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\begin{document}

%\input title
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%%%%%%%%%%%%%%%%%%55 title %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{titlepage}

\begin{center}

{\large CZECH TECHNICAL UNIVERSITY IN PRAGUE}\\


Faculty of Nuclear Sciences and Physical Engineering\\

\vspace{30ex}

{\Large\bf DIPLOMA THESIS}\\

\vspace{10ex}

{\Large  SEMICLASSICAL ANALYSIS OF TIME DEPENDENT \\
\vskip5pt QUANTUM SYSTEMS}\\


\vspace{9ex}

{\Large Ond\v{r}ej Lev }\\

\vspace{2ex}

Supervisor: Prof. Ing. Pavel \v{S}\v{t}ov\'{\i}\v{c}ek, DrSc.\\
Consultant: Prof. Pierre Duclos, Universit\'e de Toulon et du Var, France
\vspace{20ex}

{\Large Trutnov}\\

\vspace{3ex}

{\Large 5.5. 2002}

\end{center}

\end{titlepage}


\newpage


\begin{titlepage}
\vspace*{100ex}
\noindent
I declare that I wrote this diploma thesis independently using the listed references.
\vskip 10ex
\hskip 65ex
Ond\v{r}ej Lev        
\end{titlepage}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%% Preface %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Preface}$\\$
During the last three years when I have been dealing with mathematical 
physics, I encountered six functional analytic methods:\\
1) KAM algorithm \cite{DuclosStovicek}; \\
applications of \\
2) WKB in Feynman integral and Geometric asymptotics  
\cite{schu}, \cite{GS} \\
3) Dyson expansion and \\
4) Egorov theorem \cite{br}, \cite{Lev} \\
on quantum ball;\\
5) Egorov theorem for time dependent case \cite{Lev}, \\
6) Matrix elements investigated with WKB.\\
I decided to include in this thesis three of them. 
Matrix elements investigated with WKB because it is my last work, 
Egorov theorem for time dependent case because new improvements
appeared and finally KAM, although not semiclassical method it was
inspiration for starting semiclassical analysis of matrix elements.
As a consequence of this decision this thesis
consists of three more or less independent parts.\\
The last part contains the results of paper \cite{D2} and application
of this improved KAM technique  on two models.
The quantity $V_{knm}$ defined by (\ref{Vknm_eq_int0T}), in fact
matrix element of perturbation $V$ in eigen-basis of unperturbed
Floquet Hamiltonian,  is crucial for \mbox{Theorem \ref{MainTheorem}}
to be applicable.\\
Section \ref{Egorov} gives a new  statement of Egorov theorem for time dependent
systems, based on ideas of \cite{br}, \cite{Lev} and Joachim Asch, who
showed direct approach to time dependent generalization. Also
assumptions on the form and growth of time dependent Hamiltonian
are weaken.\\
The heart of thesis, section \ref{WKB}, contains application of WKB
(although nonstandard one, \mbox{see \ref{Intro}.} Introduction) 
in investigation of asymptotic behavior of 
\[
W_{n,\;l}:=<W\,\Psi_n,\;\Psi_l> ,\
\]
for large $n,\;l$, where $\Psi_n$ is the normalised Dirichlet
$L^2(\R_+)$ solution of
\[
(-\Delta+x^\alpha)\Psi_n(x)=E_n\;\Psi_n(x)\;,\ \;\alpha>1 .\
\]
Asymptotic behavior is important for finiteness of quantities like 
$\sup_{n}\sum_l W_{n,\;l}$ or $\sup_{n}\sum_l
\frac{W_{n,\;l}}{E_n-E_l}$, which are important in KAM, Dyson or
adiabatic analysis of time dependent systems.\\
The result of this method is precise as far as computations of leading terms
of normalization
constant and the diagonal of matrix $W_{n,\;l}$ are concerned,
 with the numerical
values being derived, but less precise for offdiagonal elements, which
are only estimated from above (see Theorem \ref{sum}). Unfortunately, there is
no sufficient condition yet implying that $W$ is a Shur-Holmgren operator
($\sup_{n}\sum_l W_{n,\;l} <\infty$).\\ 
This is the beginning of the work which started 3 month ago
within consultations with \mbox{prof. Pierre Duclos} and Dr. Michel Vittot
during my stage in CPT, Marseille and continued in Prague within meetings with
prof. Pavel \v{S}\v{t}ov\'{\i}\v{c}ek, my supervisor.\\
 Possible
continuation could be an improvement of estimate (\ref{offdiag}),
computing further terms in (\ref{C_E}), (\ref{diagelements}), a
generalization to asymptotically homogeneous potentials, taking Neumann
condition instead of Dirichlet and using  it to treat the true symmetric well
on the whole line.
%\\ \\
\newpage
$\\$
{\bf Acknowledgments}
$\\$
I would like to thank to Dr. Joachim Asch (CPT, Marseille) for
consultations and ideas in Egorov theorem problematic,\\ 
prof. Pierre Duclos and Dr. Michel Vittot for consultations and ideas in
WKB problematic and their care during my stage in CPT, Marseille.\\
Finally, I would like to thank very much to prof. Pavel 
\v{S}\v{t}ov\'{\i}\v{c}ek
for his perfect leadership during all three years.


