\magnification = \magstep1
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\font\big=cmbx10 scaled \magstep2
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{\big
\centerline{DELOCALIZATION FOR }
\smallskip
\centerline{SLOWLY DECAYING POTENTIALS}
\smallskip
\centerline{RANDOMLY DEPENDING ON TIME}
}  %ende big
\vskip 2.5 true cm
\centerline{\bf by}
\medskip
\centerline{\medium Sergei E. Cheremshantsev*}
 \vskip 2 true cm
 * On leave from: Leningrad Department of Steklov mathematical
Institute, St.-Petersburg, Russia.
 \smallskip

 \vskip 2 true cm
 \centerline{\bf Abstract}
 \medskip
 \noindent
We consider the one-dimensional Schrodinger equation with
slowly decaying potential which depends on time in a random
manner. \par
We prove that solutions of this equation are a.s. delocalized
as t tends to infinity and give the bound from below for the
rate of propagation.
\vskip 3 true cm
\noindent
SFB 237 -- Preprint Nr. 130  \hfil\break
Institut f\"ur Mathematik   \hfil\break
Ruhr-Universit\"at-Bochum
\smallskip\noindent
October 1991                                     % Monat und Jahr eintragen
$$
\buildrel {\hrulefill {                         }} \over
{                            }   $$
\eject
\magnification=\magstep1
\centerline{\bf DELOCALIZATION FOR SLOWLY }
\centerline {\bf DECAYING POTENTIALS }
\centerline {\bf RANDOMLY DEPENDING ON TIME }
\bigskip
\centerline
{\bf by }
\centerline {\bf Sergei E. Cheremshantsev*}
\smallskip
\centerline {\bf Fakultat fur Mathematik }
\centerline {\bf Ruhr-Universitat }
\centerline {\bf D4630 Bochum 1}
\centerline {\bf Germany }
\bigskip
\centerline {*On leave from: Leningrad Department
of Steklov Mathematical
Institute,}
\centerline {St.-Petersburg, Russia.}
\bigskip
\centerline
{\bf 1.\hskip 4pt Introduction}
\bigskip
\par
Theory of quantum systems with time-dependent Hamiltonians
has been actively developed in the last ten years.
Time-periodic interactions, in particular, were treated in many
papers.
A new direction in this field is studying of quantum systems with
Hamiltonians depending on time in a random manner. There are here
only few rigorous mathematical results.
\par
When considering quantum systems with time-dependent Hamiltonians
$H(t)$ (random or not) one can classify them according to the
character of spectrum of $H(t)$ for a fixed $t$. Three situations
are possible:
\par
1. $H(t)$ have absolutely continuous and, possibly, discrete spectrum.
\par
2. $H(t)$ have only discrete spectrum.
\par
3. $H(t)$ have dense pure point spectrum.
\par
If $H$ does not depend on time, the spectral properties of $H$
determine the behaviour of solutions
of nonstationary equation
as ${t\to\infty}$.
For time-dependent
Hamiltonians this is in general not true. The behaviour
of solutions of the Schrodinger equation
\par
$$i{{{\partial }  \Psi}\over{{\partial }t}}=H(t)\Psi$$
is not directly related with spectral properties of $H(t)$ for a fixed
$t$. However, this classification turns out to be useful.
\par
In the case where $H(t)$ has the form
\par
$$H(t)=-\Delta_{x}+V(t,x),$$
the case 1 corresponds to $V(t,x)$, sufficiently fast decaying as
$\vert x\vert\to\infty$; $V(t,x)\to +\infty$ in the case 2, and
  $V(t,x)\sim $ const or slowly decays as $\vert x\vert\to\infty$
 in the case 3
(Anderson potentials).
\par
The case 1 for $V(t,x)$ randomly
depending on time by some Markov process was
studied by Pillet [1,2] and Cheremshantsev [3,4] for different kinds
of Markov processes and conditions on $V$. The main results
are asymptotic completeness and absence of bound states for a.e.
path of the Markov process.
\par
The \ problem \ of \ second \ kind \ for \ random $H(t)$
was investigated by Guarneri [5],
Combescure [6] and some other authors [7,8]. The typical
result is a growth of energy as $t\to\infty$.
\par
As to the case 3, it seems that so far there were no rigorous
mathematical results. What we suggest below is an attempt
to observe for a
relatively simple model possible nontrivial effects caused by
random dependence on time.
\par
Let us consider one-dimensional
Schrodinger equation with potential whose sign
randomly depends on time:
$$i{\partial \Psi\over \partial t}=
-{d^2 \over dx^2} \Psi +q(x)\omega (t)\Psi,\hskip 5pt
\Psi\vert_{t=0}=\Psi_{0}(x),\hskip 5pt
 t\ge0 \eqno(1) $$
\par
Here $\Psi_{0}\in L_{2}({\bf R})$, $q$ is some bounded real function,
$\omega :[0,+ \infty ) \to \{1,-1\}$.
 Suppose that $\omega (t)$ is some path of Markov process with
two states 1 and -1
and transition probabilities
$$p(\tau ,i,j)=1/2(1-\exp (-2p \tau )),\hskip 5pt
 i \not= j, \eqno(2)$$
$$p(\tau ,i,i)=1/2(1+\exp (-2p \tau )), \eqno(3)$$
where $i,j= \pm 1$, $p>0, \tau >0$.
Let $\omega (0)=1$ (one can make this
assumption without loss of generality)
and $ \mu $ be the corresponding probability measure on
the space of paths $X$.
 It is known [9] that
$\mu $-a.e. $\omega $
has only a finite number of jumps on $[0,t]$, so in fact
$$\Psi (t,\omega)=U_{(-1)^{n}}(t-t_n)U_{(-1)^{n-1}}(t_n-t_{n-1})
...U_1(t_1-t_0)\Psi_0,\eqno(4) $$
where $0=t_0<t_1(\omega )<t_2(\omega )<...<t_n(\omega )\le t$,
$\omega (t)=(-1)^{n+1}$ for $t\in [t_{n-1},t_n)$,
$U_1(t)$ and $U_{-1}(t)$ are propagators for operators
$-{d^2 \over dx^2}+q(x)$ and $-{d^2 \over dx^2}-q(x)$ respectively.
