On the Simon-Spencer Theorem
This paper presents a generalization of the classical result by B. Simon and T. Spencer on the absence of absolutely continuous spectrum for the continuous one-dimensional Schr odinger operator with an unbounded potential.
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nasplib_isofts_kiev_ua-123456789-1064962025-02-09T21:44:19Z On the Simon-Spencer Theorem Gordon, A. Holt, J. Laptev, A. Molchanov, S. This paper presents a generalization of the classical result by B. Simon and T. Spencer on the absence of absolutely continuous spectrum for the continuous one-dimensional Schr odinger operator with an unbounded potential. 2008 Article On the Simon-Spencer Theorem / A. Gordon, J. Holt, A. Laptev, S. Molchanov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 1. — С. 108-120. — Бібліогр.: 10 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106496 en Журнал математической физики, анализа, геометрии application/pdf Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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This paper presents a generalization of the classical result by B. Simon and T. Spencer on the absence of absolutely continuous spectrum for the continuous one-dimensional Schr odinger operator with an unbounded potential. |
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Article |
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Gordon, A. Holt, J. Laptev, A. Molchanov, S. |
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Gordon, A. Holt, J. Laptev, A. Molchanov, S. On the Simon-Spencer Theorem Журнал математической физики, анализа, геометрии |
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Gordon, A. Holt, J. Laptev, A. Molchanov, S. |
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Gordon, A. |
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On the Simon-Spencer Theorem |
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On the Simon-Spencer Theorem |
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On the Simon-Spencer Theorem |
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On the Simon-Spencer Theorem |
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On the Simon-Spencer Theorem |
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on the simon-spencer theorem |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2008 |
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https://nasplib.isofts.kiev.ua/handle/123456789/106496 |
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On the Simon-Spencer Theorem / A. Gordon, J. Holt, A. Laptev, S. Molchanov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 1. — С. 108-120. — Бібліогр.: 10 назв. — англ. |
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Журнал математической физики, анализа, геометрии |
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AT gordona onthesimonspencertheorem AT holtj onthesimonspencertheorem AT lapteva onthesimonspencertheorem AT molchanovs onthesimonspencertheorem |
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2025-12-01T02:08:17Z |
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2025-12-01T02:08:17Z |
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1850269920300892160 |
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Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 1, pp. 108�120
On the Simon�Spencer Theorem
A. Gordon
University of North Carolina at Charlotte, Charlotte, NC 28223, USA
E-mail:aygordon@uncc.edu
J. Holt
University of South Carolina Lancaster, Lancaster, SC 29721, USA
E-mail:jholt@gwm.sc.edu
A. Laptev
Imperial College London, London SW7 2AZ, UK
E-mail:a.laptev@imperial.ac.uk
S. Molchanov
University of North Carolina at Charlotte, Charlotte, NC 28223, USA
E-mail:smolchan@uncc.edu
Received October 20, 2007
This paper presents a generalization of the classical result by B. Simon
and T. Spencer on the absence of absolutely continuous spectrum for the
continuous one-dimensional Schr�odinger operator with an unbounded po-
tential.
Key words: Schr�odinger operator, localization, Simon�Spencer theorem.
Mathematics Subject Classi�cation 2000: 81Q10, 47E05, 34L40.
Dedicated with great respect to V. Marchenko and L. Pastur
1. Introduction
The fundamental paper by B. Simon and T. Spencer (see [9]) has played
an essential role in our understanding of localization phenomena. For the lattice
Schr�odinger operator, the main result of this paper is quite transparent and can
be formulated in the following form:
Theorem 1.1. Let h = � + V (x), x � 0 be the lattice Schr�odinger operator
on l2(Z+) with the boundary condition (0) = 0. If
lim sup
x!1
jV (x)j =1;
then
P
ac
(h) = ;.
c
A. Gordon, J. Holt, A. Laptev, and S. Molchanov, 2008
On the Simon�Spencer Theorem
R e m a r k. Due to general results (see [5]) concerning compact perturbations
of h,
P
ac
(h) = ; for any boundary condition of the form (�1) cos � � (0)
sin � = 0 with � 2 [0; �):
The result cannot be improved. There are many examples of operators h with
bounded potentials V (x) whose spectra are either purely absolutely continuous,
or contain a rich absolutely continuous component. For instance, for periodic
V the spectrum
P
(h) of h is purely absolutely continuous. This statement is
physically nontrivial for energies in the range of the potential V .