\newpage

%\input diag 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%% diag %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Behavior of matrix elements investigated with WKB}\label{WKB}

\subsection{Introduction} \label{Intro}
$\\$
WKB analysis is well-known method in quantum mechanics, which allows
us to construct wave functions in semiclassical regime. This method
exists also in mathematics, under name Liouville-Green method. For
details for further discussion one can consult \cite{Olver} Chapters 6 and
11.\\
Let us consider equation 
\[
(-\Delta+x^\alpha-E)\Psi(x)=0
\]
in $E\to\infty$ regime. Using scaling trick (\ref{scaling}) this
equation becomes equivalent to
\[
(-\hbar^2\Delta+x^\alpha-1)\Phi(x)=0 \; ,\
\]
where $\hbar\to 0$.\\
Standard WKB yields the solution
\be
\Phi(x)=&&\frac{1}{\sqrt[4]{1-x^\alpha}}
\sin\left(\frac{1}{\hbar}\int^x_0\sqrt{1-y^\alpha}\; dy \right)\;
\;\; {\rm for} \;x<1\\
&&\frac{1}{\sqrt[4]{x^\alpha-1}}
\exp\left(-\frac{1}{\hbar}\int^x_1\sqrt{y^\alpha-1}\; dy \right)\;
\;\; {\rm for} \;x>1 \; ,\
\ee
which is diverging in turning point $1$.\\
The problem is that standard WKB is in fact unitary transformation
between the solutions of
\[
(-\hbar^2\Delta+V)\Phi=0 
\]
and
\[
(-\hbar^2\Delta+C)\zeta=0 \;.\ 
\]
Constant $C$ may be chosen $+1$, where $V>0$, or $-1$, where $V<0$,
i.e. in classically forbidden or permitted region. The solution is there
exponentially decaying or oscillating respectively. The divergence of
standard WKB solution is caused by the fact that the unitary mapping
between solutions of equations 
\[
(-\hbar^2\Delta+V)\Phi=0 
\]
and 
\[
(-\hbar^2\Delta+v)\zeta=0 \; ,\ 
\]
where $v$ is some function,
conserves the sign of $V$.\\
Because our problem $V(x)=x^\alpha-1$ changes the sign, it looks more
naturally trying to map the equation
\[
(-\hbar^2\Delta+x^\alpha-1)\Psi(x)=0 
\]
to
\[
(-\hbar^2\Delta+x-1)\zeta(x)=0 \; .\ 
\] 
If we proceed this mapping, Airy functions $\Ai$ and $\Bi$
appear. Because we are interested in $L^2(\R_+)$ solutions, we exclude
$\Bi$. Now our new solution has no singularity!
Other advantage is that error is estimated uniformly on $\R_+$. We
see that this approach is convenient for our problematic. \\
Now if we ask for Dirichlet boundary condition at origin, we will
obtain Bohr-Sommerfeld quantization condition and Energy $E$  become labeled
by a natural number $n$.\\
So at the moment we have two Dirichlet $L^2(\R_+)$ solutions 
and we would like to compute matrix elements
\[
\frac{<W \,\Psi_n, \, \Psi_l  > }
{\sqrt{ < \Psi_n ,\, \Psi_n  >< \Psi_l ,\, \Psi_l  >}} ,\
\]
for some perturbation $W$. 
This expression looks very complicated, but numerically calculated
examples $\alpha=1,\,2$ showed us the way, the most important thing is
the behavior in the neighborhood of the origin.
\\
The section is divided into several parts: computation of
normalization constant, diagonal and offdiagonal matrix elements. 
The approach is given in details, what is interesting, is that in fact
only mathematical tool, which is used is integration by parts. The
results are formulated in the beginning of each subsection and at the
end of this section is given summary of all results, stated as Theorem 
\ref{sum}.
$\\$
$\\$
{\bf Convention:} 
We will use (for simplicity) the letter $K$ for  all positive
constants
 which depend only on $\alpha$, although they may vary.\\
This section is dealing only with $\alpha>1$ case. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%% Results due to  F.W.J. Olver  %%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	


