\documentclass[reqno]{amsart}
\usepackage[mathscr]{eucal}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 55, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2005/55\hfil Zero modes of the Pauli operator]
{On the Aharonov-Casher formula for different self-adjoint extensions
of the Pauli operator with singular magnetic field}
\author[M. Persson\hfil EJDE-2005/55\hfilneg]
{Mikael Persson}
\address{Mikael Persson \hfill\break
Department of Mathematical Sciences\\
Division of Mathematics\\
Chalmers University of Technology\\
and G\"oteborg University\\
412 96 G\"oteborg, Sweden}
\email{mickep@math.chalmers.se}
\urladdr{http://www.math.chalmers.se/$\sim$mickep}
\date{}
\thanks{Submitted February 17, 2005. Published May 24, 2005.}
\subjclass[2000]{81Q10, 35Q40, 47F05}
\keywords{Schr\"odinger operators; spectral analysis}
\begin{abstract}
Two different self-adjoint Pauli extensions describing a
spin-1/2 two-dimensional quantum system with singular
magnetic field are studied. An Aharonov-Casher type formula
is proved for the maximal Pauli extension and the possibility
of approximation of the two different self-adjoint extensions
by operators with regular magnetic fields is investigated.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{example}[thm]{Example}
\newtheorem*{remark}{Remark}
\newcommand{\gnorm}[1]{\left\lvert\mspace{-1.7mu}\left\lvert\mspace{-1.7mu}\left\lvert#1%
\right\rvert\mspace{-1.7mu}\right\rvert\mspace{-1.7mu}\right\rvert}
\section{Introduction} \label{sec:intro}
Two-dimensional spin-$1/2$ quantum systems involving magnetic
fields are described by the self-adjoint Pauli operator. One
interesting question about such systems is the appearance of zero
modes (eigenfunctions with eigenvalue zero). Aharonov and Casher
proved in \cite{ac} that if the magnetic field is bounded and
compactly supported, then zero modes can arise, and the number of
zero modes is simply connected to the total flux of the magnetic
field. Since then, Aharonov-Casher type formulas have been proved
for more and more singular magnetic fields in different settings,
see \cite{cfks,gegr,lali,mi}. Recently they were proved for
measure-valued magnetic fields in \cite{ervo} by Erd\H{o}s
and Vougalter.
We are interested in the Pauli operator when the magnetic field
consists of a regular part with compact support and a singular
part with a finite number of Aharonov-Bohm (AB) solenoids
\cite{ab}. The Pauli operator for such singular magnetic fields,
defined initially on smooth functions with support not touching
the singularities, is not essentially self-adjoint. Thus there are
several ways of defining the self-adjoint Pauli extension,
depending on what boundary conditions one sets at the AB
solenoids, see \cite{adte,dast,exstvy,gest,gest2}. Different
extensions describe different physics, and there is a discussion
going on about which extensions describe the real physical
situation.
There are two possible approaches to making the choice of the
extension: trying to describe boundary conditions at the
singularities by means of modelling actual interaction of the
particle with an AB solenoid, or considering approximations of
singular fields by regular ones, see \cite{bopu,ta}. We are going
to study the maximal extension introduced in \cite{gegr}, called
the Maximal Pauli operator, and compare it with the extension
defined in \cite{ervo}, that we will call the EV Pauli operator.
These two extensions were recently studied in \cite{rosh} in the
presence of infinite number of AB solenoids, and it was proved
that a magnetic field with infinite flux gives an
infinite-dimensional space of zero modes for both extensions.
When studying the Pauli operator in the presence of AB solenoids
one must always keep in mind the possibility to reduce the
intensities of solenoids by arbitrary integers by means of
singular gauge transformations. In Section~\ref{sec:def} we define
both extensions via quadratic forms. The Maximal Pauli operator
can be defined directly for arbitrary strength of the AB fluxes,
while the EV Pauli operator is defined via gauge transformations
if the AB intensities do not belong to the interval $[-1/2,1/2)$.
The EV Pauli operator has the advantage that the Aharonov-Casher
type formula in its original form holds even for singular AB
magnetic fields. However, as we show, it does not satisfy another
natural requirement, that the number of zero modes is invariant
under the change of sign of the magnetic field. This absence of
invariance exhibits itself only if both singular and regular parts
of the field are present. This justifies our attempt to study the
Maximal Pauli operator which lacks the latter disadvantage. The
price we have to pay for this is that our Aharonov-Casher type
formula has certain extra terms.
For the Dirac operators with strongly singular magnetic field the
question on the number of zero modes was considered in
\cite{hiog}. The definition of the self-adjoint operator
considered there is close to the one in Erd\H{o}s-Vougalter, however
it is not gauge invariant, therefore the Aharonov Casher-type
formula obtained in \cite{hiog} depends on intensity of each AB
solenoid separately.
In Section~\ref{sec:prop} we establish that the Maximal Pauli
operator is gauge invariant and that changing the sign of the
magnetic field leads to anti-unitarily equivalence. Our main
result is the Aharonov-Casher type formula for the Maximal Pauli
operator. An interesting fact is that this operator can have both
spin-up and spin-down zero modes, in contrary to the EV Pauli
operator and the Pauli operator for less singular magnetic fields,
which have either spin-up or spin-down zero modes, but not both.
In \cite{gegr} a setting with an infinite lattice of AB solenoids
with equal AB flux at each solenoid is studied, having both
spin-up and spin-down zero modes, both with infinite multiplicity.
In Section~\ref{sec:approx} we discuss the approximation by more
regular fields in the sense of Borg and Pul\'{e}, see \cite{bopu}. It
turns out that both the Maximal Pauli operator and the EV Pauli
operator can be approximated in this way. However, the EV Pauli
operator can be approximated as a Pauli Hamiltonian, while the
Maximal Pauli operator can only be approximated one component at a
time. Since different ways of approximating the magnetic field may
lead to different results, see \cite{bovo,ta}, we leave the
question if the Maximal Pauli operator can be approximated as
Pauli Hamiltonian open.
\section{Definition of the Pauli operators}
\label{sec:def}
The Pauli operator is formally defined as
\[
P=\left(\sigma\cdot\left(-i\nabla+\mathbf{A}\right)\right)^2
=\left(-i\nabla+\mathbf{A}\right)^2+\sigma_3 B
\]
on $L_2(\mathbb{R}^2)\otimes \mathbb{C}^2$. Here $\sigma=(\sigma_1,\sigma_2)$,
where $\sigma_1$, $\sigma_2$ and $\sigma_3$ are the Pauli matrices
\[
\sigma_1=
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
,\quad
\sigma_2=
\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}
,\quad \text{and}\quad
\sigma_3=
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix},
\]
where $\mathbf{A}$ is the real magnetic vector potential and
$B=\mathop{\rm curl}(\mathbf{A})$ is the magnetic field. This definition does
not work if the magnetic field $B$ is too singular, see the
discussion in \cite{ervo,so}. If $\mathbf{A}\in
L_{2,\text{loc}}(\mathbb{R}^2)$, using the notations $\Pi_k =
-i\partial_k+A_k$, for $k=1,2$, $Q_\pm= \Pi_1\pm i\Pi_2$ and
$\lambda$ for the Lebesgue measure, the Pauli operator can be
defined via the quadratic form
\begin{equation}
\label{eq:kvadform}
p[\psi]=\|Q_+\psi_+\|^2+\|Q_-\psi_-\|^2=\int|\sigma
\cdot(-i\nabla+\mathbf{A})\psi|^2d\lambda(x),
\end{equation}
the domain being the closure in the sense of the metrics $p[\psi]$ of the
core consisting of smooth compactly supported functions. With this notation,
we can write the Pauli operator $P$ as
\begin{equation}
\label{eq:paulinotion}
P=\begin{pmatrix} P_+ & 0\\ 0 & P_-\end{pmatrix}=\begin{pmatrix} Q_+^*Q_+ & 0\\
0 & Q_-^*Q_-\end{pmatrix}.