 It is important
that these two operators have no common eigenfunctions (one can
easily show it). Therefore there are no solutions looking
like $\Psi (t,x)=\exp (ih(t))\Psi _0 (x)$.
The main question is what is the behaviour of $\Psi (t,x)$ given
by (4) as
$t\to + \infty$.
\par
{\bf Definition.(see [10]). If
$$\lim_{R\to+\infty}\sup_{t\ge 0}\Vert F(\vert x \vert \ge R)
\Psi (t,x)\Vert_2=0,\eqno(5)$$
then $\Psi _0$ is called bound state. \par
If $\forall R>0$
$$\lim_{T\to +\infty}{1\over T}\int_0^Tdt\Vert F(\vert x \vert
\le R)\Psi (t,x)\Vert_2^2=0\eqno(6)$$
then $\Psi _0$ is called scattering or propagating state.}
\par
Let ${\it H}_b=\lbrace
\Psi_0: \Psi$  satisfies  (5)$\rbrace $. For
time-independent
potentials $V$
with reasonable properties it is known [11] that (5)
holds for any $\Psi_0\in {\it H}_{pp}(-{d^2 \over dx^2}+V(x))$,
 and
(6) holds for
$\Psi_0\in {\it H}_{c}(-{d^2 \over dx^2}+V(x))$
(RAGE-theorem). For time-dependent potentials, however, we cannot
be sure that (6) holds for all $\Psi_0\in {\it H}_b^{\perp}$.
\par
What kind of behaviour can we expect for the solutions of (1)
given by (4)? If $q$ decays sufficiently fast, then two operators
 $- {d^2 \over dx^2 } \pm q(x)$ have only absolutely continuous
spectrum on $(0,+ \infty )$ and complete wave operators. One can
prove by the same method as in [2] or [3] that solutions (4)
behave like free ones as $t\to + \infty $ for $\mu $ -a.e $\omega $
for any $\Psi_0 \in L_2({\bf R})$. In particular, (6) holds for any
$\Psi_0 $ and there are no bound states. Here the behaviour of
solutions (4) as $t\to + \infty $ is essentially the same (except
the absence of bound states) as the behaviour of
$U_{\pm 1} (t)\Psi_0$.
\par
More interesting is the situation where $q$ decays slowly.
It is known that in this case operators $- {d^2 \over dx^2}
\pm q(x)$ may have only pure point spectrum on $(0,+ \infty )$.
An important example is given in the paper of Kotani and
Ushiroya [12] where some class of random slowly decaying
potentials $q$ is considered. If $q(x)$ decay like $\vert x \vert
^{- \nu }, \nu \in (0,1/2)$ as $\vert x \vert \to \infty$, then a.s.
both operators $- {d^2 \over dx^2} \pm q(x)$ have only pure
point spectrum dense
on $(0, + \infty )$. That means that all the states
$U_1 (t) \Psi_0$ and
$U_{-1} (t) \Psi_0$ are localized (5) as $t\to +\infty $. One
can ask whether the solutions (4) are also localized, where
$U_1(t)$ and $U_{-1}(t)$
interchange with random $t$. The answer is
negative, solutions (4) being
 delocalized for $\mu $-a.e. $\omega $
for any $\Psi_0$. Let us formulate now the main results of the
paper.
\par
{\bf Theorem 3. Assume that for some $r \ge 1 $
$q \in L_r ({\bf R}) \cap
L_{\infty} ({\bf R})$. Set  $\beta_0= {1 \over 2r }$
for $r \ge 2$ and
$\beta_0=1/4$ for $1 \le r \le 2$. If $0<\beta <\beta_0 $
then for
$\mu $-a.e. $\omega$ for any $\Psi_0\in L_2({\bf R})$
$$\lim_{T\to +\infty}
{1\over T}\int_0^Tdt \Vert F(\vert x \vert
\le t^{\beta })
\Psi (t,x,\omega ) \Vert _2^2=0 $$
In particular, (6) holds and there are no bound states.}
\par
{\bf Theorem 4. If $\nu <2\beta_0$ then
for $\mu $-a.e. $\omega $ for any $\Psi_0 \in L_2({\bf R}),
\hskip 5pt \Psi_0 \ne 0 $
$$\liminf_{T\to +\infty}
{1\over T}\int_1^T dt {<x^2(t)> \over
 t^{ \nu }}>0 $$
where
$$<x^2(t)> \buildrel \rm def \over =
\int dx x^2 \vert \Psi (t,x,\omega ) \vert ^2 $$ }
\par
The proof of these results consist of three steps. First
of all we get
formula for the random dynamics averaged over probability measure.
Then we introduce some nonnegative functional $\Phi $ on the space
of paths $X$ and prove that its expectation is finite. From this
fact we get finally the statements of Theorems.
\par
In further consideration $\int dx $ denotes integrals over
$(- \infty , + \infty )$, $\hskip 5pt \Vert f \Vert _p $
is the norm in
$L_p({\bf R}), \hskip 5pt (f,g)_H $ -
the inner product in $H$ (on the
contrary, $(f,g) $ is an element of $H\oplus H$). We denote by
$\hat f $ the Fourier transform of the function $f$.
\bigskip
\centerline
{\bf 2. Averaging of solutions of evolution equation }
\bigskip
\par
Let us consider an evolution equation in Hilbert space $H$ :
$$i {d\phi \over dt}=A(\omega (t))\phi, \hskip 5pt
\phi \vert_{t=0}=\phi_0,
\eqno (7),$$
where $\omega$ is a path of Markov process introduced above, $A(1)$
and $A(-1)$ are selfadjoint operators in $H$. The solution of the
equation (7) is given by
$$\phi (t,\omega )= \prod_{j=0}^k U_{(-1)^j} (t_{j+1}-t_j))\phi_0
\equiv U(t, \omega )\phi_0,
\eqno(8),$$
\noindent
where $0=t_0<t_1<...<t_{k-1} \le t_k=t; \hskip 5pt
\omega (t)=(-1)^j$ for
$t\in [t_j,t_{j+1})$, $ U_1(t) \equiv \exp (-iA(1)t),$
$\hskip 5pt U_{-1}(t)
\equiv \exp (-iA(-1)t)$. We may consider (8) as a definition
of solutions of the equation (7).