\subsection{Results due to F.W.J. Olver} $\\$
Due to  Theorem 3.1 , Chapter 11 of \cite {Olver} 
the  only $L^2(\R^+)$ solution of equation 
($ E> 0 $)
\bea
\label{rice}
(-\Delta +x^\alpha  )\Psi(x)=E \Psi(x)
\eea
can be written in the form 
\bea
\label{presny}
\Psi(x)= \left( \frac{s(x)} {x^\alpha -E}\right)^\frac{1}{4} \left(
\Ai(s(x)) +\varepsilon(x)
\right) , 
\eea
where 
\be
s (x)  := && (\frac{3}{2} \int^x _{E ^{ \frac{1}{\alpha} } }
 \sqrt{y^\alpha-E} \,\, dy)^\frac{2}{3},  \,\,\,	
\rm{for} \,\,\,\,  x \geq E ^{\frac{1}{\alpha} } \\
&-& (\frac{3}{2} \int^{E ^{\frac{1}{\alpha} }} _x
\sqrt{E -y^\alpha} \,\, dy)^
\frac{2}{3},  \,\, \rm{for} \,\,\,\, x< E ^{\frac{1}{\alpha} } .\
\ee
$\varepsilon$ is an error, which is controlled by the inequality
\bea
\label{er}
\frac{|\varepsilon(x)|}{M(s(x)) }\, , \; 
\frac{|\varepsilon'(x)|}{N(s(x)) } \sqrt{\frac{s(x)}{x^\alpha-E} }
\leq \frac{1}{ G(s(x))} 
\left( \exp(\frac{8}{\pi^2} E^{-\frac{\alpha+2}{2\alpha}} 
\V(x,\infty)) -1
\right) ,\
\eea
where $\V $ is  given by the formula 
\be
\V(x,y):= \int^y _x \frac{1}{\sqrt{|\tilde {s} (z)|}} 
\left| \frac{4VV''-5V'^2 }{ 16 V^\frac{5}{2}} 
+\frac{5\sqrt{|V|}}{16|\tilde s|^3}\right|(z)\; dz= 
 \int^y _x \left|  \frac{2 {\tilde s}'(z){\tilde s}'''(z) 
-3{{\tilde s}''}{^2}(z) } {4\sqrt{|{\tilde s}(z)|} {{\tilde s}'}{^3}(z) } \right|
dz .\
\ee
We used short notation
 \bea \label{V,tilde s}
V(x):=x^\alpha -1 \, , \;\; 
\tilde s(x):=  
&&(\frac{3}{2} \int^x _1
\sqrt{y^\alpha-1} \,\, dy)^\frac{2}{3},  \;\;\;
\mbox{for}\ x \geq 1  \\ \nn
&-&(\frac{3}{2} \int^1 _x
 \sqrt{1 -y^\alpha} \,\, dy)^   \frac{2}{3}, \;\;\;
 \mbox{for}\  x< 1 .\
\eea
To investigate possible boundedness of $\V$ it is useful to notice
 that $\tilde s {\tilde s}'^2=V$.\\
If we take into account the behavior of $V$ and $\tilde s$
\[
\tilde{s}(x) \stackrel {\infty}{\sim} 
x^\frac{\alpha+2}{3} 
 \,,\ \, \,\, \tilde{s}(x) \stackrel{1}{\sim} x-1  ,\
\]
we conclude (using the fact that $\tilde s\in C^\infty$) 
that $\V$ is a bounded function on $\R^+$.
%\newpage 
$\\$
Let $c$ be negative root of the equation
\[
\Ai(c)=\Bi(c)
\]
of the smallest absolute value ($c=-0.36605$).\\
