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Inverse Scattering Aharonov-Bohm
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\begin{document}
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\title{Aharonov-Bohm Effect and Time-Dependent Inverse Scattering Theory 
 \thanks{2000 {\sc AMS} classification 35P25, 35R30  and 81U40. Research partially supported
by proyecto PAPIIT  IN105799, DGAPA-UNAM.}}
\author{ Ricardo Weder \thanks{On leave of absence from Instituto de Investigaciones
en Matem\'aticas Aplicadas y en Sistemas, Universidad Nacional Aut\'onoma de M\'exico, 
Apartado Postal 20-726,
M\'exico D.F. 01000. Fellow, Sistema Nacional de Investigadores.}\\Instituto Argentino de 
Matem\'atica, CONICET\\
Saavedra 15, 1083-Buenos Aires, Argentina \\
E-Mail: weder@servidor.unam.mx\\}
\date{}
\maketitle


\begin{center}
\begin{minipage}{5.75in}
\centerline{{\bf Abstract}}\bigskip
We study the Aharonov-Bohm effect from the point of view of time-dependent
inverse scattering theory. As this three dimensional problem is invariant under translations along the vertical axis, it reduces to a problem in
$\ER^2$. We first consider an unshielded  magnetic field that has a singular part produced by a tiny solenoid and a regular part. The wave 
function is zero at the location of the solenoid. We then consider the case where the singular part of the magnetic field is shielded inside a cylinder whose  transversal section is  compact set $K$, and
 there is also  a regular magnetic field. In this case the magnetic field inside $K$ is quite general. Actually, the only condition is  that the 
magnetic flux accross $ K$ has to be finite. Moreover, the wave function is defined in $\Omega:= \ER^2 \setminus K$ and it is zero on $\partial K$.

We prove that  in the unshielded  case  the scattering operator determines uniquely the regular magnetic field and that in the shielded case it determines uniquely the magnetic field in $\Omega$. Moreover, in the unshielded case the scattering operator determines the magnetic flux of the solenoid modulo 2 and in the shielded case it determines the magnetic flux across $K$ modulo 2. Our results follow from a reconstruction formula with error term. 
\end{minipage}
\end{center}
\newpage
\section{Introduction}\sss
In this paper we study the Aharonov-Bohm effect \cite{ab} from the point of view of time-dependent  inverse 
scattering theory. Aharonov and Bohm considered the scattering of an electron off the magnetic
field of a tiny solenoid, idealized as having infinite lenght and zero radius (scattering off a
thread of magnetic flux). We assume that the solenoid is located on the vertical axis of 
coordinates, and in consequence it is enough to consider scattering in the plane orthogonal to 
the solenoid, as the problem is invariant under translation along the vertical direction. 
The magnetic field of an unshielded solenoid at zero is given by, $ B_s \,\delta (x)\linebreak $ with $ x=(x_1,x_2) \in  \ER^2$ (this 
actually corresponds to a magnetic field in  $\ER^3$ with components $(0,0, B_s \,\delta (x))$.
We also assume that there is a regular magnetic field $B_{R} \in C^1_0 (\ER^2)$ that is 
continuously differentiable and has compact support. In particular, this allows us to take into
account that our idealized solenoid is only an approximation to a real solenoid. The total magnetic
field is the sum of the two,
\beq
B:=  B_s \,\delta (x) + B_R,
\label{1.1}
\ene
 where, of course, $B_s$ is real and $B_R$ is real valued. In order to define the 
Schr\"{o}dinger operator for an electron in the presence of $B$ we have to introduce a magnetic
potential. The magnetic potential for  $B_s \,\delta (x)$ in the Coulomb gauge is given by,
\beq
A_0:= \frac{\alpha_{\0} }{ |x|^2} \left[\begin{array}{c} -x_2 \\ x_1 \end{array}
\right], 
\label{1.2}
\ene
where, we denote $\alpha_{\0}:=\frac{B_s  }{2 \pi}$. Observe that  $\alpha_{\0}$ is the flux ( across zero) of the singular magnetic field normalized by $2\pi$. It is easily checked that 
$\nabla \times A_s:= \frac{\partial}{\partial x_1} A_{s,2}
-\frac{\partial}{\partial x_2} A_{s,1}= B_s \,\delta (x)$ and that $ \nabla \cdot A_s=0$,
with the derivatives taken in distribution sense in $\acute{{\cal D}}$.
The magnetic potential in the Coulomb gauge for $B_R$ is given by,
\beq
A_R= \frac{1}{2 \pi} \int B_R(x-y) \left[\begin{array}{c} -\hat{y}_2 \\ \hat{y}_1 
\end{array}\right]\, \frac{dy}{|y|},
\label{1.4}
\ene



where, $ \hat{y}:= \frac{y}{|y|}$. Clearly, $A_R\in C^1 \left(\ER^2, \ER^2\right)$.
The magnetic potential for $B$ in the Coulomb gauge is given by,
\beq
A_c:= A_s+ A_R.
\label{1.5}
\ene


As is well known, the magnetic potential is not uniquely defined by the magnetic field, actually there is
always the possibility of a gauge transformation. We introduce
below a general class of magnetic potentials that is convenient for our purposes.

\begin{definition}
We denote by ${\cal A}_{\0}(\alpha_{\0},B_R)$ the set of all real-valued 
$A \in C^1\left( \ER^2 \setminus 0, 
 \ER^2 \right) \cap L^1_{{\rm loc}} \left( \ER^2 , \ER^2 \right)$, with $ \nabla \times A =
 2 \pi \alpha_{\0} \,\delta (x) + B_R$, in  $\acute{{\cal D}}$. Moreover, we assume that
$A(x)=O\left(|x|^{-1}\right), |x| \rightarrow \infty$ and that,
\beq
a(r):= \sup_{|x| \geq r} |A(x)\cdot \hat{x}| \in L^1 ([0, \infty )).
\label{1.6}
\ene 
\end{definition}

We now consider the situation where a general singular magnetic field
is contained inside an infinite  cylinder, 
with axis along the vertical direction, and transversal section $K$, where $K$ is a compact    
subset of $ \ER^2 $.  The purpose of the cylinder is to shield the singular magnetic field from the incoming electrons. As we will see below, we can not hope that the scattering operator
determines uniquely the magnetic field inside $K$. In fact, we can only determine the (normalized ) flux of the magnetic field across $K$ modulo 2. This suggests that instead of
specifying the magnetic field inside $K$ we only fix the magnetic flux across $K$, normalized by $2 \pi$, that by Stoke's theorem is given by the circulation of the magnetic potential, $A$, along $ \partial K$,

\beq  
 \alpha_K := \frac{1}{2 \pi} \int_{\partial K} A(x)\cdot dx,
\label{1.8}
\ene
where we integrate in counter-clockwise sense.
Of course, we also specify a regular magnetic field
$B_R \in C^1_0 \left( \overline{\Omega}\right)$ outside of $K$.
The magnetic flux $\alpha_K$ could be produced, for example, by a finite number of
solenoids inside $K$, and also by a regular magnetic field contained inside
$K$, or by a combination of both. The considerations above suggest the definition of the following class of magnetic potentials. In what follows
we will always denote,  $\Omega:= \ER^2 \setminus K$. 

\begin{definition}
Let $K$ be a compact set such that $0 \in K$ and that its boundary,
$\partial K$, is a simple, closed, $C^1$ curve.  Then, for any  $\alpha_K \in \ER$ and any real-valued 
$B_R \in C^1_0 \left( \overline{\Omega}\right)$  we denote by 
${\cal A}_K \left( \alpha_K, B_R\right)$  the set of all real-valued 
$A \in C^1\left( \overline{\Omega}, 
 \ER^2 \right)$ with $ \nabla \times A = B_R$ and such that,
\beq
\alpha_K =  \frac{1}{2 \pi}  \int_{\partial K} A(x)\cdot dx,
\label{1.8b}
\ene
where we integrate in counter-clockwise sense.
 Moreover, we assume that
$A(x)=O\left(|x|^{-1}\right), |x| \rightarrow \infty$ and that,
\beq
a(r):= \sup_{x \in \Omega ,|x| \geq r} |A(x)\cdot \hat{x}| \in L^1 ([0, \infty )).
\label{1.9}
\ene 
\end{definition}

As we will see in Appendix A, the magnetic potential in the Coulomb gauge belongs 
to ${\cal A}_{K}(\alpha_{K},B_R)$.

In what follows we will denote by $K$ either $K:=\{0\}$ or a set  as in 
Definition 1.2.
It is an 
elementary fact that if $A^{(1)}, A^{(2)} \in {\cal A}_{K}(\alpha_{K},B_R)$, then there is a real-valued function, 
$ \lambda $, such that  if $K=\{0 \}$, \linebreak  $\lambda   \in C^{2}\left(\ER^2 \setminus 0 \right) \cap L^{\infty} \left( \ER^2\right)$  and  $ \lambda  \in C^{2}\left(\overline{\Omega} \right)\cap 
L^{\infty}\left(\overline{\Omega}\right)$ if $K$ is as in Definition 1.2,  with the property that,
\beq
A^{(2)}= A^{(1)}+ \nabla \lambda,
\label{1.9b}
\ene
in $\acute{{\cal D}}$ if $K=\{ 0\}$ and in $\Omega$ otherwise. Moreover, 
for any $\hat{x} \in S^1 , \lambda_{\infty}(\hat{x}):= 
\lim_{r \rightarrow \infty}
\lambda (r \hat{x})$ exists and it is a continuous  function on the circle.  We also denote by $\lambda_{\infty}$ the extension of this function to a continuous and homogeneous function  of degree zero, defined in 
$\ER^2 \setminus 0$, i.e., $ \lambda_{\infty}(r \hat{x}):= \lambda_{\infty}
(\hat{x}),\,  r > 0$. For a proof of this fact 
in the case of regular fields defined in $\ER^n$ see \cite{jung}.
In Appendix A  we give details of the proof for the reader's convenience.




