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magnetic Schrodinger operator, Lifshitz tail, random magnetic field, density of states
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%%%%%%%%%%%%  Title  %%%%%%%%%%%%%%%%

\title{Lifshitz tail for Schr\"odinger operator \\
with random magnetic field}
\author{Shu Nakamura \\ 
{\normalsize Graduate School of Mathematical Sciences, University of Tokyo}
\\ {\normalsize 3-8-1, Komaba, Meguro-ku, Tokyo, Japan 153-8914}
\\ {\normalsize E-mail: {\tt shu@ms.u-tokyo.ac.jp}}}
\date{}

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%     Paper from here
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\begin{document}
\maketitle

\abstract{
We study the behavior of the density states at the lower
edge of the spectrum for Schr\"odinger operators with random
magnetic fields. We use a new estimate on magnetic Schr\"odinger
operators, which is similar to the Avron-Herbst-Simon estimate but 
the bound is always nonnegative.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%  SECTION 1
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction}
In this paper, we consider the magnetic Schr\"odinger operator
without scalar potential on $\re^d$ ($d\geq 2$):
\[
H=(p-A(x))^2 \qquad \text{on} \ L^2(\re^d), 
\]
where $p=-i\pa_x$  the momentum operator, 
\[
A(x)= (A_1(x), \dots, A_d(x)) \qquad x\in\re^d
\]
is a vector potential. For the moment we suppose $A$ is a 
$C^1$-class function. We identify $A(x)$ with the 1-form
$A=\sum_{j=1}^d A_j(x) dx_j$. The magnetic field is the 2-form
defined by
\[
B(x)=dA = \sum_{i<j} B_{ij}(x) dx_i\wedge dx_j, 
\]
where $B_{ij}(x)= \pa_{x_i} A_j(x)-\pa_{x_j} A_i(x)$, and we identify
$B$ with the skew-symmetric matrix valued function $(B_{ij}(x))$ on 
$\re^d$.  It is well-known that for any closed 2-form $B$, there is a
1-form $A$ such that $B=dA$, and the corresponding Schr\"odinger
operator $H$ is uniquely determined by $B$ modulo the gauge
transformation group. 

We suppose that $B=B^\om$ is a metrically transitive random closed
2-form in the following sense: 
There exist a probability space $(\Om,\Meas,\Pro)$ and $\re^d$ acts on 
$\Om$ as a measure preserving transformation group. 
We denote the action by $\normalset{T_x}{x\in\re^d}$. 
The action is supposed to be ergodic, i.e., if $E\subset \Om$
is invariant under $\{T_x\}$ then $\Pro(E)=0$ or 1. $B=B^\om$ is
assumed to satisfy
\[
B^\om(x-y)=B^{(T_x \om)}(x), \qquad x,y\in\re^d,
\]
and it is a continuous closed 2-form almost surely. 

We denote 
\[
\La_L= \normalset{x\in\re^d}{|x_j|\leq L/2 \text{ for } j=1,\dots,d},
\qquad L>0, 
\] 
and let $H^\om_\La$ be 
the Schr\"odinger operator (with the same symbol as
above) on $\L\subset\re^d$ with, e.g.,  the Neumann boundary condition~(cf.\
Section~3).  We also denote the Lebesgue measure of $\La$ by $|\La |$.
Then it is well-known that the density of states
\[
k(E)= \lim_{L\to\infty} \frac{1}{|\La_L|} \#
\{\text{eigenvalue of }H^\om_{\La_L}\leq E\}
\]
exists for almost all $E\in\re$ and $\om\in\Om$, and it is independent of 
$\om$ (almost surely). 
Our main assumption is the following. 

