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\begin{document}

\title[Bound States of Inverse Square Potentials]{Bound States of
Discrete Schr\"odinger Operators with Super-Critical Inverse Square Potentials}


\author[D. Damanik]{David Damanik}
\address{Mathematics 253--37\\ California Institute of Technology\\
Pasadena\\ CA 91125\\ U.S.A.}
\email{\href{mailto:damanik@caltech.edu}{damanik@caltech.edu}}
\urladdr{\href{http://www.math.caltech.edu/people/damanik.html}{http://www.math.caltech.edu/people/damanik.html}}

\author[G. Teschl]{Gerald Teschl}
\address{Faculty of Mathematics\\ University of Vienna\\
Nordbergstrasse 15\\ 1090 Wien\\ Austria\\ and International Erwin Schr\"odinger
Institute for Mathematical Physics\\ Boltzmanngasse 9\\ 1090 Wien\\ Austria}
\email{\href{mailto:Gerald.Teschl@univie.ac.at}{Gerald.Teschl@univie.ac.at}}
\urladdr{\href{http://www.mat.univie.ac.at/~gerald/}{http://www.mat.univie.ac.at/\string~gerald/}}

\thanks{{\it Work supported by the National Science Foundation under Grant No.\ DMS-0500910
and the Austrian Science Fund (FWF) under Grant No.\ P17762}}

\keywords{Discrete Schr\"odinger operators, bound states, oscillation theory}
\subjclass[2000]{Primary 47B36, 81Q10; Secondary 39A11, 47B39}


\begin{abstract}
We consider discrete one-dimensional Schr\"odinger operators whose potentials decay
asymptotically like an inverse square. In the super-critical case, where there are
infinitely many discrete eigenvalues, we compute precise asymptotics of the number of
eigenvalues below a given energy $E$ as this energy tends to the bottom of the essential
spectrum.
\end{abstract}

\maketitle


\section{Introduction}

This paper is concerned with discrete one-dimensional Schr\"odinger operators in
$\ell^2(\Z_+)$, where $\Z_+ = \{ 1,2,3,\ldots\ \}$. That is,
$$
H u(n) = - \Delta u(n) + V(n) u(n), \quad \Delta u(n) = u(n+1) - 2u(n) + u(n-1),
$$
where we impose a Dirichlet boundary condition, $u(0) = 0$.

In the free case, $V \equiv 0$, the spectrum of $H = - \Delta$ is
the interval $[0,4]$. If $V(n) \to 0$ as $n \to \infty$, then $V$ is
compact and hence zero is the bottom of the essential spectrum of
$H$. We are interested in the discrete spectrum of $H$ below zero.
It consists of simple eigenvalues, so-called bound states, that can
only accumulate at zero. Thus, for $E \ge 0$, we define
$$
N_E(H) = \dim\Ran P_H((-\infty,-E]),
$$
where $P_H$ is the family of spectral projections associated with
$H$ by the spectral theorem.

It is well known that $V(n) \sim - n^{-2+\eps}$ produces finitely
many discrete eigenvalues if $\eps<0$ and infinitely many if
$\eps>0$ and so inverse square decay is critical for the existence
of infinitely many discrete eigenvalues below the essential
spectrum. Furthermore, by the discrete analogue of Kneser's theorem,
the discrete spectrum below zero of the operator $H$ with potential
$V(n) = - c n^{-2}$ is finite when $c \le \frac14$ and infinite when
$c > \frac14$. Surprisingly, this discrete analog turns out much
harder to prove than the continuum result and it is not so well
known. It was first shown by Na\u\i man \cite{n} and later
rediscovered in \cite{dhs} (see also \cite{lute} for a more general
result).

Our goal is to study the behavior of $N_E(H)$ as $E \downarrow 0$ in
the super-critical case $c>\frac14$. This question is natural since
this case describes the borderline behavior at the transition
between finitely and infinitely many bound states as pointed out in
the previous paragraph. In addition, we were also motivated by
recent results on a connection between singular spectrum embedded in
the essential spectrum and the discrete spectrum of a given
Schr\"odinger operator.

