%%%%%%%%%%%%%% The Paper %%%%%%%%%%%%%%%%%%%%
\input amstex
\documentstyle{amsppt}
%%%%%%%%%%%%%%%%%  MY Macros %%%%%%%%%%%%%%%%%%
\def\ZZ{\Bbb Z}
\def\RR{\Bbb R}
\def\znu1{\ZZ^{\nu - 1}}
\def\znu{\ZZ^{\nu}}
\def\eltwo{l^2(\znu)}
\def\eltwoz{l^2(\ZZ)}
\def\half{1\over2}
\def\third{1\over3}
\def\quarter{1\over4}
\def\tks{\thickspace}
\def\tis{\thinspace}
%%%%%%%%%%%%%%%% end of Macros %%%%%%%%%%%%%%%
\topmatter
\rightline{\bf Preprint No.: IMSc/92/48}
\vskip 1cm
\title Absolutely continuous spectrum for sparce Potentials
\endtitle
\author M.Krishna\endauthor
\address
Institute of Mathematical Sciences,  
Taramani, Madras 600113, India
\endaddress
\email krishna\@ imsc.ernet.in \endemail
\abstract
 In this paper we show that bounded perturbation of the
discrete laplacian with a potential that is sparcely
supported ( a notion made precise in the paper )
produces absolutely continuous spectrum in the
interval [-2$\nu$, 2$\nu$] for large dimension $\nu$.  We note that the
potential need not give rise to a compact operator, let alone have 
decay at infinity.
\endabstract
\endtopmatter
%%%%%%%%%%%%%%%%%% end of top matter %%%%%%%%%%%%%%%%
\TagsOnRight
\document
\head 1. Introduction\endhead
The study of the absolutely continuous spectrum is relevent in several
models associated with scattering on the one hand and the extended
states , as  is expected  in the Anderson model, etc.,
In our attemp to understand the mechanism of occurance of the absolutely
continuous spectrum in the Anderson model, we come across a situation
where a certain non compact perturbation of the discrete laplacian in 4
or more dimensions exhibits absolutely continuous spectrum if the
potential is sparcely supported in the lattice.  We may consider this
as a hierarchial model also for the purposes of understanding the
mechanism responsible for producing absolutely continuous spectrum in
the higher dimensional Anderson model.  We stress here that we only
consider a deterministic operator and show the presence of absolutely
continuous spectrum.  
\vskip 1cm
We also would like to add here that in the Anderson model the diffusive
behaviour is supposed to be responsible for the occurance of the
absolutely continuous spectrum, while in this work it is scattering that
produces the absolutely continuous spectrum.  We would like to still
present this result with the hope that it might be useful for further
work in the area.  We recall that in \cite{1} it was shown that if the
potential is i.i.d. random variable with distribution having finite
variance times a function with a  overall decay at infinity as a power
$\alpha < -1$ of the
distance from the origin, then the perturbation of the laplacian with
such potentials has absolutely continuous spectrum in [$-2\nu , 2\nu$]
almost everywhere.  In the proof of that theorem  the behaviour of the
kernel of the free evolution in time is used together with the
conditions on the potentials to show the existence of the wave operators
and thus the presence of the absolutely continuous spectrum. It was
sufficient to make crude estimates to obtain the result there. In this
paper we show that it is actually not necessary for the potential to
have decay at infinity provided it has some sparce support , in such a
way that we can still show the existence of the wave operators.  But for
technical reasons we stick to the deterministic case and consider only
single potentials.  It is clear that some random potentials of small
disorder would have sparce suport but not those assumed here. 
Therefore we cannot extend our results yet to cover them. 

\head 2. The Model \endhead
We consider the space $\eltwo$ and the laplacian $\Delta$ defined by
$$
(\Delta u)(n) = \sum\limits_{| n - i | = 1} u(i)
\tag 2.1 
$$ 
and a potential {\bf V} coming from  a real valued bounded funtion $V :
\znu \rightarrow \RR$.  We then consider the operator H = $\Delta$ + V
and note that it is self-adjoint on $\eltwo$.  We also note that
the operator $\Delta $ has spectrum [-2$\nu$ , 2$\nu$] which is purely
absolutely continuous and  H has spectrum in [ -2$\nu - {\|V\|}_\infty$
, 2$\nu + {\|V\|}_\infty$].  We make the following assumption on the
potential {\bf V}.     