\end{equation}
However, defining the Pauli operator via the quadratic form
$p[\psi]$ in~\eqref{eq:kvadform} requires that the vector
potential $\mathbf{A}$ belongs to $L_{2,\text{loc}}(\mathbb{R}^2)$,
otherwise $p[\psi]$ can be infinite for nice functions $\psi$, see
\cite{so}. If the magnetic field consists of only one AB solenoid
located at the origin with intensity (flux divided by $2\pi$)
$\alpha$, then the magnetic vector potential $\mathbf{A}$ is given
by $\mathbf{A}(x_1,x_2)=\frac{\alpha}{x_1^2+x_2^2}(-x_2,x_1)$ which
is not in $L_{2,\text{loc}}(\mathbb{R}^2)$. Here, and elsewhere we
identify a point $(x_1,x_2)$ in the two-dimensional space
$\mathbb{R}^2$ with $z=x_1+ix_2$ in the complex plan $\mathbb{C}$.
Following \cite{ervo}, we will define the Pauli operator via
another quadratic form, that agrees with $p[\psi]$ for less
singular magnetic fields. We start by describing the magnetic
field.
Even though the Pauli operator can be defined for more general magnetic fields,
in order to demonstrate the main features of the study, without extra
technicalities, we restrict ourself to a magnetic field consisting of a
sum of two parts, the first being a smooth function with compact support,
the second consisting of finitely many AB solenoids.
Let $\Lambda=\{z_j\}_{j=1}^n$ be a set of distinct points in $\mathbb{C}$
and let $\alpha_j\in\mathbb{R}\setminus\mathbb{Z}$. The magnetic field we
will study in this paper has the form
\begin{equation}
\label{eq:magnet}
B(z)=B_0(z)+\sum_{j=1}^n 2\pi\alpha_j\delta_{z_j},
\end{equation}
where $B_0\in C_0^1(\mathbb{R}^2)$. In \cite{ervo} the magnetic field is
given by a signed real regular Borel measure $\mu$ on $\mathbb{R}^2$
with locally finite total variation. It is clear that
$\mu=B_0(z)d\lambda(z)+\sum_{j=1}^n 2\pi\alpha_j\delta_{z_j}$ is such a measure.
The function $h_0$ given by
\[
h_0(z)=\frac{1}{2\pi}\int \log|z-z'|B_0(z')d\lambda(z')
\]
satisfies $\Delta h_0=B_0$ since $B_0\in C_0^1(\mathbb{R}^2)$ and
$\Delta \log|z-z_j|=2\pi\delta_{z_j}$ in the sense of distributions.
The function
\[
h(z)=h_0(z)+\sum_{j=1}^n \alpha_j\log|z-z_j|
\]
satisfies $\Delta h=B$. It is easily seen that $h_0(z)\sim\Phi_0\log|z|$
as $|z|\to\infty$, and thus the asymptotics of $e^{h(z)}$ is
\[
e^{\pm h(z)}\sim
\begin{cases}
|z|^{\pm \Phi}, & |z|\to\infty\\
|z-z_j|^{\pm \alpha_j}, & z\to z_j,
\end{cases}
\]
where $\Phi_0=\frac{1}{2\pi}\int B_0(z)d\lambda(z)$ and
$\Phi=\frac{1}{2\pi}\int B(z)d\lambda(z)=\Phi_0+\sum_{j=1}^n\alpha_j$.
We are now ready to define the two self-adjoint Pauli operators.
The decisive difference between them is the sense in which we are
taking derivatives. This leads to different domains, and, as we
will see in later sections, to different properties of the
operators. Let us introduce notations for taking derivatives on
the different spaces of distributions. Remember that
$\Lambda=\{z_j\}_{j=1}^n$ is a finite set of distinct points in
$\mathbb{C}$. We let the derivatives in $\mathcal{D}'(\mathbb{R}^2)$
be denoted by $\partial$ and the derivatives in
$\mathcal{D}'(\mathbb{R}^2\setminus\Lambda)$ be denoted by $\partial$
with a tilde over it, that is $\tilde{\partial}$. Thus, for example, by
$\partial_z$ we mean $\frac{\partial}{\partial z}$ in the space
$\mathcal{D}'(\mathbb{R}^2)$ and
by $\tilde{\partial}_z$ we mean $\frac{\partial}{\partial z}$ in the space
$\mathcal{D}'(\mathbb{R}^2\setminus\Lambda)$.
\subsection{The EV Pauli operator}
We follow \cite{ervo} and define the sesquilinear forms $\pi_+$
and $\pi_-$ by
\begin{align*}
\pi_+^h(\psi_+,\xi_+)&=4\int \overline{\partial_{\bar{z}}\left(e^{-h}\psi_+\right)}
\partial_{\bar{z}}\left(e^{-h}\xi_+\right)e^{2h}d\lambda(z),\\
\mathscr{D}(\pi^h_+)&=\left\{\psi_+\in L_2(\mathbb{R}^2):\pi^h_+(\psi_+,\psi_+)
<\infty\right\},
\end{align*}
and
\begin{align*}
\pi_-^h(\psi_-,\xi_-)&=4\int \overline{\partial_z\left(e^{h}\psi_-\right)}
\partial_z\left(e^{h}\xi_-\right)e^{-2h}d\lambda(z),\\
\mathscr{D}(\pi^h_-)&=\left\{\psi_-\in L_2(\mathbb{R}^2) :
\pi^h_-(\psi_-,\psi_-)<\infty\right\}.
\end{align*}
Set
\begin{align*}
\pi^h(\psi,\xi)&=\pi_+^h(\psi_+,\xi_+)+\pi_-^h(\psi_-,\xi_-),\\
\mathscr{D}(\pi^h)&=\mathscr{D}(\pi^h_+)\oplus\mathscr{D}(\pi^h_-)
=\big\{\psi=\begin{pmatrix}\psi_+\\ \psi_-\end{pmatrix}\in
L_2(\mathbb{R}^2)
\otimes\mathbb{C}^2:\pi^h(\psi,\psi)<\infty\big\}.
\end{align*}
Let us make more accurate the description of the domains of the
forms $\pi_\pm^h$ and $\pi^h$. For example, what is required of
a function $\psi_+$ to be in $\mathscr{D}(\pi^h_+)$? It should belong
to $L_2(\mathbb{R}^2)$, and the expression
\[
\pi_+^h(\psi_+,\psi_+)=4\int \left|\partial_{\bar{z}}\left(e^{-h}\psi_+\right)
\right|^2e^{2h}d\lambda(z)
\]
should have a meaning and be finite. This means that the
distribution $\partial_{\bar{z}}\left(e^{-h}\psi_+\right)$ actually must be a
function and its modulus multiplied with $e^h$ must belong to
$L_2(\mathbb{R}^2)$, that is
$|\partial_{\bar{z}}\left(e^{-h}\psi_+\right)|e^h\in L_2(\mathbb{R}^2)$. This
forces all the intensities $\alpha_j$ to be in the interval
$(-1,1)$, see \cite{ervo}.
Next we define the norm by
\[
\gnorm{\psi}^2_{\pi^h}=\gnorm{\psi_+}^2_{\pi_+^h}+\gnorm{\psi_-}^2_{\pi_-^h},
\]
where
\[
\gnorm{\psi_+}^2_{\pi_+^h}=\|\psi_+\|^2+\left\|\partial_{\bar{z}}\left(e^{-h}
\psi_+\right)e^h\right\|^2
\]
and
\[
\gnorm{\psi_-}^2_{\pi_-^h}=\|\psi_-\|^2+\left\|\partial_z\left(e^{h}\psi_-\right)
e^{-h}\right\|^2.