 The formal Hamiltonian is the operator,
\beq
h_A:= \frac{\left( \p -A \right)^2}{2 m},
\label{1.10}
\ene
 with domain $C^2_0 \left( \Omega \right)$, and where and $ \p := -i \nabla$ is the momentum 
operator,and  $ m> 0$ is the mass of the electron. We take Planck's constant  and the speed of light all equal to one and the  charge of the electron
equal to minus one.  
The quadratic form associated to $h$ is the following one,
\beq
q_A( \phi ,\psi ):= \left( (\p -A) \phi , (\p -A) \psi \right),
\label{1.11}
\ene
with domain  $C^1_0 \left( \Omega \right)$. The form $q_A$ is non-negative and closable.
The Hamiltonian, $H_A$, is the self-adjoint operator in $L^2 ( \Omega)$ associated to the closure 
of $q_A$ (see \cite{ka}). $H_A$ is the extension of $h_A$ with Dirichlet boundary condition on 
$\partial \Omega$. 
Under the gauge transformation (\ref{1.9b}) the Hamiltonians are unitarily equivalent:
\beq
H_{A^{(2)}}=  e^{i \lambda (x)}\, H_{A^{(1)}}\, e^{-i \lambda (x)}.
\label{1.12}
\ene


Let $J$ be the identification operator from $L^2 \left( \ER^2 \right)$ onto   
$L^2 \left( \Omega \right)$ given by multiplication by the characteristic function of $\Omega$. In the case $K= \0$ we take $J=I$.
The unperturbed Hamiltonian is given by $H_0:= \frac{\p^2}{2m}$, with domain the Sobolev space
$W_{2,2}\left( \ER^2 \right)$. The wave operators are defined as,
\beq
\w (A):= \hbox{s}-\lim_{t \rightarrow \pm \infty} e^{it H_A}\, J\, e^{-it H_0}.
\label{1.13}
\ene
We prove in Section 2 that if $A \in {\cal A}_K (\alpha_K,B_R)$, the strong limits exist and are isometric  and moreover, that the wave operators transform as follows under a change  of gauge,

\beq
\w\left( A ^{(2)}\right)= e^{i \lambda (x)}\, \w \left(A^{(1)}\right) \, 
e^{-i \lambda_{\infty}(\pm {\bf p})}.
\label{1.14}
\ene

The scattering operator is given by,

\beq
S(A):=  W_+^{\ast}\left( A \right)\,\,W_- \left( A \right).
\label{1.15}
\ene
Equation (\ref{1.14}) implies that  $S(A)$ transforms  under a change of gauge according to,
\beq
S( A^{(2)} )=  e^{i \lambda_{\infty}( {\bf p})} S(A^{(1)})e^{-i \lambda_{\infty}(- {\bf p})}.
\label{1.16}
\ene
For a similar result for the Dirac equation in all space and with regular magnetic field see \cite{jung} and \cite{jung2}.Under the general class of gauge transformations that we consider the scattering operator is 
not gauge invariant. However, it is changed  by unitary operators  that are only a function 
of momentum. As we will see below, this leaves invariant relative phases in configuration space
and hence,  makes it possible to solve the inverse problem -in a gauge invariant way- taking the scattering operator
corresponding to any magnetic potential in ${\cal A}_K(\alpha_K,B_R)$.
Observe that to define $H_A$ we only use  the values of $A$ in $\Omega$.
This means that as long as $A$ is fixed in $\Omega$, we can change the magnetic potential in the interior of $K$ without changing $H_A$. Note however that as $ A\in C^1 \left( \overline{\Omega}\right)$ the flux
$\alpha_K$ is uniquely defined by the values of $A$ in $\Omega$. This explains why we can not hope to uniquely reconstruct the magnetic field inside $K$ from the scattering operator and makes it plausible that we can reconstruct $\alpha_K$. 

 

As we said above, the only purpose of the obstacle, $K$, is to shield the incoming electron from the magnetic
field, and in order to separate the scattering effect of the magnetic potential from the one
of the obstacle, we consider asymptotic configurations that have negligible interaction
with $K$ for all times in the high-velocity limit. For this purpose, given $\hat{\v}\in S^1$,
let us denote,
\beq
 \Omega_{\uv}:= \left\{ x \in \Omega : x+ \hat{\v} \tau \in \Omega,\, \hbox{for all}\, \tau \in 
\ER \right\}.
\label{1.17}
\ene    
Given $\v \in \ER^2$ we take asymptotic configurations $\Phi \in C^{\infty}_0 ( \Omega_{\hat{\v}})$,
where $ \hat{\v}:= \v / v$, with $ v:= |\v|$. The  free evolution boosted by $\v$ is given by
 $ e^{-im \v \cdot x} \, e^{-itH_0} \, e^{im \v \cdot x}=  e^{-im v^2 t/2} 
e^{-i \p \cdot \v t} e^{-itH_0},$ and -to a good approximation- in the limit when $ v \rightarrow
\infty $ with $\hat{\v}$ fixed this can be replaced ( modulo an 
unimportant phase) by the classical translation  $e^{-it\p \cdot \v}$. Then, in the 
high-velocity limit it is a good approximation to assume that the free evolution of our 
asymptotic configuration is given  by
$e^{-it\p \cdot \v} \Phi_0 = \Phi_0 (x- \v t) $, and as 
$\Phi_0 \in C^{\infty}_0 ( \Omega_{\hat{\v}})$, it has negligible interaction with $K$ 
for all times. In the following theorem we evaluate the high-velocity asymptotics of the 
scattering operator.

\begin{theorem}
Suppose that $A \in {\cal A}_K(\alpha_K, B_R)$ and that $\Phi_0, \Psi_0 \in C^{\infty}_0\left(\Omega_{\uv}
\right)$. Let 
$ \Phi_{\v}, \Psi_{\v}$ be the boosted asymptotic configurations,
\beq 
\Phi_{\v}:= e^{im \v \cdot x} \Phi_0, \,\, \Psi_{\v}:= e^{im \v \cdot x} \Psi_0.
\label{1.18}
\ene
Then, 

\beq
 \left( S(A) \Phi_{\v}, \Psi_{\v}\right)= \left( e^{i \int_{-\infty}^{\infty} \uv\cdot A(x+ 
\uv \tau )
\, d\tau} \Phi_0, \Psi_0 \right)+ O\left(\frac{1}{|v|}\right),\, |v| \rightarrow \infty.
\label{1.19}
\ene
\end{theorem}
By Theorem 1.3, from the high-velocity limit of the scattering operator we uniquely reconstruct
the phase,

\beq
e^{i \int_{-\infty}^{\infty} \uv\cdot A(x+ \uv \tau )
 \, d\tau}, \, x \in \Omega_{\uv}.
\label{1.20}
\ene

Equation (\ref{1.9b}) implies that if $A^{(1)}, A^{(2)} \in {\cal A}_K(\alpha_K, B_R)$, then
\beq\
\int_{-\infty}^{\infty} \uv\cdot A^{(2)}(x+ \uv \tau )
 \, d\tau =  \int_{-\infty}^{\infty} \uv\cdot A^{(1)}(x+ \uv \tau )  \, d\tau + 
\lambda_{\infty}( \uv )- \lambda_{\infty}(- \uv ).
\label{1.21}
\ene
 In consequence, (\ref{1.20}) is not gauge invariant. However, the function
\beq
{\cal R}(x,y, \uv) := e^{i \int_{-\infty}^{\infty} \uv\cdot A(x+ \uv \tau )
 \, d\tau -i \int_{-\infty}^{\infty} \uv\cdot A(y+ \uv \tau )
 \, d\tau },\, \,
 x,y \in \Omega_{\uv},
\label{1.22}
\ene
 is gauge invariant and this allows us to solve the inverse scattering problem 
in a gauge invariant way. We prove in Section 2 that (\ref{1.22}) determines uniquely the 
Radon transforms,
\beq
\int_{-\infty}^{\infty}B_R (x  + \uv \tau) \, d \tau, \, x \in \Omega_{\hat{v}}.
\label{1.23}
\ene
By the support theorem for the Radon transform \cite{he} \cite{na}  $B_R$ is uniquely
determined in $\Omega$, provided that $K$ is convex. The fact   that 
the  magnetic flux, $\alpha_K$, is determined modulo 2 follows from an explicit calculation. This gives us 
the  following theorem.
\begin{theorem}
Suppose that $A^{(j)}\in {\cal A}_K\left(\alpha^{(j)}_K, B^{(j)}_R\right),
j=1,2$ and that $K$ is convex. Then, if $S\left(A^{(1)}\right)= S\left(A^{(2)}\right)$, we have that,
 $\alpha^{(1)}_K= \alpha^{(2)}_K$ modulo $2$ and that $B^{(1)}_R(x)=
B^{(2)}_R(x), x \in \Omega$. 
\end{theorem}
Observe that since (\ref{1.22}) is obtained from (\ref{1.19}) we actually only need to know the 
high-velocity limit of the scattering operator. Moreover, our proof gives a method for the 
reconstruction of $B_R$ in $\Omega$. Note that in spite of the fact that the scattering of the 
electron takes place outside of $K$, we determine the magnetic flux  in 
$K$- modulo 2- from the scattering 
operator. This is the Aharonov-Bohm effect \cite{ab} that shows that in quantum mechanics  the magnetic field acts on a charged particle -by means of the magnetic potential-
even in regions where it is identically zero.
Nicoleau has proven in \cite{ni} the following result using stationary methods. Suppose that   the magnetic field, $B$, is infinitely differentiable in $\ER^2$ and has 
compact support, and that   $K$ is compact, convex, $ 0 \in K$ and 
$\partial K$  is smooth. Then, if  $S\left(A^{(1)}\right)= S\left(A^{(2)}\right)$, with $A^{(j)}, j=1,2$, the Coulomb gauge potentials,
then the magnetic fluxs across $\ER^2$ are equal modulo $2$ and     
$B^{(1)}_R(x)=
B^{(2)}_R(x), x \in \Omega$.  The case of an unshielded solenoid with $K=\{0\}$ is not covered by the result of \cite{ni}. Note that this is actually the problem considered by  Aharonov and Bohm  \cite{ab}.
In the case where the interior of $K$ is non-empty \cite{ni} considers the 
situation where the magnetic flux across $K$ is produced by a magnetic field in
$C^{\infty}_0 \left( \ER^2 \right)$. Our result is considerably more general
in the sense that we study the case where the magnetic flux across $K$ is produced by any magnetic field inside $K$. The only restriction is that the magnetic flux across $K$ has to be finite.
We could as well consider the case where there is also a scalar potential-as is done in \cite{ni}- however, we do not pursue this direction here.

