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\begin{document}
\title{There is no Aharonov-Bohm effect in dimension three}
\author{ D. Yafaev
\\Department of Mathematics, University of Rennes,\\
Campus Beaulieu, 35042, Rennes, France,
\\ yafaev@univ-rennes1.fr}
\maketitle
\begin{abstract}
Consider the scattering amplitude $s(\omega,\omega^\prime;\lambda)$,
$\omega,\omega^\prime\in{\Bbb S}^2$, $\lambda>0$,
corresponding to an arbitrary three-dimensional short-range
magnetic field $B(x)$. The magnetic
potential $A^{(tr)}(x)$ such that ${\rm curl}\,A^{(tr)}(x)=B(x)$ and
$ =0$ decays at infinity as $|x|^{-1}$ only.
Nevertheless, we show that the structure of $s(\omega,\omega^\prime;\lambda)$
is the same as for short-range
magnetic potentials. In particular, the leading diagonal
singularity $s_0(\omega,\omega^\prime)$ of $s(\omega,\omega^\prime;\lambda)$
is the Dirac function. Thus, up to the diagonal Dirac function, the
scattering amplitude
has only a weak singularity
in the forward direction and hence
scattering is essentially of short-range nature. This is qualitatively
different from the
two-dimensional case where $s_0(\omega,\omega^\prime)$ is a linear combination
of the Dirac function and of a singular denominator, that is
the Aharonov-Bohm effect occurs. Our approach relies on a construction of
a special
gauge, adapted to a given magnetic field $B(x)$, such that
the corresponding magnetic
potential $A (x)$ is also short-range.
\end{abstract}
\section{Introduction}
There is no precise definition of the Aharonov-Bohm effect in the literature.
In the original paper \cite{AB} (see also \cite{Hen, RUI}) the
Schr\"odinger operator $H$
with magnetic potential
\[
A (x)=\alpha (-x_2,x_1) |x|^{-2},
\quad\alpha\in{\Bbb R},\quad x=(x_1,x_2)\in{\Bbb R}^2,
\]
was considered.
For such potentials, variables in the Schr\"odinger equation
can be separated in polar coordinates $(r,\theta)$. This allows to show that
the scattering matrix (SM) $S(\lambda)$, $\lambda>0$ is the energy, for the
operator $H$
has two eigenvalues $e^{i\alpha\pi}$ and
$e^{-i\alpha\pi}$ with corresponding eigenfunctions $e^{im\theta}$ ($m=0,
\pm 1, \pm 2,\ldots$
is the angular momentum) for $m\leq \alpha$ and $m\geq \alpha$,
respectively. This implies of
course that the SM is non-trivial, that is, $S(\lambda)\neq I$ ($I$ is the
identity operator).
Since the magnetic field $B(x)={\rm curl}\, A(x)=0$ for $x\neq 0$,
the fact that $S(\lambda)\neq I$ appeared to be surprising.
We do not share this point of view because, for the Aharonov-Bohm
potential, the field is not zero but
equals $2\pi \alpha \delta(x)$ where $\delta(x)$ is the
Dirac function, and a quantum particle interacts always with this field.
Nevertheless the results of \cite{AB,Hen, RUI} are quite interesting.
In fact, the Aharonov-Bohm potential is one of rare examples
where the SM can be calculated explicitly. Moreover, its properties are quite
different from the case
of short-range potentials satisfying the condition
\begin{equation} \label{eq:H1sr}
| A(x)| \leq C (1+|x|)^{- \rho },~~
\rho >1.
\end{equation}
Let us discuss properties of the SM in a more general framework.
Suppose that a magnetic field $B(x)=(0,0,B(x))$, $ x\in {\Bbb R}^2$, satisfies
the condition
\begin{equation} \label{eq:B1}
| B(x)|
\leq C (1+|x|)^{-r},\quad r>2,
\end{equation}
(for simplicity, one may assume for a moment that
$B(x)$ has compact support). Then the magnetic
potential $A^{(tr)}(x)=(A^{(tr)}_1(x),A^{(tr)}_2(x),0)$, ${\rm
curl}\,A^{(tr)}(x)=B(x)$, obeying
the transversal gauge condition
$ =0$ has asymptotics as $|x|\rightarrow\infty$
\begin{equation}\label{eq:A'}
A^{(\infty)}(x)=a(\hat{x})(-x_2,x_1) |x|^{-2},\quad \hat{x}=x/|x|,
\end{equation}
where $a$ is a smooth function on the unit circle.
Potentials with asymptotics (\ref{eq:A'}) were considered in \cite{RY, Y2}.
For any $\omega\in{\Bbb S}$, we denote by $\omega^{(\pm)}$ the points
obtained from $\omega $ by rotation at the angle $\pm\pi/2$
in the positive (counter-clockwise) direction.
Let
\[
f (\omega)= \int_{{\Bbb S}( \omega^{(-)},\omega^{(+)})} a(\theta)d\theta
=\int_{\geq 0} B(x)dx,
\]
where the first integral is taken in the positive direction over the
half-circle ${\Bbb S}( \omega^{(-)},\omega^{(+)})$ between the points
$\omega^{(-)}$ and $\omega^{(+)}$, and let
\[
\Phi = \int_{\Bbb S} a(\theta)d\theta=\int_{{\Bbb R}^2} B(x)dx
\]
be the total magnetic flux. The kernel
$s(\omega,\omega^\prime;\lambda)$ of the SM regarded as an integral operator
(the variables $\omega,\omega^\prime\in{\Bbb S}$ and $\lambda>0$) is smooth
away from the diagonal $\omega = \omega^\prime$ and its leading diagonal
singularity
is given by the function
\begin{equation}\label{eq:SAB}
s_{0} (\omega,\omega^\prime)= e^{i(f(\omega^{(-)})-f(\omega^{(+)}))/2}
\Bigl(\cos(\Phi/2)\,
\delta(\omega,\omega^\prime) + \pi^{-1} \sin(\Phi/2)\, {\rm P.V.}
\{\omega,\omega^\prime\}^{-1}\Bigr).
