Dynamical Localization for Discrete and Continuous Random Schrodinger Operators F. Germinet UFR de Mathematiques et LPTMC Universite Paris VII - Denis Diderot 75251 Paris Cedex 05 France e-mail:
[email protected] S. De Bievre UFR de Mathematiques et URA GAT Universite des Sciences et Technologies de Lille 59655 Villeneuve d'Ascq Cedex France e-mail:
[email protected] 1997 Abstract
We show for a large class of random Schrodinger operators H! on `2 ( ) and on L2 ( ) that dynamical localization holds, i.e. that, with probability one, for a suitable energy interval I and for q a positive real, sup rq (t) sup < PI (H! ) t ; jX jq PI (H! ) t > < 1: t
t
Here is a function of suciently rapid decrease, t = e?iH! t and PI (H! ) is the spectral projector of H! corresponding to the interval I . The result is obtained through the control of the decay of the eigenfunctions of H! and covers, in the discrete case, the Anderson tight-binding model with Bernouilli potential (dimension = 1) or singular potential ( > 1), and in the continuous case Anderson as well as random Landau Hamiltonians.
1 Introduction
We show for a large class of random Schrodinger operators H! on `2 ( ) and on L2 ( ) that dynamical localization holds, i.e. that, with probability one, for a 1
suitable energy interval I and q > 0, sup rq (t) sup < PI (H! ) t ; jX jq PI (H! ) t > < 1: t t
Here is a function of suciently rapid decrease, t = e?iH! t , PI (H! ) is the spectral projector of H! corresponding to the interval I . The result covers all random Schrodinger operators for which exponential localization has been proved, including operators with Bernouilli potentials in dimension 1 and random Landau Hamiltonians, for example. The strategy of the proof is as follows. First recall that exponential localization, i.e. pure point spectrum and exponentially decaying eigenfunctions, is by now a well established property of random Schrodinger operators in many situations. On the other hand, it is also known that exponential localization does not systematically entail dynamical localization [10]. The authors of [10] point out that, to obtain dynamical localization, some control is needed on the location and the size of the boxes outside of which the eigenfunctions \e ectively" decrease exponentially. This is precisely what is achieved for random Schrodinger operators in the present paper (Theorem 3.1 and Theorem 4.2). Our proof here uses the ideas of Von Dreifus and Klein [13], and in particular those of the proof of their Theorem 2.3. We proceed as follows: once exponential localization has been proved, and using the fact that the spectrum is now known to be discrete, one can exploit the result of the multi-scale analysis a second time to get better (and sucient) control on the eigenfunction decay. We rst deal with the discrete case (Sections 2 and 3). In section 2 we start by proving (along the lines of [10]) a sucient condition (see (2.1)) on the eigenfunctions of a Hamiltonian H which implies dynamical localization. In section 3 we give the proof of the announced result for the discrete Anderson model. The continuous case is dealt with in sections 4 and 5. Exponential localization for Schrodinger operators has recently been carried over to the continuum by Combes and Hislop [6] and by Klopp [19]. The case of random Landau Hamiltonians is dealt with by Combes, Hislop and Barbaroux [3] [7], by Wang [23] and by Dorlas, Macris and Pule [12]. All those papers use an adaptation to the continuous case of the multi-scale analysis originally developed for discrete Schrodinger operators ([16] [17] [13] or see [5]). This reduces the proof of exponential localization to the veri cation of two hypotheses: a Wegner estimate and an estimate allowing the \initialization" of the multi-scale analysis. Our central result here (Theorem 4.2) shows that those two hypotheses actually imply dynamical localization. We give some applications in section 5. To put our results in perspective, we recall that Del Rio, Jitomirskaya, Last and Simon [10] used bounds of Aizenman [1] to give a simple proof (avoiding the multi-scale analysis) of dynamical localization for the discrete Anderson model with a potential with bounded density. But the bounds of [1] do not seem to carry over to the continuous case, nor to Bernouilli and other singular potentials in the discrete case. To deal with these cases, we were therefore obliged to return 2
to the (rather painful) multi-scale analysis. A further application of our results to the random dimer model [15] will be given elsewhere [9].
2 Eigenfunction decay and dynamical localization In this section we give, for a class of self-adjoint operators H with pure point point spectrum, de ned on either l2( ) or L2( ), a sucient condition on the eigenfunctions (see (2.1)) that guarantees dynamical localization. Our strategy for proving dynamical localization for random Schrodinger operators is then to prove that a property much stronger than this condition is indeed satis ed (Theorem 3.1, Theorem 4.2). Let H0 be the following operator on L2 ( ):
H0 = H 1 H 2 ; where P H1 = p21 + p22 , p1 = @x1 + Bx2 =2, p2 = @x2 ? Bx1 =2, B 0, and H2 = 3 p2i , pi = @xi . One can also write H0 = (P ? A)2 , where A is the vector potential B=2(x2 ; ?x1 ), written in the symmetric gauge, associated to ! the constant magnetic eld ? B = B~ex3 .