\par
Set
$$U_N (t,\omega ) \buildrel \rm def \over =
 \prod_{j=1}^{N-1} U_{\omega (\tau_j)} (t/N), \eqno(9)$$
where $\tau_j=tj/N$. Both groups $U_1(t)$ and $U_{-1}(t)$ being
unitary and strongly continuous, it is easy to get from (8) and
(9) that
$$U(t, \omega )=s-\lim_{N\to \infty} U_N(t,\omega )\eqno(10)$$
for a.e. $\omega $ and any $t>0$.
\par
Consider the complex-valued function
$$F_N({\bf i}) \buildrel \rm def \over =
F_N(i_1,i_2,...,i_{N-1})=
(\prod_{j=1}^{N-1} U_{i_j} (t/N) \phi_0, g)_H,$$
where $i_j=\pm 1, \hskip 5pt \phi_0, g \in H$. Obviously,
$$h_N(t,\omega )\equiv (U_N(t,\omega )\phi_0, g)_H=
F_N(\omega (\tau_1), \omega (\tau_2), ...,\omega (\tau_{N-1}))$$
The functional  $h_N(t,\cdot )$ is peacewise
constant on probability space.
Therefore $h_N(t,\cdot )$ is measurable with respect
to probability measure $\mu $ and for any measurable set $\Omega $
we have
$$J_{N,\Omega }(t)\equiv \int_{\Omega } d\mu (\omega )h_N(t,\omega )=
\sum_{\bf i} F_N({\bf i})\mu (\Omega \cap I_N({\bf i}))
\eqno(11)$$
Here $I_N({\bf i})=\lbrace
\omega : (\omega (\tau_1),\omega (\tau_2),...,
\omega (\tau_{N-1}))={\bf i} \rbrace
,\hskip 5pt I_N({\bf i})\cap I_N({\bf j})=
\emptyset $ for ${\bf i}\not= {\bf j}$ and the summation
is carried over all possible ${\bf i} \in \lbrace 1,-1 \rbrace ^{N-1}$.
\par
It follows from (10) that $(U(t,\omega )\phi_0 ,g)_H$ is $\mu $-
measurable for any $t,\phi_0 ,g$ and
$$\int_{\Omega }d\mu (\omega )(U(t,\omega )\phi_0,g)_H=
\lim_{N \to \infty } J_{N,\Omega }(t)
\eqno(12)$$
\par
Define the following two operators in $H \oplus H$:
$$A(\phi_1,\phi_2) \buildrel \rm def \over =
 (A(1)\phi_1,A(-1)\phi_2),$$
$$M(\phi_1,\phi_2) \buildrel \rm def \over =
 (-p \phi_1 +p \phi_2,
-p \phi_2 +p \phi_1),$$
where $\phi_1,\phi_2 \in H, \hskip 5pt (\phi_1,\phi_2) \in
H \oplus H$.
One can easily show that $A$ and $M$
are selfadjoint in $H \oplus H$, $M$ is bounded and Re $ M \le 0$.
\par
{\bf Theorem 1. Let $\Omega_l= \lbrace \omega : \omega (t)=l \rbrace,
\hskip 5pt l=\pm 1.$ Then for any $\phi_0, g \in H$
$$\int_{\Omega_l }d\mu (\omega )(U(t,\omega )\phi_0,g)_H=
(P_l \exp (t(-iA+M))\vec \phi_0, g)_H
\eqno(13)$$
where $\vec \phi_0=(\phi_0 ,0)\in H \oplus H, \hskip 5pt
P_1(f,h)=f, \hskip 5pt
P_2(f,h)=h.$   }
\par
{\bf Proof.} It is essentially the same as in [2] or [3]. Let
$p(\tau, i,j )$ be the transition probabilities of considered
Markov process given by (2),(3). The measure can be easily
calculated:
$$\mu (\Omega_l \cap I_N({\bf i}))=\prod_{j=1}^N p(t/N, i_j,i_{j-1}),
\eqno(14)$$
where $i_0=1,\hskip 5pt i_N=l.$ It follows from (11) and (14) that
$$J_{N,l}(t) \buildrel \rm def \over =
\int_{\Omega_l } d\mu (\omega )h_N(t,\omega )=
\Biggl ( U_l(t/N)  \sum_{i_{N-1}=\pm 1 } p(t/N, l, i_{N-1})$$
$$U_{i_{N-1}}(t/N)
\sum_{i_{N-2}=\pm 1} p(t/N,i_{N-1},i_{N-2})...$$
$$\sum_{i_1=\pm 1}
p(t/N, i_2,i_1)U_{i_1}(t/N)
p(t/N,i_1,1)\phi_0 , \hskip 5pt
g \Biggr )  _H
\eqno(15)$$
Let the operator $L_N$ in $H \oplus H$ be given by
$$P_l(L_N \vec h)=U_l(t/N)(p(t/N,l,1)h_1+p(t/N,l,-1)h_{-1}),$$
where $\vec h=(h_1,h_{-1}), \hskip 5pt
l=\pm 1$. It follows from the definition
of $A$ and $M$ that
$$L_N=\exp (-i(t/N)A) \exp((t/N)M)$$
Hence one can rewrite (15) as follows:
$$J_{N,l}(t)=(P_l(L_N^N \vec \phi_0 ), \hskip 3pt g)_H,
\eqno(16)$$
By Trotter formula for dissipative operators, (12) and (16) we get
(13). The proof is completed.
\par
{\bf Corollary. The average over all $\omega $ is given by
$$\int_X d\mu (\omega )(U(t,\omega )\phi_0, g)_H=
(\exp(t(-iA+M))\vec \phi_0 , \vec g )_{H \oplus H},$$
where $\vec g =(g,g)$. }
\par
In our further consideration we shall estimate the integral
$$I_{\delta }=\int_X d\mu (\omega )K_{\delta } (\omega ),$$
where
$$K_{\delta }(\omega )=\int_{\delta }^1 d\epsilon \epsilon^{-\gamma }
\int_0^{+ \infty} dt \exp(-2\epsilon t)
(U(t, \omega )\phi_0, g(t))_H,$$
$0<\delta <1, \hskip 3pt \gamma >0, \hskip 3pt
g$ is some $H$-valued function.