\end{equation}
Here $\delta(\omega,\omega^\prime)$ is the Dirac function on the unit circle,
$\{\omega,\omega^\prime\}$ is the oriented angle between an
initial vector $\omega$ and a final vector
$\omega^\prime$ and the second term is understood in the sense of the
principal value.
Singularity (\ref{eq:SAB}) is qualitatively different from the
short-range case where
\[
s (\omega,\omega^\prime;\lambda)=
\delta(\omega,\omega^\prime) + O(|\omega-\omega^\prime|^{-2+\rho})
\]
as $\omega-\omega^\prime\rightarrow 0$. Thus, in the dimension two,
even for magnetic fields of
compact support with $\Phi\not\in 2\pi {\Bbb Z}$,
the SM contains the singular integral operator.
Hence the forward singularity
$\pi^{-2} \sin^2(\Phi/2)|\omega-\omega^\prime|^{-2}$ of the
scattering cross section $| s (\omega,\omega^\prime;\lambda)|^2$
is stronger than for short-range
magnetic potentials where it is $O(|\omega-\omega^\prime|^{-4+2\rho})$.
This fact can be taken for a definition of the
Aharonov-Bohm effect. We emphasize that the Coulomb decay of $A(x)$
cannot be gauged away even for fields $B(x)$ of compact support.
Indeed, if $A (x)=
o(|x|^{-1}) $ as $|x|\ri\infty$, then the circulation of $A(x)$ over the circle
$|x|=r$ tends to $0$ as $|x|\ri\infty$. The Stokes theorem implies that in
this case
necessarily $\Phi= 0$.
In the present paper we show that in the three-dimensional case
the scattering amplitude corresponding to a short-range
magnetic field $B(x)$ satisfying condition
(\ref{eq:B1})
has also the short-range structure, that is
the Aharonov-Bohm effect does not occur.
The difference between dimensions
$d=3$ and $d=2$ is of topological nature: if
$d=3$, then the set ${\Bbb R}^d\setminus\{0\}$ is simply connected, whereas
it is not
true for $d=2$. This allows us to construct for $d=3$ a special gauge, adapted
to a given short-range magnetic field $B(x)$, such that
the corresponding magnetic potential $A(x)$,
\begin{equation}\label{eq:potfield}
{\rm curl}\, A(x)=B(x),
\end{equation}
is also short-range, i.e., it satisfies
condition (\ref{eq:H1sr}) with $\rho=r-1>1$.
If $B(x)$ has compact support, then $A(x)$ is also of compact support.
Of course, our results remain true for all dimensions $d \geq 3$
(in the general case
one has to consider $A$ as a 1-form and $B=dA$ as a 2-form).
In the next section we recall some basic facts of scattering theory and
discuss the
behaviour of the SM with respect to gauge transformations. In Section~3 we
prove our main result
announced above. Finally, in Section~4 we show that condition (\ref{eq:B1}) is
precise, that is the Aharonov-Bohm effect occurs if $B(x)$ decays as
$|x|^{-2}$ only.
\section { Scattering by magnetic fields }
{\bf 2.1.}
Let us first discuss briefly some basic facts of scattering theory.
Consider the Schr\"odinger operator
\[
H=(i\nabla+A(x))^2,\quad x\in{\Bbb R}^d, ,\quad d\geq 2,
\]
with a real magnetic potential $A(x)=(A_1(x),\ldots,A_d(x))$
satisfying condition (\ref{eq:H1sr}).
We do not assume that the function $A(x)$ is differentiable
so that, strictly speaking, $H$ is correctly defined
as a self-adjoint operator in the space $ L_2({\Bbb R}^d)$
in terms of the corresponding quadratic form. In general,
equality (\ref{eq:potfield}) should be understood in the sense of
distributions.
Let $H_0=-\Delta$ be the ``free" operator.
Under assumption (\ref{eq:H1sr}) the wave operators
\[
W_\pm= W_\pm(H,H_0 ) = s-\lim_{t\rightarrow\pm\infty} e^{iHt}
e^{-iH_0t}
\]
exist, are unitary and possess the intertwining property
$HW_\pm=W_\pm H_0$. The scattering operator
${\bf S}=W_+^\ast W_-$ is unitary and commutes with $H_0$.
Let $\hat{f}={\cal F}f$ be the Fourier transform of $f\in L_2({\Bbb R}^d)
$, and
let the unitary operator $F :L_2({\Bbb R}^d) \rightarrow
L_2({\Bbb R}_+; L_2({\Bbb S}^{d-1}))$ be defined by the formula
\[
(F f)(\lambda;\omega)=2^{-1/2}\lambda^{(d-2)/4}\hat{f}(\lambda^{1/2}\omega),
\quad \lambda>0, \quad \omega\in {\Bbb S}^2.
\]
Since ${\bf S}H_0=H_0 {\bf S}$, we have that
\[
(F{\bf S} f)(\lambda)=S(\lambda)(Ff)(\lambda)
\]
where the unitary operator
$S(\lambda): L_2({\Bbb S}^{d-1})\rightarrow L_2({\Bbb S}^{d-1})$
is known as the scattering matrix (SM).
The scattering amplitude $s(\omega,\omega^\prime;\lambda)$,
$\omega,\omega^\prime\in{\Bbb S}^{d-1}$, is kernel of
$S(\lambda)$ regarded as integral operator.
The following assertion is well known (see, e.g.,
\cite{Y5, Y2}).
\begin{proposition}\label{reg1A}
Let condition $(\ref{eq:H1sr})$ hold. Then
the operator $T(\lambda)=S(\lambda)-I$ is compact and it belongs to the
trace class
if $ \rho>d$. If $ \rho >d+n$,
$n=0,1,2,\ldots$, then
$T(\lambda) $ is integral operator with kernel from the class
$C^n({\Bbb S}^{d-1}\times {\Bbb S}^{d-1})$.