\]
This form $\pi^h$ is symmetric, nonnegative and closed with
respect to $\|\cdot\|$, again see \cite{ervo}, and hence it
defines a unique self-adjoint operator $\mathcal{P}_h$ via
\begin{equation}
\label{eq:pdefenr}
\mathscr{D}(\mathcal{P}_h)=\{\psi\in\mathscr{D}(\pi^h) : \pi^h(\psi,\cdot)\in
\left(L_2(\mathbb{R}^2)\otimes\mathbb{C}^2\right)\}
\end{equation}
and
\begin{equation}
\label{eq:pdomenr}
(\mathcal{P}_h\psi,\xi)=\pi^h(\psi,\xi),\quad \psi\in\mathscr{D}(\mathcal{P}_h),
\xi\in\mathscr{D}(\pi^h).
\end{equation}
We call this operator $\mathcal{P}_h$ the \emph{non-reduced EV Pauli operator}.
If some intensities $\alpha_j$ belongs to $\mathbb{R}\setminus[-1/2,1/2)$,
we let $\alpha_j^*$ be the unique real number in $[-1/2,1/2)$ such that
$\alpha_j$ and $\alpha_j^*$ differ only by an integer, that is
$\alpha_j^*-\alpha_j=m_j\in\mathbb{Z}$. We define the
\emph{reduced EV Pauli operator} (or just the \emph{EV Pauli
operator}), $P_h$, to be
\begin{equation}
\label{eq:pdefe}
P_h=\exp(i\phi)\mathcal{P}_h\exp(-i\phi)
\end{equation}
where $\phi(z)=\sum_{j=1}^n m_j \arg(z-z_j)$. Hence, if there are some $\alpha_j$
outside the interval $(-1,1)$ only the reduced EV Pauli operator is well-defined. If
all the intensities $\alpha_j$ belong to the interval $[-1/2,1/2)$ then we do not
have to perform the reduction and hence there is only one definition. However, if
there are intensities $\alpha_j$ inside the interval $(-1,1)$ but outside the
interval $[-1/2,1/2)$ then we have two different definitions of the EV Pauli
operator, the direct one and the one obtained by reduction. In the next section we
will show that these two operators are not the same.
\subsection{The Maximal Pauli operator}
Let $\alpha_j\in\mathbb{R}\setminus\mathbb{Z}$. We define the forms
\begin{align*}
\mathfrak{p}_+^h(\psi_+,\xi_+)&=4\int
\overline{\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_+\right)}\tilde{\partial}_{\bar{z}}\left(e^{-h}\xi_+\right)e^{2h}d\lambda(z),\\
\mathscr{D}(\mathfrak{p}^h_+)&=\left\{\psi_+\in L_2(\mathbb{R}^2) :
\mathfrak{p}^h_+(\psi_+,\psi_+)<\infty\right\},
\end{align*}
and
\begin{align*}
\mathfrak{p}_-^h(\psi_-,\xi_-)&=4\int
\overline{\tilde{\partial}_z\left(e^{h}\psi_-\right)}\tilde{\partial}_z\left(e^{h}\xi_-\right)e^{-2h}d\lambda(z),\\
\mathscr{D}(\mathfrak{p}^h_-)&=\left\{\psi_-\in L_2(\mathbb{R}^2) :
\mathfrak{p}^h_-(\psi_-,\psi_-)<\infty\right\}.
\end{align*}
Set
\begin{align*}
\mathfrak{p}^h(\psi,\xi)&=\mathfrak{p}_+^h(\psi_+,\xi_+)+\mathfrak{p}_-^h(\psi_-,\xi_-),\\
\mathscr{D}(\mathfrak{p}^h)&=\mathscr{D}(\mathfrak{p}^h_+)\oplus\mathscr{D}(\mathfrak{p}^h_-)=\big\{\psi=\begin{pmatrix}\psi_+\\
\psi_-\end{pmatrix}\in
L_2(\mathbb{R}^2)\otimes\mathbb{C}^2:\mathfrak{p}^h(\psi,\psi)<\infty\big\}.
\end{align*}
Again, let us make clear about the domains of the forms. For a
function $\psi_+$ to be in $\mathscr{D}(\mathfrak{p}_+^h)$ it is required that
$\psi_+\in L_2(\mathbb{R}^2)$ and that $\tilde{\partial}_z(e^{-h}\psi_+)$ is a
function. After taking the modulus of this derivative and
multiplying by $e^h$ we should get into
$L_2(\mathbb{R}^2\setminus\Lambda)$, that is
$|\tilde{\partial}_{\bar{z}}(e^{-h}\psi_+)|e^h\in L_2(\mathbb{R}^2\setminus\Lambda)$.
Note that the form $\mathfrak{p}^h$ does not feel the AB fluxes at
$\Lambda$ since the derivatives are taken in the space
$\mathcal{D}'(\mathbb{R}^2\setminus\Lambda)$, and integration does
not feel $\Lambda$ either since $\Lambda$ has Lebesgue measure
zero. This enable the AB solenoids to have intensities that lie
outside $(-1,1)$.
Also, define the norm
\[
\gnorm{\psi_h}^2_{\mathfrak{p}^h}=\gnorm{\psi_+}^2_{\mathfrak{p}_+^h}
+\gnorm{\psi_-}^2_{\mathfrak{p}_-^h},
\]
where
\[
\gnorm{\psi_+}^2_{\mathfrak{p}_+^h}=\|\psi_+\|^2+\left|\left|\tilde{\partial}_{\bar{z}}
\left(e^{-h}\psi_+\right)e^h\right|\right|^2
\]
and
\[
\gnorm{\psi_-}^2_{\mathfrak{p}_-^h}=\|\psi_-\|^2+\left|\left|\tilde{\partial}_z
\left(e^{h}\psi_-\right)e^{-h}\right|\right|^2.
\]
\begin{prop} The form $\mathfrak{p}^h$ defined above is symmetric, nonnegative
and closed with respect to $\|\cdot\|$.
\end{prop}
\begin{proof} It is clear that $\mathfrak{p}^h$ is symmetric and nonnegative.
Let $\psi_n=(\psi_{n,+},\psi_{n,-})$ be a Cauchy sequence in the
norm $\gnorm{\cdot}_{\mathfrak{p}^h}$. This implies that
$\psi_{n,\pm}\to\psi_{\pm}$ in $L_{2}(d\lambda(z))$,
$\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_{n,+}\right)\to u_+$ in
$L_2(e^{2h}d\lambda(z))$ and $\tilde{\partial}_z(e^h\psi_{n,-})\to u_-$ in
$L_2(e^{-2h}d\lambda(z))$. We have to show that
$\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_{+}\right)= u_+$ and
$\tilde{\partial}_z(e^h\psi_{-})=u_-$. For any test-function $\phi\in
C_0^\infty(\mathbb{R}^2\setminus\Lambda)$,
\begin{align*}
&\left|\int\bar{\phi}\left(u_+-\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_{+}\right)\right)
d\lambda(z)\right|\\
& \leq \left|\int bar{\phi}\left(u_+-\tilde{\partial}_{\bar{z}}\left(e^{-h}
\psi_{n,+}\right)\right)\right|
+\left|\int\tilde{\partial}_{\bar{z}}(\bar\phi)e^{-h}\left(\psi_+-\psi_{n,+}
\right)\right|\\
& \leq \|\bar\phi e^{-h}\|\cdot
\left\|u_+-\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_{n,+}
\right)\right\|_{L_2(e^{2h})}
+\left\|\tilde{\partial}_{\bar{z}}(\bar\phi)e^{-h}\right\|\cdot \|\psi_+-
\psi_{n,+}\|.
\end{align*}
The above expression tends to zero as $n\to\infty$, since the
first terms in each sum is bounded (thanks to $\phi$) and the
other one tends to zero. The proof is the same for the spin down
component. This shows that $\mathfrak{p}^h$ is closed.