\begin{ass}
\begin{enumerate}
\item $B^\om$ is a metrically transitive random closed 2-form on
$\re^d$ in the above sense. 
\item There is a constant $M\in \re_+=[0,\infty)$ such that 
\[
|B^\om(x)| \leq M
\qquad \text{for any $x\in\re^d$ and almost all $\om\in\Om$,}
\]
where $|B(x)|=(\sum_{i<j} |B_{ij}(x)|^2)^{1/2}$. 
\item There exists a real-valued continuous function $\f$ on $\re_+$
such that $\f(\l)\to0$ as $\l\to\infty$, and it satisfies the following
condition: 
Let $\La_1$, $\La_2\in \Meas$ and let $f\in L^1(\Om)$, 
$g\in L^\infty(\Om)$ such that $f$ is $\Meas_{\La_1}$-measurable and 
$g$ is $\Meas_{\La_2}$-measurable, where $\Meas_\La$ denotes the
$\s$-algebra generated by $\normalset{B^\om(x)}{x\in\La}$. Then 
\[
|\Exp(fg)-\Exp(f)\Exp(g)| \leq 
\f(\text{dist}(\La_1,\La_2)) \norm{f}_{L^1}\norm{g}_{L^\infty}, 
\]
where $\Exp(\cdot)$ denotes the expectation with respect to $\Pro$, 
and $\text{dist}(\cdot,\cdot)$ is the Euclidean distance on $\re^d$. 
\item Let $C_0$ be the unit cube in $\re^d$. Then 
\[
\Exp\bigpare{\int_{C_0} |B^\om(x)|dx} >0. 
\]  
\end{enumerate}
\end{ass}

\begin{thm}
Suppose Assumption~A. Then 
\[
\limsup_{E\dto 0} \log(-\log k(E))/\log E \leq  -d/2. 
\]
\end{thm}

The Lifshitz tail (or the Lifshitz singularity) of the density of states
for Schr\"odinger operator with random potential 
have been studied by many authors~(see, e.g., the textbook of Carmona
and Lacroix~\cite{CL} or the lecture note of Kirsch~\cite{K}),
but very few results have been obtained on the Lifshitz tail for
Schr\"odinger operator with random magnetic field so far. 
The author is aware of only a work by Ueki~\cite{U} where the
Lifshitz tail is studied for Schr\"odinger operators with a class of
Gaussian random magnetic field, and a work by the author on  
2D discrete Schr\"odinger operator with Anderson type random
magnetic field~\cite{Na}. 

The strategy of the proof is similar to \cite{Na}. 
We first prove that magnetic Schr\"odinger operator is bounded from below by a
nonnegative  function, which does not vanish where $B$ does not (Section~2). 
This estimate is similar to the
well-known AHS (Avron-Herbst-Simon) estimate:
\[
H\geq \pm B_{ij}(x), \qquad x\in\re^d, i\neq j.
\]
The right hand side of the AHS estimate is not necessarily positive.
In order to obtain global lower bounds of the
operator,  we have to use a partition of unity, and thus introduce (not necessarily
positive) error terms.  Our estimate (Theorem~2) is in fact weaker than the AHS
estimate in many  situations, but the right hand side of the estimate is always
nonnegative,  and hence easier to apply in our context. 
In Section~3, we discuss Neumann decoupling and prove a simple a priori estimate
on the density of states for magnetic Schr\"odinger operators. 
We  then combine them with a result on the
Lifshitz tail of  Schr\"odinger operator with random potential to
conclude the assertion~(cf. Kirsch-Martinelli \cite{KM}, Kirsch \cite{K}). 

\textbf{Notation:} 
We denote the definition domain of an operator $A$ by $\dom(A)$. 
The quadratic form domain of $A$ is denoted by $\qdom(A)$. 
The inner product of a Hilbert space $\mathcal{H}$ is denoted by 
$\jap{\f|\g}$ ($\f,\g\in\mathcal{H}$), which is linear in the second entry. 
We denote the space of the Schwartz functions on $\re^d$ by $\schwartz(\re^d)$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%   SECTION 2
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Energy estimates for magnetic Schr\"odinger operators}