Theorem 2.1 Let H be a self-adjoint operator on H = l2( ) or L2( ) with pure point spectrum on some interval I . Let 'n be its eigenfunctions with corresponding eigenvalues En 2 I . In the case H = L2 ( ), suppose that I is compact and that H has the form H0 + V , H0 as described above, V 2 L1 ( ). Suppose moreover that
9 > 0; 0 2]0; =2[ and sites (xn ) s:t: 8 n; j'n (x)j < C e 0jxnj e? jx?xnj : (2.1) Let q > 0 and 2 H decaying exponentially at a rate > 2 0. Then there exists a constant C = C (I; ; 0; ; q) such that:
8 t 0; jj jX jq=2 e?iHt PI (H ) jj2 C :
(2.2)
This simple result relies on ideas of section 7 in [10]. Note that (2.1) says roughly that the eigenfunctions are localized inside boxes of size jxn j=2 around \centers" xn . This is stronger than exponential localization of H on I , which only means that
9 > 0 such that 8n; 9Cn > 0; j'n (x)j Cn e? jxj;
(2.3)
but weaker than what the authors of [10] called SULE (Semi-Uniformly Localized Eigenfunctions). 3
We choose to present the proof of this theorem in the continuous case, since the rst part of the proof (Lemma 2.2) is a little bit more technical in this situation. In order to prove Theorem 2.1 we need control on the growth of the jxn j in n, which is given by the following preliminary lemma: Lemma 2.2 Let H be as in the proposition and > 0. Then one can order the jxn j in increasing order, and there exists a constant C = (4 max(B; 1))?1 such that for n large enough (depending on ):
jxn j Cn1=( +) :
Proof of Lemma 2.2: Essentially we follow the ideas of the proof of Theorem 7.1 of [10]. We recall that the energies En that we consider belong to I . Let > 0 be given and let 0 > 0 so that > 0 ( ? 1). Let L > 0 be given, de ne 0
J = [0; L ], and write 2L (x) for the characteristic function of the ball of radius 2L and centered at 0. Suppose that jxn j < L. Then, for such n and for some constant C1 , one has: < 'n ; (1 ? 2L (X )) J (H0 ) 2L (X )'n > jj(1 ? 2L (X )) 'n jjL2 jj 2L (X )'n jjL 2 C e( 0+ )jxnj (1 ? 2L (X ))e? jxj L2 C1 e?( ? 0)L : (2.4) 0 ? Secondly, using 1 ? J (y) yL ; y 0, and H'n = En 'n : < 'n ; (1 ? J (H0 ))'n > (< 'n ; H'n > +jjV jj1 ) L?0 C (I; jjV jj1 )L?0 : (2.5) So, using (2.4) and (2.5), tr( 2L (X ) J (H0 ) 2L (X ))
X
nj En 2I; jxn jL
< 'n ; 2L(X ) J (H0 ) 2L (X )'n >
]fnj En 2 I; jxn j Lg 1 ? 3C1 e?( ? 0)L ? C (I; jjV jj1 )L?0
(2.6)
21 ]fnj En 2 I; jxn j Lg for L L0 = L0( ; 0 ; I; jjV jj1 ; 0 ): (2.7)
The next step is then to bound the trace class norm of the operator Q = 2L (X ) J (H0 ) 2L (X ). Let's study the case B 6= 0. Since fu21 + u22 Lg fu21 L; u22 Lg, and denoting by (2dL)(x) the characteristic function of the d-dimensional ball of radius 2L centered at 0, remark that tr(Q) tr ( 2L(X )( J (H1 ) J (H2 )) 2L (X )) (2) (X )) ( ( ?2) (X ) (H ) ( ?2) (X )) ( X ) ( H ) = tr ( (2) J 2 J 1 2L 2L 2L 2L tr (Q1 Q2 ) : (2.8) 4
The operator Q2 ?is the product of two Hilbert-Schmidt ?1 (?2)operators with respective ( ? 2) ? 1 kernel 2L (x) F g (x ? y) and J (x) F 2L (x ? y) [21], where F ?1 g denotes the inverse Fourier transform of g(x) = J s(x), with s(x1 ; :::; x ?2 ) = P ?2 2 1 xi . So, denoting by jjC jj1 the trace class norm of an operator C de ned on H, one has:
jjQ2 jj1 jj 2L (X ) J (H2 )jjHS jj J (H2 ) 2L (X )jjHS jj( (2L?2) (x)jj2L2 jj J s(x)jj2L2 (4L) ?2 L0 ( ?2) :
(2.9)
We turn now to the magnetic part Q1 . It is well known that the spectrum of our H1 consists of eigenvalues (2n + 1)B , n = 0; 1; 2; :::. The corresponding projectors are operators with kernel [20],
Pn (x; x0 ) = ei B2 x^x0 pn
p
B (x ? x0 ) ;
(2.10)
Z 1 pn ((x1 ; x2 )) = 2 eikx1 hn (k + x2 =2)hn(k ? x2 =2)dk; and hn (k) are the normalised Hermite functions. Note that we gave the expression of Pn in the in the Landau gauge as in [20]. P symmetric gauge, and notcorresponding Write now PJ = nj En 2J Pn , the projector (2) to the eigenvalues (2) belonging to J . Then Q1 = 2L (X )PJ PJ 2L (X ) , which are two HilbertSchmidt operators with the same norm (because of (2.10)). And one has, for each n,
where
2 jj (2) 2L (X )PZn jjHS
Z p 2 p B (x ? x0 ) dxdx0 n x Z x 0 2 (4BL)2 F hn (: + x22 )hn (: ? x22 ) (x1 ) L2 dx2 Zx2 2 2 = (4BL)2 hn (k + x22 ) hn (k ? x22 ) dx2 dk x ;k =
(2) 2L (x)
2
= (4BL)2 jjhn jj4L2
So, since jjhn jjL2 = 1, and using ]fn 0; (2n +1)B L0 g L0 =2B if L0 B , one has 2 2+0 : jj (2) 2L (X )PJ jjHS 8BL And then, using (2.8) and (2.9),
tr(Q) CL + ; 5
(2.11)
taking > 0 ( ? 1), and C = 4 max( P B; 1). Note here that if B = 0 then the free Hamiltonian H0 has the form 1 @i2 . Hence the analysis made previously for Q2 is valid for such a Q and (2.11) holds. In order to nish the argument, note that, together with (2.7) and (2.8), (2.11) tells us that, for L L0 , N (L) ]fnj; En 2 I; jxn j Lg is nite. Order then the eigenfunctions in such a way that jxn j increases. So, N (jxn j) = n, and if n N0 N (L0 ), one has, with L = jxn j:
jxn j Cn1=( +) , for n > N0 :
(2.12)
The main di erence between the continuous and discrete cases comes from this Lemma, in the sense that on L2 ( ) one has to control the behaviour of the eigenfunctions in the momentum variables: that was achieved through the use of J (H0 ). In the discrete case, (2.11) is replaced by the trivial equality tr( 2L ) = (4L +1) . This also explains why no speci c form for H is needed on l2 ( ). It is easy, then, to rewrite the proof of Lemma 2.2 in this case (see also [10]). 2 Proof of Theorem 2.1: Let 2 H such that, for some constant C ( ) > 0 and > 2 0, j (x)j < C ( )e?jxj. We have to bound jjX q=2 PI (H )e?iHt jj, for q > 0, t > 0. Let 0 < < inf( =2 ? 0 ; =2 ? 0 ). Then 2 P (H )e?iHt jj2 jjX q=X I j < 'n ; > j2 jjX q=2 'n jj2
nj En 2I
X
nj En 2I
Z
(C ( )C )2 e?4jxnj
Z
dx dy jyjq e?2jxje?2( ? 0?)jx?yje2( 0 +)(2jxnj?jx?xnj?jy?xnj)
1 Z Z 0 X e?4jxnj A jyjq e?2( ?2( 0?))jx?yjdy e4( 0+)jxj?2jxjdx: C3 @ nj En 2I
This last line is uniformly bounded by a constant C ( ; 0 ; I; ; ), according to Lemma 2.2 and because of the choice of . 2