\vskip 1cm

{\noindent \bf Assumption 2.1} {\sl We assume that if $\Lambda$ is a
$\nu$-dimensional cubic
region in $\znu$ centered at any point in $\znu$ ,then the volume $|S|$
of the set $ S = \{ n\in\Lambda :  V(n) \neq 0 \} $ satisfies
$$
|S| \leq | \Lambda | ^{1-\eta}
$$
for $|\Lambda|$ sufficiently large. We assume that $\eta$ satisfies
${4\over\nu} > \eta > {3\over \nu}$ .}

\vskip 1cm

Our main theorem of the paper is the following. 
\vskip 1cm
{\noindent \bf Theorem 2.2} {\sl Consider the operator H defined above on
$\eltwo$ and  suppose the potential {\bf V} satisfies the assumption
2.1, then for  $\nu \geq 4$  the absolutely continuous spectrum of H
contains the interval [-2$\nu$ , 2$\nu$].}

\vskip 1cm

At this stage we would like to mention the strategy of proof.  It turns
out that the time evolution generated by the operator $\Delta$ has decay
like $t^{-\nu\over 2}$ in $\nu$ dimensions in most of the regions of the
lattice.  However there is a thin region in the lattice where the decay
is slightly lower.  Therefore if the potential is nonvanishing in
regions which are in small volume compared to the volume of the region
under consideration, then the overall decay of the sum fo matrix
elements of the time evoluition would have integrable decay , which
would then give using standard techniques of scattering theory the
necessary result from the existence of the wave operators.  
\vskip 1cm
 
{\noindent \bf Proof.} We notice that to prove the theorem we need to 
show the existence of the limits $ \lim_{t\rightarrow\infty}
\tks$ exp(itH) exp(-it$\Delta$)f for each f $\in \eltwo$. 
This is reduced to showing by density arguments that the sequence 
$$
F(\phi,f,t) = < \phi , exp(itH) exp(-it\Delta) f>
\tag 2.2
$$
is Cauchy in t , uniformly in $\phi$ and for all elements f of $\eltwo$
of finite support.

Thus we have to show that
$$
\lim_{t,w \rightarrow \infty} \sup_{\|\phi\|=1} \{ |<\phi ,
exp(itH)exp(-it\Delta) f> \tks - \thickspace <\phi ,
exp(iwH)exp(-iw\Delta) f>|\} 
$$
$$
= 0
\tag 2.3 
$$

By writing the difference on the left hand side as an integral of the
differential one has to show that 
$$
\lim_{t \rightarrow \infty} \sup_{\|\phi\|=1} \{ 
\int\limits_w^t {d\over{ds}} < \phi , exp(isH)exp(-is\Delta) f >\} = 0
\tag 2.4
$$

Computing the integrand in the above expression and simplifying we get
that it is sufficient to show the integrability in {\bf s} of

$$
\sup_{\|\phi\|=1} \{|< \phi, exp(isH) V exp(-is\Delta) f>| \}
\tag 2.5
$$

Since by the argument above f is a vector of finite support in the above
expression , it is sufficient to consider the case when f is a vector
with support at a single point in $\znu$ in particular the element
$\delta_0$.  We shall not make use of any special properties of the
point {\bf 0} of $\znu$ , in the following.  Therefore we consider,

$$
\sup_{\|\phi\|=1} \{|< \phi, exp(isH) V exp(-is\Delta) \delta_0>| \}
\tag 2.6
$$

By introducing a partition of unity in the above expression we find that
it is equal to 

$$
\sup_{\|\phi\|=1} \{|< \phi, exp(isH) \sum_{n\in \znu
}|\delta_n><\delta_n|V exp(-is\Delta) \delta_0>| \}
\tag 2.7
$$

Since the operator exp(isH) is unitary for each fixed s , the above
quantity can be estimated ,using the Cauchy-Schwarz inequality, by  the
following quantity

$$
\sup_{\|\phi\|=1}\{\|\phi \|^2  \sum_{n \in \znu} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2 \}^{\half} 
\tag 2.8
$$

Therefore it is enough to show that the following estimate is valid
as s$\rightarrow \infty$,

$$
 \sum_{n \in \znu} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2 \thickapprox O(|s|^{-2-}) 
\tag 2.9
$$ 

To show this we need to consider the sum split into several regions in
$\znu$ depending upon the parameter {\bf s} and estimate the summand
in each of these regions .  We shall do this in the following lemmas
and complete the proof of the theorem.