\end{proof}
Hence $\mathfrak{p}^h$ defines a unique self-adjoint operator
$\mathfrak{P}_h$ via
\begin{equation}
\label{eq:paulidef}
\mathscr{D}(\mathfrak{P}_h)=\{\psi\in\mathscr{D}(\mathfrak{p}^h):\mathfrak{p}^h(\psi,\cdot)\in
\left(L_2(\mathbb{R}^2)\otimes\mathbb{C}^2\right)\}
\end{equation}
and
\begin{equation}
\label{eq:paulidom}
(\mathfrak{P}_h\psi,\xi)=\mathfrak{p}^h(\psi,\xi),\quad
\psi\in\mathscr{D}(\mathfrak{P}_h),\xi\in\mathscr{D}(\mathfrak{p}^h).
\end{equation}
We call this operator $\mathfrak{P}_h$ the \emph{Maximal Pauli operator}.
\section{Properties of the Pauli operators}
\label{sec:prop}
In this section we will compare some properties of the two Pauli
operators $P_h$ and $\mathfrak{P}_h$ defined in the previous
section. We start by showing that $\mathfrak{P}_h$ is gauge invariant
while the non-reduced EV Pauli operator $\mathcal{P}_h$ is not.
\subsection{Gauge transformations}
Let $B(z)=B_0(z)+\sum_{j=1}^n2\pi \alpha_j\delta_{z_j}$ be the
same magnetic field as before and let $\hat{B}(z)$ be another
magnetic field that differs from $B(z)$ only by multiples of
the delta functions, that is
$\hat{B}(z)-B(z)=\sum_{j=1}^n 2\pi m_j\delta_{z_j}$,
where $m_j$ are integers, not all zero. Then the
corresponding scalar potentials $\hat{h}(z)$ and $h(z)$ differ
only by the corresponding logarithms
$\hat{h}(z)-h(z)=\sum_{j=1}^nm_j\log|z-z_j|$. With
$\phi(z)=\sum_{j=1}^nm_j\arg(z-z_j)$ we get
$\hat{h}(z)+i\phi(z)=h(z)+\sum_{j=1}^nm_j\log(z-z_j)$. This
function is multivalued, however, since $m_j$ are integers, we
have
\begin{gather}
\label{eq:diffe}
\partial_{\bar{z}}\left(\hat{h}(z)+i\phi(z)\right)=\partial_{\bar{z}}
h(z)+\sum_{j=1}^nm_j\partial_{\bar{z}}\log(z-z_j),\\
\label{eq:diffm}
\tilde{\partial}_{\bar{z}}\left(\hat{h}(z)+i\phi(z)\right)=\tilde{\partial}_{\bar{z}}
h(z),\\
\label{eq:exprule}
e^{\hat{h}+i\phi}=e^h\prod_{j=1}^m(z-z_j)^{m_j}.
\end{gather}
Let us check what happens with $\mathfrak{p}^h$ when we do gauge
transforms. Let $\psi=(\psi_+,\psi_-)^t\in\mathscr{D}(\mathfrak{p}^h)$. We
should check that $e^{-i\phi}\psi$ belongs to
$\mathscr{D}(\mathfrak{p}^{\hat{h}})$, where $\phi(z)=\sum_{j=1}^n
m_j\arg(z-z_j)$ is the harmonic conjugate to $\hat{h}(z)-h(z)$. We
do this for $\mathfrak{p}_+^{\hat{h}}$. It is similar for
$\mathfrak{p}_-^{\hat{h}}$. Since $\psi_+\in\mathscr{D}(\mathfrak{p}_+^h)$ we know that
$\tilde{\partial}_{\bar{z}}(\psi_+e^{-h})\in
L_{1,\text{loc}}(\mathbb{R}^2\setminus\Lambda)$. Let us check that
$\tilde{\partial}_{\bar{z}}(\hat{\psi}_+e^{-\hat{h}})\in
L_{1,\text{loc}}(\mathbb{R}^2\setminus\Lambda)$. Again,
by~\eqref{eq:exprule} we have
\begin{align*}
\tilde{\partial}_{\bar{z}}(\hat{\psi}_+e^{-\hat{h}})&=\tilde{\partial}_{\bar{z}}
\Big(\psi_+e^{-h}\prod_{j=1}^n(z-z_j)^{-m_j}\Big)\\
&=\tilde{\partial}_{\bar{z}}(\psi_+e^{-h})\prod_{j=1}^n(z-z_j)^{-m_j}+\psi_+e^{-h}\tilde{\partial}_{\bar{z}}
\Big(\prod_{j=1}^n(z-z_j)^{-m_j}\Big),
\end{align*}
which clearly belongs to $L_{1,\text{loc}}(\mathbb{R}^2\setminus\Lambda)$.
Next we should check that
$|\tilde{\partial}_{\bar{z}}(\hat{\psi}_+e^{-\hat{h}})|e^{\hat{h}}$
belongs to $L_2(\mathbb{R}^2\setminus\Lambda)$ under the assumption that
$|\tilde{\partial}_{\bar{z}}(\psi_+e^{-h})|e^{h}$ belongs to
$L_2(\mathbb{R}^2\setminus\Lambda)$.
A calculation using \eqref{eq:diffm} and \eqref{eq:exprule} gives
\begin{equation} \label{eq:ltwo}
\begin{aligned}
&\left|\tilde{\partial}_{\bar{z}}\left(e^{-\hat{h}}\hat{\psi}_+\right)\right|
e^{\hat{h}}\\
&=
\left|\tilde{\partial}_{\bar{z}}\left(e^{-\hat{h}-i\phi}\psi_+(z)\right)\right|e^{\hat{h}}\\
&=
\Big|\left(\tilde{\partial}_{\bar{z}}(-h(z))\psi_++\tilde{\partial}_{\bar{z}}\psi_+(z)\right)
e^{-h}\prod_{j=1}^n(z-z_j)^{-m_j}\Big|e^{h}\prod_{j=1}^n|z-z_j|^{m_j}\\
&=\left|\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_+\right)\right|e^{h}.
\end{aligned}
\end{equation}
Hence $\psi_+\in\mathscr{D}(\mathfrak{p}_+^h)$ implies
$\hat{\psi}_+=e^{-i\phi}\psi_+\in\mathscr{D}(\mathfrak{p}_+^{\hat{h}})$. In the
same way it follows that $\psi_-\in\mathscr{D}(\mathfrak{p}_-^h)$ implies that
$\hat{\psi}_-=e^{-i\phi}\psi_-\in\mathscr{D}(\mathfrak{p}_-^{\hat{h}})$. Thus
$e^{-i\phi}\mathscr{D}(\mathfrak{p}^h)\subset\mathscr{D}(\mathfrak{p}^{\hat{h}})$.
In the same
way we can show that
$e^{i\phi}\mathscr{D}(\mathfrak{p}^{\hat{h}})\subset\mathscr{D}(\mathfrak{p}^h)$,
and thus we
can conclude that
$e^{-i\phi}\mathscr{D}(\mathfrak{p}^h)=\mathscr{D}(\mathfrak{p}^{\hat{h}})$.
From the calculation in~\eqref{eq:ltwo} and a similar calculation
for the spin-down component $\psi_-$ it also follows that
\begin{align*}
\mathfrak{p}^{\hat{h}}\left(e^{-i\phi}\psi,e^{-i\phi}\psi\right)
& = 4\int\left|\tilde{\partial}_{\bar{z}}\left(e^{-\hat{h}-i\phi}\psi_+\right)\right|^2
e^{2\hat{h}}+\left|\tilde{\partial}_z\left(e^{\hat{h}-i\phi}\psi_-\right)\right|^2
e^{-2\hat{h}}d\lambda(z)\\
& =
4\int\left|\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_+\right)\right|^2e^{2h}+\left|\tilde{\partial}_z
\left(e^{h}\psi_-\right)\right|^2e^{-2h}d\lambda(z)\\
& = \mathfrak{p}^h(\psi,\psi).