3 The discrete Anderson model We consider the self-adjoint operator H!
H! = ? + V! ; where is the discrete Laplacian on `2 ( ) and V! (! 2 ) is a random potential, the (V! (x))x2 being i.i.d. random variables. Their common probability 6
measure is assumed to be non degenerate, i.e. not concentrated on a single point. The conditions that we impose on are:
R
if = 1 : 9 > 0 ; j v j d(v) < 1; if 2 : is -Holder continuous:
(3.1)
Let us recall how the disorder () of a -Holder continuous measure is de ned:
()?1 = > inf0 sup jb ? aj? ([a; b]): jb?aj 0. If = 1, de ne ? (I ) inf f (E ); E 2 I g, where (E ) is the Lyapunov exponent at energy E . Suppose ? = (I ) > 0. If > 1, pick ? > 0. Suppose the disorder () is taken suciently high. Then, almost surely, H! has pure point spectrum on I , and there exist centers xn;! associated to the eigenfunctions 'n;! with energy En;! 2 I such that: 8 0 2]0; ?[, there exists a constant C (!; ; 0 ) such that
8x 2 ; j'n;! (x)j C (!; ; 0 )e 0 jxn;! j e? 0jx?xn;! j :
(3.2)
Evidently, one can also write a \low energy" version of this theorem. As an immediate consequence of Theorem 2.1 and Theorem 3.1 we have: Corollary 3.2 Let H! be as in Theorem 3.1, and PI (H! ) the spectral projection on the compact interval I . Then for q > 0 and 2 `2( ) decaying exponentially with rate > 0, jj jX jq PI (H! )e?iH! t jj2 is bounded uniformly in t almost surely. Comparing (3.2) to (2.3) and to (2.1), one notices that now the size of the boxes in which the eigenfunctions \live" can grow at most as jxn;! j . One expects this can be improved to a polynomial bound (but no more - see [10]). Supposing has a bounded density with compact support, the polynomial bound follows from the proof of Theorem 7.6 in [10]. The proof of Theorem 3.1 is based on the ideas of [13], and in particular on the proof of Theorem 2.3 in [13]. The strategy is the following: since the hypotheses of Theorem 3.1 imply exponential localization, we know that there exist \centers" xn;! where the eigenfunction 'n;! is maximal, and one can then exploit the result of the multi-scale analysis a second time to improve the control of the decay of the eigenfunctions. As already pointed out, this proof has the advantage of yielding the result for singular potentials and in particular for Bernouilli potentials in dimension 1. In addition, the proof extends to continuous random Schrodinger operators, as shown in sections 4 and 5. 7
To make this paper self-contained, we start by recalling the elements from [13] that we need. First of all, L(x) denotes the cube of side L=2 centered in x and @ L(x) its boundary. HL (x);! is the restriction of the operator H! to the cube L(x) with Dirichlet boundary conditions, and GL (x) (E; :; :) is its resolvent. Given L0 > 1, 2]1; 2[, we de ne Lk (k 2 ) recursively via Lk+1 = L k . Given in addition an integer b 2, we de ne
Ak+1 (xo ) = 2bLk+1 (xo )n2Lk (xo ): Note that we do not indicate the dependence of Lk and Ak on L0 ; and b since these quantities will at any rate be xed later on. We further need the following de nition:
De nition 3.3 Let > 0 and an energy E 2 be given. A cube L(x) is said to be ( ; E )-regular if E 62 (HL (x) ) and if for all y 2 @ L(x); jGL (x) (E; x; y)j e? L=2: Otherwise L (x) will be called ( ; E )-singular.