For the continuous Hamiltonian, the corresponding result is not so strong,
and the result depends on the existence of very high �peaks" in the potential
function V .
Theorem 1.2. Let H+ = � 00 + V be a 1-D Schr�odinger operator on
L2(R+) with the Dirichlet boundary condition (0) = 0 and V (x) � 0. If there
exist sequences fxngn�0, fhngn�0 and fÆngn�0 of positive numbers with
xn, hn ! 1 for which V (x) � hn on [xn; xn + Æn] and Æn
p
hn ! 1, thenP
ac
(H+) = ;.
Theorem 1.2 does not cover the physically signi�cant class of �Æ-like" poten-
tials. We can expect that for potential functions of the type
V (x) =
X
n�1
hnÆ(x� xn) or V (x) =
X
n�1
hn
In(x)
Æn
(here In represents the indicator function of the interval [xn; xn + Æn]) for which
xn � xn�1 ! 1, hn ! 1 and Æn ! 0, the corresponding Hamiltonian H+ will
have no absolutely continuous component. However, Th. 1.2 cannot be used at all
to prove this for the Æ-potential shown above, and requires a strong assumption
in the second case, namely
p
hnÆn !1: Our goal is to prove the following result
generalizing Th. 1.2 in several directions:
Theorem 1.3. Let H be a one-dimensional Schr�odinger operator on L2(R)
de�ned by
H = � d2
dx2
+ V (x): (1)
Assume that V (x) � 0 and that
lim sup
jxj!1
x+1Z
x
V (z) dz =1: (2)
Then
P
ac
(H) = ;.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1 109
A. Gordon, J. Holt, A. Laptev, and S. Molchanov
A similar statement is true for the half axis case.
Theorem 1.4. Let H� be a 1-D continuous Schr�odinger operator on L2(R+)
with the boundary condition (0) cos � � 0(0) sin � = 0, � 2 [0; �). Assume that
V (x) � 0 and that
lim sup
x!1
x+1Z
x
V (z) dz =1: (3)
Then
P
ac
(H�) = ;.
R e m a r k 1. Of course, (3) implies that lim sup
x!1
R
x+!
x
V (z) dz =1 for
any ! > 0:
R e m a r k 2. All �nal or �nearly �nal" results in spectral theory contain
local L1 norms of the potential. We remind the reader of the following results
(see [2]) of M. Birman and A. Molchanov. M. Birman proved that the spectrum
of H is bounded from below if and only if
lim sup
x!1
x+1Z
x
V�(s) ds <1; (4)
where V�(x) = max(0;�V (x)). Moreover, if
lim
x!1
x+1Z
x
V�(s) ds = 0; (5)
then the negative spectrum is purely discrete (possibly with an accumulation
point at 0). Additionally, if V � 0, then condition (5) is also necessary.
Another result was given by A. Molchanov: if V � 0, then the spectrum of H
is purely discrete if and only if for any ! > 0
lim
jxj!1
x+!Z
x
V (s) ds =1:
In the second part of the paper, we will present an example showing that the
condition
lim sup
x!1
x+1Z
x
V (z) dz =1;
together with self-adjointness ofH, cannot guarantee the absence of the absolutely
continuous spectrum. In fact, in this example the absolutely continuous spectrum
110 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1
On the Simon�Spencer Theorem
will coincide with [0;1): The key feature of this example will be the presence of
very deep wells which tend to destroy the repulsive e�ects caused by high positive
peaks.