\par
{\bf Lemma 1. Suppose that  $g$ is strongly continuous and
$$\Vert g(t) \Vert _H \le C \exp(\nu t),\hskip 5pt
  \nu < 2\delta$$ Then
$K_{\delta }(\omega )$
is $\mu $-integrable and
$$I_{\delta }=\int_{\delta }^1 d\epsilon
\epsilon^{-\gamma} \int_0^{+\infty }dt \exp(-2\epsilon t)
(\exp(t(-iA+M))\vec \phi_0, \vec g(t) )_{H \otimes H},$$
where $\vec g(t) =(g(t),g(t))$.}
\par
The $\hskip 6pt$
result $\hskip 6pt$
follows $\hskip 6pt $
from $\hskip 6pt $
the statement of $\hskip 6pt$
Theorem 1, $\hskip 6pt $
continuity
of the function
$\epsilon^{-\gamma }\exp(-2\epsilon t)(U(t,\omega )
\phi_0, g(t))_H$ with respect to $\epsilon, t$ for a.e. $\omega $
and the proof of Lemma 1.2 in [3].
\bigskip
\centerline
{\bf 3. Finiteness of the main functional.}
\bigskip
\par
Define the following nonnegative functional on the paths:
$$\Phi (\omega )=\int_0^1 d\epsilon \epsilon^{-\gamma }
\int_0^{+\infty } dt \exp(-2\epsilon t) f(t,\omega ),
\eqno(17)$$
where
$$f(t, \omega )=\int dx F(\vert x \vert \le t^{\beta })
\vert \Psi (t,x,\omega ) \vert ^2,$$
$\gamma, \beta \in (0,1), \hskip 5pt
0 \le f \le \Vert \Psi_0 \Vert _2 ^2,\hskip 5pt
\Psi$ is a solution of the equation (1). If the integral in (17)
diverges, we set $\Phi =+ \infty $. Our aim is to prove that for
some set of $\Psi_0 $
$$J \buildrel \rm def \over =
\int_X d\mu (\omega )\Phi (\omega )<+ \infty $$
To do it, we first consider the following functionals:
$$\Phi_{\delta }(\omega )\buildrel \rm def \over =
 \int_{\delta }^1
 d\epsilon \epsilon ^{-\gamma } \int_0^{+\infty } dt \exp(-2\epsilon
t)f(t,\omega ),$$
where $0<\delta <1$. It is obvious that
$$\Phi (\omega )=\lim_{\delta \to +0} \Phi_{\delta }(\omega )
\eqno(18)$$
for $\mu $-a.e. $\omega$. Let us introduce the function
$$N(t,x,y,\omega )\buildrel \rm def \over =
 \Psi (t,x,\omega ) \cdot
\buildrel {\hrulefill {          }} \over {\Psi (t,y,\omega )}$$
According to Lemma 1.1 of [3] we can write the equality
$$f(t,\omega )=\lim_{n\to \infty}(N(t,\omega ),
R_n(t))_{L_2({\bf R}^2)}
\equiv \lim_{n \to \infty} f_n(t,\omega ),
\eqno(19)$$
where $R_n(t,x,y)=F(\vert x \vert \le t^{\beta })l_n(x-y)$,
$l_n \in C_0^{\infty }({\bf R}), l_n(x) \ge 0,
 \int l_n(x)dx=1$ and
$$\lim_{n\to\infty } \int dx F(\vert x \vert \ge \nu )l_n(x)=0$$
for any $\nu >0$.One can see [3] that
$$\vert f_n(t,\omega ) \vert \le C
\eqno(20)$$
uniformly on $n,t,\omega $. From (19)-(20) for any
$\delta >0, \omega \in X$
$$\Phi_{\delta }(\omega )=\lim_{n\to \infty } \int_{\delta }^1
 d\epsilon \epsilon ^{-\gamma } \int_0^{+\infty } dt \exp(-2\epsilon
t)f_n(t,\omega )$$
$$\equiv \lim_{n\to\infty } \Phi_{\delta ,n}
(\omega )
\eqno(21)$$
by dominated convergence theorem.