Let the condition
\[
|\partial^\alpha A(x)| \leq C_\alpha (1+|x|)^{- \rho-|\alpha| }
\]
hold for some $\rho \in (1,d)$ and all multi-indices $\alpha$. Then
the operator $T(\lambda) $ has integral kernel which is a
$C^\infty$-function off the diagonal
$\omega=\omega^\prime$ and is bounded by $C|\omega-\omega^\prime|^{-d+
\rho }$
as $\omega^\prime\rightarrow\omega$.
\end{proposition}
\bigskip
{\bf 2.2.}
Let us now discuss the behaviour of the SM with respect to gauge
transformations
defined by the formula
\begin{equation}\label{eq:GT1}
\tilde{H}=e^{i\phi}H e^{-i\phi}=(i\nabla+\tilde{A}(x))^2
\end{equation}
where $\phi$ is a differentiable function
and the magnetic potential
\begin{equation}\label{eq:GT}
\tilde{A}(x)=A(x)+{\rm grad}\,\phi(x).
\end{equation}
Of course ${\rm curl}\,\tilde{A}(x)={\rm curl}\,A(x) $.
We are particularly interested in functions $\phi(x) $ which are
asymptotically homogeneous of degree
zero. More precisely, we assume that
$\phi(x)=\phi_0( x)+ \phi_1(x)$ where $\phi_0( x)=\phi_0(\hat{x})$
and $ \phi_1(x)=o(1)$ as $ |x|\rightarrow\infty$.
Let us find a relation between the wave operators
$ W(H,H_0)$ and $ W(\tilde{H},H_0)$.
\begin{proposition}\label{GIW}
Let the wave operators
$W_\pm(H,H_0)$ exist, and
let $\phi $ be an asymptotically homogeneous function of degree zero with
homogeneous part $ \phi_0 $. Then the wave operators $ W_\pm(\tilde{H},H_0)$
also exist and
\begin{equation}\label{eq:GT2}
W_\pm(\tilde{H},H_0)=e^{i\phi (x)}W_\pm(H,H_0){\cal F}^*
e^{-i\phi_0(\pm\xi)}{\cal F}.
\end{equation}
\end{proposition}
Indeed, since
$$
(\exp(-iH_0t)f)(x)=e^{i|x|^2/(4t)}(2it)^{-d/2}\hat{f}(x/(2t))+o(1),
$$
we have that
\begin{eqnarray*}
(e^{-i\phi}\exp(-iH_0t)f)(x)&=&e^{i|x|^2/(4t)}(2it)^{-d/2} e^{-i\phi_0(\pm
x/(2t))}
\hat{f}(x/(2t))+o(1)
\\ &=&e^{i|x|^2/(4t)}(2it)^{-d/2}\hat{f}^{(\pm)}(x/(2t))+o(1),\quad
t\rightarrow\pm\infty,
\end{eqnarray*}
where
$\hat{f}^{(\pm)}(\xi)=e^{-i\phi_0(\pm\xi)}\hat{f}(\xi)$
and the remainder $o(1)$ tends to $0$ in $L_2({\Bbb R}^d)$
as $ t\rightarrow \pm\infty$.
This is equivalent to the relation
\[
e^{-i\phi}\exp(-iH_0t)f=\exp(-iH_0t)f^{(\pm)}+o(1),
\]
which, in view of
definition (\ref{eq:GT1}), implies that
\begin{eqnarray*}
W_\pm(\tilde{H},H_0)f =
\lim_{t\rightarrow\pm\infty} e^{i\tilde{H}t} e^{-iH_0t}f=
\lim_{t\rightarrow\pm\infty} e^{i\phi}e^{iHt} e^{-i\phi}e^{-iH_0t}f
\\
= \lim_{t\rightarrow\pm\infty}e^{i\phi} e^{iHt} e^{-iH_0t}f^{(\pm)}=
e^{i\phi}W_\pm(H,H_0)f^{(\pm)}.
\end{eqnarray*}
This proves (\ref{eq:GT2}).
As an immediate consequence of Proposition~\ref{GIW}, we obtain
a relation between the corresponding scattering operators and matrices.
\begin{proposition}\label{GI}
Under the assumptions of Proposition~$\ref{GIW}$,
the scattering operators for the pairs $H_0 ,H $ and $H_0 ,\tilde{H} $
are related by the equation
\begin{equation}\label{eq:GT3}
{\cal F}{\bf S}(\tilde{H},H_0){\cal F}^* = e^{i\phi_0(\xi)}{\cal
F}{\bf S}(H,H_0) {\cal F}^* e^{-i\phi_0(-\xi)}.
\end{equation}
The SM $S(\lambda)=S(H,H_0;\lambda)$ and
$\tilde{S}(\lambda)=S(\tilde{H},H_0;\lambda)$ satisfy
for all $\lambda>0$ the relations
\begin{equation}\label{eq:GT1S}
\tilde{S}(\lambda) =e^{i\phi_0(\omega) }
S(\lambda)e^{ -i\phi_0(-\omega)}
\end{equation}
or, in terms of the scattering amplitudes,
\begin{equation}\label{eq:GT1SA}
\tilde{s}(\omega,\omega';\lambda)=e^{i\phi_0(\omega)-i\phi_0(-\omega')}
s(\omega,\omega';\lambda).
\end{equation}
\end{proposition}
We emphasize that relations (\ref{eq:GT3})-(\ref{eq:GT1SA}) for
the scattering operators and
matrices (but not (\ref{eq:GT2}) for the wave operators) depend only on the
asymptotics $\phi_0$ of the function $\phi$. Probably, formulas
(\ref{eq:GT2}),
(\ref{eq:GT3}) appeared first in the paper \cite{RUI} in the case $d=2$ under
some assumptions on $A$. Actually, these formulas do not require any
assumptions at all.
Formulas (\ref{eq:GT1S}) or (\ref{eq:GT1SA}) show that the SM is not
determined by the magnetic
field $B(x)={\rm curl}\,A(x) $ only although we have an explicit formula
which connects the SM in
different gauges. This seems to contradict the following mental experiment.