Let us sketch this connection briefly. For operators of the form $H
= -\Delta + V$ on the whole line, that is, acting in $\ell^2(\Z)$,
it was shown by Killip and Simon that the absence of bound states
implies that the potential vanishes identically \cite{ks}. In other
words, $\sigma(H) \subseteq [0,4]$ implies $V \equiv 0$. Damanik et
al.\ found a simple variational proof of this statement that also
allowed them to extend the result to operators in $\ell^2(\Z^2)$
\cite{dhks}. It does not hold in dimensions three or higher, nor on
the half-line, as shown in \cite{dhks}. The counterexamples in
$\ell^2(\Z^d)$, $d \ge 3$, are given by sparse non-decaying
potentials, whereas a possible counterexample in $\ell^2(\Z_+)$ is
given by $V(n) = (-1)^n/n$. Damanik and Killip then showed that, on
the half-line, the absence of bound states implies the absence of
embedded singular spectrum. In fact, absence of embedded singular
spectrum follows already when there are finitely many bound states
\cite{dk}. This raised the natural question of what can be said when
there are infinitely many bound states. Is it true that sufficiently
fast decay, for example finiteness of the moments
$$
M_\gamma = \sum_{n \ge 1} \left( \mathrm{dist}(E_n,[0,4])
\right)^\gamma,
$$
implies the absence of embedded singular spectrum? Here, $\{E_n\}_{n
\ge 1}$ denote the discrete eigenvalues of $H$ outside the essential
spectrum $[0,4]$. Note that the Damanik-Killip result can be phrased
as follows: $M_0 < \infty$ implies that $\sigma_{\mathrm{sing}}(H)
\cap [0,4] = \emptyset$. Damanik et al.\ then found an example with
$2$ as an embedded eigenvalue (and hence $\sigma_{\mathrm{sing}}(H)
\cap [0,4] \not= \emptyset$) and $M_\gamma < \infty$ for every
$\gamma > 0$ \cite{dks}!\footnote{It was later realized by Damanik
and Remling that $M_\gamma < \infty$ does indeed put some
restrictions on the embedded singular spectrum; see \cite{dr}. For
example, if $M_\gamma < \infty$ for every $\gamma > 0$, then the
embedded singular spectrum is supported on a set of zero Hausdorff
dimension.} This example has the potential $V(n) = (1 + \varepsilon)
(-1)^n/n$ for some small $\varepsilon > 0$. The core of the proof
consists of using methods developed in \cite{dhks} and \cite{dhs} to
compare this sign-indefinite potential with two sign-definite ones,
both of the form $- c_j n^{-2}$, for different constants $c_1, c_2 >
1/4$. This clearly motivates a study of the bound state structure of
super-critical inverse square potentials. The estimates for their
bound states derived in \cite{dks} were sufficient to prove the
result mentioned above, but it is desirable to have a more complete
understanding; in particular since in the continuous case the bound
state structure has been studied in detail earlier. Namely, Kirsch
and Simon considered operators of the form $- \frac{d^2}{dx^2} + V$
and carried out an investigation of $N_E(H)$ for super-critical
inverse square potentials \cite{kisi} (see also Schmidt \cite{schm}
for extensions). They proved that the bound states approach the
essential spectrum geometrically. More precisely, they showed that
\begin{equation}\label{ksres}
\lim_{E\downarrow 0} \frac{N_E(-\Delta+V)}{-\ln(E)} = \frac{1}{2\pi}
\sqrt{c-\tfrac{1}{4}},
\end{equation}
where $V(x) = -\frac{c}{1 + x^2} + W(x)$ and $W$ is such that
$N_0(-\Delta + \gam W)<\infty$ for all $\gam\in\R$.\footnote{This is
the half-line version of the Kirsch-Simon result. On the whole line,
one has to multiply the right-hand side of \eqref{ksres} by two;
compare \cite[Theorem~1]{ks}.} Let us also note that in the
continuous case, a study of inverse square potentials is further
motivated by the fact that they arise in the partial wave analysis
of rotationally symmetric partial differential operators.

We will prove the discrete analogue of the Kirsch-Simon result in
this paper, that is, geometric approach of the bound states to the
essential spectrum.