\vskip 1cm

{\noindent \bf Lemma 2.3} {\sl Consider the number $\beta$ as in Proposition
A.3 and consider the region $S_s = \{ n \in \znu : |n_i| > \beta s ,\tks
for some \tks i = 1,..,\nu\}$.  Then the following estimate is valid.

$$
 \sum_{n \in S_s} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2 \thickapprox O(|s|^{-2-}) 
\tag 2.10
$$

as $s \rightarrow \infty$.  } 

{\noindent \bf Proof.} To evaluate the sum in (2.10) , we note that
$S_s = \cup_i \{ n \in \znu : |n_i| > \beta |s| \} = \cup_i S_s (i)$.
Therefore on each of these regions we have the estimate,coming
from Propositions A.2 and A.3 , 

$$
 |<\delta_n , exp(-is\Delta)\delta_0 >|^2 \leq exp(-\alpha(\beta)|n_i|)
\prod_{j \neq i} |J_{n_j} (2s)|^2  
\tag 2.11
$$

Therefore we have the estimate
$$ 
 \sum_{n \in S_s} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2 
$$
$$
\leq \sum_{i=1}^{\nu} \sum_{S_s (i)}  
|V(n)|^2  exp(-\alpha(\beta)
|n_i|) \tks \prod_{j \neq i} |J_{n_j} (2s)|^2  
\tag 2.12
$$
Now V(n) is  a bounded function , $|n_i| > \beta |s|$ and $n_j , j \neq
i $ is in $\ZZ$ .  Therefore the exponential term in $|n_i|$ gives a
decay in s of the required order for the sum since  
$J_{n_j}(2s)$ is in$ \eltwoz$ with $\sum_{n_j} |J_{n_j} (2s)|^2$ uniformly
bounded in s.

Now it remains to show that in the region $\znu - S_s $
the decay at infinity in s required by equation (2.9) is valid. 
Infact in the above Lemma we did not make use of the Assumption 2.1 at
all. It is in the other region that we need the asssumption as will be
seen in the proof of the following lemma.
\vskip 1cm
{\noindent \bf Lemma 2.4} {\sl Consider the region $\znu - S_s $,
$S_s $ as in Lemma 2.3.  Then in this region the following estimate
is valid.

$$
 \sum_{n \in \znu - S_s} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2 \thickapprox O(|s|^{-2-}) 
\tag 2.13
$$

as $|s| \rightarrow \infty$.}
\vskip 1cm
{\noindent \bf Proof.} The region $\znu - S_s$ is nothing but the region
$\{ n \in \znu : |n_i| \leq \beta |s| , \forall i = 1,..,\nu \}$, let us
call it $S_s^c$.  We split the region $S_s^c$ further into the following
regions.
$$
S_s^c =  \cup_{k=0}^{\nu} S_s^c (k) 
$$
and
$$
S_s^c (k) = S_s^c \cap \{|n_i - 2s| < \sqrt{s}, i = 1,..,k\}\cap
\{|n_i -2s| \geq \sqrt{s} , i = k+1,..,\nu\}
\tag 2.14   
$$  

where k = 0 is to be understood as the absence of the second factor.
We shall show that in each of the above regions we have the estimate
calimed in the lemma for the left hand side of equation (2.13). 
All the constants that occur in the following estimates would be
independent of s, but may depend on the dimension $\nu$ , the point
{\bf 0} of $\znu$ , on $\|V\|$ and the exponent $\eta$ of
assumption 2.1. Now let us consider first the region $S_s^c (0)$.
Then we have, as $s \rightarrow \infty$, by Propositions A.2 and A.4 (1),

$$
 \sum_{n \in  S_s^c (0)} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2  \leq C \tks Vol\{n \in S_s^c (0): V(n)
\neq 0 \} |s|^{-\nu}
\tag 2.15
$$

where  Vol (A) denotes the cardinality of the set A. Now by assumption
2.1, the Vol in the above inequality is estimated as 

$$
Vol ( n \in S_s^c (0) : V(n) \neq 0 ) \leq \{Vol (S_s^c (0))\}^{1-\eta}
\leq Const. |s|^{\nu(1-\eta)}
\tag 2.16
$$

where the Const. is independent of s.  Combining equations (2.15) and
(2.16) we see that 

$$ 
 \sum_{n \in  S_s^c (0)} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2  \leq C \tks  |s|^{-\nu\eta}
\tag 2.17
$$