We prove Theorems 1.3 and 1.4 by extending to this case the time-dependent method for inverse
scattering theory \cite{ew1}, \cite{ew2}. 
This method  was previously 
extended to Schr\"{o}dinger operators with magnetic fields in \cite{ar1}, \cite{ar2} and  \cite{ar3} and to Dirac operators in \cite{jung} and \cite{jung2}.These papers study the problem  $\ER^n$ ( i.e. when $K= \emptyset$ ) and with regular magnetic fields. 

Since the asymptotic configurations that have negligible
interaction with the obstacle $K$ have compact support in configuration space we have to introduce
an energy cut-off and this requires the  extension of the propagation estimate of \cite{enss} 
-that was used in \cite{ew1}, \cite{ew2}- to the case where the momentum cut-off has support in a 
ball whose radious increases as the velocity tends to infinity ( see Lemma 2.1). 


 The Aharonov-Bohm effect has been
extensively studied. In \cite{ru} Ruijsenaars analyses in great detail the direct scattering 
problem -with time-dependent and stationary methods- in the case where $B_R \equiv 0$ and $K$ is either $\0$ or 
a circle. This problem has an explicit 
solution. The paper \cite{ru} also contains an extensive review of the literature up to 1982. For reviews of the physics literature see \cite{op} and \cite{pt}, and for related results see  \cite{ja}. In the mathematics literature the Aharonov-Bohm effect was studied from the point of view of semiclassical analysis in \cite{ls} and \cite{br}. In \cite{helf} this effect was studied  in eigenvalues. 
   

 
\section{The Proofs}\sss 
We first prove a propagation estimate that is an extension  of the estimate of \cite{enss}-that 
was used in  \cite{ew1}, \cite{ew2}- to the case where the  momentum cut-off has support  in a 
ball with  center $m \v $ and whose radius increases as $v ^{\rho}, \rho > 0$,  
$ v \rightarrow \infty$. We  denote by $  F( x \in {\cal M})$\ the operator of 
multiplication by the characteristic function of ${\cal M}$. For any Borel function $f$
let us define the operator $f(\p)$ by functional calculus, or equivalently,  as $f(\p):= {\cal F}^{-1} f(\cdot ) {\cal F}$, where  ${\cal F}$ 
denotes the Fourier transform.
Remark that under translation in configuration or momentum  space generated, respectively,  
by $\p$ and $x$ we obtain,

\beq
e^{i \p\cdot \v t} \, f(x)\, e^{-i \p \cdot \v t }= f(x+ \v t),
\label{2.1}
\ene
\beq
 e^{-i m \v \cdot x} \, f(\p)\, e^{i m \v \cdot x }= f(\p+m \v ),
\label{2.2}
\ene
and in particular,
\beq
 e^{-i m \v \cdot x} \, e^{-it H_0}\, e^{i m \v \cdot x }= e^{-im v^2 t/2} e^{-i \p \cdot \v t} 
e^{-itH_0}.
\label{2.3}
\ene
For any $\v \in \ER^n , n =1,2,\cdots $, let us denote by $B_{m \eta}(m \v)$ the open ball
in $\ER^n$, of center $m \v$ and radius $m \eta$.  In the case where $\v=0$ we use the notation $B_{m \eta}$.
We designate by $\hat{f}$ the Fourier transform of $f$,
$\hat{f}(p):= 1 /(2 \pi)^{n/2} \int e^{-ip\cdot x}\, f(x)\, dx$.


\begin{lemma}
For any $f \in C^{\infty}_0( B_{m \eta})$, for some $ \eta > 0$,  and any  $ l=1,2, \cdots $, there is a constant 
$C_l$ such that the following estimate holds:
\beq 
\left\| F( x \in \tilde{{\cal M}})\, e^{-it H_0}\, f\left( \frac{{\p} - m  {\v}}{v^{\rho}} \right) 
F( x \in {\cal M}) \right\| \leq C_l  (1+r v^{\rho}+ \eta \,v^{2 \rho}|t|)^{-l},
\label{2.4}
\ene

for every $ {\bf v} \in \ER^n, \, t \in \ER$ and any measurable sets  
${\cal M}$ and ${\tilde{\cal  M}}$  in $\ER^n$
such  that
 $r:={\rm dist}\left( \tilde{{\cal M}}, {\cal M}+ {\bf v}t\right) -\eta v^{\rho } |t| \geq 0$.
\end{lemma}

\noindent {\it Proof:}\,\,by (\ref{2.1})-(\ref{2.3}) it is enough to prove the estimate,

\beq 
\left\| F( x \in \tilde{{\cal M}})\, e^{-it H_0}\, f\left( \frac{\p}{v^{\rho}} \right) 
F( x \in {\cal M}) \right\| \leq C_l \, (1+r v^{\rho}+ \eta \,v^{2 \rho}|t|)^{-l} ,
\label{2.5}
\ene

provided that, $ r:={\rm dist} \left( \tilde{{\cal M}}, {\cal M}\right)-\eta v^{\rho } |t| \geq 0$.
We have that,
\beq
\left\| F( x \in \tilde{{\cal M}})\, e^{-itH_0}\, f\left(\frac{\p}{v^{\rho}}\right) \, F(x\in {{\cal M}}) \phi \right\|^2
\leq \frac{1}{(2 \pi)^{n} } \int_{\tilde{{\cal M}}} dx \int_{{\cal M}} dy \int_{{\cal M}}
dz \,\tilde{f}_t (x-y)\,\phi (y)\, \overline{\tilde{f}_t (x-z) \,\phi(z)},
\label{2.6}
\ene 
where,

\beq
\tilde{f}_t (x):= \frac{1}{(2 \pi)^{n/2} } \int e^{ip\cdot x} e^{-i p^2 t/2m } \, f(p/ v^{\rho})
\, dp.
\label{2.7}
\ene

But, as $ 2 |\phi (y)|\,|\phi (z)|\leq |\phi (y)|^2 +|\phi (z)|^2 $,

\beq
\left\| F( x \in \tilde{{\cal M}})\, e^{-itH_0}\, f\left(\frac{\p}{v^{\rho}}\right) \, F(x\in {\cal M}) \phi \right\|^2
\leq  \frac{1}{(2 \pi)^{n}} \left[ \int_{|x|\geq r + \eta  v^{\rho} |t|} |\tilde{f}_t (x)|\, dx \right]^2
\, \|\phi \|^2.
\label{2.8}
\ene

Finally, by the Corollary to Theorem XI.14 in  \cite{rs}, for any  $N=1,2, \cdots ,$ there is a 
constant $C_N$ such that,

\beq
 |\tilde{f}_t (x)| \leq C_N \, v^{n \rho} \, (1+|x| v^{\rho}+  v^{2 \rho}|t|)^{-N},
\label{2.9}
\ene
for $x \geq \eta v^{\rho} |t|$. The lemma follows inserting (\ref{2.9}) in (\ref{2.8}).


\begin{corollary}
For any $ Q \in \ER^2, \,  f \in C^{\infty}_0 \left( B_{m \eta}\right),\,
  0 \leq \rho < 1$ and  for any  $l=1,2, \cdots ,$ there is a 
constant $C_l$ such that,
\beq
 \left\| F(| x-Q- \v t |> |\v t|/4)\, e^{-itH_0} \, f\left(\frac{\p-m\v}{v^{\rho}} \right) 
F(| x-Q |\leq |\v t |/ 8) \right\| \leq C_l (1+ |\v t|)^{-l},
\label{2.10}
\ene
 for $ v > (8 \eta)^{1/(1-\rho)}$.
\end{corollary}

\noindent {\it Proof :}\,\, the corollary follows from Lemma 2.1 with
$\tilde{{\cal M}}=\{ |x-Q- \v t | > |\v t|/4 \}, \, 
{\cal M}= \{|x-Q| \leq |\v t|/8 \}$. Observe that  $r:={\rm dist}\left( \tilde{{\cal M}}, {\cal M}+ {\bf v}t\right) -\eta v^{\rho } |t| \geq 
|\v t|/8 -\eta v^{\rho } |t| \geq 0$ if $ v > (8 \eta)^{1/(1-\rho)}$.