\par
Consider two operators in $L_2({\bf R}^2)$:
$$A(j)=-\partial _{xx} + \partial _{yy} +(q(x)-q(y))j,
\hskip 4pt j=\pm 1 $$
It is not hard to
show (see [3]) that these operators are selfadjoint on the
domain
$$D=\lbrace f: f\in L_2({\bf R}^2),
\int ds dk \vert \hat f (s,k) \vert
^2 (\vert k^2-s^2 \vert ^2 +1) <+\infty \rbrace $$
for bounded $q$. One can see also that $N$ satisfies to the
equation
$$i {\partial N \over \partial t}=A(\omega (t))N,
\hskip 6pt
N \vert _{t=0} =\Psi_0(x)   \cdot
\buildrel {\hrulefill {       }} \over {\Psi_0(y)}$$
and all the conditions of Theorem 1 and Lemma 1 are
satisfied. Hence, functionals
$\Phi_{\delta ,n}$ are $\mu $-integrable
for all $\delta ,n$ and
$$J(\delta ,n)\buildrel \rm def \over =
 \int_X d\mu (\omega )\Phi_{\delta ,n}(\omega )=
$$
$$\int_{\delta }^1
 d\epsilon \epsilon ^{-\gamma } \int_0^{+\infty } dt \exp (-2\epsilon
t)(\exp(t(-iA+M)) \vec \phi_0 , \vec R_n(t))_{_{H \oplus H}},
\eqno(22)$$
where $\vec \phi_0 =(\Psi_0 (x)  \cdot
\buildrel {\hrulefill {       }} \over {\Psi_0(y)}
, \hskip 5pt 0), \hskip 5pt \vec R_n(t)=(R_n(t),R_n(t)),
\hskip 5pt H=L_2({\bf R}^2)$. From (20)-(22)
by dominated convergence
theorem we get for all $\delta \in (0,1)$
$$J(\delta )\buildrel \rm def \over =
 \int_X d\mu (\omega )\Phi_{\delta }(\omega )
=\lim_{n \to \infty }J(\delta ,n)
\eqno(23)$$
\par
Let us estimate the function
$$Q_n(\epsilon )\buildrel \rm def \over =
 \int_0^{+\infty } dt \exp (-2\epsilon
t)(\exp(t(-iA+M)) \vec \phi_0 , \vec R_n(t))_{_{H \oplus H}},
\eqno(24)$$
By Plancherel Theorem
$$Q_n(\epsilon )={ 1 \over 2\pi } \int_{-\infty }^{+\infty }
(\vec L_1 (\lambda , \epsilon),
\vec L_2 (\lambda , \epsilon ))_{_{H \oplus H}},
\eqno(25)$$
where
$$\vec L_1(\lambda , \epsilon )= \int_0^{+\infty }
dt \exp (-\epsilon
t-it\lambda )(\exp(t(-iA+M)) \vec \phi_0=$$
$$-i(A+iM+\lambda -i\epsilon )^{-1}\vec \phi_0,
\hskip 5pt
\vec L_2(\lambda , \epsilon )=
(L_2(\lambda , \epsilon ), L_2(\lambda , \epsilon )),$$
$$L_2(x,y, \lambda ,\epsilon )=l_n(x-y) \int_0^{+\infty }
dt \exp(-\epsilon t
-it\lambda )F(\vert x \vert \le t^{\beta })=$$
$$l_n(x-y)(\epsilon +i\lambda )^{-1} \exp(-i\vert x \vert ^{1/\beta }
\lambda -\epsilon \vert x \vert ^{1/\beta })
\eqno(26)$$
It follows from (26) and $\int dx l_n(x)=1$ that
$$\hat L_2(s,k,\lambda , \epsilon )\buildrel \rm def \over =
 \int dx dy \exp(-isx-iky)
L_2(x,y,\lambda , \epsilon ) $$
can be estimated as
$$\vert \hat L_2(s,k,\lambda , \epsilon ) \vert
 \le
\vert \epsilon +i\lambda \vert ^{-1}
\int dx dy l_n(x-y) \exp (-\epsilon \vert x \vert ^{1/\beta })$$
$$\le C(\beta )\epsilon ^{-\beta }\vert \epsilon +i\lambda \vert ^
{-1} \eqno(27)$$
uniformly in $s,k,\lambda ,\epsilon ,$ where
$C(\beta )=\int dx \exp(- \vert x \vert ^{1/\beta })$.
\par
Consider two functions:
$$\eta (\lambda ,\epsilon )=(P_1+P_{-1})(A+iM+\lambda
-i\epsilon )^{-1} \vec \phi_0,$$
$$\theta (\lambda ,\epsilon )=(P_1-P_{-1})(A+iM+\lambda
-i\epsilon )^{-1} \vec \phi_0,$$
$\eta (\cdot ,\cdot ,\lambda ,\epsilon ),\hskip 3pt
\theta (\cdot ,\cdot ,\lambda , \epsilon )\in
L_2({\bf R}^2)$. It follows from (25)-(27) and Plancherel Theorem
that
$$\vert Q_n(\epsilon ) \vert \le C(\beta )\epsilon ^{-\beta }
\int_{-\infty }^{+\infty }d\lambda \vert \epsilon +i\lambda \vert
^{-1} \int \int ds dk \vert \hat \eta  (s,k,\lambda ,\epsilon )
\vert , \eqno(28)$$
where $\eta $ is taken in Fourier represantation. One can see from
the definition of $A$ and $M$ that $\eta ,\theta $ satisfy
in space represantation the following system of equations:
$$(-\partial_{xx}+\partial_{yy} +\lambda -i\epsilon)\eta +
(q(x)-q(y))\theta =
\Psi_0 (x)  \cdot
\buildrel {\hrulefill {       }} \over {\Psi_0(y)}
\eqno(29)$$
$$(-\partial_{xx}+\partial_{yy} +\lambda -i\epsilon -2ip)\theta +
(q(x)-q(y))\eta =
\Psi_0 (x)  \cdot
\buildrel {\hrulefill {       }} \over {\Psi_0(y)}
\eqno(30)$$
\par
The following Lemma plays very important role in further
consideration. Let
$\lambda \in {\bf R}, \hskip 5pt \epsilon \in (0,1),\hskip 4pt p>0$.
\par
{\bf Lemma 2. The following uniform estimate holds
$$\Vert \theta (\cdot , \cdot , \lambda ,\epsilon )\Vert
_{L_2({\bf R}^2)} \le C
(\epsilon p )^{-1/2} \Vert \Psi_0 \Vert_2^2$$
\par
Proof.} One can rewrite (29)-(30) as
$$(B_{\lambda } -i\epsilon +iM)\vec \phi =\vec \phi_0,
\eqno(31)$$
where $\vec \phi =(\eta ,\theta ), \hskip 5pt \vec \phi_0=
(\Psi_0 (x)  \cdot
\buildrel {\hrulefill {       }} \over {\Psi_0(y)},\hskip 5pt
\Psi_0 (x) \cdot
\buildrel {\hrulefill {       }} \over {\Psi_0(y)}),$
$$B_{\lambda }\vec \phi =(
(-\partial_{xx}+\partial_{yy} +\lambda )\eta +
(q(x)-q(y))\theta, \hskip 5pt
(-\partial_{xx}+\partial_{yy} +\lambda )\theta +
(q(x)-q(y))\eta ),$$
$M \vec \phi =(0, \hskip 3pt -2p \theta )$.