Note that this de nition is !-dependent, but we follow the usual practice by not indicating this. For xo 2 ; Ek (xo ) is de ned to be the following set:
f!j 9 E 2 I; 9 x 2 Ak+1 (xo ); Lk (xo ) and Lk (x) are ( ; E ) ? singularg: Finally, we recall a well known identity. Let x 2 , E 62 (HL (x)), and ' 2 `2( ) so that H' = E' be given, then:
'(x) =
X
(y;y0 )2@ L (x)
GL (x) (E; x; y)'(y0 ):
(3.3)
Here (with some abuse of notation) (y; y0 ) 2 @ L means y and y0 are nearest neighbours with y 2 L (x) and y0 62 L(x). @ +Lk (x) will denote the points y0 just outside Lk (x). In order to prove Theorem 3.1, we start with the following three lemmas:
Lemma 3.4 Let p > and 2]1; 2 ? 2=(p + 2 )[ and b > 1 be given. Assume the hypotheses of Theorem 3.1 are satis ed. Then for any 2]0; ?[ there exists L0 = L0 (p; ; ; b; ()) > 1 such that:
8 k 2 ; 8 x 2 ; (Ek (x)) (2bL(kL+1)+p 1) : k
Lemma 3.5 Let > 0 be xed. There exists a constant L(; ) so that, if H is a Schrodinger operator and ' 2 `2 ( ) an eigenvector of H with eigenvalue E , and if x 2 satis es j'(x )j = supfj'(x)j; x 2 g, then L(x ) is ( ; E )singular, for all L L (; ). 8
Lemma 3.6 If ; s , and are some positive constants, then 8 2]0; 1[ there exists L(; ; ) such that, if L L(; ; ):
8 x; xo 2 ; s L ?1 e? L=2
xo j jL=x?2+1
e? jx?xoj :
(3.4)
The rst lemma follows immediately from the Appendix and from Theorem 2.2 of Von Dreifus and Klein [13] and constitutes the core of the proof of exponential localization in [13]. We will prove an analog of it for continuous random Schrodinger operators in the Appendix. The second lemma says roughly that if ' is an eigenvector of H with eigenvalue E , then E must be close to the spectrum of HL (x) provided L is big enough and L (x) is centered on a maximum of j'n;! j. It is quite simple, but central for what follows. Obviously the third lemma doesn't need a proof. We have stated it separately in order to make clear later on that L(; ; ) only depends on the model parameters and not on the particular eigenfunction we consider. Note that L(; ; ) behaves like (1= ) at a positive power. Proof of Lemma 3.5: Let ' be as in the lemma: ' 2 `2( ), so x exists. Suppose that L(x ) is ( ; E ) ? regular, and apply the identity (3.3) at the point x . Then for some y0 2 @ +L (x):
j'n;! (x )j s L ?1 e? L=2j'n;! (y0 )j s L ?1 e? L=2j'n;! (x )j;
(3.5)
where s is a constant depending only on the dimension. Now let L ( ; ) be a positive real such that s L ?1e? L=2 < 1 for L L . Then, for such L, (3.5) is impossible, and L(x ) cannot be ( ; E )-regular any more, that is: L(x ) is ( ; E )-singular for L L ( ; ). 2 Proof of Theorem 3.1: Under the hypotheses of the theorem, H! has ? a:s: exponential localization on I (see [4] and [13]). This means that there exists
0 , ( 0 ) = 1 so that for all ! 2 0 , c (H! ) \ I = ; and for all eigenvalue En;! 2 I , the corresponding eigenfunction 'n;! is `2 and satis es (2.3). The aim is therefore to control the constant Cn;! of (2.3) and more precisely to show that xn;! can be chosen so that this Cn;! grows slower than an exponential in jxn;! j . In order to prove Theorem 3.1, we wish to use equation (3.3) repeatedly, and on a scale Lk for suitably large k, to estimate the value of 'n;! (x) when x belongs to Ak+1 (xn;! ) for suitably chosen xn;! . To do this, one has to work \outward" from x 2 Ak+1 (xn;! ) to the boundary of Ak+1 (xn;! ), making sure that the boxes of size Lk to which one applies (3.3) are regular. After these preliminaries, let's start the proof properly speaking, which will consist of three steps. Firstly, let I; ? > 0 and 0 be as in the theorem. Pick > 0 and 2] 0 ; ?[. 9
First step: Let p > , 2]1; 2 ? 2=(p + 2 )[ and b > 1 be given. With the Lk , k 0, de ned in Lemma 3.4, consider [ Ek (xo ): Fk = jxo j(Lk+1)1=
Lemma 3.4 then implies that for some constant C (; ; b) (Fk ) C (; ; b)L?k p+ (1+1=): P (F ) < 1. Hence, since p can be chosen larger than 2 (1 + 1=), one has 1 k=0 k The Borel-Cantelli lemma then implies that:
1 0 [ @mlim Fk A = 0; !1 km
so that the set
1 = f! 2 j9 k~1 = k~1 (!; ; p; ) such that 8 k k~1 ; ! 62 Fk g has full measure. This ends the probabilistic part of the proof. Note that the choice of puts a lower bound on p. This in turn forces the disorder to be high via Lemma 3.4. Second step: Now pick an ! in ( 0 \ 1 ), which will be kept xed throughout the rest of the proof. Let En;! 2 I , 'n;! its eigenfunction, and let xn;! be a point where j'n;! (x)j is maximal. Note that such a point exists since ! 2 0 and therefore 'n;! 2 `2 ( ). Let k1 = k1 (!; ; p; ; xn;! ) = max(k~1 ; k0 (; xn;! )); (3.6) where for all y 2 , the integer k0 (; y) is de ned as follows: k0 (; y) = minfk 0 such that jyj < Lk+1 g: (3.7) Hence 8k k1 ; ! 62 Ek (xn;! ). Indeed, if k k1 , then ! 62 Fk and jxn;! j < Lk+1 . This implies that ! 62 Ek (xn;! ), and consequently that 8k k1 , 8E 2 I and 8y 2 Ak+1 (xn;! ) either Lk (xn;! ) or Lk (y) is ( ; E ) ? regular. We can then apply Lemma 3.5 to conclude that there exists an integer k2 = max(k~2 , k0 (; xn;! )) where k~2 depends on the same parameters as k~1 and not on n, so that 8k k2 ; 8En;! 2 I; 8y 2 Ak+1 (xn;! ); Lk (y) is ( ; En;! ) ? regular: Last step: We now nish the argument along the lines of the proof of Theorem 2.3 in [13] ; ! is still xed in ( 0 \ 1 ). For k k2 and for x 2 Ak+1 (xn;! ), Lk (x) being ( ; En;! ) ? regular, we can apply relation (3.3): j'n;! (x)j s Lk ?1 e? Lk =2 j'n;! (x0 )j; 10