Finally, we will consider the Hamiltonian H with the potential
V (x) =
X
n�1
hnÆ(x� xn) (6)
and prove the following theorem:
Theorem 1.5. Let Æ > 0 and H� be the operator on L2(R+) de�ned by
H� (x) = � 00(x) + V (x) (x) with the boundary condition (0) cos � � 0(0)
sin � = 0, where V (x) is de�ned by (6) with hn = n.
(a) If xn� xn�1 > (n!)2+Æ, then the spectrum of H� is purely singular contin-
uous for any boundary phase � 2 [0; �):
(b) If xn � xn�1 < (n!)2�Æ, then the spectrum of H� is pure point for a.e.
� 2 [0; �):
2. A Few Lemmas and the Proof of Theorem 1.3
Following the strategy of Simon and Spencer ([9]), we �rst want to study the
following problem: let
H � H+ I = � d2
dx2
+ V (x) + 1; (7)
where V (x) � 0 for all x 2 R. Suppose that
LZ
�L
V (s) ds = A >> 1: (8)
We want to estimate (for the energy parameter � = 0) the trace norm jjH�1 �
H�1
x0
jj1 of the di�erence of the resolvents of the operators H and Hx0
. Here, Hx0
is the operator given by the di�erential expression �d2=dx2 + V (x) + 1 with the
Dirichlet boundary condition (x0) = 0 at some point x0 2 [�L;L]:
Lemma 2.1. The kernel H�1
�
(x; y) of the resolvent operator (H ��)�1 at the
point � = 0 has the following representation:
H�1
0 (x; y) � R(x; y)
= Ex
8<
:
1Z
0
exp
0
@�
tZ
0
(1 + V (bs)) ds
1
A Æy(bt) dt
9=
; :
Here, bt is the Brownian motion with the generator L = d2=dx2.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1 111
A. Gordon, J. Holt, A. Laptev, and S. Molchanov
This is one of the well-known forms of the Feynman�Kac formula (see [6])
connecting the Schr�odinger operator (outside its spectrum) with the Brownian
motion. The expression Æy(bt) dt takes the form Æy(bt) dt = d�y(t), where �y(t)
is the local time of bs at the point y.
Lemma 2.2. If x; y < x0 or x; y > x0, then
Rx0(x; y) = Ex
8<
:
�0Z
0
exp
0
@�
tZ
0
(1 + V (bs)) ds
1
A Æy(bt) dt
9=
; ; (9)
where �0 is the time of the �rst arrival of the Brownian motion bt at the point x0,
that is, �0 = minft : bt = 0g.
R e m a r k. Of course, �0 <1 with probability one.
A similar result is true in a more general situation.
Lemma 2.20. Let Rx0;X be the resolvent (again for � = 0) of the operator H
de�ned by the expression (7) and Dirichlet boundary conditions at x0 and at each
point of a discrete set X � R (x0 =2 X). Then
Rx0;X(x; y) = Ex
8<
:
�0^�XZ
0
exp
0
@�
tZ
0
(1 + V (bs)) ds
1
A Æy(bt) dt
9=
; : (10)
Here, both x and y belong to one of the intervals �i, where f�i : i = 1; 2; : : :g is
the partition of R by the point x0 and the points of X. The random moment �X
is de�ned by
�X = minft : bt 2 Xg:
R e m a r k. Since �0 � �0^�X , from (9) and (10) it follows that Rx0;X(x; x) �
Rx0(x; x) on each interval�i. This monotonicity property will be used in the proof
of Th. 1.3.