We consider a deterministic Schr\"odinger operator 
$H=(p-A(x))^2$ with the magnetic field $B=dA$. We fix a constant $r>0$. 
For $x\in\re^d$ and $1\leq i<j\leq d$, we denote
\begin{align*}
D_{ij}(x) &= \bigset{y\in\re^d}{y_k=x_k \text{ for $k\neq i,j$}, \\
&\qquad\qquad \qquad(x_i-y_i)^2+(x_j-y_j)^2\leq r^2},\\
\c_{ij}(x)&= \pa D_{ij}(x).
\end{align*}
We define $b_{ij}(x)\in \torus=\re/(2\pi \ze)$ by
\[
b_{ij}(x) \equiv \int_{D_{ij}(x)} B_{ij}(y) dy_i dy_j
\pmod{2\pi\ze}.
\]
We identify $\torus$ with $[-\pi,\pi)$ and set
\[
W(x)= \frac{1}{(d-1)4\pi^3 r^3} \sum_{i<j} \int_{\c_{ij}(x)} 
|b_{ij}(y)|^2 dy.
\]

\begin{thm}
$H\geq W$ in the operator sense, i.e., 
\[
\jap{\g| H\g} \geq \int_{\re^d} W(x)|\g(x)|^2 dx, \qquad \g\in \dom(H).
\]
\end{thm}

\begin{rem}
Since $|b_{ij}(x)|\leq \pi$, we have 
\[
0\leq W(x)\leq \frac{1}{(d-1)4\pi^3 r^3}\cdot \frac{d(d-1)}{2} 2\pi r \cdot
\pi^2 = \frac{d}{4r^2}, 
\qquad x\in\re^d.
\]
\end{rem}

\begin{proof}
We denote the velocity operator by 
\[
v_j = p_j - A_j(x), \qquad j=1,\dots,d.
\]
We parameterize $\c_{ij}(x)$ as follows: For $t\in [0,2\pi r)$, we set
\[
y_k(x;t) =\begin{cases}
x_k  \quad &\text{if $k\neq i,j$}, \\
x_i+r\cos(t/r) \quad &\text{if $k=i$} ,\\
x_j+r\sin(t/r) \quad & \text{if $k=j$}.
\end{cases}
\]
For $\g\in\schwartz(\re^d)$, we compute
\begin{equation}\label{eq-2-1}
E_x(\g) = \int_{\c_{ij}(x)} |\dot{y}\cdot (v\g)|^2 dy,
\end{equation}
where $\dot y(x;t) =\frac{d}{dt} y(x;t)$. If we denote 
\[
\tilde\g(t) = \g(y(x;t)), \qquad \tilde A(t) = \dot{y}(x;t)\cdot A(y(x;t)), 
\]
then the right hand side of (\ref{eq-2-1}) is 
\[
\int_0^{2\pi r} \bigabs{(-i\pa_t -\tilde A(t)) \tilde \g(t)}^2 dt = 
\jap{\tilde \g|h\tilde \g}
\]
where $h$ is an operator on $L^2([0,2\pi r))$ defined by
\[
h\f(t) = (-i\pa_t -\tilde A(t))^2\f(t), 
\qquad \f\in \dom(h)= H^2(\re/(2\pi r\ze)), 
\]
i.e., $\f\in\dom(h)$ satisfies the periodic boundary condition. 
Then using the gauge transform $G(t)=\exp(i\int_0^t \tilde A(s) ds)$, we can
easily observe that $h$ is unitarily equivalent to $\tilde h$ defined by
\begin{align*}
 & \tilde h \f(t)= -\pa_t^2 \f(t), \\
&\f\in\dom(\tilde h) = \bigset{\f\in H^2((0,2\pi r))}{\f(2\pi r)= e^{ib}\f(0)},
\end{align*}
where $b\equiv \int_0^{2\pi r} \tilde A(t)dt \pmod{2\pi \ze}$. 
Now we note
\begin{align*}
\int_0^{2\pi r} \tilde A(t) dt &= \int_{\c_{ij}(x)} \dot{y}\cdot A(y(x;t)) dt \\
&= \int_{D_{ij}(x)} B_{ij}(y) dy_i dy_j \equiv b_{ij}(x) \pmod{2\pi \ze}
\end{align*}
by the Stokes formula. Thus $b=b_{ij}(x)$. 
The spectrum of $\tilde h$ is easily computed, and we learn
\[
\s(\tilde h) = \biggset{ \biggpare{\frac{b}{2\pi r}+\frac{n}{r}}^2}{n\in\ze}.
\]
In particular, 
\[
\inf\s(h)=\inf\s(\tilde h) = \biggpare{\frac{b_{ij}(x)}{2\pi r}}^2 =
\frac{1}{4\pi^2r^2}|b_{ij}(x)|^2. 
\]
This implies
\[
E_x(\g) = \jap{\tilde\g| h \tilde \g} 
\geq \inf\s(h)\norm{\tilde\g}^2 = 
\frac{1}{4\pi^2r^2}|b_{ij}(x)|^2 \norm{\tilde\g}^2.
\]
Integrating this inequality over $x\in\re^d$, we have
\begin{align}
\int_{\re^d} E_x(\g)dx &\geq \frac{1}{4\pi^2 r^2}
\int_{\re^d}\int_0^{2\pi r} |b_{ij}(x)|^2 |\g(y(x;t))|^2 dt dx\nonumber\\
&= \frac{1}{4\pi^2 r^2}
\int_{\re^d}\int_0^{2\pi r} |b_{ij}(y(x';t'))|^2 |\g(x')|^2 dt' dx'
\label{eq-2-2}
\end{align}
where $x'=y(x,t)$ and $t'\equiv t+\pi r \pmod{2\pi\ze}$. 