Suppose that a quantum
particle interacts with a magnetic field.
Note that it is exactly a field but not a potential
which can be created by our hands. However, to
calculate the SM, we have to introduce a magnetic potential
and then solve the Schr\"odinger equation. Thus, the SM
depends on a potential. So it
appears that a particle itself
chooses a gauge convenient for it. There could be (at least) two possible
explanations of this
seeming contradiction. The first is that the scattering amplitude
$s(\omega,\omega^\prime;\lambda)$ cannot be measured
experimentally although it is widely believed to be possible. From this
point of view only the
scattering cross section $|s(\omega,\omega^\prime;\lambda)|^2$ can be
practically found which is
compatible with (\ref{eq:GT1SA}). Another point of view is that
experimental devices used for observation of a quantum particle are not
harmless and fix some
specific gauge.
On the other hand, for a given field,
the SM is stable with respect to short-range perturbations
of a potential.
\begin{proposition}\label{SMsr}
Let the wave operators
$W_\pm(H,H_0)$ exist.
Suppose that $ {\rm curl}\, A (x)= {\rm curl}\, \tilde{A} (x)$ and
\begin{equation}\label{eq:SMsr1}
\tilde{A} (x)-A (x)=O(|x|^{-\rho}), \quad \rho>1,
\end{equation}
as $|x|\rightarrow \infty$.
Then the wave operators $ W_\pm(\tilde{H},H_0)$
also exist and the scattering operators and matrices
for the pairs $H_0 ,H $ and $H_0 ,\tilde{H} $
coincide.
\end{proposition}
Indeed, according to Propositions~\ref{GIW} and \ref{GI}, it suffices to show
that $A (x)$ and $ \tilde{A} (x)$ are related by equality (\ref{eq:GT})
where the function $\phi(x)$ has a limit as $|x|\rightarrow \infty$.
Let us define $\phi(x)$ as a curvilinear integral
\[
\phi(x)= \int_{\Gamma_x} <\tilde{A} (y)- A (y), dy>
\]
taken between $0$ and a variable point $x$. By the Stokes theorem,
this integral does not depend on a
choice of $\Gamma_x$ so that equality (\ref{eq:GT}) holds.
Moreover, choosing $\Gamma_x$ as the piece of straight line connecting $0$
and $x=r\omega$,
$ \omega\in{\Bbb S}^{d-1}$, and using (\ref{eq:SMsr1}), we see that the
limit of
$\phi(r\omega)$ as $r \rightarrow \infty$ exists. It remains to show that
this limit does not
depend on $ \omega\in{\Bbb S}^{d-1}$. Again by the Stokes theorem,
\begin{equation}\label{eq:CurvB}
\phi(r\omega_2) - \phi(r\omega_1)=
\int_{{\Bbb S}_r[\omega_1,\omega_2]} <\tilde{A} (y)- A (y), dy>,
\end{equation}
where ${\Bbb S}_r[\omega_1,\omega_2]$ is the arc of the circle
centered at the origin and passing
through the points $ r\omega_1$ and $ r\omega_2$.
Condition (\ref{eq:SMsr1}) implies that integral (\ref{eq:CurvB})
tends to $0$ as $r \rightarrow \infty$.
\section{Main result}
{\bf 3.1.}
In this section we suppose that the dimension $d=3$.
Our results remain true for all $d\geq 3$ but not for $d=2$.
Let $B(x)=(B_1(x),B_2(x),B_3(x))$ be a magnetic field
such that ${\rm div}\,B(x)=0$ and
estimate (\ref{eq:B1}) is satisfied. Recall that a magnetic potential
$A^{(tr)}(x)=(A_1^{(tr)}(x),A_2^{(tr)}(x),A_3^{(tr)}(x))$ satisfying the
conditions
\begin{equation}\label{eq:curl}
{\rm curl}\, A^{(tr)}(x)=B(x), \quad =0
\end{equation}
can be constructed by the formula
\begin{equation} \label{eq:BB2}
A_1^{(tr)}(x)=\int_0^1\Bigl( B_2(sx)x_3- B_3(sx) x_2\Bigr)s ds.
\end{equation}
Expressions for components $ A_2^{(tr)}(x)$ and $ A_3^{(tr)}(x)$ are
obtained by
cyclic permutations of indices in (\ref{eq:BB2}).
It follows that $A^{(tr)}(x)$ admits the representation
\begin{equation} \label{eq:B3}
A^{(tr)}(x)=A^{(\infty)}(x)+A^{(reg)}(x),
\end{equation}
where
\begin{equation} \label{eq:Binf}
A_1^{(\infty)}(x)=|x|^{-2}\int_0^\infty\Bigl( B_2(s\hat{x})x_3-
B_3(s\hat{x}) x_2\Bigr)s ds,
\end{equation}
\begin{equation} \label{eq:Breg}
A_1^{(reg)}(x)=-|x|^{-2}\int_{|x|}^\infty\Bigl( B_2(s\hat{x})x_3-
B_3(s\hat{x}) x_2\Bigr)s ds.
\end{equation}
Thus,
$A^{(\infty)}(x)$ is a homogeneous function of degree $-1$
and $ A^{(reg)}(x)=
O(|x| ^{- \rho}) $
with $ \rho=r-1 >1$ as $|x|\ri\infty$.
According to formulas (\ref{eq:curl}) and (\ref{eq:B3})
${\rm curl}\,A^{(\infty)}(x)=O(|x| ^{- r}) $.
Since $r>2$ and ${\rm curl}\, A^{(\infty)}(x)$ is a homogeneous function of
degree $-2$,
this implies that
\begin{equation}\label{eq:reg1}
{\rm curl}\,A^{(\infty)}(x)=0, \quad x\neq 0.
\end{equation}
Similarly,
\begin{equation}\label{eq:WWX}
=0.
\end{equation}
\bigskip
{\bf 3.2.}
Given a magnetic field satisfying condition (\ref{eq:B1}), we construct now
a magnetic potential satisfying condition (\ref{eq:H1sr}).