Lemma 2.3. If x; y < x0 or x; y > x0, then
R(x; y)�Rx0(x; y) = Ex
8<
:exp
0
@�
�0Z
0
(1 + V (bs)) ds
1
A
9=
;
� Ex0
8<
:
1Z
0
exp
�
�
Z
t
0
(1 + V (bs)) ds
�
Æy(bs) ds
9=
;
= �(x)R(x0; y); (11)
112 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1
On the Simon�Spencer Theorem
where +(x) is used in (11) for x; y < x0 and �(x) for x; y > x0. Here, +(x)
is the solution of H = 0 on (�1; x0], such that +(x0) = 1 and +(x) ! 0
as x ! �1. Similarly, �(x) is the solution of H = 0 on [x0;1) satisfying
�(x0) = 1 and �(x) ! 0 as x ! 1. Such solutions exist and are unique.
Furthermore, these solutions are positive, monotone and convex over the intervals
(�1; x0] and [x0;1), respectively.
R e m a r k. It is easy to see that +(x) � e�jx�x0j for x � x0 and
�(x) � e�jx�x0j for x � x0:
P r o o f. We have
R(x; y)�Rx0(x; y) = Ex
8<
:
1Z
0
exp
0
@�
tZ
0
(1 + V (bs)) ds
1
A Æy(bt) dt
9=
;
� Ex
8<
:
�0Z
0
exp
0
@�
tZ
0
(1 + V (bs)) ds
1
A Æy(bt) dt
9=
;
= Ex
8<
:
1Z
�0
exp
0
@�
�0Z
0
(1 + V (bs)) ds
1
A
� exp
0
@�
tZ
�0
(1 + V (bs)) ds
1
A Æy(bt) dt
9=
; : (12)
Using the strong Markov property for the stopping time �0, we then have
R(x; y)�Rx0;X(x; y) = Ex
8<
:exp
0
@�
�0Z
0
(1 + V (bs)) ds
1
A
9=
;
� E b�0
8<
:
1Z
0
exp
0
@�
uZ
0
(1 + V (bs)) ds
1
A Æy(bu) du
9=
;
= R(x0; y)Ex
8<
:exp
0
@�
�0Z
0
(1 + V (bs)) ds
1
A
9=
; :
(13)
We have used the obvious relation b�0 = x0: The elliptic form of the Feynman�Kac
formula gives for
u(x) = Ex
8<
:exp
0
@�
�0Z
0
(1 + V (bs)) ds
1
A
9=
;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1 113
A. Gordon, J. Holt, A. Laptev, and S. Molchanov
the equation (for x < x0 or x > x0)
u00 � (1 + V )u = 0; u(x0) = 1;
i.e., u(x) is equal to +(x) for x � x0, or �(x) for x � x0.
As with Lem. 2.2, Lem. 2.3 can be generalized.
Lemma 2.30. The di�erence R(x; y)�Rx0;X(x; y) is given by the expression
R(x; y)�Rx0;X(x; y)
= R(x0; y)Ex
8<
:exp
0
@�
�0^�XZ
0
(1 + V (bs))Æy(bs) ds
1
A
9=
; ;
where x; y belong to the same interval �i:
R e m a r k. Lemmas 2.3 and 2.30 contain fundamental information about
R(x; y) �Rx0;X(x; y) and R(x; y)�Rx0(x; y). Both of these di�erences are non-
negative and increase if we replace the potential V by a smaller function, say
by the truncated potential V (x)I�(x), where I� is the indicator function of an
arbitrary interval �, or remove extra Dirichlet boundary conditions imposed at
points of the set X. In particular,
R(x; y)�Rx0(x; y) � R(x; y)�Rx0;X(x; y)
(compare with the remark following Lem. 2.20).