On the one hand, as explained above, this result is relevant to the
investigations and results described in the previous paragraph that
led the authors of \cite{dks} to ask if the Kirsch-Simon result has
a discrete analogue; compare \cite[Sect.~2]{dks}. On the other hand,
the proof of Kirsch and Simon uses some arguments that do not carry
over directly to the discrete case: They scale the spatial variable
and use exact solvability of the Euler differential equation.
Spatial scaling is not possible in the discrete case and, while
there exists a discrete Euler equation, it is not symmetric.


\begin{theorem}\label{main}
Suppose
$$
V(n) = - \frac{c}{n^2} + W(n), \qquad c> \frac{1}{4},
$$
where $W$ is a decaying sequence such that $N_0(-\Delta + \gam W)<\infty$ for all
$\gam\in\R$. Then
\begin{equation}\label{limit}
\lim_{E\downarrow 0} \frac{N_E(-\Delta+V)}{-\ln(E)} = \frac{1}{2\pi}
\sqrt{c-\tfrac{1}{4}}.
\end{equation}
\end{theorem}
\noindent\textit{Remarks.} (i) We say that a sequence $W$ is decaying if $W(n) \to 0$
as $n \to \infty$.\\
(ii) The hypothesis on $W$ is satisfied, for example, if
$\sum_{n>0} n |W(n)|< \infty$. See \cite[Thm.~5.10]{tjac}.\\
(iii) An analogous result holds on the whole line. This follows quickly from the half-line result
since the whole-line case can be reduced to the half-line case by Dirichlet decoupling.\\
(iv) For perturbations of the form $V(n) = \frac{c}{n^2} + W(n)$, an analogous result
holds near the top of the essential spectrum.



\section{Proof of Theorem~\ref{main}}


As a preparation we state the discrete analog of Proposition~5 from \cite{kisi}.
The proof is analogous.

\begin{lemma} \label{lemNE}
Let $V$, $W$ be decaying sequences. Then for every $E>0$ and $0<\eps<1$ we have
\begin{align*}
N_E\left(-\Delta + V + W\right) & \le N_E\left(-\Delta +\tfrac{1}{1-\eps} V\right) +
N_E\left(-\Delta + \tfrac{1}{\eps} W\right),\\
N_E\left(-\Delta + V + W\right) & \ge N_E\left(-\Delta + (1-\eps) V\right) - N_E
\left(-\Delta - \tfrac{1-\eps}{\eps} W\right).
\end{align*}
\end{lemma}
\noindent
Now we come to the proof of our main theorem. We start with $$V_c(n)= - \frac{c}{n^2}
$$ and replace it by $V_{E,c}$ which is just $V_c - E$ on $\{ n : V_c(n) \le -E\}$
and equal to $V$ otherwise. To investigate the asymptotics of $N_E(-\Delta + V_{E,c})$ we
split our domain into two parts by cutting at $\sqrt{\frac{c}{E}}$. For the first part,
we will compute the asymptotics of $N_E$ directly. The remaining part does not contribute
to $N_E$. Then we use Lemma~\ref{lemNE} to show that $N_E$ has the
same asymptotics for $V_{E,c}$ and $V= V_c + W$.

\begin{lemma}\label{lem3}
We have
$$
\lim_{E\downarrow 0} \frac{N_E(-\Delta+V_{E,c})}{-\ln(E)} = \frac{1}{2\pi}
\sqrt{c-\tfrac{1}{4}}.
$$
\end{lemma}

\begin{proof}
We first decompose $-\Delta+V_{E,c}$ into two parts by imposing an additional Dirichlet
boundary condition at $\floor{\sqrt{\frac{c}{E}}}$. Since this constitutes a rank-one
resolvent perturbation it will not affect the limit. By the choice of our cut point, the
part with $n>\floor{\sqrt{\frac{c}{E}}}$ does not contribute and by oscillation theory
(see e.g.\ \cite{tosc} or \cite[Ch.~4]{tjac}) it
suffices to count the number of sign flips of some solution of $(-\Delta + V_{E,c}) u =
-E u$ on $(1,\sqrt{\frac{c}{E}})$, that is, the number of sign flips of some solution of
$(-\Delta + V_c) u = 0$ on $(1,\sqrt{\frac{c}{E}})$.