\end{align*}
Hence we can conclude that if $\psi\in\mathscr{D}(\mathfrak{P}_h)$ and
$\xi\in\mathscr{D}(\mathfrak{p}^h)$ then
$e^{-i\phi}\psi\in\mathscr{D}(\mathfrak{P}_{\hat{h}})$ and
$e^{-i\phi}\xi\in\mathscr{D}(\mathfrak{p}^{\hat{h}})$. If we denote by $U_\phi$ the
unitary
operator of multiplication by $e^{i\phi}$, then we get
\[
(\mathfrak{P}_h\psi,\xi)=\mathfrak{p}^h(\psi,\xi)
=\mathfrak{p}^{\hat{h}}(U_\phi^*\psi,U_\phi^*\xi)
=(\mathfrak{P}_{\hat{h}}U_\phi^*\psi,U_\phi^*\xi)
=(U_\phi \mathfrak{P}_{\hat{h}}U_\phi^*\psi,\xi),
\]
and hence $\mathfrak{P}_h$ and $\mathfrak{P}_{\hat{h}}$ are unitarily equivalent.
We have proved the following proposition.
\begin{prop}
Let $B$ and $\hat{B}$ be two singular magnetic fields as in \eqref{eq:magnet},
with difference $\hat{B}-B=\sum_{j=1}^n 2\pi m_j \delta_{z_j}$, where $m_j$
are integers, not all equal to zero. Then their corresponding Maximal
Pauli operators defined by~\eqref{eq:paulidef} and~\eqref{eq:paulidom}
are unitarily equivalent.
\end{prop}
To see that $\mathcal{P}_h$ is not gauge invariant it is enough to look at an
example. Note that this operator is defined only for intensities belonging
to the interval $(-1,1)$. Let $n=1$, $z_1=0$, $\alpha_1=-1/2$ and $m_1=1$,
so the two magnetic fields are $B(z)=B_0(z)-\pi\delta_0$ and
$\hat{B}(z)=B_0(z)+\pi\delta_0$.
The scalar potentials are given by $h(z)=h_0(z)-\frac12\log|z|$ and
$\hat{h}(z)=h_0(z)+\frac12\log|z|$ respectively, where $h_0(z)$ is a
smooth function with asymptotics $\Phi_0\log|z|$ as $|z|\to\infty$.
We should show that $\mathscr{D}(\pi^{\hat{h}})$ is not given by
$e^{-i\phi}\mathscr{D}(\pi^h)$, where $\phi(z)=\arg(z)$. Then it follows that
$\pi^h$ and $\pi^{\hat{h}}$ do not define unitarily equivalent operators.
Let $\psi_+\in\mathscr{D}(\pi_+^h)$. This means, in particular, that
$\partial_{\bar{z}}(\psi_+e^{-h})$ belongs to the space
$L_{1,\text{loc}}(\mathbb{R}^2)$.
Now let $\hat{\psi}_+=e^{-i\phi}\psi_+$. Then, according to~\eqref{eq:exprule}
we get
\[
\partial_{\bar{z}}(\hat{\psi}_+e^{-\hat{h}})=\partial_{\bar{z}}(\psi_+e^{-\hat{h}-i\phi})=\partial_{\bar{z}}\left(\frac{\psi_+e^{-h}}{z}\right)=\partial_{\bar{z}}(\psi_+e^{-h})\frac1z+\psi_+e^{-h}\pi\delta_0
\]
which is not in $L_{1,\text{loc}}(\mathbb{R}^2)$ since it is a distribution
involving $\delta_0$ (for non-smooth $\psi_+$ it is not even well-defined).
Thus we have $\mathscr{D}(\pi_+^{\hat{h}})\neq e^{-i\phi}\mathscr{D}(\pi_+^h)$ and
hence $\mathscr{D}(\pi^{\hat{h}})\neq e^{-i\phi}\mathscr{D}(\pi^h)$ so $\pi^h$ and
$\pi^{\hat{h}}$ are not defining unitarily equivalent operators $\mathcal{P}_h$
and $\mathcal{P}_{\hat{h}}$.
\subsection{Zero modes}
\label{sec:zero}
When studying spectral properties of the operator $\mathfrak{P}_h$ it is
sufficient to consider AB intensities $\alpha_j$ that belong to the
interval $(0,1)$, since the operator is gauge invariant. See the discussion
after the proof of Theorem~\ref{thm:AC} for more details about what happens
when we do gauge transformations.
\begin{lemma}
\label{lemma:asymptot}
Let $c_j\in\mathbb{C}$ and $z_j\in\mathbb{C}$, $j=1,\ldots,n$,
where $z_j\neq z_i$ if $j\neq i$ and not all $c_j$ are equal to zero. Then
\begin{equation}
\label{eq:asympt}
\sum_{j=1}^n \frac{c_j}{z-z_j}\sim |z|^{-l-1},\quad |z|\to\infty,
\end{equation}
where $l$ is the smallest nonnegative integer such that
$\sum_{j=1}^n c_j z_j^l\neq 0$.
\end{lemma}
\begin{proof}
If $|z|$ is large in comparison with all $|z_j|$ we have
\begin{align*}
\sum_{j=1}^n \frac{c_j}{z-z_j} &= \frac{1}{z}\sum_{j=1}^n \frac{c_j}{1-z_j/z}\\
&= \sum_{k=0}^\infty \left(\sum_{j=1}^n c_jz_j^k\right)\frac{1}{z^{k+1}}\\
&= \left(\sum_{j=1}^n c_jz_j^l\right)\frac{1}{z^{l+1}}+O(|z|^{-l-2})
\end{align*}
and thus $\sum_{j=1}^n \frac{c_j}{z-z_j}\sim |z|^{-l-1}$ as $|z|\to\infty$.
\end{proof}
\begin{remark} \rm
We note that $l$ in Lemma~\ref{lemma:asymptot} may never be greater than $n-1$.
Indeed, if $l\geq n$ then we would have the linear system of equations
$\{\sum_{j=1}^nc_jz_j^k=0\}_{k=0}^{n-1}$. But the determinant of this system
is $\prod_{i>j}(z_i-z_j)\neq 0$, and this would force all $c_j$ to be zero.
Note also that for $l< n$ we have a linear system of $l$
equations $\{\sum_{j=1}^nc_jz_j^k=0\}_{k=0}^{l-1}$ with $n$
unknowns $c_j$, and that the $l\times n$ matrix $\{z_j^k\}$ has
rank $l$.
\end{remark}
\begin{thm}\label{thm:AC}
Let $B(z)$ be the magnetic field \eqref{eq:magnet} with all
$\alpha_j\in(0,1)$, and let $\mathfrak{P}_h$ be the Pauli operator
defined by~\eqref{eq:paulidef} and \eqref{eq:paulidom} in
Section~\ref{sec:def} corresponding to $B(z)$. Then
\begin{equation}
\dim\ker \mathfrak{P}_h=\left\{n-\Phi\right\}+\left\{\Phi\right\},
\end{equation}
where $\Phi=\frac{1}{2\pi}\int B(z)d\lambda(z)$, and $\{x\}$
denotes the largest integer strictly less than $x$ if $x>1$ and
$0$ if $x\leq 1$. Using the notations $Q_\pm$ introduced in
Section~\ref{sec:def}, we also have
\begin{equation}
\dim\ker Q_+=\left\{n-\Phi\right\}\quad \text{and}\quad \dim\ker Q_-
=\left\{\Phi\right\}.
\end{equation}
\end{thm}
\begin{proof}
We follow the reasoning originating in \cite{ac}, with necessary
modifications. First we note that $(\psi_+,\psi_-)^t$ belongs to
$\ker \mathfrak{P}_h$ if and only if $\psi_+$ belongs to $\ker Q_+$ and
$\psi_-$ belongs to $\ker Q_-$, which is equivalent to
\begin{equation}
\tilde{\partial}_{\bar{z}}\left(e^{-h}\psi_+\right)=0\quad \text{and}\quad
\tilde{\partial}_z\left(e^{h}\psi_-\right)=0.