 The functions $G, M, N$ are
defined  by prescriptions
\be
 M:=\sqrt{2\Ai \Bi} \;\;\;\;\;& {\rm for}  \hskip10pt
 x\geq c \; ,\ \;\; & M:= \sqrt{\Ai^2+ \Bi^2}  \;\;\;\;\; {\rm for}
\hskip10pt  x\leq c \\
 G:=\sqrt{\frac{\Bi}{\Ai}}   \;\;\;\;\; & {\rm for} \hskip10pt
 x\geq c \; ,\ \;\; & G:= 1   \;\;\;\;\; {\rm for} \hskip10pt   x\leq c \\
 N:=\sqrt{\frac{{\Ai'}^2 \Bi^2+\Ai^2 {\Bi'}^2 }{\Ai \Bi}  }
 \;\;\;\;\; &{\rm for} \hskip10pt  x\geq c \; ,\ \;\; &
N:=\sqrt{{\Ai'}^2 +{\Bi'}^2 }  \;\;\;\;\; 
{\rm for} \hskip10pt  x\leq c \\
\ee
By graphical analysis with Mathematica P. Duclos and M. Vittot
obtained the following bounds of Airy functions and their derivatives,
which are valid on $\R$.
\bea \nn
|\Ai| \leq 2\frac{1}{\sqrt\pi A\; \omega} &,& \;\;
|\Bi|\leq 2\frac{A}{\sqrt\pi \omega} \\ 
\label{aibound}
|\Ai'| \leq \frac{\omega}{\sqrt\pi A } &,& \;\;
|\Bi'|\leq \frac{A\; \omega}{\sqrt\pi}
\eea
Auxiliary functions $A$ and $\omega$ are defined by
\[
A(s):= e^{\frac{2}{3} {s_+} ^\frac{3}{2}} \; ,\ \hskip10pt
\omega(s):= |s|^\frac{1}{4} +\frac{1}{2}
\]
$s_+:=\max(s,0)$.\\
Putting the definitions of $G, M,  N,$ the estimates above (\ref{aibound})
and the estimates of
error (\ref{er}) together we obtain
\be
e ^{\frac{2}{3} s_+(x)^\frac{2}{3} } \left(|s(x)|^\frac{1}{4} +
\frac{1}{2}  \right) |\varepsilon(x)| \leq
\frac{2} {\sqrt\pi} 
\left( \exp(\frac{8}{\pi^2} E^{-\frac{\alpha+2}{2\alpha}} 
\V(x,\infty)) -1
\right) 
\ee
\be
e ^{\frac{2}{3} s_+(x)^\frac{2}{3} } \left(|s(x)|^\frac{1}{4} +
\frac{1}{2}  \right)^{-1}  |\varepsilon'(x)| \leq
\frac{2} {\sqrt\pi} \sqrt{\frac{x^\alpha-E}{s(x)} } 
\left( \exp(\frac{8}{\pi^2} 
E^{-\frac{\alpha+2}{2\alpha}} \V(x,\infty)) -1
\right)  .\
\ee
\\
Finally using the boundedness of $\V$  on $\R^+$,  we 
conclude the following estimates of error
\bea
\label{error}
&& |\varepsilon(x) | \leq  K 
 E^{-\frac{\alpha+2}{2\alpha} } |s(x)|^{-\frac{1}{4}}
 e^{- \frac{2}{3}s_+(x)^\frac{3}{2} } \\
\label{error'}
&& |\varepsilon'(x) | \leq  K  E^{-\frac{\alpha+2}{2\alpha} }
 \sqrt{\frac{x^\alpha-E}{s(x)} }  
\left(\frac{1}{2}+|s(x)|^{\frac{1}{4}} \right)
 e^{- \frac{2}{3}s_+(x)^\frac{3}{2} }	
\eea
	