In the following paragraph and in Lem. 2.4, we denote by V�(x) the truncated
potential V�(x) = V (x)I�(x), where � is the interval [�L;L]. Let H� be the
operator
H� = �d2=dx2 + 1 + V�(x)
and R� = H�1
� :
Using the functions �(x) given by Lem. 2.3, in which V (x) is replaced by
V�(x); we can construct the resolvent kernel R�(x0; x), i.e., the L
2 solution of
the problem H�R� = �Æx0 , namely
R�(x0; x) =
�
c +(x) if x < x0
c �(x) if x > x0
; (14)
where the constant c is such that
c( 0+(x0)� 0�(x0)) = 1: (15)
It also follows that
c =
~ 0+(x0)
~ +(x0)
�
~ 0�(x0)
~ �(x0)
;
114 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1
On the Simon�Spencer Theorem
where ~ �(x) are arbitrary solutions of H� = 0, exponentially decaying at �1,
respectively. For example, we can de�ne ~ +(x) = ex for x < �L and ~ �(x) = e�x
for x > L: From (14) it then follows (see [9]) that
jjH�1
� �H�1
�;x0
jj1 = Tr
�
H�1
� �H�1
�;x0
�
=
1Z
�1
(R�(x; x)�R�;x0(x; x)) dx
= c
0
@ x0Z
�1
+(x)
2 dx+
1Z
x0
�(x)
2 dx
1
A :
Let us note that 0 � +(x) � 1 and that +(x) � ex for x < �L, so that
x0Z
�1
+(x)
2 dx �
�LZ
�1
e2x dx+
x0Z
�L
1 dx � 1
2
+ 2L: (16)
Similarly,
1Z
x0
�(x)
2 dx � 1
2
+ 2L: (17)
Now, we are ready to prove the central technical result.
Lemma 2.4. For an appropriate x0 2 [�L;L]
jjH�1
� �H�1
�;x0
jj1 = Tr(R� �R�;x0) �
c(L)p
A
; (18)
where A =
R
L
�L V (s)ds � 1 and c(L) is some constant depending only on L:
P r o o f. Let us introduce the phase function
z(x) = ~ 0(x)= ~ (x);
where ~ (x) is the solution of H = 0 satisfying the boundary conditions (�L) =
0(�L) = 1: Then z(x) = 1 for x 2 (�1;�L], since ~ (x) = ex on this interval.
The function z(x) satis�es the usual Riccati equation
z0(x) = (1 + V (x)I�(x))� z(x)2; z(�L) = 1; (19)
where I� is the indicator of the interval � = [�L;L]. After integration, (19)
becomes
z(x) = 1 +
xZ
�L
(1 + V (s)I�(s)) ds�
xZ
�L
z(s)2 ds: (20)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1 115
A. Gordon, J. Holt, A. Laptev, and S. Molchanov
Put M = maxx2� z(x). Since z(x) is continuous, there is a minimal point x0 2
[�L;L] for whichM = z(x0): Then, (20) impliesM � z(L) � 1+2L+A�2LM2,
which in turn gives M + 2LM2 � A. It follows that
M � b(L)
p
A
if A � 1, where b(L) > 0 depends only on L. Now, since 0�(x) < 0 and
�(x) > 0, by (15) it follows that 1 = c( 0+(x0) � 0�(x0)) � c 0+(x0), and
therefore putting c(L) = 1=b(L),
c � 1
0+(x0)
=
~ +(x0)
~ 0+(x0)
=
1
z(x0)
=
1
M
� c(L)p
A
:
Now we are ready to prove Th. 1.3.
P r o o f o f T h e o r e m 1.3. For �xed L > 0, let �n = [yn � L; yn + L],
n 2 Z, be a sequence of disjoint intervals for which yn ! �1 as n! �1, and
X
n2Z
1p
An
<1; (21)
where An =
R
�n
V (s) ds: Using Lem. 2.4, one can �nd a point x0;n 2 �n for
which
jjH�1
�n
�H�1
�n;x0;n
jj1 �
c(L)p
An
: (22)
Here, H�n = �d2=dx2 + 1 + V (x)I�n(x) and H�n;x0;n
is the same operator,
but with Dirichlet boundary condition added at x0;n 2 �n: Now, let us return
to the operator H de�ned by (1), and consider the resolvents (H + 1)�1 and
(HX + 1)�1, where HX is the operator H with Dirichlet boundary conditions
at the countable system of points X = fx0;ng. Using the fact that both H and
Hx0;X
are nonnegative, it follows from the monotonicity argument (see the remark
following Lem. 2.30) that
jj(H+ 1)�1 � (HX + 1)�1jj1 �
X
n
c(L)p
An
<1:
By the Kato�Birman theorem (applicable since � = �1 is outside the spectrum
of both operators, H and HX , see ([8])), it follows that
P
ac
(H) =
P
ac
(Hx0;X
).