Both operators $B_{\lambda }$ and $M$
are selfadjoint in $L_2({\bf R}^2)
\oplus L_2({\bf R}^2)$. From (31) we have
$$\Vert \vec \phi_0 \Vert ^2 =(
(B_{\lambda }-i\epsilon +iM) \vec \phi ,
(B_{\lambda }-i\epsilon +iM) \vec \phi )_{_{H \oplus H}}=$$
$$\Vert (B_{\lambda }+iM) \vec \phi \Vert_{H \oplus H } ^2 +
\epsilon^2 \Vert \vec \phi \Vert_{H \oplus H} ^2 -
2\epsilon (M \vec \phi, \vec \phi )_{_{H \oplus H}}\ge $$
$$-2 \epsilon (M \vec \phi , \vec \phi )_{_{H \oplus H}}=
4p \epsilon \Vert \theta \Vert _H^2.$$
The proof is completed.
\par
We shall need also the following result of technical caracter.
\par
{\bf Lemma 3. Let $1<\alpha <2, \hskip 5pt 0<\epsilon <1,
\hskip 5pt \lambda \in {\bf R} $. Consider the integral
$$I(s,\lambda ,\epsilon )=\int_{-\infty }^{+\infty }
{ dk \over \vert s^2-k^2-\lambda \vert ^{\alpha }+
\epsilon ^{\alpha }}$$
The estimate holds
$$\int ds (I(s,\lambda , \epsilon ))^{2/\alpha } \le
C(\alpha )\epsilon ^{2/\alpha -2} \vert \lambda \vert
^{1/2-1/\alpha }$$
uniform in $\lambda ,\epsilon $.
\par
Proof. } From $\vert k^2-s^2+\lambda \vert ^{\alpha }\ge
\vert k^2-\vert s^2-\lambda \vert \vert ^{\alpha }$ after
the change of variable $t=k \vert s^2- \lambda \vert ^{-1/2}$
we have
$$I(s,\lambda , \epsilon ) \le C \vert s^2-\lambda \vert ^
{1/2-\alpha }\int_0^{+\infty } { dt \over
\vert t^2-1 \vert ^{\alpha }+\nu ^{\alpha }}\hskip 3pt
, \eqno(32)$$
where $\nu =\epsilon \vert s^2-\lambda \vert^{-1}$. Consider
two cases.
\par
1. $\nu \ge 1.$ From
$$ \sup_{\nu \ge 1} \sup_{t \ge 0}
{\vert t^2-1 \vert ^{\alpha }+\nu ^{\alpha }
\over
\vert t-1 \vert ^{2 \alpha }+\nu ^{\alpha }  } <+ \infty $$
we obtain
$$I(s, \lambda ,\epsilon )\le C \vert s^2- \lambda \vert ^{1/2-
\alpha } \int_{- \infty }^{+ \infty } d \tau
( \vert \tau \vert ^{2 \alpha  } + \nu ^{\alpha } )^{-1} \le $$
$$ C  \vert s^2- \lambda \vert ^{1/2 - \alpha } \nu ^{1/2- \alpha }
= C \epsilon ^{1/2 - \alpha }
\eqno(33)$$
\par
2. $\nu \le 1.$ Then
$$ T( \nu ) \buildrel \rm def \over =
 \int_0^{+ \infty } dt
(\vert t^2-1 \vert ^{\alpha }+\nu ^{\alpha })^{-1} =
\int_0^{+ \infty }dt
(\vert t^2-1 \vert ^{\alpha }+\nu ^{\alpha })^{-1} \cdot $$
$$(F( \vert t-1 \vert \ge 1/2)+
(F( \vert t-1 \vert \le 1/2) )
\equiv T_1 (\nu )+T_2(\nu )$$
Further,
$$T_1(\nu ) \le \int_0^{+\infty }dt F( \vert t-1 \vert \ge 1/2)
\vert t^2 -1 \vert ^{-\alpha }=C<+\infty ,
\eqno(34)$$
since $\alpha >1$. Finally,
$$T_2(\nu )\le C \int_{1/2}^{3/2} dt
(\vert t^2-1 \vert ^{\alpha }+\nu ^{\alpha })^{-1}
\le C \int_{-1/2}^{1/2} d \tau
(\vert \tau  \vert ^{\alpha }+\nu ^{\alpha })^{-1}
\le C \nu ^{1- \alpha },
\eqno(35)$$
since $1< \alpha <2$. From (32),(34)-(35) we have for
$\nu \le 1$
$$ I(s,\lambda , \epsilon ) \le C \epsilon ^{1-\alpha }
\vert s^2 - \lambda \vert ^{-1/2}
\eqno(36)$$
Now using (33) and (36) estimate the integral
$$L \buildrel \rm def \over =
\int ds (I(s,\lambda ,\epsilon ))^{2/\alpha }\le
C \int ds (F(\vert s^2 -\lambda \vert \le \epsilon )\epsilon ^
{2/ \alpha (1/2 - \alpha ) } +$$
$$ F( \vert s^2 - \lambda \vert \ge \epsilon )
(\epsilon ^{1-\alpha }\vert s^2 - \lambda \vert ^{-1/2} )^
{2/ \alpha })\equiv L_1+L_2 $$  From
$\vert s^2- \vert \lambda \vert \vert \le
\vert s^2 - \lambda \vert $ we have
$$L_1 \le C \int ds F( \vert s^2 -\vert \lambda \vert
\vert \le \epsilon ) \epsilon ^{1/\alpha -2}$$
\par
If $\vert \lambda \vert \le \epsilon $ then
$$L_1 \le C \epsilon ^{1/2} \epsilon ^{1/\alpha -2}
\le C \epsilon ^{2/\alpha -2} \vert \lambda \vert
^{1/2 -1/\alpha }
\eqno(37)$$
because $1/2-1/\alpha <0 $.
\par
If $\vert \lambda \vert \ge \epsilon $ then
$$L_1 \le C \epsilon ^{1/\alpha -2} ((\vert \lambda \vert + \epsilon )
^{1/2}-(\vert \lambda \vert - \epsilon )^{1/2})\le $$
$$C \epsilon ^{1/\alpha -1} (\vert \lambda \vert + \epsilon )^{-1/2}
\le C \epsilon ^{2/\alpha -2} \vert \lambda \vert
^{1/2- 1/\alpha }
\eqno(38)$$
Finally, for all $\lambda $
$$L_2 \le C \epsilon ^{2/\alpha -2} \int ds
\vert s^2 -\lambda \vert ^{-1/\alpha }\le $$
$$C \epsilon ^{2/\alpha -2} \int ds \vert s^2 -
\vert \lambda \vert \vert  ^{-1/\alpha } \le
C \epsilon^{2/\alpha -2} \vert \lambda \vert ^{1/2-1/\alpha }
\eqno(39)$$  From
(37)-(39) follows the statement of Lemma.