\bull
In the proofs of Lemmas 2.3 and 2.4 below we first use a special shifted
transversal gauge and we then transfer the result to a general gauge.
Similar ideas were used previously in \cite{ar3} and \cite{jung}, respectively, in the cases of the Schr\"{o}dinger and the Dirac equation in all space with regular magnetic field.
\begin{lemma}
The wave operators $\w(H_A)$ exist and are isometric for all  $A \in {\cal A}_K(\alpha_K,B_R)$. Moreover, if $A^{(1)}, A^{(2)} 
\in {\cal A}_K(\alpha_K,B_R)$ with $A^{(2)}= A^{(1)} + \nabla \lambda$, 
then,
\beq
\w\left( A ^{(2)}\right)= e^{i \lambda (x)}\, \w \left(A^{(1)}\right) \, 
e^{-i \lambda_{\infty}(\pm {\bf p})}.
\label{2.11}
\ene
\end{lemma}

\noindent {\it Proof :} \, \,Let $\chi \in C^{\infty}(\ER^2)$ satisfy
$\chi (x)=0$ for  $x$ in a neigborhood of $K$ and $ \chi (x)=1$ for $|x| \geq M$, with $M$
large enough. Then as $(1- \chi (x)) (H_0 + 1)^{-1}$ is a compact operator,

\beq
W_{\pm} (A)\phi = \hbox{s}-\lim_{t \rightarrow \pm \infty} e^{it H(A)} \chi (x) e^{-it H_0}\phi,
\label{2.12}
\ene
for all $\phi \in D(H_0)$, and by density for all $\phi \in L^2$.
We prove the existence  of $W_+ (A)$. The case of $W_-(A)$ follows in the
same way. It is enough to prove that the strong limits (\ref{2.12})  exist
for all $\Phi_{\v}$ with $\hat{\Phi}_{\v} \ \in C^{\infty}_0(B_{m\eta}(m\v))$, for all $\v, \eta$ with 
$  v > 8 \eta$, since the set of all these functions is dense in $L^2 ( \ER^2 )$.
Note that $\Phi_{\v}= e^{im \v\cdot x} \Phi_0$ with 
$\hat{\Phi}_{0}\in C^{\infty}_0 \left(B_{m\eta} \right)$. Let us take a 
gauge $A^{(f,Q)}$, as in 
(\ref{a.6}) in Appendix A, with $f,Q$ such that,

\beq
 A^{(f,Q)}(x)\, F(|x-Q-  \uv \tau |\leq |\tau|/4 )=0, \, \tau \in \ER.
\label{2.13}
\ene

By Duhamel's formula,
\beq
W_+(A^{(f,Q)})  \Phi_{\v} = \chi (x)  \Phi_{\v} + 
i \int_0^{\infty} \, dt\, e^{it H_{A^{(f,Q)}}}\, V(x,\p)\, 
e^{-itH_0}\,  \Phi_{\v} ,
\label{2.14}
\ene
where,
\beq
V(x,\p):= \chi (x)\left[-\frac{A^{(f,Q)}\cdot \p}{m}+ 
\frac{\left(-\p \cdot A^{(f,Q)}  \right)+  \left(A^{(f,Q)} \right)^2}{2m}
\right] 
- \frac{1}{2m}\left( (\Delta \chi)+ 2i (\nabla \chi)\cdot \p 
+2 A^{(f,Q)}\cdot (\p \chi)\right).
\label{2.15}
\ene
  
Let $g \in C^{\infty}_0 \left(B_{m \eta}\right)$ satisfy, $g\equiv 1$ on the support of 
$\hat{\Phi}_0 $. Then, By (\ref{2.13})
$$
\left\| \chi  A^{(f,Q)} e^{-itH_0}\,(\p - m \v ) \,\Phi_{\v} \right\| \leq C\left[ \left\| 
\chi A^{(f,Q)} 
F(| x-Q- \v t |> 
|\v t|/4) e^{-itH_0} \, g(\p-m\v )\right. \right.
$$
\beq
\left. \left.
(\p- m \v ) F(| x-Q |\leq |\v t |/ 8) \right\| \| \Phi_0 \| 
 +\left\| 
F(| x-Q |> |\v t|/8)\,  \ \Phi_0 \right\|\right].
\label{2.16}
\ene
 
By (\ref{2.10}) with $\rho =0 $,
\beq
 \left\| \chi A^{(f,Q)} F(| x-Q- \v t |> |\v t|/4) e^{-itH_0} \, g(\p-m\v ) (\p -m \v ) 
F(| x-Q |\leq |\v t |/ 8) \right\| \leq C_l (1+ |\v t|)^{-l}.
\label{2.17}
\ene
As $ \Phi_0$ belongs to the Schwartz class,
\beq
\left\| F(| x-Q |> |\v t|/8)\, \Phi_0 \right \| \leq C_l (1+ |\v t|)^{-l},\, l=1, 2, \cdots ,
\label{2.18}
\ene


and it follows that,
\beq
\left\| \chi A^{(f,Q)} e^{-itH_0}\,(\p -m \v ) \,\Phi_{\v} \right\| \leq C_l (1+ |\v t|)^{-l}.
\label{2.19}
\ene
In a similar way we prove that,

\beq
\left\| \chi A^{(f,Q)} e^{-itH_0}\,m \v  \,\Phi_{\v} \right\| \leq C_l \, v \, (1+ |\v t|)^{-l},
\label{2.20}
\ene
and that,


\beq
\left\| \chi \left(( \nabla \cdot \A)+ \left(\A\right)^2\right)\,  e^{-itH_0} \,\Phi_{\v} \right\| 
\leq C_l (1+ |\v t|)^{-l}.
\label{2.21}
\ene

Moreover,

$$
\left\| (\Delta \chi (x))\, e^{-itH_0} \,\Phi_{\v} \right\| \leq C\left[ \left\| 
F(| x- \v t |> 
|\v t|/4 )\, e^{-itH_0} \, g(\p-m\v ) F(| x|\leq |\v t |/ 8) \right\|
\|\Phi_0 \| \right.
$$
\beq
 +\left. \left\| (\Delta \chi (x))\,
F(| x- \v t |\leq 
|\v t|/4) \, e^{-itH_0} \, g(\p-m\v ) F(| x |\leq |\v t |/ 8) \right\|
\|\Phi_0\|  
 +\left\| 
F(| x |> |\v t|/8)\,  \Phi_0 \right\|\right].
\label{2.22}
\ene
Since $ \Delta \chi$ has compact support, 
$$
 \left\| (\Delta \chi (x))\,
F(| x- \v t |\leq
|\v t|/4) e^{-itH_0} \, g(\p-m\v ) F(| x |\leq |\v t |/ 8) \right\| \leq 
$$
\beq
 C \left\| (\Delta \chi (x))\,
F(| x- \v t |\leq
|\v t|/4) \right\| \leq  C  \left\| (\Delta \chi (x))\,
F(|x| \geq 
3|\v t|/4) \right\| \leq   C_l (1+ |\v t|)^{-l},
\label{2.23}
\ene
and we prove as above that,
\beq
\left\| (\Delta \chi (x))\, e^{-itH_0} \,\Phi_{\v} \right\| \leq C_l (1+ |\v t|)^{-l}.
\label{2.24}
\ene
In the same way we obtain  that

\beq
\left\|  (\nabla \chi)\cdot \p \, e^{-itH_0}\,\Phi_{\v} \right\| \leq C_l \, v \,(1+ |\v t|)^{-l},
\label{2.25}
\ene
and that,
\beq
\left\|   A^{(f,Q)}\cdot (\p \chi )\, e^{-itH_0}\,\Phi_{\v} \right\| \leq C_l  \,(1+ |\v t|)^{-l},
\label{2.25b}
\ene

By (\ref{2.15}),(\ref{2.19}),(\ref{2.20}),(\ref{2.21}),(\ref{2.24}),
(\ref{2.25})and (\ref{2.25b}),


\beq
\left\| V(x,\p) e^{-itH_0}\, \Phi_{\v} \right\| \leq C_l \, v \, (1+ |\v t|)^{-l},\, 
l= 1,2, \cdots,
\label{2.26}
\ene
and this implies that the integral in the right-hand side of (\ref{2.14}) is absolutely
convergent. In particular, $W_+(A^{(f,Q)}) \Phi_{\v}$ exists. Let $\lambda $ satisfy ( see (\ref{1.9b}) and Remark
3.1 in Appendix A), $A= A^{(f,Q)} + \nabla \lambda$. 
Since, $ H_{ A} = e^{i \lambda (x)}\, H_{A^{(f,Q)}} \,e^{-i \lambda (x)}$, we have that 
for $\phi \in D(H_0)$,
\beq
 W_+ (A) \phi_{\v} = e^{i \lambda (x)}\, \hbox{s}-\lim_{t \rightarrow \infty} e^{it H_{A^{(f,Q)}}}  
e^{-i \lambda (x)}\, J \, e^{-itH_0} \phi_{\v} =  e^{i \lambda (x)}\, \hbox{s}-\lim_{t \rightarrow \infty} 
e^{it H_{A^{(f,Q)}}}  \,J\, e^{-i \lambda_{\infty} (x)} e^{-itH_0} \phi_{\v},
\label{2.27}
\ene
where we used the fact that by (\ref{a.3}) if $K$ is contained inside the
open ball of center zero and radius $M$, then   $ | \lambda (x) - \lambda_{\infty}(x)| \leq \int_{|x|}^{\infty}
a(r)\, dr$, for $|x|\geq M$, and then by Rellich's selection theorem the multiplication  
operator 
$e^{-i\lambda (x)}- 
e^{-i\lambda_{\infty} (x)}$ is compact from $D(H_0)$ into $L^2$.
As, $e^{it H_0}\, x \, e^{-itH_0}= x+ \p t/m$, it follows by functional calculus that, $e^{it H_0}\, \lambda_{\infty}(x) \, e^{-itH_0}= 
\lambda_{\infty}(x+ \p t/m)$, and as 
 $ \lambda_{\infty}$ is homogeneous of degree zero, $ e^{itH_0}  \lambda_{\infty}(x)
 e^{-itH_0}= \lambda_{\infty}(\frac{m x}{t} + \p ), t > 0$. 
Finally, as $\hbox{s}-\lim_{t \rightarrow \infty} \lambda_{\infty}(mx/t + 
{\p} )
 = \lambda_{\infty}(\p)$,
and by (\ref{2.27}),