We proceed from the magnetic potential $A^{(tr)}$ in the transversal gauge.
Let $ A ^{(\infty)}$ be function (\ref{eq:Binf}).
Then we define (cf. the proof of Proposition~\ref{SMsr}) the function
$U(x)$ for $ x\neq 0$ as a curvilinear integral
\begin{equation}\label{eq:Curv}
U(x)= \int_{\Gamma_{x_0,x}} < A^{(\infty)}(y), dy>
\end{equation}
taken between some fixed point $x_0 \neq 0$ and a variable point $x$. It
is required that
$0 \not\in\Gamma_{x_0,x}$,
so that, in view of (\ref{eq:reg1}) and the Stokes theorem, $U(x)$ does
not depend on a choice of
$\Gamma_{x_0,x}$. Here it is used that the set ${\Bbb R}^3\setminus\{0\}$
(and ${\Bbb
R}^d\setminus\{0\}$ for all $d\geq 3$) is simply connected. Clearly,
\begin{equation}\label{eq:GG}
A^{(\infty)} (x)={\rm grad}\, U(x).
\end{equation}
Moreover, the function $U(x)$ is homogeneous of degree $0$. Indeed, if
$x_2=\gamma x_1$,
$\gamma>1$, then we can choose $\Gamma_{x_0,x_2}= \Gamma_{x_0,x_1}\cup
(x_1,x_2)$ where
$ (x_1,x_2)$ is the piece of straight line connecting $x_1$ and $x_2$.
If $y \in(x_1,x_2)$, then according to (\ref{eq:WWX})
$=0$.
Hence $U(x_1)=U (x_2)$. We use definition (\ref{eq:Curv})
away from some neighbourhood of the point $x=0$, say, for $|x|\geq 1/2$, and
extend $U(x)$ as a differentiable
function to all $x\in{\Bbb R}^3$. For example, we can choose a function
$\eta\in C^\infty ({\Bbb R}^3)$ such that $\eta(x)=1$ for $|x|\geq 1/2$,
$\eta(x)=0$ in a neighbourhood of $x=0$ and then replace $U(x)$ by
$\eta(x) U(x)$.
Let us now set
\begin{eqnarray}\label{eq:GGx}
A(x) &=& A^{(tr)}(x)-{\rm grad}\, ( \eta(x) U(x))
\nonumber\\
&=& A^{(reg)}(x) + (1-\eta(x)) A^{(\infty)}(x) - U(x) {\rm grad}\,\eta(x),
\end{eqnarray}
so that $A(x)=A^{(reg)}(x)$ for $|x|\geq 1/2$ and $A(x)=A^{(tr)}(x)$
in a neighbourhood of $x=0$.
Thus, we have the following result.
\begin{proposition}\label{pot}
Suppose that ${\rm div}\,B(x)=0$ and
that condition $(\ref{eq:B1})$
holds. Let the magnetic potential $A(x)$ be defined by formula
$(\ref{eq:GGx})$ where $A^{(\infty)}(x)$, $A^{(reg)}(x)$ and $U(x)$
are functions $(\ref{eq:Binf})$, $(\ref{eq:Breg})$ and
$(\ref{eq:Curv})$, respectively.
Then $A(x)$ satisfies condition $(\ref{eq:potfield})$
and estimate $(\ref{eq:H1sr})$. Moreover, $A(x)$ has compact support
if $B(x)$ has compact support.
\end{proposition}
By the proof of Proposition~\ref{pot} we could have proceeded from the
magnetic potential
$A^{(c)}(x)$ satisfying the Coulomb gauge condition ${\rm div}\, A^{(c)}(x)=0$.
This is however less convenient. We emphasize that
Proposition~\ref{pot} is definitely not true for $d=2$ if the magnetic flux
$\Phi\neq 0$.
Let us denote by ${\cal A}(B)$ the class of
magnetic potentials satisfying (\ref{eq:potfield})
and (\ref{eq:H1sr}) for some $\rho>1$.
This class is non-empty according to Proposition~\ref{pot}.
If $A\in{\cal A}(B)$, then, for an arbitrary function
$\phi (x)$ such that ${\rm grad}\, \phi (x)=
O(|x|^{- \rho}) $, potential (\ref{eq:GT}) also belongs to this class.
However according to Proposition~\ref{SMsr} the SM
for the pair $H_0=-\Delta$, $H=(i\nabla+A(x))^2$ does not depend on the
choice of
$A\in{\cal A}(B)$ and, thus, is determined by
the magnetic field $B(x)$ only. We say that this SM $S(\lambda)=S(\lambda;B)$
is the SM for the field $B(x)$.
Comparing Propositions~\ref{reg1A} and \ref{pot}, we arrive at our main result.
\begin{theorem}\label{main}
Let a magnetic field $B(x)$ be such that
${\rm div}\,B(x)=0$ and
condition $(\ref{eq:B1})$ holds, and let
a magnetic potential $A\in{\cal A}(B)$. Then the wave operators
for the pair $H_0=-\Delta$, $H=(i\nabla+A(x))^2$
exist, are unitary and the SM
$ S(\lambda) $ for the magnetic field $B(x)$ is a unitary
operator for all $ \lambda>0 $.
The operator $T(\lambda)=S(\lambda)-I$ is compact and it belongs to the
trace class
if $ r>4$. If $r>4+n$,
$n=0,1,2,\ldots$, then
$T(\lambda) $ is integral operator with kernel from the class
$C^n({\Bbb S}^2\times {\Bbb S}^2)$. If
\[
|\partial^\alpha B(x)| \leq C_\alpha (1+|x|)^{- r-|\alpha| }
\]
for some $r \in (2,4)$ and all multi-indices $\alpha$, then
the operator $T(\lambda) $ has integral kernel which is a
$C^\infty$-function off the diagonal
$\omega=\omega^\prime$ and is bounded by $C|\omega-\omega^\prime|^{-4+ r }$
as $\omega^\prime\rightarrow\omega$.