where x0 2 @ +Lk (x) is chosen so that
j'n;! (x0 )j = supfj'n;! (y)j; y 2 @ +Lk (x)g:
As long as x0 is still in Ak+1 (xn;! ) we can use (3.3) again, so that we can repeat this step at least d(x; @Ak+1 (xn;! ))=(Lk =2 + 1) times. Hence for k k2 and x 2 Ak+1 (xn;! ),
j'n;! (x)j s Lk ?1 e? Lk =2
k+1 (xn;! )) d(x; @A L =2 + 1 k
:
Comparing this to (3.4), we see that in order to get a useful estimate we need a lower bound on d(x; @Ak+1 (xn;! )). For that purpose we cover with new annular regions A~k+1 (xn;! ) de ned as follows. Pick 2]0; 1[ so that > 0 and choose the integer b introduced at the beginning of the proof so that b > (1 + )=(1 ? ). Then set A~k+1 (xn;! ) = [2bLk+1=(1+)] (xn;! )n2Lk =(1?) (xn;! ) Ak+1 (xn;! ): Note that if x 2 A~k+1 (xn;! ) then d(x; @Ak+1 (xn;! )) jx ? xn;! j. Hence, repeating (3.3) jx ? xn;! j times, one has that for all k k2 and for all x 2 A~k+1 (xn;! ), xn j ?1 ? Lk=2 Ljkx?=2+1 j'n;! (x)j s Lk e ;
or, applying Lemma 3.6, and choosing 2]0; 1[ such that 0 = we conclude that there exists an integer k3 = max(k~3 ; k0 (; xn;! )) where k~3 depends again not on n, such that: 8 k k3 ; and 8 x 2 A~k+1 (xn;! ); j'n;! (x)j e? 0 jx?xnj: (3.8)
But now note that for all x 2 , and provided jx ? xn;! j > L0=(1 ? ) there exists a k so that x 2 A~k+1 (xn;! ). This means that there exists an integer k = max(k~4 ; k0 (; xn;! )), k~4 depending once again not on n, such that (3.8) holds for all x 2 satisfying jx ? xn;! j > Lk . Hence, using that j'n;! (x)j 1 for all x 2 :
8x 2 ; j'n;! (x)j C (!; ; 0 )e 0 Lk e? 0jx?xn;! j :
(3.9)
So far, we have only proved that the eigenfunctions decay exponentially, but we are now in a position to control the n-dependence of the constant e 0 Lk as follows. Note that the only n-dependence of k comes from k0 (; xn;! ). Suppose supfjxn;! j; En;! 2 I g < 1, then k can be chosen n-independently, so that we actually obtain a uniform localization (called ULE in [10]), and a fortiori 11
condition (2.1) of Theorem 2.1. But Lemma 2.2 contradicts this rst possibility. So, in fact, supfjxn;! j; En;! 2 I g = 1, and, for n suciently large, one has: k = k0 (; xn;! ) i.e., with (3.7),
Lk jxn;! j :
Inserting this in (3.9) yields the announced result.
2
4 The continuous case In this section, our goal is to obtain the analog of Theorem 3.1 and Corollary 3.2 for continuous random Schrodinger operators. The result is stated in Theorem 4.2 below. In section 5, we will present some models where the hypotheses of this theorem are satis ed. We consider random Schrodinger operators on L2 ( ) of the following type ( 1): X (4.1) H! = H0 + i (!)u(x ? i): i2
Here i) H0 = (ir? A)2 + Vper , where A is a vector potential of a constant magnetic ! eld ? B =rot(A), and Vper a periodic potential. ii) The variables i (!), i 2 are independent and identically distributed, with common distribution . iii) The function u(x) belongs to C02 ( ), with supp u [?R; R] . To state the hypotheses, we need to recall some notations and simple facts. We introduce jxj = maxfjxi j; i = 1; :::; g, and denote by L (x) the cube L(x) = fy 2 j jy ? xj < L=2g: Moreover > 0 being xed (independently of L), ~ L(x) is the subset ~ L(x) = fy 2 L(x) such that L=2 ? < jx ? yj < L=2g; ~L;x will denote its characteristic function.We denote furthermore by L;x a function in C 2 ( ) with support in L(x) and satisfying, 0 L;x 1, L;x 1 on the subcube L (x)n~ L (x) so that r L;x lives in ~ L (x). Note that we will often drop the ! or x-dependence of the objects introduced in order to alleviate the notations. We furthermore de ne local Hamiltonians HL (x);! as follows. When A = 0 and Vper = 0, HL (x);! is the restriction of the operator H! to the cube L(x) with Dirichlet boundary conditions. When A 6= 0 or Vper 6= 0,
HL (x);! = H0 +
X
i2L \
12
i (!)u(x ? i):
We denote by WL;x the rst order di erential operator WL;x [H0 ; L;x] and write RL (x) (E ) the resolvent of HL (x);! . We then always have the geometric resolvent equation: if l L and if E 62 (HL ) and E 62 (Hl ) then
l RL (E ) = Rl (E ) l + Rl (E )Wl RL (E ):
(4.2)
De nition 4.1 Let > 0 and an energy E 2 be given. A cube L(x) is said to be ( ; E )-regular if E 62 (HL (x) ) and if: jj 4;xRL (x) (E )WL;x jj e? L=2: Otherwise L (x) will be called ( ; E )-singular.
We now state the result. Given a compact interval I , and reals 0 > 0; p > ,
L0 ; L~ > 1, we introduce Hypothesis [H1]( 0; I; p; L0):
(8 E 2 I; L0 is ( 0 ; E ) ? regular) > 1 ? L1p : 0
Hypothesis [H2](I; L~ ) (Wegner): there exists CW so that for all I~ I1 fE j d(E; I ) < 1g and L > L~ , (Tr(EL (0) (I~))) < CW jI~j jLj: Note that, using Chebyshev's inequality and [H2](I; L~ ), one also has, for L > L~ , 0 < < 1 and E 2 I , (d(E; (HL (!) )) < ) < CW jL j:
(4.3)
This is the so-called \Wegner Estimate". We will need both [H2] and (4.3) for the proof of Proposition 4.3.