\beq
 W_+ ( A ) \phi_{\v} =  e^{i \lambda (x)}\, \hbox{s}-\lim_{t \rightarrow \infty} e^{it H_{A^{(f,Q)}}} J 
e^{-itH_0}\,\,e^{itH_0}e^{-i \lambda_{\infty} (x)} e^{-itH_0} \phi_{\v}
=  e^{i \lambda (x)}\, W_+ \left(A^{(f,Q)} \right)\,e^{-i \lambda_{\infty} (\p)}\phi_{\v}.
\label{2.28}
\ene
This proves the existence of  $W_+ ( A )$. Note that the same argument proves (\ref{2.11})
for $W_+ (A )$.
In the case of $W_-(A)$ we proceed as above using that 
for $t < 0$, $ \lambda_{\infty}(x + {\p} t /m )= \lambda_{\infty}(-\frac{mx}{t} - \p )$. The fact that the wave operators are isometric is immediate because $e^{it H(A)}$ and $ e^{-itH_0}$ are unitary and 
$(1- \chi_{\Omega}) (H_0 + 1)^{-1}$ is a compact operator.

\begin{lemma}
For any  $A\in {\cal A}_K\left( \alpha_K, B_R\right)$, and every $\Phi_{0} \in 
C^{\infty}_0\left(\Omega_{\uv}
\right)$, 
\beq
\left\| \left(e^{-im \v \cdot x} W_{\pm}(A)\, e^{im \v \cdot x} - 
e^{-i \int_{0}^{\pm \infty} \uv\cdot A(x+ \uv \tau)\, d\tau } 
\right)\Phi_{0}\right\| = O\left(1/ v \right),
\, v \rightarrow \infty.
\label{2.29}
\ene
\end{lemma} 

\noindent {\it Proof:}\,\,We give the proof for $W_+(A)$. The 
case of $W_-(A)$ follows in the same way. let $A^{(f,Q)}$ be the 
gauge chosen in the proof of Lemma 2.3 . We first prove that,

\beq
\left\| \left( e^{-im \v \cdot x} \, W_{+}\left(A^{(f,Q)}\right)\,e^{im \v \cdot x}  -
 e^{-i \int_{0}^{ \infty} \uv\cdot A^{(f,Q)}(x+\uv \tau )
\, d\tau } 
\right)\Phi_{0}\right\| = O\left(1/ v\right),
\, v \rightarrow \infty.
\label{2.30}
\ene

As $\Phi_0 \in C^{\infty}_0\left( \Omega_{\uv} \right)$,
$\hbox{distance} \left( K, \{x: x=y+ \uv \tau, y \in \hbox{support}\, 
\Phi_0, \tau \in \ER \}\right) > 0$.
 Hence, there is a function $\chi \in C^{\infty}(\ER^2)$ such that $ \chi \equiv 0$ for $ x$ in 
a neighborhood of $K$ 
and $ \chi \equiv 1$,  for $ x\in \{x: x=y+ \uv \tau, y \in \hbox{support}\, \Phi_0, \tau \in \ER \}
\cup \{x : |x| \geq M \}$, for some $M$ large enough. We designate, $\tilde{A}(x):= \chi (x)
A^{(f,Q)}(x)$.
Let us denote, 

\beq
 H_1:= \frac{1}{v} e^{-im \v \cdot x}
\, H_0 \, e^{im \v \cdot x},\, \hbox{and}\, H_2:= \frac{1}{v} e^{-im \v \cdot x}
\, H \left(A^{(f,Q)} \right) \, e^{im \v \cdot x}.
\label{2.30b}
\ene



 Observe that,
\beq
 e^{-im \v \cdot x} \, W_{+}\left(A^{(f,Q)}\right)\, e^{im \v \cdot x}= 
\hbox{s}-\lim_{t\rightarrow \infty}
e^{it H_2} \,\chi (x) \, e^{-it H_1}.
\label{2.31}
\ene
Moreover, as $ \tilde{A}(x+ \uv \tau)= A^{(f,Q)}(x+\uv \tau ), x \in \hbox{support} \,\Phi_0$,

\beq
\left( e^{-im \v \cdot x} \,W_{+}\left(A^{(f,Q)}\right) \, e^{im \v \cdot x} -
 e^{-i \int_{0}^{ \infty} \uv\cdot A^{(f,Q)}(x+\uv \tau )
\, d\tau }\right)\Phi_{0}=\hbox{s-}\lim_{t \rightarrow \infty}\left[  e^{it H_2} \,\chi (x) \, e^{-it H_1}
- e^{-i \int_{0}^t \uv\cdot \tilde{A}(x+\uv \tau )
\, d\tau } 
\right]\Phi_{0}.
\label{2.32}  
\ene

Let $g \in C^{\infty}_0 (\ER^2 )$ satisfy, $g(k)=1, |k| \leq 1$ and $g(k)=0, |k| \geq 2$.
Then, as $\hat{\Phi}_0$ belongs to the Schwartz space, we have that for every integer $N$,
\beq
\left\| \left[g\left(\frac{\p}{v^{\rho}}\right) -1\right] \, \Phi_0 \right\|_{L^2(\ER^2)}= 
O\left(1/ \,v^{\rho N}\right), v \rightarrow \infty ,
\label{2.33}
\ene
where $\rho > 0$. Then,  it is enough to prove that,

\beq
\left\|\lim_{t \rightarrow \infty}\left[  e^{it H_2} \,\chi (x) \, e^{-it H_1}
- e^{-i \int_{0}^t \uv \cdot \tilde{A}(x+\uv \tau )
 \, d\tau } 
\right]\tilde{\Phi} \right\|=  O\left(1/ v \right),
\, v \rightarrow \infty,
\label{2.34}
\ene
 
with $\tilde{\Phi}:= g\left(\frac{\p}{v^{\rho}}\right)\, \Phi_0$,
and $ 0 < \rho < 1$.

By Duhamel's formula,
$$
\left(e^{it H_2} \,\chi (x)\, e^{-it H_1} -\chi (x) e^{-i \int_{0}^t \uv \cdot\tilde{A}(x+\uv \tau )
\, d\tau }\right) \tilde{\Phi}= \int_0^t \, d\tau e^{i \tau H_2}i \left[
H_2 e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv
\mu) d\mu} \chi (x) \right.
$$
\beq
 \left. - e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv \mu ) d\mu} \chi(x)
( H_1 -\uv \cdot\tilde{A}
(x+(t-\tau)\uv))\right] e^{-i\tau H_1}\, \tilde{\Phi}.
\label{2.35}
\ene
It follows from the identity \cite{ar1}, \cite{ar2},
\beq
(\p-\tilde{A}(x)) e^{-i \int_0^{t} \uv \cdot\tilde{A}(x+ \uv \tau) d\tau} =
e^{-i \int_0^{t} \uv \cdot\tilde{A}(x+ \uv \tau) d\tau} \, (\p- b(x,t)),
\,\, b(x,t):= \tilde{A}(x +\uv t)+ \int_0^t (\uv \times \tilde{B})(x+  \uv \mu ) \, d \mu,
\label{2.36}
\ene
with $(\uv \times \tilde{B})_i:=\sum_{j=1}^2(\tilde{F}_{ij})\uv_j, \, \tilde{F}_{ij}:= 
\frac{\partial}{\partial x_i}
\tilde{A}_j - \frac{\partial}{\partial x_j}
\tilde{A}_i$, and a simple computation that, 

\beq
\left(H_2 \, e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv
\mu) d\mu} \chi (x) - e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv \mu) d\mu}\chi(x)
[ H_1 -\uv \cdot\tilde{A}
(x+(t-\tau)\uv)]\right)= T_1+T_2+T_3,
\label{2.37}
\ene
where,
$$
T_1:= \frac{1}{v}\, e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv
\mu)d\mu} \frac{-1}{2m}\left\{ \chi (x)\left[ \p \,\, b(x,t-\tau)+b(x,t-\tau)\,\p 
-\left( b(x, t-\tau )\right)^2 \right] \right.
$$
\beq
\left.
+ (\Delta \chi (x))- 
2  (\p \chi (x))\cdot \p + 2 \tilde{A}(x)\cdot (\p \chi (x)) \right\},
\label{2.38}
\ene 

$$
T_2:= \frac{1}{v}\, e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv
\mu) d\mu}\left\{\frac{\chi (x)}{2m}\left[-\left( \p\cdot (\A(x)-\tilde{A}(x))\right)+ |\A(x)|^2-|\tilde{A}(x)|^2\right]
\right.
$$
\beq
\left. -\frac{\chi (x)}{m} (\A(x)-\tilde{A}(x))\cdot (\tilde{A}(x)+\p
- b(x, t- \tau)) - \frac{1}{m} (\p \chi (x))\cdot 
(\A(x)-\tilde{A}(x)) \right\},
\label{2.39}
\ene
\beq
T_3:= e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv
\mu )d\mu}\left[(\p \chi (x))\cdot \uv- \chi (x) (\
A^{(f,Q)}(x)-\tilde{A}(x))\cdot \uv\right].
\label{2.40}
\ene
Equation (\ref{2.2}) implies that,
$$
T_1 \, e^{-i\tau H_1}\, \tilde{\Phi}=  
\frac{1}{v}\,e^{-im \v \cdot x}\, e^{-i \int_0^{t-\tau} \uv \cdot\tilde{A}(x+ \uv
\mu)d\mu} \frac{-1}{2m}\left\{\chi (x)\left[(\p-m \v) \,\, b(x,t-\tau)+b(x,t-\tau)\,
(\p -m \v )
\right.\right.
$$
\beq
\left.\left.
- \left( b(x, t - \tau )\right)^2 \right] 
 + (\Delta \chi (x))- 
2(\p \chi (x))\cdot (\p -m \v )+2 \tilde{A}(x)\cdot (\p \chi (x)) \right\} \, e^{-iH_0 \tau /v} e^{im \v \cdot x} \tilde{\Phi}.
\label{2.41}
\ene 
Furthermore, by (\ref{2.1}), for $ 0 \leq \tau \leq t$,  
$$
\left\| \tilde{A}(x+ \uv (t-\tau)) F\left(|x-Q- \uv \tau |\leq \tau /4 \right) \right\|=
\left\|e^{i\p\cdot \uv (t-\tau)} \tilde{A}(x) F\left(|x-Q- \uv t |\leq \tau /4 \right) 
 e^{-i\p\cdot \uv (t-\tau)}     \right\|
$$
\beq
\leq \left\| \tilde{A}(x) F\left(|x-Q- \uv t |\leq t  /4 \right) \right\|=0,
\label{2.42}
\ene
where in the last equality we used (\ref{2.13}).
We prove in the same way that,
\beq
 \tilde{B}(x+ \uv \mu) F\left(|x-Q- \uv \tau |\leq 	\tau /4 \right) = 0,
\label{2.43}
\ene
for $ \mu , \tau \geq 0$ and  then,
\beq
b(x,t- \tau )\, F\left(|x-Q- \uv \tau |\leq \tau /4 \right) = 0, \hbox{for}
\, \tau \in (0, t ).
\label{2.44}
\ene