\par
Now we are able to estimate $\Vert \hat \eta (\cdot , \cdot ,
\lambda , \epsilon ) \Vert _{L_1({\bf R}^2)}$.
It follows from (29) that
$$ \hat \eta (s,k,\lambda ,\epsilon )= (s^2-k^2+\lambda -i
\epsilon )^{-1} (
\hat \Psi_0(s)  \cdot
\buildrel {\hrulefill {      }}\over
{\hat \Psi_0(-k) }
-U_1 +U_2) \buildrel \rm def \over =
\hat \eta _0 - \hat \eta _1 + \hat \eta _2
 \eqno(40)$$
where
$$U_1(s,k,\lambda ,\epsilon )=\int dx \exp (-isx)
q(x) \theta _1 (x,k, \lambda , \epsilon ),$$
$$U_2(s,k,\lambda ,\epsilon )=\int dy \exp (-iky)
q(y) \theta _2 (s,y, \lambda , \epsilon ),$$
$$\theta _1(x,k)= \int dy \exp(-iky) \theta (x,y),\hskip 5pt
\theta _2(s,y)= \int dx \exp(-isx) \theta (x,y).$$
\par
Set $\alpha =(1/r+1/2)^{-1}$ for $r > 2$ and
$\alpha =1+ \nu , \nu \in (0,1/3)$ for $r \in [1,2]$,
where $q \in L_r({\bf R}) \cap L_{\infty }({\bf R})$.
Suppose that $\hat \Psi_0 \in L_2({\bf R}) \cap L_{\mu }({\bf R}),
\hskip 5pt \alpha ^{-1}+ \mu ^{-1} =1$.
By Lemma 3 and Holder
inequality
$$ \Vert \hat \eta _0 \Vert _{L_1({\bf R}^2)}\le C
\Vert \hat \Psi_0 \Vert _{\mu }
\Vert \hat \Psi_0 \Vert _2 \hskip 2pt
\epsilon ^{1/\alpha -1}
\vert \lambda \vert ^{1/4-1/2\alpha }
\eqno(41)$$
and
$$ \Vert \hat \eta_2 \Vert _{L_1({\bf R}^2)} \le C I(s, \lambda ,
\epsilon )^{1/\alpha } \Vert U_2(s, \cdot , \lambda ,\epsilon )
\Vert _{\mu },
\eqno(42)$$
where  $I(s, \lambda , \epsilon )$ is
integral introduced in Lemma 3. Further,
by Hausdorff-Young inequality and
Holder inequality
$$\Vert U_2 (s, \cdot , \lambda , \epsilon ) \Vert _{\mu } \le
C \Vert q(\cdot ) \theta_2(s, \cdot , \lambda ,\epsilon ) \Vert
_{\alpha } \le C \Vert q \Vert _{\gamma } \Vert \theta_2(s, \cdot ,
\lambda , \epsilon ) \Vert _2,
\eqno(43)$$
where $\gamma ^{-1}=-1/2 + \alpha^{-1}, \gamma \in (2, +\infty )$
One can easily see that $\Vert q \Vert_{\gamma }<+\infty $.
On the basis of Lemma 3, Lemma 2, Plancherel Theorem
and (42)-(43) one can write
the estimate
$$\Vert \hat \eta _2 \Vert _{L_1({\bf R}^2)} \le C
 \epsilon ^{1/\alpha -1}
\vert \lambda \vert ^{1/4-1/(2 \alpha )}
\Vert \theta_2 (\cdot ,\cdot ,\lambda ,\epsilon ) \Vert
_{L_2({\bf R}^2)} \le C \epsilon ^{1/\alpha -3/2}
\vert \lambda \vert ^{1/4-1/(2 \alpha )}
\eqno(44)$$
The norm $\Vert \hat \eta _1 \Vert $
can be estimated as (44) in the same manner. From (41)
and (44) we obtain the following estimate for $Q_n(\epsilon )$:
$$ \vert Q_n(\epsilon ) \vert \le C(\beta )\epsilon
^{-\beta +1/\alpha -3/2 } \int d \lambda (\vert \lambda \vert
+\epsilon )^{-1} \vert \lambda \vert ^{1/4 -1/(2 \alpha )}=$$
$$C_1  \epsilon ^{-\beta +1/(2\alpha )-5/4},
\eqno(45)$$
because $-1<1/4-1/(2\alpha )<0$. This inequality is uniform in
$n, \epsilon $.
\par
Substituting (45) in (22) we get the estimate for
$J(\delta ,n)$:
$$\vert J(\delta ,n) \vert \le C \int_{\delta }^1 d \epsilon
\epsilon ^{-\gamma -\beta +1/(2\alpha ) -5/4}
\eqno(46)$$
Let $\gamma +\beta <1/(2\alpha )-1/4$. It follows from (46) that
$$\vert J(\delta ,n) \vert \le C <+\infty
\eqno(47)$$
uniformly on $\delta \in (0,1), \hskip 5pt
n \in {\bf N}$. From (23) and (47) we have
$$\vert J(\delta ) \vert \le C <+\infty
\eqno(48)$$
uniformly in $\delta $. Functionals $\Phi_{\delta }(\omega )$
being nonnegative, by Fatou Lemma we get from (18) and (48):
$$J=\int_X d \mu (\omega ) \Phi (\omega ) <+\infty
\eqno(49)$$
\bigskip
\centerline
{\bf 4. Proof of Theorems. }
\bigskip
\par
Now we can prove the results formulated in the introduction. First
of all, one can easily see from the definition of $\Phi (\omega )$
that
$$\Phi (\omega ) \ge  C
\int _1^{+\infty } dt t^{\gamma -1}
\int dx F( \vert x \vert \le t^{\beta } ) \vert \Psi (t,x, \omega )
\vert ^2 \equiv \Xi (\omega ),
\eqno(50)$$
where $C>0$. From (49) and (50) we obtain the following statement.