But the operator HX is the orthogonal sum of the operators Hn, where Hn =
�d2=dx2 + V (x) on the interval [x0;n; x0;n+1] with Dirichlet boundary conditions
at the endpoints. Since each Hn has purely discrete spectrum, the spectrum of
HX is pure point. Therefore
P
ac
(H) = ;:
116 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1
On the Simon�Spencer Theorem
3. A Few Examples
The �rst example will show that the presence of a strong positive part of the
potential cannot guarantee the absence of the absolutely continuous spectrum of
H, even when H is essentially selfadjoint.
Example 3.1. Let xn; hn !1, Æn ! 0, xn+1�xn !1 and H = �d2=dx2+
V (x); where
V (x) =
1X
n=1
hn(I[xn�Æn;xn] � I[xn;xn+Æn]): (23)
Note that, from the conditions of Ex. 3.1,
xnZ
xn�1
V (x) ds = Ænhn !1
and that H is essentially selfadjoint. This follows by results due to P. Hartman
and M. Eastham (see [4, 1]) giving the essential self-adjointness of H, without any
assumption on V other than that V (x) � 0 on some in�nite disjoint sequence of
intervals of �xed length.
Theorem 3.2. If in Ex. 3.1
P
n
h2
n
Æ3
n
<1, then
P
ac
(H) = [0;1).
It will be helpful to consider the monodromy matrix M� in the generalized
Pr�ufer representation, that is, M�(a; b) is the matrix satisfying"
(b)
0(b)p
�
#
=M�(a; b)
"
(a)
0(a)p
�
#
:
Lemma 3.3. Let VÆ;h = h(I[�Æ;0] � I[0;Æ]) and M�(�Æ; Æ) be the monodromy
matrix in the generalized Pr�ufer representation for the problem H = �d2=dx2 +
VÆ;h(x) on [�Æ; Æ]: Let � be a �xed interval on the positive energy axis and suppose
� 2 �. Then, with the assumption Æ << 1 and h >> 1,
jjM�(�Æ; Æ) � Ijj � ch2Æ3:
P r o o f. Assume that h >> � > 0 and let �h;Æ(�) =
p
h� �Æ. Let us write
an explicit formula for M�(�Æ; 0) and M�(0; Æ): Simple calculations show that
M�(�Æ; 0) =
0
@ cosh�h;Æ(�)
p
�Æ
�h;Æ(�)
sinh�h;Æ(�)
�h;Æ(�)p
�Æ
sinh�h;Æ(�) cosh�h;Æ(�)
1
A
=
1 +O(hÆ2) O(Æ)
hÆp
�
+O(h2Æ3) 1 +O(hÆ2)
!
(24)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1 117
A. Gordon, J. Holt, A. Laptev, and S. Molchanov
while
M�(0; Æ) =
0
@ cos�h;Æ(�)
p
�Æ
�h;Æ(�)
sin�h;Æ(�)
��h;Æ(�)p
�Æ
sin�h;Æ(�) cos�h;Æ(�)
1
A
=
1 +O(hÆ2) O(Æ)
� hÆp
�
+O(h2Æ3) 1 +O(hÆ2)
!