\end{equation}
This means exactly that $f_+(z)=e^{-h}\psi_+(z)$ is holomorphic and
$f_-(z)=e^h\psi_-(z)$ is antiholomorphic in $z\in\mathbb{R}^2\setminus\Lambda$.
It is the change in the domain where the functions are holomorphic that
influences the result.
Let us start with the spin-up component $\psi_+$. The function
$f_+$ is allowed to have poles of order at most one at $z_j$,
$j=1,\ldots,n$, and no others, since $e^h\sim |z-z_j|^{\alpha_j}$
as $z\to z_j$ and $\psi_+=f_+e^h$ should belong to
$L_2(\mathbb{R}^2)$. Hence there exist constants $c_j$ such that
the function $f_+(z)-\sum_{j=1}^n\frac{c_j}{z-z_j}$ is entire.
From the asymptotics $e^h\sim |z|^{\Phi}$, $|z|\to\infty$, it
follows that $f_+-\sum_{j=1}^n\frac{c_j}{z-z_j}$ may only be a
polynomial of degree at most $N=-\Phi-2$. Hence
\[
f_+(z)=\sum_{j=1}^n\frac{c_j}{z-z_j}+a_0+a_1z+\ldots a_Nz^N,
\]
where we let the polynomial part disappear if $N<0$. Now, the
asymptotics for $\psi_+$ is
\[
\psi_+(z)\sim |z|^{-l-1+\Phi}+|z|^{N+\Phi},\quad |z|\to\infty,
\]
where $l$ is the smallest nonnegative integer such that
$\sum_{j=1}^n c_jz_j^l\neq0$. To have $\psi_+$ in
$L_2(\mathbb{R}^2)$ we take $l$ to be the smallest nonnegative
integer strictly greater than $\Phi$. Remember also from the
remark after Lemma~\ref{lemma:asymptot} that $l\leq n-1$. We get
three cases. If $\Phi<-1$, then all complex numbers $c_j$ can be
chosen freely, and a polynomial of degree $\{-\Phi\}-1$ may be
added which results $\{n-\Phi\}$ degrees of freedom. If
$-1\leq\Phi0$. Let us begin with the spin-up component
$\psi_+$. This time, the holomorphic $f_+=e^{-h_2}\psi_+$ may not
have any poles since then $\psi_+$ would not belong to
$L_2(\mathbb{R}^2)$, and $f_+(z)=e^{-h_2}\psi_+(z)\to0$ as
$|z|\to\infty$, so we must have $f_+\equiv 0$, and thus
$\psi_+\equiv 0$. For $\psi_-(z)$ to be in $L_2(\mathbb{R}^2)$ it
is possible for $f_-$ to have a pole of order $1$ at the origin.
Hence there exist a constant $c$ such that $f_-(z)-c/\bar{z}$ is
antiholomorphic in the whole plane. The function $f_-(z)\to0$ as
$|z|\to\infty$ since the total intensity $\Phi_2>0$. This implies,
by Liouville's theorem, that $f_-(z)\equiv c/\bar{z}$, so the
solution space to $\mathfrak{P}_{h_2}\psi=0$ is spanned by
$\psi(z)=(0,e^{-h_2}/\bar{z})$.
Finally, we determine the solutions to $\mathfrak{P}_{h_3}\psi=0$.
Now $\Phi_3=\frac{1}{2\pi}\int B_3(z)d\lambda(z)=9/4$. Consider
the spin-up part $\psi_+$. For $\psi_+$ to be in
$L_2(\mathbb{R}^2)$ our function $f_+$ may have a pole of order no
more than two at the origin. As before, there exist constants
$c_1$ and $c_2$ such that $f_+(z)-c_1/z-c_2/z^2$ is entire and its
limit is zero as $|z|\to\infty$, and thus $f_+(z)\equiv
c_1/z+c_2/z^2$. Again, both $c_1$ and $c_2$ must vanish for
$\psi_+$ to be in $L_2(\mathbb{R}^2)$ (otherwise we would not stay
in $L_2$ at infinity). Thus $\psi_+\equiv 0$. On the other hand,
the function $f_-$ may not have any poles (these poles would push
$\psi_-$ out of $L_2(\mathbb{R}^2)$), so it is antiholomorphic in
the whole plane. It also may grow no faster than $|z|^{5/4}$ as
$|z|\to\infty$, and thus $f_-$ has to be a first order polynomial
in $\bar{z}$, that is $f_-(z)=c_0+c_1\bar{z}$. Moreover for
$\psi_-$ to be in $L_2(\mathbb{R}^2)$ it must have a zero of order
$1$ at the origin, and thus $f_-(z)=c_1\bar{z}$. We conclude that
the solutions to $\mathfrak{P}_{h_3}\psi=0$ are spanned by
$(0,\bar{z}e^{-h_3})^t$.
\end{example}
A natural property one should expect of a reasonably defined Pauli
operator is that its spectral properties are invariant under the
reversing the direction of the magnetic field: $B\mapsto -B$. The
corresponding operators are formally anti-unitary equivalent under
the transformation $\psi\mapsto \bar{\psi}$ and interchanging of
$\psi_+$ and $\psi_-$.
\begin{example} \label{ex:BminusBEV} \rm
The number of zero modes for $P_h$ is not invariant under
$B(z)\mapsto-B(z)$, which we should not expect since the interval
$[-1/2,1/2)$ is not symmetric. We check this by showing that the number
of zero modes are not the same. To see this, let $B(z)=B_0(z)+\pi\delta_0$,
where $B_0$ has compact support and
$\Phi_0=\frac{1}{2\pi}\int B_0(z)d\lambda(z)=\frac{3}{4}$. Then $B$ has
to be reduced since the AB intensity at zero is $1/2\not\in[-1/2,1/2)$.
After reduction we get the magnetic field
$\hat{B}(z)=B_0(z)-\pi\delta_0$, and we can apply
Theorem~\ref{thm:ACEV}. Let $\hat{\Phi}=\frac{1}{2\pi}\int
\hat{B}d\lambda(z)=\frac14$. Thus the number of zero modes for
$P_h$ is $0$. Now look at the Pauli operator $P_{-h}$
defined by the magnetic field $B_-(z)=-B(z)=-B_0(z)-\pi\delta_0$.
This magnetic field is reduced and thus we can apply
Theorem~\ref{thm:ACEV} directly. The total intensity is
$\Phi_-=\frac{1}{2\pi}\int -B(z)d\lambda(z)=-\frac{5}{4}$, so the
number of zero modes for $P_{-h}$ is $1$. If $B$ has several
AB fluxes then the difference in the number of zero modes of
$P_h$ and $P_{-h}$ can be made arbitrarily large.
\end{example}
\begin{remark} \rm
If there are only AB solenoids then the EV Pauli operator
preserves the number of zero modes under $B\mapsto -B$, so the
absence of signflip invariance can be noticed only in the
presence of both AB and nice part.