\par
{\bf Theorem 2. Let $q \in
L_r({\bf R}) \cap L_{\infty }({\bf R}), r \ge 1$. Set
$\alpha =(r^{-1}+1/2)^{-1}$  for $r >2$ and $1<\alpha <4/3$ for
$r \in [1,2]$. If $ \gamma ,\beta \in (0,1), \hskip 5pt
0<\gamma +\beta <
1/(2\alpha )-1/4$ and $\Psi_0 \in
L_2({\bf R}) \cap L_{\alpha }({\bf R})$,
then
$$\int_X d \mu (\omega ) \Xi (\omega )<+\infty $$ }
\par
{\bf Corollary 1. For any $\Psi_0$ satisfying condition of the Theorem,
for a.e. $\omega $ $\Xi (\omega )<+\infty $.}
\par
Let $\beta_0 = {1 \over 2r}$ for $r \ge 2 $ and $\beta_0 =1/4$
for $r \le 2$.
\par
{\bf Corollary 2.
If $0< \beta < \beta_0 $, then for any
$\Psi_0 \in L_1({\bf R}) \cap L_2({\bf R})$
for a.e. $\omega $
$$\lim_{T \to +\infty } {1 \over T} \int_0^T dt \int
dx F(\vert x \vert \le t^{\beta }) \vert \Psi (t,x, \omega )
\vert ^2 =0
\eqno(51)$$ }
\par
The result follows from the estimate
$${1 \over T} \int_0^T dt f(t) \le {1 \over T }
\Biggl ( \int_0^T dt t^{\gamma -1} f^2(t) \Biggr
 )^{1/2}
\Biggl ( \int_0^T dt t^{1-\gamma } \Biggr
)^{1/2}\le $$
$${C \over T } T^{1- \gamma /2} \Biggl  (
\int_0^T dt t^{\gamma -1}
f(t) \Biggr
 )^{1/2} \to 0 $$
as $T \to +\infty $, where $0 \le f(t) \le \Vert \Psi_0 \Vert ^2$.
\par
{\bf Theorem 3. There exists a set of full measure $\Omega \subset
X$ such that (51) holds for any $\omega \in \Omega $ for all
$\Psi_0 \in L_2({\bf R})$.
\par
Proof. } It follows from Corollary 2 that there exists a dense in
$L_2({\bf R})$ countable set $M$ such that (51) holds for
$\Psi_{0,n} \in M $ if $\omega \in X \backslash W_n, \hskip 5pt
\mu (W_n)=0, \hskip 5pt
n=1,2,... $ Define $W= \cup  W_n, \hskip 5pt
\mu (W)=0$. Then (51) holds for
any $\Psi_0 \in M $ and any $\omega \in X \backslash W$. From
the estimate
$${1 \over T }\int_0^T dt \int dx F(\vert x \vert \le t^{\beta })
\vert \Psi (t,x, \omega ) \vert ^2 \le \Vert \Psi_0 \Vert _2^2$$
we get (51) for any $\Psi_0 \in L_2({\bf R}),\hskip 3pt \omega \in
X \backslash W$. The proof is completed.
\par
{\bf Lemma 4. If $0< \nu< 1/\alpha -1/2$ then for any
$\Psi_0 \in L_2({\bf R}) \cap L_{\alpha }({\bf R}),
\hskip 5pt \Psi_0 \ne 0,$
$$\int_1^{+\infty }dt {<x^2(t)> \over t^{1+\nu }}
=+\infty $$
for a.e. $\omega $.
\par
Proof. } From
the definition of $<x^2(t)>$ we have
$$ <x^2(t)>= \int dx x^2 \vert \Psi (t,x, \omega ) \vert ^2 \ge
t^{2\beta }\int dx F(\vert x \vert \ge t^{\beta })
\vert \Psi (t,x, \omega ) \vert ^2=$$
$$t^{2\beta }(\Vert \Psi_0 \Vert _2^2 -
\int dx F( \vert x \vert \le t^{\beta })\vert \Psi (t,x, \omega )
\vert ^2 )
\eqno(52)$$
By Corollary 1
$$\int_1^T dt <x^2(t)>t^{\gamma -1 -2\beta } \ge
\Vert \Psi_0 \Vert_2^2 \int_1^T dt t^{\gamma -1}-$$
$$\int_1^T dt t^{\gamma -1}\int dx F(\vert x \vert \le
t^{\beta }\vert \Psi (t,x, \omega )\vert ^2 \ge C_1 T^{\gamma }-
C_2 \to +\infty $$
as $T \to +\infty $. Taking $\gamma $ sufficiently small, we get the
statement of the Lemma.
\par
{\bf Theorem 4. If $0< \nu <2 \beta_0 $ then for a.e. $\omega $
for any $\Psi_0 \in L_2({\bf R}), \hskip 5pt \Psi_0 \ne 0, $
$$\liminf_{T \to +\infty } {1 \over T} \int_1^T dt <x^2(t)>
t^{-\nu } >0
\eqno(53)$$
\par
Proof. } From (52) and Theorem 3
$${1 \over T} \int_1^T dt <x^2(t)> t^{-2\beta }
\ge {T-1 \over T } \Vert \Psi_0 \Vert
_2^2 -{1 \over T } \int_1^T dt
\int dx F( \vert x \vert \le t^{\beta })\vert \Psi (t,x, \omega )
\vert ^2 \to \Vert \Psi_0 \Vert _2^2$$
as $T \to +\infty $ that proves (53).
\par
\bigskip
\centerline {\bf Acknowledgements. }
\bigskip
\par
I would like to thank Professor Sergio Albeverio for hospitality
at the Institut fur Mathematik
Ruhr-Universitat Bochum. \par
I am grateful to Alexander von Humboldt Stiftung for financial
support.
\bigskip
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\bigskip
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\par
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\par
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\end