\end{theorem}
Let
\[
\Sigma_{diff}(\omega ;\omega_0,\lambda)=(2\pi)^2
\lambda^{-1} | s (\omega ,\omega_0;\lambda)|^2, \quad\omega\neq\omega_0,
\]
be the scattering cross section for an incident direction $\omega_0$ of a
beam of particles and a
direction of observation $\omega$.
\begin{corollary} \label{B6}
If estimate $(\ref{eq:B1})$ is satisfied for $r>4$, then
$\Sigma_{diff}(\omega ;\omega_0,\lambda)$ is a bounded function of
$\omega\in{\Bbb S}^2$. If condition $(\ref{eq:B1})$ is satisfied
for some $r \in (2,4)$ and all multi-indices $\alpha$, then
\[
\Sigma_{diff}(\omega ;\omega_0,\lambda)=
O(|\omega -\omega_0|^{-8+2r})
\]
as $\omega \rightarrow\omega_0$.
\end{corollary}
Using formula (\ref{eq:GGx}) and applying
Proposition~\ref{GI} to the function $\phi(x)=U(x)$, we obtain also the
following
\begin{proposition} \label{B6tr}
Suppose that a magnetic field $B(x)$
satisfies the assumptions of Theorem~$\ref{main}$.
Let $ S^{(tr)}(\lambda) $ be the SM for the pair $H_0=-\Delta$,
$H=(i\nabla+A^{(tr)}(x))^2$.
Set
\[
u(\omega)=U(\omega)-U(-\omega)
\]
where the function $U(x)$
is defined by formula $(\ref{eq:Curv})$.
Let $S_0$ be the operator of
multiplication by the function $\exp(iu(\omega) )$.
Then all the results of
Theorem~$\ref{main}$ are true for the operator
$T^{(tr)}(\lambda)= S^{(tr)}(\lambda)-S_0$.
\end{proposition}
Let us introduce the circulation
\begin{equation}\label{eq:EABAB}
I(x,\omega)= \int_{-\infty}^\infty
dt,\quad |\omega|=1,
\quad x\neq 0,
\end{equation}
of $A^{(\infty)}(x)$ over the straight line $x+ t\omega$, $t\in{\Bbb R}$.
We shall consider this function only on the set $=0$.
Integral (\ref{eq:EABAB}) is convergent by virtue of transversal
condition (\ref{eq:WWX}).
It is easy to see that $I(x,\omega)$ is a homogeneous function of degree
zero in the variable $x$.
Under assumption (\ref{eq:reg1}) the function $I(x,\omega)$ does not
actually depend on $x$
and equals $u(\omega)$.
Indeed, it follows from (\ref{eq:GG}) that
\begin{eqnarray}\label{eq:GG2}
I(x,\omega)=\int_{-\infty}^\infty <{\rm
grad}\,U(x+t\omega),\omega>d t
= \lim_{T\rightarrow\infty}\int_{-T}^T \frac{d}{dt}U(x+t\omega) dt
\nonumber\\
=\lim_{T\rightarrow\infty}(U(x+T\omega)-U(x-T\omega)) =U(\omega)-U(-\omega).
\end{eqnarray}
\bigskip
{\bf 3.3.}
As a concrete example, let us consider a toroidal solenoid
${\bf T}$ in the space ${\Bbb R}^3$ symmetric with respect to rotations
around the $x_3$-axis
(which does not intersect ${\bf T}$).
We assume that the section of ${\bf T}$ by a half-plane passing through
the $x_3$-axis is convex
and has a smooth boundary
$\partial{\bf T}$ but is not necessarily a disc. Suppose
(which looks quite realistic) that
\[
B(x_1,x_2,x_3)= \alpha (x_1^2+x_2^2)^{-1}(-x_2, x_1,0), \quad
\alpha={\rm const},
\]
inside of ${\bf T}$ and is zero outside. Then ${\rm div}\,B(x)=0$ and
the current ${\rm curl}\,B(x)=0$ if $x\not\in\partial{\bf T}$.
Of course, Theorem~\ref{main} applies to this field and hence
$ S(\lambda)-I$ is integral operator with kernel from the class
$C^\infty({\Bbb S}^2\times {\Bbb S}^2)$.
Let us illustrate our construction on this example.
In particular, we shall find the SM $S^{(tr)}(\lambda)$.
First, we construct the potential $A^{(tr)}(x)$ by formula (\ref{eq:BB2}).
Let the half-line $L_z$ consist of points $s (1+z^2)^{-1/2}(1,0,z)$ for all
$s\in{\Bbb
R}_+$. Denote by $z_1$ and $z_2$ the values of $z$ for which the
half-line $L_z$
is tangent to $\partial{\bf T}$ and, for $z\in[z_1,z_2]$,
denote by $\varkappa_\pm(z)$, $\varkappa_+(z)\geq
\varkappa_-(z)$, the values of $s$ for which $L_z$ intersects $\partial{\bf
T}$.
Set $z=z(x)=x_3 (x_1^2+ x_2^2)^{-1/2}$. Taking into account the rotational
symmetry, we
see that a point $s(x_1,x_2,x_3)\in{\bf T}$ if and only if
$s|x|(1+z^2)^{-1/2}(1,0,z)\in{\bf T}$ or $\varkappa_-(z)\leq s|x| \leq
\varkappa_+(z)$. Thus, integral (\ref{eq:BB2}) equals zero (and hence
$A^{(tr)}(x)=0$)
if $z\not\in (z_1,z_2)$ and it is actually taken over the set
$[0,1]\cap [\varkappa_-(z)|x|^{-1},
\varkappa_+(z)|x|^{-1}]$ if $z \in (z_1,z_2)$.