We prove as in Lemma 2.3, using (\ref{2.2}),(\ref{2.10}), (\ref{2.30b}), (\ref{2.38}) and (\ref{2.44})
that, for each $l=1,2, \cdots $, there are constants $C_l$ such that,
\beq
\left\| T_1 \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\| \leq \frac{C_l}{v} \, (1+ |\tau |)^{-l}.
\label{2.45}
\ene 
For this purpose observe that,
\beq
e^{im \v \cdot x}\, \tilde{\Phi}= g\left(\frac{\p - m \v}{v^{\rho}}\right)
\, e^{im \v \cdot x}\, \Phi_0.
\label{2.46}
\ene
We prove in the same way that,
\beq
\left\| T_2 \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\| \leq \frac{C_l}{v} \, 
(1+ |\tau |)^{-l}.
\label{2.47}
\ene 
Let us denote,

\beq
a(x):=|(\p \chi(x))|+ |\chi (x)| \left|\A (x)- \tilde{A}(x)\right|.
\label{2.47.1}
\ene
Then,

\beq
\left\| T_3 \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\|
\leq \left\| a(x) \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\|.
\label{2.47.2}
\ene
We prove as above that,
\beq
\left\| a(x) \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\| \leq C_l \, 
(1+|\tau| )^{-l}.
\label{2.47.3}
\ene

Moreover, by (\ref{2.1}), (\ref{2.3}) and as $a(x+\uv \tau )\, \Phi_0(x)=
0$,
\beq
a(x )\, e^{-i\tau H_1}\, \tilde{\Phi}  = a(x)\, e^{-i\tau H_1}\, (\tilde{\Phi}
-\Phi_0 )+ e^{-i( \p \cdot \uv \tau + m v \tau / 2 )} 
a(x+ \uv \tau )   \left(e^{-i H_0 \tau /v}-I \right)\, \Phi_0.  
\label{2.47.4}
\ene
Hence,
\beq
\left\|a(x) \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\| \leq C \frac{(1+|\tau|)}{v},
\label{2.47.5}
\ene
where we used (\ref{2.33}). By (\ref{2.47.3}) and (\ref{2.47.5}),
\beq
\left\| a(x) \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\| \leq C_{\delta ,l} \,\frac{1}{v^{\delta}}
(1+|\tau| )^{-l}, l=1,2,\cdots , 0 \leq \delta < 1.
\label{2.47.6}
\ene
Let us denote,
\beq
 I(\v):=\int \, \gamma (\v, \tau ) \, d \tau,
\label{2.47.7}
\ene
where,
\beq
\gamma (\v ,\tau ):=\left[ \left\| a(x) \,e^{-i H_1 \tau }
\tilde{\Phi} \right\|^2 + \epsilon \, v^{-4} (1+|\tau |)^{-4}\right]^{1/2},
\label{2.47.8}
\ene
with $ \epsilon > 0$.
It follows from (\ref{2.47.6}) that $I(\v) \ < \infty$ and that
$ \lim_{ v \rightarrow  \infty} I(\v)= 0$. Equations  (\ref{2.1}) and (\ref{2.3}) imply  that,
\beq
\left\| a(x) \,e^{-i H_1 \tau }
\tilde{\Phi} \right\|^2 = \left\| a(x+ \uv \tau )) \,e^{-i H_0 \tau /v }
\tilde{\Phi} \right\|^2.
\label{2.47.9}
\ene
 Hence,
\beq
\left|\frac{\partial}{\partial v} \gamma(\v ,\tau)\right|
\leq C \, \left[ \frac{|\tau|}{v^2}  \left\| a( x + \uv \tau )\, e^{-i H_0 \tau /v}
H_0 \,\tilde{\Phi} \right\|+ v^{-3}\, (1+|\tau|)^{-2} \right].
\label{2.47.10}
\ene
But, as
\beq
\left\| a( x + \uv \tau )\, e^{-i H_0 \tau /v}
H_0 \,\tilde{\Phi} \right\|= \left\| a( x )\, e^{-i H_1 \tau }
 H_0 \,\tilde{\Phi} \right\|,
\label{2.47.11}
\ene
we prove as above that,

\beq
\left\| a( x + \uv \tau )\, e^{-i H_0 \tau /v}
H_0 \,\tilde{\Phi} \right\| \leq C_l (1 + |\tau |)^{-l},
\label{2.47.12}
\ene
and we obtain,
\beq
\left|\frac{\partial}{\partial v} \gamma(\v ,\tau)\right|
\leq \leq C\, v^{-2}\,
(1+|\tau |)^{-2}, v \geq 1.
\label{2.48}
\ene
This proves the following estimate,
\beq
\left|\frac{\partial}{\partial v} I(\v)\right| \leq C \, v^{-2}
\label{2.49},
\ene
and then,
\beq
I(\v)= \left| \int_{v}^{\infty} \frac{\partial}{\partial s} I(s \uv )
 \, ds \right| \leq C v^{-1}.
\label{2.50}
\ene
By  (\ref{2.47.2}), (\ref{2.47.7}), (\ref{2.47.8}) and (\ref{2.50}),

\beq 
\int \, \left\| T_3 \, e^{-i\tau H_1}\, \tilde{\Phi}  \right\|
\, d \tau \leq I(\v) \leq C v^{-1}, v \geq 1.
\label{2.51}
\ene

 Equation (\ref{2.34}) follows from (\ref{2.35}),(\ref{2.37}), 
(\ref{2.45}), (\ref{2.47}) and (\ref{2.51}). Equation (\ref{2.30}) is a 
consequence of (\ref{2.32}), (\ref{2.33})and (\ref{2.34}).
By (\ref{2.2}), (\ref{2.11}) and as $\lambda_{\infty}$ is homogeneous of 
degree zero,

$$
 \left(e^{-im \v \cdot x} W_{+}(A)\, e^{im \v \cdot x} - 
e^{-i \int_{0}^{ \infty} \uv\cdot A(x+ \uv \tau)\, d\tau } 
\right)\Phi_{0} = e^{i \lambda (x)} \left( e^{-im \v \cdot x} \, W_{+}\left(A^{(f,Q)}\right)\,
e^{im \v \cdot x}  e^{-i \lambda_{\infty}(\uv + \p /mv)}\right.
$$
 \beq
\left.
 - e^{-i \lambda_{\infty}(\uv)}e^{-i \int_{0}^{ \infty} \uv\cdot A^{(f,Q)}(x+\uv \tau )
\, d\tau } 
\right)\Phi_{0},
\label{2.52}
\ene
where we used that $A= A^{(f,Q)}+ \nabla \lambda$.
As $ \left|\lambda_{\infty}(\uv + p /mv)- \lambda_{\infty}(\uv)\right| \leq C \, |p| / v$ for
$|p|/ mv  \leq 1/2$ and since $\hat{\Phi}_0$ belongs to the Schwartz space,


\beq
\left\|\left(e^{-i \lambda_{\infty}(\uv + \p /mv)} -
  e^{-i \lambda_{\infty}(\uv)}\right) \Phi_0 \right\| \leq \frac{C}{v}.
\label{2.53}
\ene
Equation (\ref{2.29}) follows from (\ref{2.30}), (\ref{2.52}) and (\ref{2.53}).

\noindent {\it Proof of Theorem 1.2:}\,\,The theorem follows from  (\ref{1.15}) and Lemma 2.4.

\noindent {\it Proof of Theorem 1.3 :}
By (\ref{1.4}) and (\ref{A.1})
\beq
A_R (x)= \frac{\alpha_{\Omega} }{ |x|^2} \left[\begin{array}{c} -x_2 \\ x_1 \end{array}
\right] + \tilde{A}_R (x),
\label{2.54}
\ene
where,
$\alpha_{\Omega}:= \frac{1}{2 \pi}\int_{\Omega} B_R (y)\, dy$, and
with, $\tilde{A}_R(x)= O\left(\frac{1}{|x|^{2}}\right), |x| \rightarrow \infty$.

