\hatv)\,d\xi}
\left(S^D-e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}\right)\Phi_\v,
\Psi_\v\right) \nn\\
&& \!\!\!\!\!\!\mbox{}-\int_{0}^\infty
\left(e^{-i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}
K_+(\p,\x+\tau\hatv)e^{i\int_{-\infty}^\infty \hatv\A(\x+\xi\hatv)\,d\xi}
\Phi_0,\Psi_0\right)\,d\tau\nn\\
&& \!\!\!\!\!\!\label{exactformulaD} \mbox{}-\int_{-\infty}^0
(K_-(\p,\x+\tau\hatv)\Phi_0,\Psi_0)\,d\tau \\
&& \!\!\!\!\!\!\mbox{}-\int_{-\infty}^\infty \left(\left\{V^l(\x+\tau\hatv)-
V^l(\tau\hatv)\right\}\Phi_0,\Psi_0\right)d\tau \nn\\
& = & \!\!\!\!\!\int_{-\infty}^\infty (V^s(\x+\tau\hatv)\Phi_0,\Psi_0)\,d\tau,
 \nn
\end{eqnarray}
which is the reconstruction formula (\ref{ffVsD}) in Theorem \ref{th4}.
The injectivity already has been shown in the short--range case above.
Again the error term is bounded by
\beq \label{errorboundD}
v^{-1} \|v R^D_-\Phi_0\|\,\|v R^D_+\Psi_0\|
\eeq
and therefore of order ${\cal{O}}(v^{-1})$.\\
A short computation for the commutator of $\exp(-i\int_{-\infty}^\infty
\hatv\A(\x+\xi\hatv)\,d\xi)S^D$
with the first component of the momentum $\p_1$ gives (\ref{ffV}).
Then the total potential $V$ can be computed from (\ref{ffV}) (see
\cite[Proof of Theorem 1.2]{EW 95}).
Hence Theorem \ref{th4} is proved. \qed


\section{Generalizations}
\setcounter{equation}{0}


For magnetic fields with compact support we can construct suitable
gauges for the vector potential. In this case we do not need decay assumptions
of the vector potential for proving the reconstruction formulas.
In the general case the behaviour of the magnetic field inside an arbitrary,
but fixed ball has no influence for the formulas. It suffices that the
regularized magnetic field and the regularized current density outside
the ball decays as $|\y|^{-(2+\delta)},\,\delta>0$.
See \cite{AP2} for details.


\section*{Acknowledgement}
I thank Professor V. Enss and Professor R. Weder for their encouragement and
their helpful conversations and Deutsche Forschungsgemeinschaft for
their financial support.


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Uspekhi Mat. Nauk {\bf 42}(3), 93-152 (1987), translation in Russ. Math.
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\end{thebibliography}}

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