Theorem 4.2 Let > 0. Suppose that for some interval I and reals 0 > 0,
p > 2 (1+1=), the hypotheses [H1]( 0 ; I; p; L0) and [H2](I; L~ ) hold for L0 > L~ large enough. Then with probability one there exist points xn;! , associated to eigenfunctions 'n;! with energy En;! 2 I , so that: 8 2]0; 0 [ and for some constant C! = C (!; ; ; 0 ; I; L0), one has, for all x 2 , j'n;! (x)j C! e jxn;! j e? jx?xn;!j : (4.4) Moreover, if q > 0 and 2 L2 ( ) decays exponentially with mass > 0, then, with probability 1, there exists a constant C ;! such that,
jj jX jq PI (H (!))e?iH (!)t jj2 C ;! : 13
Analysing the proofs of the previous two sections, one sees that the only missing ingredient for the proof of Theorem 4.2 is an analog of Lemma 3.4, stated as Proposition 4.3 below. Indeed, the arguments of sections 2 and 3 are readily transcribed to the continuous case, provided one makes the following adaptation. First, one replaces equation (3.3) by the following equality: if ' 2 L2 (R ) satis es H' = E' for some E , then for L(x) so that E 62 (HL (x) ):
4;x' = 4;xRL (x) (E )WL;x ': Secondly, one de nes x ('), the analog of x in Lemma 3.5, as follows. If ' belongs to L2 ( ) and H' = E', then sup
y24
(Z
4 (y)
) Z
j'(x)j2 dx =
4 (x ('))
j'(x)j2 dx:
So, writing xn;! x ('n;! ), and rede ning the annular Ak+1 (xn;! ) as 2bLk+1 (x)n2Lk +2R (xn;! ), one obtains the bound written in (4.4), but for jj 4;x'n;! jj rather than for j'n;! (x)j. Then to get the pointwise estimate (4.4) apply Theorem 2.4 of [8], or decompose jje jxn;!j e? jx?xn;!j 'n;! jj2L2 on boxes of size 4 and centered on 4k, k 2 , and apply Theorem IX.26 of [21]. It therefore remains to state and prove the analog of Lemma 3.4. Following [13] let's denote by R( ; L; x; y) the set
R( ; L; x; y) f! 2 j 8 E 2 I; L(x) or L(y) is ( ; E ) ? regularg: As in the discrete case, for all 2]1; 2[ and L0 > 1 we de ne the sequence (Lk )k2 by Lk+1 = L k. Proposition 4.3 For any 2]0; 0[, p > and 2]1; 2?2=(p+2 )[ there exist L = L( ; I; ) such that if [H1]( 0 ; I; p; L0) and [H2](I,L~ ) hold for L0 > L , L0 > L~ , then for all k 0:
jx ? yj > Lk + 2R =) (R( ; Lk ; x; y)) > 1 ? 12p : Lk
The proof of Proposition 4.3 follows upon adapting the arguments of the proof of Theorem 2.2 in [13] to the continuum. Various authors [6, 19] have written up versions of the multi-scale analysis for continuous Schrodinger operators, but, to our knowledge, the version we need is not available in the literature. Nethertheless, nobody seems to doubt that any such argument can be carried over from the discrete to the continuous case. Since multi-scale arguments are in addition to this painful, we have chosen to put the proof of Proposition 4.3 in the appendix, while making an e ort to give a clear, complete and relatively simple argument. 14
5 Applications We brie y indicate two applications of Theorem 4.2 to an Anderson model on [6] and to random Schrodinger operators with a magnetic eld.
The Anderson tight-binding model
Here the free Hamiltonian H0 of equation (4.1) is ?. We suppose, following [6], that u(x) > 3=2 (x); where 3=2 is the characteristic function of the cube 3=2 (0). Putting together Proposition 4.5 and Theorem 5.1 of [6], one has immediately from Theorem 4.2: Theorem 5.1 Suppose that has a L1 density g() with support [0; max] and disorder 0 = jjgjj?11 > 0 then, for energy EA > 0 xed and disorder 0 high enough, or for disorder 0 > 0 xed and energy EA low enough, the conclusions of Theorem 4.2 hold on [0; EA ].
The Landau Hamiltonian
Here the Hamiltonian has the general form described in equation (4.1), i.e.
A 6= 0 and Vper = 0. Although the result is still valid in arbitrary dimension
under further assumptions (see [2]) we prefer to state the application in the wellknown two dimensional version [3] [7] [23]. In that case, the vector potential A is given by A = B2 (x2 ; ?x1 ); with B > 0. Recall that the spectrum of the free Landau Hamiltonian H0 consists of a sequence of eigenvalues En (B ) = (2n + 1)B; n 2 : p We suppose that u > 0, supp u B (0; 1= 2), and that there exist C0 and r0 > 0 such that ujB(0;r0) > C0 . We suppose that the common measure (of the i (!)) has a bounded density function g 2 C02 (), g being even and positive for almost every 2 supp g. Under those assumptions it is well-known that M0 = supfjV! (x)j; x; !g < +1. Let's de ne the following bands: I0 (B ) = [?M0 ; B ? 0 (B )]; In+1 (B ) = [En (B ) + n (B )); En+1 (B ) ? n(B )] with some n (B ) > 0. It follows from [7] and Theorem 4.2: Theorem 5.2 Let V be as described above. Then for B high enough, there exist some n (B ) = O(B ?1 ) such that the conclusions of Theorem 4.2 hold on each interval In (B ), n 2 . A similar result holds for the model treated by Wang [23], where u can be negative and its support is included in B (0; r), 0 < r < 1. 15
A Appendix We turn to the proof of Proposition 4.3. Let us point out that our de nition of a ( ; E ) ? regular box (De nition 4.1) di ers slightly from the one in [6]. This di erence will allow us to free ourselves from most of the diculties due to the use of an auxiliary lattice in [6] and [19]. We need the following two concepts:
De nition A.1 Let r > 0. Two boxes 1 and 2 will be called r non-overlapping i d(1 ; 2 ) > 2r. Note that, if 1 and 2 are R non ? overlapping, then, since supp u [?R; R] , two events depending respectively on the i with i in 1 and in 2 are necessarily independent.
De nition A.2 Let 2]0; 1[ be given. A box L will be called non resonant at energy E (we'll write E ? NR) if jjRL (E )jj 2eL This means, in other words, that d(E; (HL )) > (1=2)e?L .