We designate,
$\Omega^{\pm}_{\uv}:= \{ x \in \Omega_{\uv}: \pm x \cdot \tilde{\v} > 0 \}$, where
$\tilde{\v}:= ( \uv_2, -\uv_1)$. By  explicit calculation ( see (\ref{1.2}),(\ref{1.5}), (\ref{2.54}) and (\ref{A.7})
we see that,

\beq
L_c(x, \uv ):=\int_{-\infty}^{\infty}\uv\cdot A_c(x+ \uv \tau )
 \, d\tau = \pm \alpha \pi + \int_{-\infty}^{\infty} \uv\cdot \tilde{A}_R(x+ \uv \tau )
 \, d\tau, \, x \in \Omega^{\pm}_{\uv}, 
\label{2.55}
\ene
with $\alpha := \alpha_K - \alpha_A  + \alpha_{\Omega}$,  where
in the case $K=\0$ we take  $\alpha_A = 0$. Moreover,
\beq
\int_{-\infty}^{\infty}\uv\cdot \tilde{A}_R(x+ \uv \tau )
 \, d\tau = O\left(  \frac{1}{|\xp |}\right), \, |\xp | \rightarrow \infty , 
\label{2.56}
\ene
where, $x_{\perp}:= x - (x\cdot \uv) \uv$. In consequence, for any $A \in {\cal A}_K\left(\alpha_K, B_R\right)$,


\beq
L(x, \uv ):=\int_{-\infty}^{\infty}\uv\cdot A(x+ \uv \tau )
 \, d\tau =   \pm \alpha \pi + \int_{-\infty}^{\infty} \uv\cdot \tilde{A}_R(x+ \uv \tau )
 \, d\tau + \lambda_{\infty}(\uv)- \lambda_{\infty}(-\uv),   \, x \in \Omega^{\pm}_{\uv}, 
\label{2.56b}
\ene
where we used (\ref{1.21}) with $A^{(1)}= A_c$. Then, by (\ref{2.56}),

\beq
\lim_{|x_{\perp}| \rightarrow \infty }
 L(x,\uv )= \pm   \alpha \pi + \lambda_{\infty}(\uv)- 
\lambda_{\infty}(-\uv).
\label{2.57}
\ene
In the limit above, $ x \in  
\Omega^{\pm}_{\uv}$. By (\ref{1.19}),
 from $S\left( A\right)$ we uniquely reconstruct
$e^{i L(x, \uv)},\,  x \in \Omega_{\uv}$ and it follows from (\ref{2.57})
that if $S(A^{(1)})= S(A^{(2)})$,
\beq
 \alpha^{(1)} \pi+\lambda^{(1)}_{\infty}(\uv)- \lambda^{(1)}_{\infty}(-\uv)=
   \alpha^{(2)} \pi+\lambda^{(2)}_{\infty}(\uv)- \lambda^{(2)}_{\infty}(-\uv)+
2 \pi k(\uv),
\label{2.57b}
\ene
where the function $k(\uv)$ takes integers values. But as 
$\lambda^{(j)}_{\infty}(\uv), j=1,2$ are continuous for $\uv$ on the unit circle, the function $k$ is a constant independent of $\uv$. Hence, (\ref{2.57b}) taken at $\uv$ and at $-\uv$ implies that 
 $\lambda^{(1)}_{\infty}(\uv)-\lambda^{(1)}_{\infty}(-\uv)= \lambda^{(2)}_{\infty}(\uv)-\lambda^{(2)}_{\infty}(-\uv) $ and we have that $\alpha^{(1)}= \alpha^{(2)}$ modulo $2$.

Let us denote,

\beq
I(x , y,\uv):= \int_{-\infty}^{\infty} \left( \uv\cdot (A(x+ \uv \tau )
 - \uv\cdot A(y+ \uv \tau )
 \right)\, d\tau, \, \, x,y \in \Omega_{\uv}.
\label{2.58}
\ene 

Observe that,
\beq
{\cal R}(x,y, \uv) = e^{i I(x,y,\uv )}.
\label{2.59}
\ene

Using  the fact that $I(x,y, \uv)$  is gauge invariant
and that $B_R \in C^1_0 \left(\overline{\Omega} \right)$, we observe from (\ref{1.4}), (\ref{A.1}) and (\ref{2.55}) that $I(x,y, \uv) \in C^1\left( \Omega_{\uv} \times \Omega_{\uv} \right)$. Then, ${\cal R}(x,y, \uv)\in C^1\left( \Omega^{\pm}_{\uv} \times \Omega^{\pm}_{\uv} \right)$ 
and  
\beq
\frac{\partial}{\partial \xp} {\cal R}(x,y, \uv)
= i {\cal R}(x,y, \uv)  \frac{\partial}{\partial \xp} I(x, y, \uv),
\, x, y \in \Omega_{\uv}.
\label{2.61}
\ene
This implies that from the high-velocity limit of the scattering operator we uniquely reconstruct \linebreak 
$ \frac{\partial}{\partial \xp} I(x ,y, \uv), x,y \in \Omega_{\uv}$.
 But it follows  from the dominated convergence theorem, taking  derivatives under the 
integral sign, using (\ref{1.21}) and the gauge (\ref{a.6}), in Appendix A, with appropriate $f$ 
and $Q$,( see \cite{jung}) that,

\beq
 \frac{\partial}{\partial \xp} I(x ,y,\uv)= \frac{\partial}{\partial \xp}
\int_{-\infty}^{\infty} \left( \uv\cdot (A(x+ \uv \tau ) \right) d \tau =
\int_{-\infty}^{\infty} B_R(x+ \uv \tau )
 \, d\tau ,  \, x,y \in \Omega_{\uv}.
\label{2.62}
\ene
By the support theorem for the Radon transform \cite{he}, \cite{na}, if $K$ is convex, $B_R$ is uniquely determined in $\Omega$ by the Radon transforms,

\beq
\int_{-\infty}^{\infty} B_R(x+ \uv \mu )
 \, d\mu ,  \, x \in \Omega_{\uv}.
\label{2.63}
\ene
In consequence, if $S\left( A^{(1)}\right)= S\left( A^{(2)}\right),
A^{(j)} \in {\cal A}_K\left(\alpha^{(j)}_K, B^{(j)}_R \right), j=1,2$, we have that,
$B^{(1)}_R(x)= B^{(2)}_R(x), x \in \Omega$.
Once $B_R$ is uniquely reconstructed, $\alpha_A$ and $\alpha_{\Omega}$ are uniquely known, and then $\alpha_K$ is  determined modulo $2$. 
\bull




 Note that in the case where $ K=\{0\}$ we have that,
  $\Omega_{\uv}= \{x \in \ER^2 : \xp \ne 0\}$. Hence, in this case we know the Radon transform 
of  $B_R$ for  all lines in  $\ER^2$, because the integrals in (\ref{2.63}) are continuous functions of $\xp$.

We can reconstruct $\alpha$ from the gauge invariant quantity
${\cal R}(x,y, \uv)$ as follows. By (\ref{2.57}),

\beq
\lim_{|x_{\perp}| \rightarrow \infty , |y_{\perp}| \rightarrow \infty}
{\cal R}(x,y, \uv)  = e^{2i \alpha \pi}.
\label{2.63b}
\ene
In the limit above, $ x \in  
\Omega^{+}_{\uv}, y \in \Omega^{-}_{\uv} $. From (\ref{2.63b})we 
reconstruct $\alpha$ modulo 1. 





 
Lemma 2.4 implies -by a density argument- that, for every $ \Phi \in
L^2\left(\ER^2 \right)$ with $\hbox{support} \Phi \in  \Omega_{\uv}$, 

\beq
\hbox{s-}\lim_{v \rightarrow \infty} e^{-im \v \cdot x}\, \w\left(A \right) \, e^{i m \v \cdot x} \Phi  = e^{-i \int_{0}^{\pm \infty} \uv\cdot 
A(x+ \uv \tau)\, d\tau } \Phi. 
\label{2.64}
\ene
Moreover, as the wave operators are isometric, and the limits in (\ref{2.64}) are unitary, we also have strong convergence of the adjoint of the wave operators, as the following argument shows. Let us denote,
$ W_{\pm ,\v}:=  e^{-im \v \cdot x}\, \w (A) \, e^{im \v \cdot x }$,
$ W_{\pm ,\v}^{\ast}:=  e^{-im \v \cdot x} \, \a (A)\, e^{im \v \cdot x }$  and
$L_{\pm}:= \int_{0}^{\pm \infty} \uv\cdot 
A(x+ \uv \tau)\, d\tau $. 
Then,
 
\beq
\left\|\left( W_{\pm ,\v}^{\ast} -  e^{i L_{\pm}}\right) \Phi \right\|=
\left\|W_{\pm ,\v}^{\ast} \left( e^{-i L_{\pm}} - W_{\pm ,\v} \right) 
e^{i L_{\pm}}  \Phi \right\| \leq
\left\|   \left( e^{-i L_{\pm}} - W_{\pm ,\v} \right) e^{i L_{\pm} }  \Phi \right\| \rightarrow 0, \, \hbox{as} \, v \rightarrow \infty ,
\label{2.65}
\ene
and it follows that,

\beq
\hbox{s-}\lim_{ v \rightarrow \infty} e^{-im \v \cdot x}\, \a\left(A \right) \, e^{i m \v \cdot x} \,\Phi = e^{i \int_{0}^{\pm \infty} \uv\cdot 
A(x+ \uv \tau)\, d\tau } \,\Phi. 
\label{2.66}
\ene
Hence, for all   $ \Phi \in
L^2\left(\ER^2 \right)$ with $\hbox{support} \Phi \in  \Omega_{\uv}$, 
\beq
\hbox{s-}\lim_{ v \rightarrow \infty} e^{-im \v \cdot x}\, S\left(A \right) \, e^{i m \v \cdot x} \,\Phi = e^{ i \int_{-\infty}^ {\infty} \uv\cdot 
A(x+ \uv \tau)\, d\tau } \,\Phi . 
\label{2.67}
\ene


\section{Appendix A}\sss
In this Appendix we prove, for the reader's convenience, some elementary results on gauge transformations, and on special gauges with support in cones. See also \cite{ar3} and \cite{jung}
 on these points. We recall that the classes of allowed magnetic potentials were given  in Definitions 1.1 and 1.2. In the case were the obstacle $K$ is as in Definition 1.2 we define,
$$
A_R(x):= \frac{1}{2 \pi} \int \tilde{B}_R(x-y) \left[\begin{array}{c} -\hat{y}_2 \\ \hat{y}_1 
\end{array}\right]\, \frac{dy}{|y|}=
$$
\beq
\frac{1}{2 \pi} \int \tilde{B}_R(y) \frac{1}{|x-y|^2} \left[\begin{array}{c} -(x-y)_2 \\ (x-y)_1 
\end{array}\right]\, dy, 
\label{A.1}
\ene
where, $\tilde{B}_R(x)= B_R(x), x \in \Omega$, and $\tilde{B}_R(x)=0$
if $ x \in K$. For any $ x \in \Omega$, let us denote $\delta := dist(x, K) >0$. Let $\phi \in C^{\infty}_0 \left( \ER^2 \right)$ satisfy $ \phi (x)
= 1, |x|\leq \delta /3$ and $ \phi (x)= 0, $ for $ |x|\geq \delta /2$.
We define,