Therefore $A^{(tr)}(x)=0$ if $|x|\leq \varkappa_-(z)$,
\[
A^{(tr)}_1(x)= \alpha x_1 x_3
(x_1^2+x_2^2)^{-1}\int_{\varkappa_-(z)/|x|}^1 ds =\alpha x_1 x_3
(x_1^2+x_2^2)^{-1}(1-\varkappa_-(z)/|x|)
\]
if $\varkappa_-(z)\leq |x|\leq \varkappa_+(z)$ and
\[
A^{(tr)}_1(x)= \alpha x_1 x_3
(x_1^2+x_2^2)^{-1}\int_{\varkappa_-(z)/|x|}^{\varkappa_+(z)/|x|} ds
=\alpha x_1 x_3 (x_1^2+x_2^2)^{-1}|x|^{-1}(\varkappa_+(z)-\varkappa_-(z) )
\]
if $ |x|\geq \varkappa_+(z)$. Components $A^{(tr)}_2(x)$ and
$A^{(tr)}_3(x)$ can be found quite
similarly. In particular,
%%%% we see that the first derivatives of $A^{(tr)}(x)$ are bounded and
\begin{equation} \left.\begin{array}{lcl}
A^{(\infty)}_1(x)&=& x_1 x_3 (x_1^2+x_2^2)^{-1} |x|^{-1} g
(z),
\nonumber\\
A^{(\infty)}_2(x)&=& x_2 x_3 (x_1^2+x_2^2)^{-1} |x|^{-1} g (z),
\nonumber\\
A^{(\infty)}_3(x)&=&- |x|^{-1} g (z),
\end{array}\right\}
\label{eq:lagrA}\end{equation}
where
\[
g (z)= \alpha(\varkappa_+ (z) - \varkappa_- (z) ).
\]
Clearly, $g(z)$ is a bounded function, $\pm g(z)> 0$ if $\pm\alpha > 0$ for
$z \in(z_1,z_2)$ and $g(z)=0$ for $z\not\in (z_1,z_2)$.
The function $g(z)$ can be calculated explicitly if the section of ${\bf
T}$ by a half-plane
passing through the $x_3$-axis is a disc. Suppose that this disc has radius
$r$, its center belongs to the plane $x_3 =0$ and the distance from the
center to the $x_3$-axis
is $l$, $l>r$. Then the equation of $\partial{\bf T}$ is
\[
((x_1^2 +x_2^2)^{1/2}-l)^2+x_3^2=r^2.
\]
The numbers $(1+z^2)^{-1/2}\varkappa_\pm (z)$ are defined as the roots of
this equation
for $x_1$ if $x_3=zx_1$, $x_2=0$, so that
\[
\varkappa_\pm (z)=(1+z^2)^{-1/2}(l\pm (r^2-(l^2-r^2)z^2)^{1/2})
\]
and hence
\[
g(z)=2\alpha(1+z^2)^{-1/2} (r^2-(l^2-r^2)z^2)^{1/2}.
\]
In particular, $-z_1=z_2=r(l^2-r^2)^{-1/2}$ for this function.
Returning to the general case and taking into account (\ref{eq:lagrA}),
we see that a function
$U(x)$ satisfying (\ref{eq:GG}) can be constructed by the explicit formula
\[
U(x)=G(x_3 (x_1^2+ x_2^2)^{-1/2}),
\]
where
$$
G^\prime (z)=-g(z)(z^2+1)^{-1/2}.
$$
According to Proposition~\ref{B6tr}, up to an integral operator with
$C^\infty$-kernel, the SM $S^{(tr)}(\lambda)$ is the operator
$S_0$ of multiplication by the function $\exp (iu(\omega))$,
where
\begin{equation}\label{eq:quI}
u(\omega)= q(\omega_3 (1- \omega_3^2)^{-1/2}) ,\quad
\omega =(\omega_1,\omega_2,\omega_3)\in {\Bbb S}^2,
\end{equation}
and
\[
q(z)=G(z)-G(-z)=-\int_{-z}^z g(t)(t^2+1)^{-1/2} dt.
\]
Clearly, $q(-z)=-q(z) $, $ q( z) $ is an increasing (decreasing)
function if $\alpha<0$ ($\alpha>0$) and it
is a constant, $ q( z)=q_0 $, if
$z\geq \max\{ -z_1,z_2\}$.
Let us finally show that $ -q_0 $ equals the
magnetic flux $\Phi_s$ through a section of toroidal solenoid ${\bf T}$.
Let $\omega_0=(0,0,1)$, $=0$, $x_0\neq 0$, and let the
function $I$
be defined by formula (\ref{eq:EABAB}). Then it follows from (\ref{eq:GG2})
and (\ref{eq:quI})
that $ q_0=u(\omega_0) =I(x_0,\omega_0)$. By the Stokes theorem,
$\Phi_s$ equals
the circulation of the potential
$A^{(tr)}(x)$ over the closed contour formed by the $x_3$-axis and the
line $x_0 +\omega_0 t$,
where $|x_0|$ is sufficiently large and $t$ runs from $ \infty$ to
$-\infty$.
We remark that the integrals over pieces of lines connecting
these infinite parallel lines as well as the
integral over the $x_3$-axis are equal to zero. Therefore, by
(\ref{eq:EABAB}),
\[
\Phi_s=-I(x_0,\omega_0)=-q_0.
\]
Note that smooth potentials satisfying (\ref{eq:lagrA}) for $|x|$ large
were considered in
\cite{R} by the method of \cite{Y3}.
\section{Magnetic potentials with Coulomb decay}
Here we consider arbitrary magnetic potentials
$ A(x) $, $x\in{\Bbb R}^2$, with Coulomb decay at infinity
satisfying, at least asymptotically, the transversal gauge condition.
For such potentials, the magnetic field $B(x)={\rm curl}\,A(x)$
decays, in general, as $|x|^{-2}$ at infinity,
so that the assumptions of the previous section are
not satisfied. We shall show that in this case the SM
contains a singular integral operator and hence the Aharonov-Bohm effect
occurs.
Recall the following result obtained in \cite{Y2}.
\begin{theorem}\label{SMsing}
Suppose that a magnetic potential
$ A(x) $ is a $C^\infty$-function and admits the representation
\[
A(x)=A^{(\infty)}(x)+A^{(reg)}(x),
\]
where $A^{(\infty)}(x)$ is a homogeneous
function of degree $-1$
satisfying transversal condition
$(\ref{eq:WWX})$
and $ \partial^\alpha A^{(reg)}(x)=
O ( |x| ^{-\tilde{\rho}-|\alpha|})$, $ \tilde{\rho}>1$, as $|x|\ri\infty$
for all $\alpha$.