Now we estimate the left hand side of (\ref{eq-2-2}). 
We use the same change of variables as above to learn
\begin{align*}
 &\int_{\re^d} E_x(\g)dx= \int_{\re^d} \int_0^{2\pi r} 
\bigabs{\dot{y}(x;t)\cdot \g(y(x;t))}^2 dtdx \\
&\qquad= \int_{\re^d} \int_0^{2\pi r} 
\bigbra{\sin^2(t/r)|v_i\g(x)|^2 
+\cos^2(t/r)|v_j\g(x)|^2} dtdx\\ 
&\qquad= \pi r \int_{\re^d}\bigbra{|v_i\g(x)|^2 + |v_j\g(x)|^2} dx.
\end{align*}
Combining this with (\ref{eq-2-2}), we obtain
\begin{multline*}
\int_{\re^d}\bigbra{|v_i\g(x)|^2 + |v_j\g(x)|^2} dx \\
\geq \frac{1}{4\pi^3 r^3} \int_{\re^d}\int_0^{2\pi r} 
|b_{ij}(y(x;t))|^2 |\g(x)|^2 dtdx.
\end{multline*}
Then we sum up this estimate over $1\leq i<j\leq d$ to conclude
\[
(d-1)\sum_{i=1}^d |v_i\g(x)|^2 
\geq \frac{1}{4\pi^3 r^3} \int_{\re^d} \sum_{i<j} |b_{ij}(y(x;t))|^2 |\g(x)|^2 dtdx.
\]
and this implies the assertion of Theorem~2. 
\end{proof}

\begin{example}
Let $H$ be the Schr\"odinger operator on $\re^2$ with constant magnetic field
with $B>0$. We choose $r>0$ so that $b=b_{12}=\pi$, i.e., 
$r=B^{-1/2}$. Then we have $W(x)=B/2$, and hence $H\geq B/2$. It is well-known,
however, that $\inf\s(H)=B$, and the estimate is not optimal. 
\end{example}

The following estimate (or, strictly speaking, its generalization) will be useful in
the proof of Theorem~1. 