Unfortunately, $(-\Delta + V_c) u = 0$ is not explicitly solvable, but
$$
\ti{u}_c(n)= \sqrt{n} \exp\left(\I \sqrt{c-\tfrac{1}{4}} \ln(n)\right)
$$
solves $(-\Delta + \ti{V}_c) \ti{u}=0$ with
the complex-valued potential
$$
\ti{V}_c(n) = \frac{\Delta \ti{u}_c(n)}{\ti{u}_c(n)} = -\frac{c}{n^2} + O(\frac{1}{n^3}).
$$
Moreover, it is straightforward to check
(cf.~\cite[Lemma~7.10]{tjac}, resp.~\cite{lute}) that $-\Delta u +
V_c u = 0$ has a solution $u_c$ which asymptotically looks like
$\ti{u}_c(n)$. Taking the real part of $u_c$, we see that the number
of sign flips behaves to leading order like $-\frac{1}{2\pi}
\sqrt{c-\frac{1}{4}} \ln(E)$.
\end{proof}
\noindent
Let us prove the upper bound in \eqref{limit}. By Lemma~\ref{lemNE},
\begin{align*}
N_E(-\Delta + V_c + W) &= N_E(-\Delta + (V_c - \chi_{(1-\eps)E,c}) +
N_E(\chi_{(1-\eps)E,c} + W))\\
&\le N_E(-\Delta + V_{E,c/(1-\eps)}) + N_E(-\Delta + \tfrac{1}{\eps} (\chi_{(1-\eps)E,c}
+ W)),
\end{align*}
where $\chi_{E,c}= E \chi_{(0,\sqrt{c/E})}$ and $\chi_\Omega$ is the characteristic function of the
set $\Omega$. Using
$$
N_E(-\Delta + \tfrac{1}{\eps} (\chi_{(1-\eps)E,c} + W)) \le N_0(-\Delta + \tfrac{1}{\eps}
W),
$$
the assumption on $W$, and Lemma~\ref{lem3}, we see that
$$
\limsup_{E\downarrow 0} \frac{N_E(-\Delta+V_c+W)}{-\ln(E)} \le \frac{1}{2\pi}
\sqrt{\tfrac{c}{1-\eps}-\tfrac{1}{4}}
$$
for every $0<\eps<1$, that is,
\begin{equation}\label{upper}
\limsup_{E\downarrow 0} \frac{N_E(-\Delta+V_c+W)}{-\ln(E)} \le \frac{1}{2\pi}
\sqrt{c-\tfrac{1}{4}}.
\end{equation}
It remains to show the lower bound in \eqref{limit}. By
Lemma~\ref{lemNE},
\begin{align*}
N_E(-\Delta + V_c + W) &= N_E(-\Delta + (V_c - \chi_{E/(1-\eps),c}) +
N_E(\chi_{E/(1-\eps),c} + W))\\
&\ge N_E(-\Delta + V_{E,(1-\eps) c}) - N_E(-\Delta - \tfrac{1-\eps}{\eps}
(\chi_{E/(1-\eps),c} + W)).
\end{align*}
Observe that it suffices to show that the second
summand does not contribute to the limit. Invoking Lemma~\ref{lemNE} a second time we
have
$$
N_E(-\Delta - \tfrac{1-\eps}{\eps} (\chi_{E/(1-\eps),c} + W)) \le N_E(-\Delta -
\tfrac{1}{\eps} \chi_{E/(1-\eps),c}) + N_E(-\Delta - \tfrac{1-\eps}{\eps^2} W).
$$
The second term is bounded for fixed $\eps$ as $E\downarrow 0$ by assumption and it
remains to investigate the first one. As before we impose a Dirichlet boundary condition at
$\floor{\sqrt{\frac{c(1-\eps)}{E}}}$ and we need to count the sign flips of the solution
of $-\Delta u - \frac{E}{\eps(1-\eps)}u = -E u$ on $(0, \sqrt{\frac{c(1-\eps)}{E}})$.
Since this equation is explicitly solvable we obtain
$$
N_E(-\Delta - \tfrac{1}{\eps} \chi_{E/(1-\eps),c}) = \sqrt{c(1-\eps-\tfrac{1}{\eps})} + O(E).
$$
Hence
$$
\liminf_{E\downarrow 0} \frac{N_E(-\Delta+V_c+W)}{-\ln(E)} \ge
\frac{1}{2\pi} \sqrt{(1-\eps) c-\tfrac{1}{4}}
$$
for every $0 < \eps < 1$ and thus, \be \label{lower}
\liminf_{E\downarrow 0} \frac{N_E(-\Delta+V_c+W)}{-\ln(E)} \ge
\frac{1}{2\pi} \sqrt{c-\tfrac{1}{4}}. \ee Combining \eqref{upper}
and \eqref{lower}, we obtain the assertion of the theorem.
\hfill\qedsymbol

\bigskip

\noindent{\bf Acknowledgments.} G.T.\ gratefully acknowledges the
extraordinary hospitality of the Department of Mathematics at
Caltech, where this work was done.

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\end{document}