Let $\Pi_\omega$ be the plane orthogonal to $\omega$, and let
the function $I(x,\omega)$ be defined by $(\ref{eq:EABAB})$.
Set
\[
p^{(av)}(\omega)=(2\pi)^{-1}\int_{{\Bbb S}_\omega}
\exp(i I(\psi, \omega ))d\psi,
\quad {\Bbb S}_\omega ={\Bbb S}^2\cap
\Pi_\omega,
\]
\begin{equation}\label{eq:DD1}
q ( \omega,\varphi)= -(2\pi)^{-2} \int_{{\Bbb S}_\omega }
\Bigl( \exp(i I (\psi, \omega ))- p^{(av)}(\omega)\Bigr)
(<\psi,\varphi>-i0)^{-2}d \psi
\end{equation}
and
\[
s_0(\omega ,\omega^\prime)
=p^{(av)}(\omega)\delta(\omega,\omega^\prime)+
{\rm P.V.} q(\omega, \omega^\prime-\omega ).
\]
Then for arbitrary $\lambda>0$ the scattering amplitude satisfies the estimate
\[
|s(\omega,\omega';\lambda)-s_0
(\omega,\omega')|\leq C|\omega-\omega^\prime|^{-2+\nu},
\]
where $\nu= \tilde{\rho} -1$ if $\tilde{\rho}\in (1,2)$ and $\nu=1$ if
$\tilde{\rho}\geq 2$.
\end{theorem}
We emphasize that the function $q( \omega,\varphi)$ is homogeneous in $\varphi$
of degree $-2$ and satisfies the condition
\[
\int_{{\Bbb S}_\omega}
q (\omega,\varphi)d\varphi=0.
\]
Therefore the integral operator $Q$
with kernel $q(\omega,\omega^\prime-\omega)$ is well defined (as a
bounded operator in
$L_2({\Bbb S}^2)$) in the sense of the principal value by the formula
\[
(Q f)(\omega)=
\lim_{\varepsilon\to 0}\int_{ |\omega' - \omega|>\varepsilon}
q (\omega, \omega' - \omega) f(\omega')d\omega'.
\]
Theorem~\ref{SMsing} implies
\begin{corollary} \label{B6as}
If $\omega\neq\omega^\prime$ but $\omega-\omega^\prime\rightarrow 0 $, then
\[
s(\omega ,\omega^\prime ;\lambda)= q(\omega,
\omega^\prime-\omega ) + O(|\omega -\omega^\prime|^{-2+\nu}).
\]
\end{corollary}
\begin{corollary} \label{section}
If $\omega\rightarrow\omega_0$, then
\[
\Sigma_{diff}(\omega ;\omega_0,\lambda)=(2\pi)^2
\lambda^{-1} |q (\omega_0, \omega_0-\omega ) |^2
+ O(|\omega -\omega_0|^{-4+\nu}).
\]
\end{corollary}
Note that the order of singularity $|\omega - \omega_0 |^{-4}$
here is the same as for electric Coulomb potentials.
If, under the assumptions of Theorem~\ref{SMsing}, ${\rm curl}\,
A(x)=o(|x|^{-2})$ as
$|x|\rightarrow\infty$, then necessarily condition (\ref{eq:reg1}) is satisfied
and hence function (\ref{eq:Curv}) is correctly defined. It follows from
equality (\ref{eq:GG2}) that in this case
the function $I(x,\omega)$ does not depend on $x$ and, thus,
$q (\omega,\varphi)=0$. This fact agrees of course with the results of the
previous section
for magnetic fields satisfying condition
(\ref{eq:B1}).
Nevertheless $q (\omega,\varphi)$ is non-trivial in the general case.
Let us consider two concrete examples
of potentials $A^{(\infty)}(x)$ homogeneous of degree $-1$ and satisfying
transversal condition (\ref{eq:WWX}).
The corresponding
fields ${\rm curl}\, A(x)$ decay only as $|x|^{-2}$ at infinity,
so that the conclusion of
Theorem~\ref{main} is violated.
We define the first of these potentials by the equation
\[
A^{(\infty)}(x)=|x|^{-3}(\alpha_1 x_2 x_3,\alpha_2 x_3 x_1 , \alpha_3 x_1
x_2),
\quad x=(x_1,x_2,x_3)\in{\Bbb R}^3,
\]
where $\alpha_j$ are constants and
$\alpha_1+\alpha_2 + \alpha_3 =0$.
An easy calculation shows that
function (\ref{eq:EABAB}) equals
\begin{equation}\label{eq:EX2}
I(x,\omega)= 2 |x|^{-2} (\alpha_1\omega_1 x_2 x_3+\alpha_2 \omega_2 x_3 x_1+
\alpha_3\omega_3 x_1 x_2).
\end{equation}
As another example,
we choose a modification of the Aharonov-Bohm potential
\[
A^{(\infty)}(x)=\alpha |x|^{-2}(- x_2, x_1,0).
\]
In this case
\begin{equation}\label{eq:EXA21}
I(x,\omega)= \pi\alpha |x|^{-1} ( \omega_2 x_1-\omega_1 x_2).
\end{equation}
Functions (\ref{eq:EX2}) and (\ref{eq:EXA21}) depend on the variables
$x_1, x_2, x_3$ so that function
(\ref{eq:DD1}) is non-trivial (see \cite{Y2}, for details).
In both these examples the SM contain singular integral operators
and hence the Aharonov-Bohm effect occurs.
Thus, a long-range behaviour (as $b(\hat{x}) |x|^{-2}$)
of a magnetic field in dimension $3$
plays the same role as the topological obstruction
in dimension $2$ if the flux $\Phi\not\in
2\pi{\Bbb Z}$.
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\end{document}