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%% Bohr-Sommerfeld %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	

\subsection{Bohr-Sommerfeld} $\\$
If we want (\ref{presny}) to be Dirichlet solution on $\R^+$, i.e. 
\b \label{Psi(0)}
\Psi(0)=0
\e
we have to satisfy the condition
\be
%\label{dirichlet}
\Ai(s(0))+\varepsilon(0)=0
\ee
Using the well-known asymptotics of $\Ai$
\bea
&& \label{ai-infty}
\Ai(-z)=z^{-\frac{1}{4}} \frac{1}{\sqrt\pi}\cos(\frac{2}{3}z^\frac{2}{3} -
 \frac{\pi}{4} ) \left(1+\0 (z^{-\frac{3}{2}} ) \right) \, \\
&& \label{Ai+infty}
\Ai(z)=\frac{z^{-\frac{1}{4}}} {2\sqrt\pi} e^{-\frac{2}{3} z^\frac{3}{2} }
\left( 1 +\0( z^{-\frac{3}{2} }) 
\right)
\eea
and (\ref{error}), we obtain 
\be
&& \frac{1}{\sqrt{\pi} |s(0)|^\frac{1}{4}} \cos \left(
\frac{2}{3} |s(0)|^\frac{3}{2} -\frac{\pi}{4} \right) 
\left( 1+\0(E^{-\frac{\alpha+2}{2\alpha } })
\right)
 = -\varepsilon(0)   \\ && \iff
\sin \left(
\frac{2}{3} |s(0)|^\frac{3}{2} +\frac{\pi}{4} \right) =
-\sqrt{\pi} |s(0)|^\frac{1}{4} \varepsilon(0) \left( 1+\0(E^{-\frac{\alpha+2}{2\alpha } })
\right) \\&&
\iff 
\frac{2}{3} |s(0)|^\frac{3}{2} +\frac{\pi}{4} = \pi n + \0(E^{-\frac{\alpha+2}{2\alpha } })
 \,
\ee
which yields Bohr-Sommerfeld quantization condition 
\bea
\label{bohr}
\int^{E^\frac{1}{\alpha}} _0 \sqrt{E-x^\alpha} \; dx = \pi(n-\frac{1}{4} ) +
 \0(E^{-\frac{\alpha+2}{2\alpha } }),
\eea
or  equivalently	
\be
\int^{E^\frac{1}{\alpha}} _0 \sqrt{E-x^\alpha}\; dx = \pi(n-\frac{1}{4} ) +
 \0(\frac{1}{n}).
\ee
\begin{remark}
This formula, well-known as the Bohr-Sommerfeld quantization
condition, may be seen as the defining relation between quantum
mechanical Energy $E$ and a natural number $n$.\\
What is interesting is that this formula is here rigorously proven for
the class of homogeneous potentials $x^\alpha$, using Liouville-Green
method.\\
This result can be similarly proven to much more general class of potentials.
\end{remark}

$\\$ To satisfy (\ref{Psi(0)}), we define the Energy behavior by 
\b \label{E(n)}
E_n= \left(\frac{\pi}{\int^1_0  \sqrt{1-x^\alpha} dx } (n-\frac{1}{4})
\right)^\frac{2\alpha}{\alpha+2}
 + \0\left((\frac{1}{n})^\frac{2\alpha}{\alpha+2}\right).
\e


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%% Normalization constant and diagonal elements %%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\subsection{Normalization constant and diagonal
elements}
$\\$
In this section we will compute the leading terms of normalization
constant of the solution (\ref{presny}) and of diagonal matrix
elements of some regular perturbation with respect to 
\mbox{functions (\ref{presny}).}\\
Firstly we will formulate the result of this subsection, which will be
proved at the end of it. 