\beq
A^{(1)}_R(x):= \frac{1}{2 \pi} \int \tilde{B}_R(y)\, \phi(|x-y|) \frac{1}{|x-y|^2} \left[\begin{array}{c} -(x-y)_2 \\ (x-y)_1 
\end{array}\right]\, dy, 
\label{A.2}
\ene
and



\beq
A^{(2)}_R(x):= \frac{1}{2 \pi} \int \tilde{B}_R(y)\,(1- \phi(|x-y|)) \frac{1}{|x-y|^2} \left[\begin{array}{c} -(x-y)_2 \\ (x-y)_1 
\end{array}\right]\, dy. 
\label{A.3}
\ene
Since for all $ \acute{x}$ in a small neighborhood of $x$ and for all 
$y \in \ER^2$, with $|y|\leq \delta /2$ we have that, $ \acute{x}-y \in \Omega$, it follows that,


\beq
\frac{\partial}{\partial x_j}A^{(1)}_R(x)= \frac{1}{2 \pi} \int_{\Omega}  
\left( \frac{\partial}{\partial x_j}  B_R(x-y) \right) \phi (y) \left[\begin{array}{c} -\hat{y}_2 \\ \hat{y}_1 
\end{array}\right]\, \frac{dy}{|y|}.
\label{A.4}
\ene
Moreover, as $ |z| \geq \delta /3$ for all
$z$ in the support of   $(1- \phi (z))$,

\beq
\frac{\partial}{\partial x_j}A^{(2)}_R(x)= \frac{1}{2 \pi} \int  
  \tilde{B}_R(y)  \left( \frac{\partial}{\partial x_j}
\, 
(1-\phi (|x-y|)) \frac{1}{|x-y|^2} \left[\begin{array}{c} -{(x-y)_2} \\ (x-y)_1 
\end{array}\right]\right)\, dy.
\label{A.5}
\ene
By (\ref{A.4}) and (\ref{A.5}), 
$A_R \in C^1 \left( \overline{\Omega}\right)$. It follows from Gauss theorem
that for any $\phi \in C^{\infty}_0 \left( \Omega \right)$
\beq
- \int_{\Omega} dx \left[ A_{R,2} \frac{\partial}{\partial x_1} \phi (x)
-A_{R,1} \frac{\partial}{\partial x_2} \phi (x)\right] =
\int_{\Omega} dx \, B_R(x) \phi (x) \, dx.
\label{A.6}
\ene
This implies that, $\nabla \times A_R(x) = B_R (x), x \in \Omega$.

We prove in a similar way that, $ \nabla  \cdot A_R (x)=0, x \in \Omega $.
The magnetic potential in the Coulomb gauge, in the case were $K$ is as in Definition 1.2, is defined as follows,

 \beq
A_c:= \frac{(\alpha_{K}-\alpha_{R}) }{ |x|^2} \left[\begin{array}{c} -x_2 \\ x_1 \end{array}
\right] + A_R(x), 
\label{A.7}
\ene

with, $\alpha_{R}:= \frac{1}{2 \pi}\int_{\partial K} A_R(x)\cdot dx$,
where we integrate in counter-clockwise sense.
It is easyly checked that, $ A_c \in {\cal A}_K\left( \alpha_K , B_R\right)$. 

\begin{remark}
Suppose that $A^{(1)}$ and $A^{2}\in {\cal A}_K \left(\alpha_K, B_R\right)$. Then, if $K=\0$, there is a real-valued function, $\lambda \in
C^{2}(\ER^2\setminus 0 )\cap L^{\infty}(\ER^2)$, such that $ A^{(1)}= A^{(2)}
+ \nabla \lambda,\,\,\hbox{in}\,\, \acute{{\cal D}}(\ER^2)$. Furtheremore,
if $K$ is as in Definition 1.2 there is a real-valued function, $\lambda$, in 
$C^{2}(\overline{\Omega} )\cap L^{\infty}\left(\overline{ \Omega }\right)$ such that, $ A^{(1)}= A^{(2)}
+ \nabla \lambda,\,\,\hbox{in}\,\, \Omega$. 
 Moreover, in both cases,
$ \lambda_{\infty}(\hat{x}):= \lim_{r \rightarrow \infty} \lambda (r x)$\, exists and it
is a continuous function on the circle, $S_1$. We also denote by $\lambda_{\infty}$ the extension of this function to a continuous and homogeneous function  of degree zero, defined in 
$\ER^2 \setminus 0$, i.e., $ \lambda_{\infty}(r \hat{x}):= \lambda_{\infty}
(\hat{x}),\,  r > 0$.\end{remark}
\noindent {\it Proof :}\, \,
We denote, $A:= A^{(1)}-A^{(2)}$. Then $\nabla \times A =0$. Suppose first that $K=\0$. Let $\phi \in C^{\infty}_0(\ER^2)$ 
satisfy
$\int \phi (x) dx =1$, and let us denote $\phi_n:=  \phi (x/n)$. We define, $A_n:= \phi_n * A$.
Then, $A_n \in  C^{\infty},  \nabla \times A_n =0$, and

\beq
\lim_{n \rightarrow \infty} \left\|A-A_n \right\|_{L^{\infty}(D)}=0,
\label{a.1}
\ene
for any compact set $D \in \ER^2\setminus 0$. It follows from Stoke's theorem that for any simple closed
$C^1$ curve, $ {\cal C}$, that does not go through zero,
\beq
\int_{{\cal C}} A \cdot dl= \lim_{n \rightarrow \infty} \int_{{\cal C}} A_n \cdot dl=
\lim_{n \rightarrow \infty}\int_{{ \cal C}_{int}} \nabla \times A_n \,dx =0.
\label{a.2}
\ene
When $K$ is as in Definition 1.2, Stoke's theorem implies
that (\ref{a.2}) holds for any simple closed $C^1$ curve in $\Omega$.
For this purpose observe that,
\beq
\int_{\partial K} A(x)\cdot dx =0.
\label{a.2.2}
\ene
 
For any fixed $ x_0 \in \Omega $, we denote,
\beq
\lambda(x):= \int_{{\cal C}^{x}_{x_0}} A \cdot dl,
\label{a.3}
\ene
where ${\cal C}^{x}_{x_0}$ is any simple, $C^1$ curve in $\Omega$ that goes from $x_0$ to $x$. Then, when $K=\0$, $ \lambda \in C^2(\ER^2 \setminus 0)$
and when $K$ is as in Definition 1.2, 
$\lambda \in C^2\left(\overline{\Omega}\right)$. Furthermore,$ \nabla  \lambda = A$.
Since, $a(r):= \sup_{|x| \geq r} |A(x)\cdot \hat{x}|
 \in L^1\left((0, \infty )\right)$, we have that, $ \lambda \in L^{\infty}(\overline{\Omega})$. Moreover,
$\lambda_{\infty}(\hat{x}):= \lim_{r \rightarrow \infty} \lambda (r\hat{x})$ exists, and  it is a continuous 
function on $S_1$.

\bull
We state some special gauges with support in a cone. Let us take polar coordinates, $(r, \theta),
r > 0, \theta_0 \leq \theta < \theta_0 + 2 \pi$ for any $ 0 \leq \theta_0 < 2 \pi$. Let 
$ f(\theta )\in C^1 (S^1 )$ satisfy, $f (\theta )=0$ for $\theta \leq \theta_0 + \epsilon /2$,
and for $ \theta \geq \theta_0 + 2 \pi - \epsilon /2$, and moreover, $f(\theta )= \theta $ for
 $\theta_0 + \epsilon  \leq \theta \leq \theta_0 +2 \pi - \epsilon$,
for some $ \epsilon > 0$. We define,
\beq
A^{(f)}:= \frac{(\alpha_{K}-\alpha_{R}) }{ |x|^2} \left[\begin{array}{c} -x_2 \\ x_1 \end{array}
\right]- (\alpha_{K}-\alpha_{R}) \nabla f(\theta).
\label{a.4}
\ene   
Then, $A^{(f)}$ has support in a cone with center zero, axis $ e^{i \theta_0}$,
and opening angle $ \epsilon $. 

For any $Q =(q_1,q_2) \in \ER^2$ let $A^{(Q)}$ be 
the following shifted transversal magnetic potential for $B_R$,

\beq
A^{(Q)}:= \left[\begin{array}{c}q_2 - x_2 \\ x_1- q_1 \end{array}\right]
\,\, \int_{0}^1 \tilde{B}_R (x \mu + (1-\mu ) Q)\, \mu \, d\mu,
\label{a.5}
\ene
where if $K=\0, \tilde{B}_R = B_R$ and if $K$ is as in Definition 1.2,
$\tilde{B}_R(x)= B_R(x), x \in \Omega$, and $\tilde{B}_R(x)=0, x \in K$.
If we  take $Q$ far enough from the support of $B_R$ the support of $A^{(Q)}$ will be contained
in a cone with vertex $Q$ and with a small opening angle. Furthermore, 
\beq
A^{(f,Q)}:= A^{(f)}+ A^{(Q)}  \in {\cal A}_K\left(\alpha_K, B_R\right)  ,
\label{a.6} 
\ene
where if $K=\0$ we take $\alpha_R=0$, and  we can always choose $\theta_0 $ and $Q$ in such a way that  $A^{(f)}+ A^{(Q)}$ has support
in a cone with small opening angle and vertex $Q$. 

\noindent Acknowledgement

\noindent I thank G. Lozano, F. Nicoleau and F. Schaposnik for useful remarks.

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