Now $\eta$ being bigger than ${3\over{\nu}}$, by assumption 2.1, the
required estimate follows for the sum in this region.  As for the other
region we shall estimate for a general {\bf k }.  We have by
Propositions A.2 and A.4 (1),(2) that 

$$
 \sum_{n \in  S_s^c (k)} |V(n)|^2 |<\delta_n ,
exp(-is\Delta)\delta_0 >|^2 
$$
$$
 \leq C \tks Vol\{n \in S_s^c (k): V(n)
\neq 0 \} |s|^{-(\nu - k)} |s|^{-{2k\over3}}
\tag 2.18
$$

and the estimate for the Vol in the above inequality for each k is done
as follows. We consider a region $S_s^c (k)$ and cover it by disjoint
cubic regions of sides $\sqrt{s}$.  Then , since in each of such cubic
regions the set of points where V does not vanish scales as $(1 - \eta)$
of the volume,  and since there are about $|s|^{{\half}(\nu - k)}$ such
cubes covering the region $S_s^c (k)$ we have the estimate that
for each k = 1,..,$\nu$,

$$
Vol ( \{ n \in S_s^c (k) : V(n) \neq 0 \} ) \leq  Const.
|s|^{({\nu\over2})(1 - \eta)} |s|^{ {(\nu - k)}\over2}
\tag 2.19
$$ 

On the otherhand for some values of k the above estimate could turnout to
be a gross over estimate of the actual volume of the points where V is
non zero.  Therefore we also consider the estimate that 

$$
Vol( \{ n \in S_s^c (k) : V(n) \neq 0 \}) \leq Vol ( \{ n \in S_s^c :
V(n) \neq 0 \}) \leq  Const. |s|^{\nu(1 - \eta)}
\tag 2.20
$$ 

>From equations (2.18) - (2.20) , it is clear that the decay as $ s
\rightarrow \infty$ claimed in lemma is valid if , for each k =
1,..,$\nu$ we have that 

$$
min \{{(\nu - k)\over2} + \nu{(1 - \eta)\over2} , \nu(1 - \eta) \} +
(-\nu + k - {2k\over3}) < -2 
\tag 2.21
$$

This inequality simplifies to  the following

$$
-\eta - max \{  \eta , {k\over\nu} \} + {2\over3}{k\over\nu} <
-{4\over\nu} 
\tag 2.22
$$

This equation is satisfied for each k = 1,..,$\nu$ if $\eta$ is chosen
so that ${4\over\nu} > \eta > {3\over\nu}$, when $\nu \geq 4$.
Thus under the assupmtion 2.1 , the lemma is proved. 

\vskip 1cm

We shall give an example of a potential V satisfying assumption 2.1 ,
below.
\vskip 1cm
{\noindent \bf Example.} Let us consider $\ZZ^4$ and the set S defined
by 
$$
S = \{ n \in \ZZ^4 : \forall i = 1,..,4 , n_i = k_i^5 \tks for some \tks
k_i \in \ZZ \}
$$

Now consider any real valued function V which takes non zero values on S
and is zero in $\ZZ^4 - S$.  It is not difficult to check that such a V
satisfies the assumption 2.1 , with $\eta$ = ${4\over5}$.

One can construct in a similar fashion a host of example of functions in
any dimension bigger than 3 .  It is also clear that if V is bounded but
does not vanish on any point of S , it cannot give rise to a compact
oprator in $\eltwo$. Of course it goes without saying that if the
potential consits of two parts one satisfying the assumptions given in
assumption 2.1 and the other having a decay at infinity as in \cite{1},
the Theorem 2.2 will be valid for the sum.

\vskip 1cm
%%%%%%%%%%%%%%%%% Appendix  %%%%%%%%%%%%%%%
\head Appendix \endhead
In the appendix we collect a few propositions that are standard and
which are used in the paper.
\vskip 1cm
{\noindent \bf Proposition A.1} {\sl If A and B are two bounded self-adjoint
operators on a Hilbert space and suppose the limits ${slim}_{t\rightarrow
\infty}$ exp(iAs)exp(-iBs)$E_{ac}$(B) exists.  Then \par $\sigma_{ac}(B)
\subset \sigma_{ac}(A)$}
\vskip 1cm
The above theorem is standard and the proof can be found in any standard
work on scattering theory , for example \cite{2} \cite{3}. 