Remark that the commutator WL does not appear in De nition A.2 as it did in De nition 4.1. In fact one can replace WL with the characteristic function ~L;x de ned above (see [6] for the Anderson case and [7] Lemma 5.1 and lines (5.27 - 5.29) if A 6= 0) as follows. There exists a constant C (; I ) with:
jj 4;x RL (x) (E )WL;x jj C (; I )jj 4;xRL (x) (E )~ L;x jj:
(A.1)
Remark then that this last bound tells us that L is E ? NR =) jj 4 RL (E )WL jj 2C (; I )eL
(A.2)
An essential ingredient of the proof of Proposition 4.3 is the following deterministic lemma.
Lemma A.3 Let L = l with 2]1; 2[ and x 2 ; l > 12R. Denote by s the number of faces of a cube in dimension . Assume that for some
?
> 27l + 4( ? 1)ln(l=)) + 4ln(2s ) =l; with and de ned previously, and for some energy E i) L (x) is E ? NR; ii) Each box of size 4j(l+R), j=1,2,3, centered in x + l and contained in L(x)
is E-NR; iii) Among all the ( ; E ) ? singular boxes of size l contained in L(x), there are no more than three that are two by two R non-overlapping.
16
Then L (x) is ( 0 ; E ) ? regular with
0 = 1 ? l 27?1
?
?1 ? l (1? ) ) ? ln 2s C l(l=) ;
2
(A.3)
with C = C (; I ) de ned in (A.1).
In [6], Combes and Hislop have proved a simpler version of this result. In fact, they have adapted to the continuous case a simpli ed version of [13] which is contained in chapter IX of [5] (see also [14]). But, as in the discrete case, this simpli ed version does not seem to suce to obtain the results of this paper, since we need to obtain regular boxes at any size with good probability, uniformly in a compact interval of energy, and no longer at some xed energy E . So we turn to [13] and adapt it to the continuous case. Proof of Lemma A.3: The aim is to bound jj 4;xRL(x)(E ) WL;xjj. Using rst inequality (A.1), we are reduced to control jj 4;x RL (x) (E )~ L;xjj. This is achieved in (A.12) below. We recall that > 1 has been chosen small, so, without loose of generality one can suppose l > 3. In order to achieve our goal, we will recursively construct inside L(x) a chain of n boxes l (vk ), k = 0; :::; n ? 1, being most of the time ( ; E ) ? regular, and starting at v0 x. At each step of this process, we will use the geometric resolvent equation as follows. Let l0 > 3 and consider any box l0 (z ) L(x) with d(l0 (z ); ~ L(x)) > 0. For E 62 (Hl0(z) ) [ (HL (x) ), the resolvent identity (4.2) gives:
4;z RL (x) (E )~ L;x = 4;z Rl0(z) (E )Wl0 ;z RL (x) (E )~ L;x: The support of Wl0 ;z can be covered by a family of boxes 4 (v) L(x), indexed by points v that satisfy jv ? z j = l0 =2 and so that the sum over all the corresponding characteristic functions 4;v is equal to 1 on SuppWl0 ;z (Note that s (1+ l0=3) ?1 < s (l0 =) ?1 such boxes suce). Clearly, there exists one of those v for which
jj 4;z RL (x) (E )~ L;xjj s (l0 =) ?1 jj 4;z Rl0(z) (E )Wl0 ;z jj jj 4;v RL (x) (E )~ L;xjj: (A.4) Suppose now in addition that l0 (z ) is ( ; E ) ? regular. Then we immediately have:
jj 4;z RL (x) (E )~ L;x jj s (l0 =) ?1 e? l0 =2 jj 4;v RL (x) (E )~ L;x jj:
(A.5)
Apply now this argument to l (x), and set v0 = x; v1 = v. Repeat the process as long as l (vk ) is ( ; E ) ? regular and stays away from the boundary of L(x). Clearly, there exists some k 0 so that for all 0 k < k , l (vk ) 17
is ( ; E ) ? regular and l (vk ) L (x); d(l (vk ); ~ L (x)) > 0, whereas one of these conditions fails for l (vk ). As a result, if k > 0, we have jj 4;x RL (x)(E )~ L;x jj (s (l0 =) ?1 )k e? k l=2 jj 4;vk RL (x)(E )~ L;x jj: (A.6) If k = 0, this equation holds trivially. The important point here is that we gained a factor e? kl=2 : if there were no ( ; E ) ? singular boxes l inside L(x), this would end the proof. Indeed, in that case, the process could only end when l (vk ) gets too close to the boundary of L (x), implying k (L=l), so that (A.6) immediately yields the result upon using hypothesis (i). Of course, there may be ( ; E ) ? singular boxes in L (x) and we now use hypothesis (iii) of the lemma to control the case in which the above process stops because l (vk ) is ( ; E ) ? singular and vk is at a distance greater than 12(l + R) from the boundary of L(x). Using hypothesis (iii) and drawing a few pictures one easily convinces oneself that one can pack all the singular boxes of size l in t 3 slightly bigger and disjoint boxes li L (x), centered in x + l , and so that each box l (z ), where z belongs to the edge of one of those li , is ( ; E ) ? regular. More P precisely, the li are taking on one of the values 4j (l + R), 1 j 3, they satisfy ti=1 li 12(l + R) l0 , and the two following facts are simultaneously true: 0 d(z; @ (x)) l=2 1 L (1) B @ z 2 L(x)n [r li CA =) (l(z) is ( ; E ) ? regular) : (A.7) i=1
S