: (25)
With the fact that M�(�Æ; Æ) =M�(0; Æ)M�(�Æ; 0), it follows from (24) and (25)
that
jjM�(�Æ; Æ) � Ijj = O(h2Æ3); (26)
which implies jjM�;n � Ijj � Ch2
n
Æ3
n
; where M�;n = M�(xn � Æn; xn + Æn). Now
M�(0; xn+ Æn) = OnM�;n � � �O2M�;2; O1M�;1; where Oi are appropriate orthogo-
nal matrices, and from (26) it follows that
jjM�(0; xn + Æn)jj �
nY
k=1
(1 + Ch2
k
Æ3
k
) � exp
nX
k=1
Ch2
k
Æ3
k
!
<1:
It is known that the existence of a sequence xn, for which the monodromy
matrix is uniformly bounded from above for all energies in a �xed interval �,
implies the absolute continuity of the spectrum in this interval (see [7]). We have
proved that
P
ac
(H) � [0;1). In fact, it is easy to prove that
P
ac
(H) = [0;1).
The second example is related to the one above. We will use here and in Ex. 3.4
the following observation: let H = �d2=dx2 + hÆ0(x). Then, in the generalized
Pr�ufer representation
M�(0�; 0+) =
1 0
hp
�
1
!
: (27)
Example 3.4. Let V (x) be the potential de�ned by
V (x) =
X
n
hn(Æ(x � xn)� Æ(x� xn � Æn));
where hn; xn !1 and Æn ! 0. Let H� be de�ned on L2(R+) by H� = �d2=dx2+
V (x) with the boundary condition (0) cos � � 0(0) sin � = 0 with � 2 [0; �).
From (27), an explicit formula for M�(xn � 0; xn + Æn + 0) can be obtained,
namely
M�(xn � 0; xn + Æn + 0)
=
1 0
hnp
�
1
!
O�(xn � Æn � 0; xn + 0)
1 0
� hnp
�
1
!
; (28)
118 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1
On the Simon�Spencer Theorem
where
O�(xn � Æn � 0; xn + 0) =
cos
p
hn � �Æn
1p
hn��
sin
p
hn � �Æn
�phn � � sin
p
hn � �Æn cos
p
hn � �Æn
!
: (29)
From (28) and (29), one can deduce that
jjM�jj = 1 +O(hnÆ
2
n
);
and hence if X
n
Æ2
n
hn <1;
then
P
ac
(H�) = [0;1) for any � 2 [0; �):
Example 3.5. Let V (x) =
P
hnÆ(x�xn); where hn; xn > 0 and hn; xn !1.
Let H� be de�ned on L2(R+) by H� = �d2=dx2+V (x) with the boundary condition
(0) cos � � 0(0) sin � = 0 with � 2 [0; �).
Since hn !1, it follows immediately from Th. 1.4 that
P
ac
(H) = ;. We can
estimate the norm by
jjM�(0; xn + 0)jj =
nY
k=1
1 0
hkp
�
1
!
� c(�)n
nY
i=1
hi
and in general jjM�(0; x + 0)jj =
Qn
k=1
1 0
hkp
�
1
!
� c(�)n(x)
Qn(x)
i=1 hi with
n(x) = #fxijxi � xg: Now,
1Z
0
dx
jjM�(0; x)jj2
�
X
i
xi � xi�1
h21h
2
2 � � � h2n(xi)c(�)2n(xi)
:
For fast increasing distances xi � xi�1 and �xed hi the last series diverges, from
which it follows (see [10])
P
pp
(H) = ;. In this particular case the spectrum is
purely singular continuous. It is probably the simplest example of an operator
with purely singular continuous spectrum (compare [3] and [10]). In fact, if hn = n
and xn = (n!)2+Æ for Æ > 0, then
P
(H�) =
P
sc
(H): One can prove also that for
xn = (n!)2�Æ with Æ > 0, the spectrum of H� is pure point for a.e � 2 [0; �): This
proves Th. 1.5 formulated in the introduction.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1 119
A. Gordon, J. Holt, A. Laptev, and S. Molchanov
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120 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 1
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