\begin{cor}
Let $W$ as above, and let $H_0=p^2$ be the free Schr\"odinger operator. Then
\[
\inf\s(H)\geq \frac12 \inf\s(H_0+W).
\]
\end{cor}

\begin{proof}
By Kato's inequality (\cite{RS} Theorem~X.33), we have 
\[
\jap{\g|H\g} \geq \jap{|\g|\,|H_0|\g|}, \qquad \g\in \schwartz(\re^d).
\]
Combining this with Theorem~1, we learn
\[
\jap{\g|H\g} \geq \frac12\jap{|\g|\,|(H_0+W)|\g|},
\]
and hence 
\begin{align*}
\inf\s(H) &= \inf_{\g\neq 0} \frac{\jap{\g|H\g}}{\norm{\g}^2} \\
&\geq \frac12 \inf_{\g\neq 0} \frac{\jap{|\g|\,|(H_0+W)|\g|}}{\norm{\g}^2}
\geq \frac12 \inf\s(H_0+W).
\end{align*}
(Note that the last
inequality is in fact equality, since the ground state of Schr\"odinger operator is
nonnegative.) 
\end{proof}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
%   SECTION 3
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Proof of Theorem~1}

As in the last section, we fix $A(x)$ and consider a deterministic magnetic
Schr\"odinger operator $H=(p-A)^2$. Let $\La\subset\re^d$ be an open set. 
Then the Neumann Hamiltonian $H_\La$ on $L^2(\La)$ is defined by
\begin{align*}
& H_\La \g(x) = (p-A(x))^2\g(x), \qquad \g\in\dom(H_\La), \\
& \qdom(H_\La) = \dom((H_\La)^{1/2}) \\
&\qquad = \bigset{\g\in H_{loc}^1(\La)}{(p_j-A_j(x))\g\in L^2(\La), j=1,\dots,d}.
\end{align*}
Note that $\qdom(H_\La)$ is not necessarily the same as $H^1(\La)$ if $\La$ is
unbounded. If $\tLa\subset \La$ are open sets in $\re^d$ such that
$|\La\setminus\tLa|=0$, then we can regard $L^2(\La)=L^2(\tLa)$, and $H_\La$ and 
$H_\tLa$ may be considered as operators on the same Hilbert space. Since 
$\qdom(H_\tLa)\supset \qdom(H_\La)$ and the symbols are the same, we learn
\[
H_\tLa \leq H_\La \qquad \text{on $L^2(\La)$}.
\]
Hence, in particular, by the min-max principle (\cite{RS} Theorem~XIII-1) 
we have
\[
\frac{1}{|\La|} \#\{\text{eigenvalues of }H_\La\leq E\}
\leq \frac{1}{|\La|} \#\{\text{eigenvalues of }H_\tLa\leq E\}
\]
for $E\in\re$ if $|\La|$ is finite.  We denote
\[
C(L;x) =\bigset{y\in\re^d}{x_j< y_j<x_j+L, j=1,\dots,d}, 
\]
for $L>0$, $x\in\re^d$. In particular, we write $\La_L= C(L,0)$. 
We will need the following a priori bound on the number of eigenvalues. 

\begin{lem}
Suppose that $|B(x)|\leq M$ with $M<\infty$. Then there exists a constant 
$C>0$ depending only on $d$ and $M$ such that 
\[
\frac{1}{|\La_L|} \#\{\text{\rm eigenvalues of }H_{\La_L}\leq E\} 
\leq C(1+E^{d/2})
\]
for any $E\geq 0$ and for any integer $L>0$. 
\end{lem}

\begin{proof}
Let $\La=\La_L$, and let 
\[
\tLa= \bigcup \bigset{C(1,n)}{n\in\ze^d, 0\leq n_j<L, j=1,\dots,d}.
\]
Then $\tLa\subset\La$ and $|\La\setminus \tLa|=0$, and hence it suffices to
estimate
\begin{equation}\label{eq-3-1}
\frac{1}{|\La_L|} \#\{\text{e.v.'s of }H_{\tLa}\leq E\} 
= \L^{-d} \sum_{\substack{0\leq n_j<L\\j=1,\dots,d}} \#\{\text{e.v.'s of
}H_{C(1,n)}\leq E\} 
\end{equation}