(2) if z belongs to the edge of one of the li ; then z 62 ri=1 li : (A.8) So, if l (vk ) is ( ; E ) ? singular, there exists i 2 f1; ::; tg so that l (vk ) li . De ne a new family of points v on the boundary of li , such that the boxes 4 (v) cover supp Wli . Then use the equivalent of (A.4) with l0 (z ) = li and z = vk . This produces some vk +1 that belongs to the edge of li , and consequently (A.7)-(A.8) implies that l (vk +1 ) is ( ; E ) ? regular, provided d(vk +1 ; @ L(x)) > l=2. By checking how vk +1 is positioned with respect to vk one sees that the latter condition is satis ed because vk is at least at a distance 12(l + R) from @ L(x)). Use now Hypothesis (ii) and apply once again (A.4) to l (vk +1 ), to obtain that for some vk +2 on the edge of l (v), with jvk ? vk +2 j li :
jj 4;vk RL (x)(E )~ L;x
jj 2s2
l0l ?1 2
e? l=2+li jj 4;vk +2 RL (x) (E )~ L;x jj:
(A.9) Using the preliminary condition on , this leads to (A.10) jj 4;vk RL (x) (E )~ L;xjj jj 4;vk +2 RL (x) (E )~ L;x jj: This is the way in which we get past a singular box l (vk ) far from the edge of L (x). We have now completely described the recursive construction of the 18
vk and it is clear that the process grinds to a standstill only when, for some n, l (vn ) is too close to @ L(x). From (A.10) one sees that, when meeting ? l=2
a singular box, we do not gain a factor exp , so that we have to assure ourselves this does not happen to often before the process ends. We therefore need to count how many of the boxes l (vk ); 0 k n are regular. Since jvk+1 ? vk j = l=2 in that case and jvk+2 ? vk j li if not, it is not hard to see that the process cannot stop before n = n = n1 + 2t where
Pt l # i=1 i ; l=2
"
n = L=2 ? 1
or
[L=l] ? 27 n1 L=l: Hence, using (i) of the lemma, one has
(A.11)
jj 4;x RL (x) (E )~ L;xjj s (l=) ?1 e? l=2
n1
2eL :
(A.12)
Putting together relations (A.1) and (A.12) and using (A.11) as well as the de nition of 0 stated in the lemma leads to the desired result. 2 Proof of Proposition 4.3: Take 2]1; 2 ? 2=(p + 2 )[ and 2]0; 0[. Let L Lk+1 = L k, and use equation (A.3) to produce a sequence of exponents
k 2]0; 0 ]. It will be enough to show that a) 8k 0, k 0 ; b) 8k 0 : jx ? yj > Lk+1 + 2R =) (R( k ; Lk+1 ; x; y)) > 1 ? 1=L2kp+1. To prove (a), choose L0 > 0: the sequence ( k )k0 produced by repeatedly using (A.3) decreases, so for all k 0, k+1 k 0 . Then, using equation (A.3), it is clear there exists L = L( ; 0 ; ; ; ) so that, if L0 > L , the sequence ( k )k0 satis es 0
I;Lthe hypotheses of Lemma A:3 with =
2p k and L = Lk+1 are satisfied for either point x or y > 1 ? 1=Lk+1:(A.13) Firstly, since for all k 0, k , provided L0 is large enough, one has:
8k 0; k > L1 27L k + 4( ? 1)ln(Lk =)) + 4ln(2s ) k
19
Let's now de ne Il = \fE 2 ; d(E; I ) e?l =2g and 0 (Hl ) = (Hl ) \ Il . It is easy to estimate the probability that the distance between the respective spectrum of two R non-overlapping boxes l1 and l2 , (l1 ; l2 > L~ , is greater than , 0 < < 1. Using rst (4.3) and then Hypothesis [H2], one has, with some abuse of notations: (d(0 (Hl1 ); 0 (Hl2 )) < )
Z
X
l1 E 2I \0 (Hl2 )
?d(0 (H ); E ) < d! 2 l1
CW jl1 j (Tr(El2 (Il ))) CW2 (jI j + 1)jl1 j jl2 j (A.14) = CW;I jl1 j jl2 j: Hence, for all k, and writing for convenience L Lk+1 and l = Lk : if jx ? yj > L + 2R, it follows from (A.14) that (9 u 2 (x + l ) \ L(x); v 2 (y + l ) \ L(y) and l1 ; l2 = L or 4j (l + R)l; j = 1; ::; 3 with l1 (u) L(x) and l2 (v) L(y); (A.15) with d(0 (Hl1 ); 0 (Hl2 )) < ) CW;I (L=l)2 jL j2 : But consider this elementary exercise in logic: let Ai and Bj , i and j = 1; ::; J be 2J intervals, then (8 i; j = 1; :::; J; d(Ai ; Bj ) > ) () (8 E 2 8 i; j = 1; :::; J; (d(E; Ai ) > =2 or d(E; Bj ) > =2)) (8 i = 1; :::; J; d(E; Ai ) > =2) : () 8 E 2or; either (8 j = 1; :::; J; d(E; Bj ) > =2) This, combined with inequality (A.15) and = e?Lk , gives for all k 0, and if jx ? yj > Lk+1 + 2R that
8 E 2 I; (i) and (ii) of Lemma A:3 with =
2p+1 and L = Lk+1 are satisfied for either point x or y > 1 ? 1=Lk+1 : (A.16)
k
Let's nish the proof: for L0 large enough, Hypothesis [H1]( ; I; p; L0) gives (A.13) at rank 0. Suppose it is true at rank k: points (i) and (ii) of Lemma A.3, with = k and L = Lk+1 , are satis ed for either points x or y, with probability evaluated line (A.16). Now, (for any E 2 I; (iii) of Lemma A:3 holds) = 1 ? (9 E 2 I s:t: there are at least 4 R non ? overlapping ( k ; E ) ? singular boxes Lk contained in Lk+1 (x)) 1 ? (9 E 2 I s:t: there are at least 2 R non ? overlapping 20
( k ; E ) ? singular boxes Lk contained in Lk+1 (x))2 ! 2 2 ( L =L + 1) k +1 k ; (A.17) 1? 2p
Lk
where we obtained the last inequality using the recurrence hypothesis. Hence, since < 2 ? 2=(p + 2 ), combining (A.16), (A.17), there exists a constant L = L (p; ; 0 ; ) such that if L0 > L and jx ? yj > Lk+1 + 2R: (R( ; L ; x; y)) > 1 ? 1 : k
k+1
L2kp+1
Use now that k > , and Proposition 4.3 is proved.
2
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