\end{remark}
\begin{example} \rm
\label{ex:BminusBMP} The number of zero modes for $\mathfrak{P}_h$ is
invariant under $B(z)\mapsto -B(z)$. Since it is clear that the
number of zero modes is invariant under $z\mapsto \bar{z}$ we
look instead at how the Pauli operators change when we do
$B(z)\mapsto \hat{B}(z)=-B(\bar{z})$. If we set $\zeta=\bar{z}$ we
get $\hat{B}(\zeta)=-B(z)$ and the scalar potentials satisfy
$\hat{h}(\zeta)=-h(z)$. Assume that
$\psi=(\psi_+(z),\psi_-(z))^t\in\mathscr{D}(\mathfrak{p}^h)$. Then
\begin{align*}
&\mathfrak{p}^{h(z)}\left(\begin{pmatrix}\psi_+(z)\\ \psi_-(z)\end{pmatrix},
\begin{pmatrix}\psi_+(z)\\ \psi_-(z)\end{pmatrix}\right) \\
&=4\int \left|\tilde{\partial}_{\bar{z}}(\psi_+(z)e^{-h(z)})\right|^2e^{2h(z)}
+\left|\tilde{\partial}_z(\psi_-(z)e^{h(z)}\right|^2e^{-2h(z)}d\lambda(z)\\
&=4\int\left|\tilde{\partial}_\zeta(\psi_+(\bar{\zeta})e^{\hat{h}(\zeta)}\right|^2
e^{-2\hat{h}(\zeta)}+\left|\tilde{\partial}_{\bar{\zeta}}(\psi_-(\bar{\zeta})
e^{-\hat{h}(\zeta)}\right|^2e^{2\hat{h}(\zeta)}d\lambda(\zeta)\\
&=\mathfrak{p}^{\hat{h}(\bar{z})}\left(\begin{pmatrix}\psi_-(z)\\
\psi_+(z)\end{pmatrix},\begin{pmatrix}\psi_-(z)\\
\psi_+(z)\end{pmatrix}\right)
\end{align*}
Hence we see that $(\psi_+,\psi_-)^t$ belongs to
$\mathscr{D}(\mathfrak{P}_{h(z)})$ if and only if $(\psi_-,\psi_+)^t$ belongs
to $\mathscr{D}(\mathfrak{P}_{\hat{h}(\bar{z})})$ and then
$\mathfrak{P}_{\hat{h}(\bar{z})}=\mathfrak{P}_{h(z)}V$ where
$V:L_2(\mathbb{R}^2)\otimes\mathbb{C}^2\to
L_2(\mathbb{R}^2)\otimes\mathbb{C}^2$ is the isometric operator
given by $V((\psi_+,\psi_-)^t)=(\psi_-,\psi_+)^t$. Hence it is
clear that $\mathfrak{P}_{\hat{h}(\bar{z})}$ and $\mathfrak{P}_{h(z)}$ have
the same number of zero modes.
\end{example}
\begin{example} \rm
In the previous example we saw that changing the sign of the
magnetic field results in unitarily equivalent Maximal Pauli
operators. This implies that the number of zero modes for the
Maximal Pauli operators corresponding to $B$ and $-B$ are the
same. This, however, can be seen directly from the Aharonov-Casher
formula in Theorem~\ref{thm:AC}. To be able to apply the theorem
to $-B=-B_0-\sum_{j=1}^n 2\pi\alpha_j\delta_j$ we have to do gauge
transformations, adding $1$ to all the AB intensities, resulting
in $\hat{B}=-B_0+\sum_{j=1}^n 2\pi(1-\alpha_j)\delta_j$. Now
according to Theorem~\ref{thm:AC} the number of zero modes of
$\mathfrak{P}_{-h}$ is equal to
\[
\dim\ker\mathfrak{P}_{-h}=\{\hat{\Phi}\}+\{n-\hat{\Phi}\}=\{n-\Phi\}
+\{\Phi\}=\dim\ker\mathfrak{P}_h,
\]
where we have used that $\hat{\Phi}=\frac{1}{2\pi} \int
\hat{B}d\lambda(z)=n-\Phi$.
\end{example}
\section{Approximation by regular fields}
\label{sec:approx}
We have mentioned that the different Pauli extensions depend on
which boundary conditions are induced at the AB fluxes. Let us now
make this more precise. Since the self-adjoint extension only
depends on the boundary condition at the AB solenoids it is enough
to study the case of one such solenoid and no smooth field. For
simplicity, let the solenoid be located at the origin, with
intensity $\alpha\in(0,1)$, that is, let the magnetic field be
given by $B=2\pi\alpha\delta_0$. We consider self-adjoint
extensions of the Pauli operator $P$ that can be written in the
form
\[
P=\begin{pmatrix}P_+ & 0\\ 0 & P_-\end{pmatrix}=
\begin{pmatrix}
Q_+^*Q_+ & 0\\
0 & Q_-^*Q_-,
\end{pmatrix}
\]
with some explicitly chosen closed operators $Q_\pm$. It is
exactly such extensions $P$ that can be defined by the quadratic
form~\eqref{eq:kvadform}. A function $\psi_+$ belongs to
$\mathscr{D}(P_+)$ if and only if $\psi_+$ belongs to $\mathscr{D}(Q_+)$ and
$Q_+\psi_+$ belongs to $\mathscr{D}(Q_+^*)$, and similarly for $P_-$.
With each self-adjoint extension $P_\pm=Q_{\pm}^*Q_{\pm}$ one can
associate (see \cite{dast,exstvy,gest,ta}) functionals
$c_{-\alpha}^\pm$, $c_{\alpha}^{\pm}$, $c_{\alpha-1}^\pm$ and
$c_{1-\alpha}^\pm$, by
\begin{gather*}
c_{-\alpha}^{\pm}(\psi_\pm) = \lim_{r\to 0}r^\alpha
\frac{1}{2\pi}\int_{0}^{2\pi}\psi_\pm d\theta,\\
c_{\alpha}^{\pm}(\psi_\pm) = \lim_{r\to 0}r^{-\alpha}
\left(\frac{1}{2\pi}\int_{0}^{2\pi}\psi_\pm d\theta-r^{-\alpha}c_{\alpha}^{\pm}
(\psi_\pm)\right),\\
c_{\alpha-1}^{\pm}(\psi_\pm) = \lim_{r\to 0}r^{1-\alpha}\frac{1}{2\pi}
\int_{0}^{2\pi}\psi_\pm e^{i\theta} d\theta, \\
c_{1-\alpha}^{\pm}(\psi_\pm) = \lim_{r\to 0}r^{\alpha-1}
\left(\frac{1}{2\pi}\int_{0}^{2\pi}\psi_\pm e^{i\theta}d\theta-r^{\alpha-1}
c_{1-\alpha}^{\pm}(\psi_\pm)\right).
\end{gather*}
such that $\psi_\pm\in\mathscr{D}(P_\pm)$ if and only if
\begin{equation}
\psi_\pm\sim c_{-\alpha}^\pm r^{-\alpha}+c_{\alpha}^\pm r^{\alpha}
+c_{\alpha-1}^\pm r^{\alpha-1}e^{-i\theta}+c_{1-\alpha}^\pm
r^{1-\alpha}e^{-i\theta}+O(r^\gamma)
\end{equation}
as $r\to 0$, where $\gamma=\min(1+\alpha,2-\alpha)$ and $z=re^{i\theta}$.
Any two nontrivial independent linear relations between these
functionals determine a self-adjoint extension. In order that the
operator be rotation-invariant, none of these relations may
involve both $\alpha$ and $1-\alpha$ terms simultaneously.
Accordingly, the parameters
$\nu_0^\pm=c_\alpha^\pm/c_{-\alpha}^\pm$ and
$\nu_1^\pm=c_{1-\alpha}^\pm/c_{\alpha-1}^\pm$, with possible
values in $(-\infty,\infty]$, are introduced in \cite{bopu}, and
it is proved that the operators $P_\pm$ can be approximated by
operators with regularized magnetic fields in the norm resolvent
sense if and only if $\nu_0^\pm=\infty$ and
$\nu_1^\pm\in(-\infty,\infty]$ or if
$\nu_0^\pm\in(-\infty,\infty]$ and $\nu_1^\pm=\infty$.
Before we check what parameters the Maximal and EV Pauli operators
correspond to, let us in a few words discuss how the approximation
in \cite{bopu} works.