\vskip 1cm
{\noindent \bf Proposition A.2} {\sl Let m,n $\in \znu $and $\Delta$ be as in
(2.1). Then we have
$$
<\delta_n , exp(-i\Delta s)\delta_m> = \prod\limits_{i=1}^{\nu}
(-i)^{(n-m)_i} J_{(n-m)_i} (2s) 
\tag A.1
$$
where $J_k (s)$ is the Bessel function of integral order k.}
\vskip 1cm
The expression for the matrix elements of the free evolution in terms of
the Bessel functions follows from direct verification , see for example
the Appendix of \cite{1}. 
\vskip 1cm
{\noindent \bf Proposition A.3} {\sl Consider $J_n (x)$,$n\in \ZZ$ and $x>0$.
Then $\exists \tks \beta > 0$ and $\alpha(\beta) >0$ such that
$$
|J_n (x)| \leq C exp( - \alpha(\beta) | n |)
\tag A.2
$$
whenever $ |n| > \beta |x|$.}
\vskip 1cm
{\noindent \bf Proof.} We have the following relation for the Bessel
function of integral order.
$ J_n (x) = (-1)^n J_{-n}(x) $
for n a positive integer. Therefore we consider only positive n in
the following.  Now the  power series representation of $J_n
(x)$ , we have that
$$
J_n (x) = \sum\limits_{m=0}^{\infty} {{(-1)^m ({\half} x
)^{n+2m}}\over{m!(n+2m)!}}
$$
>From this it follows that
$$
|J_n (x)| \leq {{{\half}|x|^n}\over{n!}} \tks exp({\quarter}
{{|x|^2}\over{(n+1)}} )
$$
Now  we have the Stirling's approximation for the factorial n, when
$n\rightarrow \infty$, as 
$$
n! \thickapprox {1\over{\sqrt{2\pi n}}}\tks  exp(-n) n^n
$$
Hence, for some constant C ,
$$
|J_n (x)| \leq C \tks \sqrt{2\pi n} \tks exp\{ n [ \log ({{|x|e}\over{2n}}) +
{{|x|^2}\over{4n(n+1)}}]\}
$$
Therefore if $ |n| > {{e^2 |x|}\over{2}} $ we can easily check that
$$
|J_n (x)| \leq C \tks \sqrt{2\pi n}\tks exp\{\thickspace  (-1 +
 {1\over{e^4}})n\tks \} \leq C
\tks exp(- \alpha(\beta)n)
$$
for $\beta = {\half} e^2$ and an appropriate $\alpha$.
\vskip 1cm
{\noindent \bf Proposition A.4} {\sl Consider $J_n (x)$ for x and n both
large.  Then we have the following behaviour as $x \rightarrow \infty $
, in the regions indicated below.
\TagsOnLeft
$$
 If \tks |x-n| \geq |x|^{\half} ,\tks then \tks
 |J_n (x)| \leq C |x|^{-\half}.
\tag 1
$$
$$
 If \tks |x-n| \leq |x|^{\half} ,\tks  then \tks
 |J_n (x)| \thickapprox O(|x|^{-\third}).
\tag 2
$$
\TagsOnRight
In both the cases the error involved would be $|x|^{-1}$.}
\vskip 1cm
The above proposition is well known classically and is proved using the
stationary phase techniques and the details can be found in Watson \cite{4}. 
\vskip 1cm

%%%%%%%%%%%%%%%%%%% End of Appendix %%%%%%%%%%%
%%%%%%%%%%%%% References %%%%%%%%%%%%%%%%%%%%%%
\Refs

\ref
\key{1}
\by{Krishna M}
\paper{Anderson model with decaying randomness}
\jour{Proc.Indian.Acad.Sci}
\issue{4}
\yr{1990}
\pages{220-240}
\endref

\ref
\key{2}
\by{Amrein W O, Jauch J M, Sinha K B }
\book{Scattering theory in quantum mechanics}
\publ{Benjamin:Reading}
\yr{1977}
\endref

\ref
\key{3}
\by{Reed M and Simon B.}
\book{Methods of modern mathematical physics III, Scattering theory}
\publ{Academic Press}
\publaddr{NewYork}
\yr{1978}
\endref

\ref
\key{4}
\by{Watson G N}
\book{A Treatiese on the theory of Bessel functions}
\publ{Oxford University Press}
\ed{Second}
\yr{1962}
\endref


\endRefs
%%%%%%%%%%%%% End of References %%%%%%%%%%%%%%%%%%%%%%

\enddocument