We note that the number of eigenvalues are independent of the choice of gauge.
Hence we may choose $A(x)$ on each $C(1,n)$ so that $|A_j(x)|\leq M$
$(j=1,\dots,d)$. Then 
\begin{align*}
H_{C(1,n)} &= p^2-A\cdot p - p\cdot A +A^2 \\
&= \frac12 p^2 +\frac12 (p-2A)^2 -A^2 \\
&\geq \frac12 p^2 -dM^2.
\end{align*}
Hence we learn
\begin{align}
\#\{\text{e.v.'s of } H_{C(1,n)}\leq E\} 
&\leq \#\{\text{e.v.'s of }  (\tfrac12 p^2 -dM^2)\leq E\} \nonumber \\
&\leq c_1(E+dM)^2 \leq c_2(1+E^{d/2}) \label{eq-3-2}
\end{align}
with some $c_1,c_2>0$. The lemma now follows from (\ref{eq-3-1}) and 
(\ref{eq-3-2}). 
\end{proof}

We also need a generalization of the results of the last section for the Neumann 
Hamiltonian. Let $r>0$ and $W(x)$ as in the last section.  For $\La\subset \re^d$,
we set
\[
W_\La(x) = 
\begin{cases} W(x) \qquad & \text{if dist$(x,\La^c)>r$}, \\
0 &  \text{if dist$(x,\La^c)\leq r$}.
\end{cases}
\]
Then, by exactly the same argument as in the last section, we can prove the
following estimates:

\begin{lem}
$H_\La\geq W_\La$ as operators on $L^2(\La)$. Hence, in particular, 
\[
\inf\s(H_\La) \geq \frac12 \inf\s(H_{0,\La}+W_\La), 
\]
where $H_{0,\La}$ denotes the free Schr\"odinger operator with the Neumann
boundary condition on $\La$. 
\end{lem}

\begin{proof}[Proof of Theorem~1]
Here we suppose that $B=B^\om$ satisfies Assumption~A. Let $L>0$ and we set
\[
k_L(E)= \frac{1}{|\La_L|} \Exp(\#\{\text{e.v.'s of } H^\om_L\leq E\}), 
\qquad E\in\re,
\]
where $H^\om_L$ is the Neumann Hamiltonian on $\La_L$. Then we can show 
\begin{equation}\label{eq-3-3}
k(E)\leq k_L(E), \qquad \text{for any $E$ and $L$},
\end{equation}
using exactly same argument as in the case of Schr\"odinger operator with random
potential~(cf. \cite{K} Chapter~7, Corollary~2). Now we apply the argument of
Kirsch-Martinelli~\cite{KM}. It follows from Assumption~A-(iii) that $W(x)$
satisfies Assumption~B-ii) of \cite{KM}.  By Assumption~A-(iv), we have 
\[
\Exp\bigpare{\int_{C_0} |W(x)|dx} >0
\]
and hence 
\[
\Exp(|\normalset{x\in C_0}{W(x)=0}|) <1.
\]
Then we learn from the argument of Section~3 of \cite{KM} that there are $c_1$,
$c_2>0$ such that if $L\sim c_1 E^{-1/2}$ then 
\[
\Pro\bigpare{\inf\s(H_{0,\La_L}+W_{\La_L})\leq E} 
\leq \exp(-c_2 E^{-d/2}), \qquad E\sim 0.
\]
We note that the support of $W-W_{\La_L}$ is contained in a neighborhood of
$\pa\La_L$ with the width $r$, and the volume is $O(L^{d-1})$. Hence the
difference of $W$ and $W_{\La_L}$ does not affect the large deviation argument.
This and Lemma~5 imply 
\[
\Pro\bigpare{\inf\s(H^\om_L)\leq E} \leq \exp(-c_3 E^{-d/2}) 
\]
with $c_3>0$. Combining this with Lemma~4, we have
\begin{align*}
k_L(E) &\leq C(1+E^{d/2})\cdot \Pro\bigpare{\inf\s(H_L)\leq E} \\
&\leq \exp(-c_4 E^{-d/2})
\end{align*}
with $c_4>0$ if $E$ is sufficiently small. Theorem~1 follows from this and
(\ref{eq-3-3}). 
\end{proof}


%%%%%%%%%%%%%% REFERENCES %%%%%%%%%%%%%%%%%%%%

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\end{thebibliography}


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