The vector magnetic potential $\mathbf{A}$ is approximated with the
vector potential
\[
\mathbf{A}_R(z)=
\begin{cases}
\mathbf{A}(z) & |z|>R\\
0 & |z|\varepsilon}g(z)\overline{-2i\frac{\partial}{\partial
\bar{z}}(e^{-h}\phi_+(z))e^h} d\lambda(z)\\
& =\lim_{\varepsilon\to0}\int_{|z|>\varepsilon}-2i\frac{\partial}{\partial
z}(g(z)e^h)e^{-h}\overline{\phi_+(z)}d\lambda(z)\\
&\quad+ \lim_{\varepsilon\to0}\varepsilon\int_0^{2\pi}g(\varepsilon
e^{i\theta})\overline{\phi_+(\varepsilon
e^{i\theta})}e^{-i\theta}d\theta\\
& = \langle
Q_-g,\phi_+\rangle+\lim_{\varepsilon\to0}\frac{\varepsilon}{2}\int_0^{2\pi}g(\varepsilon
e^{i\theta})\overline{\phi_+(\varepsilon
e^{i\theta})}e^{-i\theta}d\theta
\end{align*}
Hence, for $g$ to belong to $\mathscr{D}(Q_+^*)$ it is necessary and sufficient that
\[
\lim_{\varepsilon\to0}\varepsilon\int_0^{2\pi}g(\varepsilon e^{i\theta})
\overline{\phi_+(\varepsilon e^{i\theta})}e^{-i\theta}d\theta=0
\]
for all $\phi_+\in\mathscr{D}(p_+)$, and thus for $Q_+\psi_+$ to belong to
$\mathscr{D}(Q_+^*)$ it is necessary and sufficient that
\[
\lim_{\varepsilon\to0}\varepsilon\int_0^{2\pi}\left(\frac{\partial}{\partial
\bar{z}}(e^{-h}\psi_+)e^h\right)
\Big|_{z=\varepsilon e^{i\theta}}\overline{\phi_+(\varepsilon e^{i\theta})}
e^{-i\theta}d\theta=0
\]
for all $\phi_+\in\mathscr{D}(p_+)$. We know that $\psi_+$ has asymptotics
$\psi_+\sim c_{-\alpha}^+r^{-\alpha}+c_{\alpha}^+r^{\alpha}
+c_{\alpha-1}^+r^{\alpha-1}e^{-i\theta}+c_{1-\alpha}^+
r^{1-\alpha}e^{-i\theta}+O(r^\gamma)$ and that$\frac{\partial}{\partial
\bar{z}}=\frac{e^{i\theta}}{2}
\left(\frac{\partial}{\partial r}+\frac{i}{r}\frac{\partial}{\partial\theta}
\right)$ in polar coordinates. A calculation gives
\[
\varepsilon\frac{\partial}{\partial
\bar{z}}(e^{-h}\psi_+)e^he^{-i\theta}\Big|_{z=\varepsilon e^{i\theta}}
\sim -2\alpha c_{-\alpha}^+\varepsilon^{-\alpha}+2(1-\alpha)c_{1-\alpha}^+
\varepsilon^{1-\alpha}e^{-i\theta}+O(r^\gamma),
\]
hence we must have
\begin{equation}
\label{eq:spinupkrav}
\lim_{\varepsilon\to0}
\int_0^{2\pi}\left(-2\alpha c_{-\alpha}^+\varepsilon^{-\alpha}+2(1-\alpha)
c_{1-\alpha}^+\varepsilon^{1-\alpha}e^{-i\theta}\right)\overline{\phi_+
(\varepsilon e^{i\theta})}d\theta=0
\end{equation}
for all $\phi_+\in\mathscr{D}(p_+)$. A similar calculation for the spin-down
component yields
\begin{equation}
\label{eq:spindownkrav}
\lim_{\varepsilon\to0}
\int_0^{2\pi}\left(2\alpha c_{\alpha}^-\varepsilon^{\alpha}+2(\alpha-1)
c_{\alpha-1}^-\varepsilon^{\alpha-1}e^{i\theta}\right)
\overline{\phi_-(\varepsilon e^{i\theta})}d\theta=0.
\end{equation}
To calculate what parameters $\nu_0^\pm$ and $\nu_1^\pm$ the
Maximal and EV Pauli extensions correspond to, it is enough to
study the asymptoics of the functions in the form core.
Let us first consider the Maximal Pauli extension. Functions on
the form \break
$(\phi_0^+c/z)e^h$ constitute a form core for
$\mathfrak{p}^h_+$, where $\phi_0$ is smooth. Hence there are elements in
$\mathscr{D}(\mathfrak{p}^h_+)$ that asymptotically behave as $r^\alpha$ and
also elements with asymptotics $r^{\alpha-1}e^{-i\theta}$.
According to~\eqref{eq:spinupkrav} this means that $c_{-\alpha}^+$
and $c_{1-\alpha}^+$ must be zero. Similarly, the elements that
behave like $r^{-\alpha}$ and elements that behave like
$r^{1-\alpha}e^{i\theta}$ constitute a form core for $\mathfrak{p}^h_-$,
which by~\eqref{eq:spindownkrav} forces $c_{\alpha}^-$ and
$c_{\alpha-1}^-$ to be zero. The parameters $\nu_0^\pm$ and
$\nu_1^\pm$ are given by
$\nu_0^+=c_{\alpha}^+/c_{-\alpha}^+=\infty$,
$\nu_1^+=c_{1-\alpha}^+/c_{\alpha-1}^+=0$,
$\nu_0^-=c_{\alpha}^-/c_{-\alpha}^-=0$ and
$\nu_1^-=c_{1-\alpha}^-/c_{\alpha-1}^-=\infty$. We see that the
spin-up component can be approximated as in (II), while the
spin-down component can be approximated as in (I).
Let us now consider the EV Pauli extension, and study the case
when $\alpha\in(0,1/2)$. The case $\alpha<0$ follows in a a
similar way. A form core for $\pi^h_+$ is given by $e^h\phi_0$
where $\phi_0$ is smooth, see \cite{ervo}. These functions have
asymptotic behavior $r^\alpha$. From~\eqref{eq:spinupkrav} follows
that $c_{-\alpha}^+$ must vanish. However, $\psi_+$ belonging to
$\mathscr{D}(Q_+)$ must also belong to $\mathscr{D}(\pi^h_+)$ and since the
functions in the form core for $\pi^h_+$ behave as $r^{\alpha}$
or nicer, we see that the term
$c_{\alpha-1}^+r^{\alpha-1}e^{-i\theta}$ gets too singular to be
in $\mathscr{D}(Q_+)$ if $c_{\alpha-1}^+\neq 0$, and hence
$c_{\alpha-1}^+$ must be zero.
Similarly, a form core for $\pi^h_-$ is given by $e^{-h}\phi_0$,
with $\phi_0$ smooth. Functions in this form core have asymptotic
behavior $r^{-\alpha}$ or $r^{-\alpha+1}e^{i\theta}$ which forces
$c_{\alpha}^-$ and $c_{\alpha-1}^-$ to be zero.
Hence the parameters $\nu_0^\pm$ and $\nu_1^\pm$ are given by
$\nu_0^+=c_{\alpha}^+/c_{-\alpha}^+=\infty$,
$\nu_1^+=c_{1-\alpha}^+/c_{\alpha-1}^+=\infty$,
$\nu_0^-=c_{\alpha}^-/c_{-\alpha}^-=0$ and
$\nu_1^-=c_{1-\alpha}^-/c_{\alpha-1}^-=\infty$.
We conclude that the spin-up part of the EV Pauli operator can be
approximated in either of the ways (I) or (II), while the
spin-down part can be approximated in way (I).
\begin{remark} \rm
From the calculations above it follows that the EV Pauli operator
can be approximated as a Pauli Hamiltonian in the sense of
\cite{bopu}, while the Maximal Pauli operator cannot be
approximated as a Pauli Hamiltonian, since the spin-up and
spin-down components are approximated in different ways.
Since AB is defined up to a singular gauge transformation and
regular fields can not be transformed in this way it is unclear
which additional physical requirements or principles can decide on
which way of approximation is the most physically reasonable.
\end{remark}
\subsection*{Acknowledgements} I would like to thank my Ph. D. thesis
supervisor Professor Grigori Rozenblum for introducing me to this
problem and for giving me all the support I needed. I would also
like to thank the referee for pointing out a mistake and giving
very helpful comments.
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