\subsubsection{{\underline{Result}}} 
\begin{theorem} \label{3.1}
Let $\alpha>1$ and a function $\Psi$ be defined by (\ref{presny}). This means 
that $\Psi$ is the only $L^2(\R+)$ solution of
equation 
\be
(-\Delta +x^\alpha  )\Psi(x)=E\; \Psi(x) .\
\ee
Then for normalization constant holds 
\bea \label{C_E}
C^{-2} _E :=<\Psi,\;\Psi>=\int^\infty _0 |\Psi|^2=
\frac{E ^\frac{2-\alpha}{2\alpha} } {2\pi}
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}} +\0 (E
^\frac{16-10\alpha}{18\alpha}) .\
\eea\\
Suppose moreover that a perturbation $W$ satisfies 
the following decay properties \\
\mbox{($ <x>=\sqrt{1+x^2}$).}
\bea
\label{asumW}
W, W' \in L^1(\R^+)   ,\
<x>^rW \in L^\infty(\R^+) \, ,\ \,\,\, \rm{for\,\, some} \,\,
\, r>1 
\eea   
Then it holds true that
\bea \label{Ediag}
\frac{<W\,\Psi,\,\Psi>}{<\Psi,\, \Psi>}=
E^{-\frac{1}{\alpha}}\left(\int^1 _0
\frac{dx}{\sqrt{1-x^\alpha}}\right)^{-1}
\left( 
\int_{\R_+}W +\0\left(E^{-\min\{\frac{\alpha+2}{18\alpha}
, \frac{r-1}{\alpha}  \} } \right)
\right)
\eea
If we want $\Psi$ to satisfy Dirichlet boundary condition
(\ref{Psi(0)}), then Energy has to be restricted by the Bohr-Sommerfeld
quantization condition (\ref{E(n)}) and then the quantity
(\ref{Ediag}) becomes a diagonal matrix element
\bea \label{diagelements}
\frac{<W\,\Psi_n,\,\Psi_n>}{<\Psi_n,\, \Psi_n>}=
\left(\frac{\pi(n-\frac{1}{4}) }{\int^1_0  \sqrt{1-x^\alpha} dx } 
\right)^{-\frac{2}{\alpha+2}}
\left(\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}}\right)^{-1}
\left( \int_{\R_+}W +
\0\left(n^{-t }
 \right)  \right) ,\
\eea
where $t:=\min\{\frac{1}{9}, \frac{2(r-1)}{\alpha+2}\} >0$
\end{theorem}
\begin{corollary}
A straightforward consequence of (\ref{C_E}) is that the
 quantum density of probability
\[
p_Q(x):=\frac{1}{\sqrt{\int_{\R+} |\Psi|^2 }}|\Psi(x)|^2
\]
tends to the classical one
\[
p_C(x):=\frac{1}{\int^1_0 (1-y^\alpha)^{-\frac{1}{2}} dy}
\frac{\chi_{[0,1]}(x)}{\sqrt{1-x^\alpha}}
\]
in $L^1(\R_+)$ limit, or equivalently
\[
\int_{\R_+} |p_Q-p_C| \stackrel{E\to\infty} {\longrightarrow} 0
\]
\end{corollary}
\begin{Remark}
It is interesting that both integrals in (\ref{diagelements}) 
are important classical quantities. (This is the reason, why we didn't
write their explicit numerical values, although they are known.)\\
The first one
\[
\int^1_0  \sqrt{1-x^\alpha} dx
\]
is the  volume of area $\{(q,p)|E(q,p)\leq 1\}$ in phase space. $\\$
The second one 
\[
\int^1_0 \frac{dx}{ \sqrt{1-x^\alpha} }
\]
is proportional to the period of the classical motion with Energy $E=1$.
\end{Remark}
$\\$
Because there is only one index $n$ in this subsection, which 
labels the Energy $E_n$ according to the Bohr-Sommerfeld condition 
(\ref{E(n)}),
 we will use  scaling to replace $E_n$, 
which is assumed to be large, by the small parameter
$\hbar$.\\
The only reason of this scaling is the 
manipulation with $\hbar$ becomes more comfortable.
It is of course possible to pass through everything
without  scaling, like in the part dedicated to  offdiagonal
elements where it is necessary, because of the presence of both
indeces $n,l$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%% Scaling %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsubsection{\underline {Scaling}} $\\$
Consider equation
\[	
(-\Delta +x^\alpha-E)\Psi(x)=0 .\
\]
For any $\lambda$ this is equivalent to the equation
\[
(-\lambda^{-2-\alpha} \Delta + x^\alpha-\lambda^{-\alpha}E
)\Psi(\lambda x)=0
\]
We define the scaling by the relations 
\be
\lambda^{-\alpha}E:=1 \;, \hskip10pt
\lambda^{-2-\alpha}:=\hbar^2 \;, \hskip10pt 
\Phi(x):= \lambda^\frac{1}{2}\Psi(\lambda x)
\ee
which means that our problem (\ref{rice}) is transformed to 
\bea \label{scaling}
(-\hbar^2\Delta+x^\alpha-1)\Phi(x)=0 \; , \hskip10pt  
E=\hbar ^{-\frac{2\alpha}{2+\alpha}} \; , \hskip10pt
\Phi(x)= \hbar^{-\frac{\beta}{2}}\Psi(\hbar^{-\beta}x)\; , \hskip10pt
\beta:=\frac{2}{\alpha+2}\;  .\
\eea
We see that investigation of  the behavior with respect
to the large parameter $E$ is equivalent to the one with respect to
 small parameter $\hbar$.\\ 
By rescaling (\ref{presny}), (\ref{error}) we conclude that
exact $L^2(\R+)$ solution of equation
\bea 
\label{eqpsi}
(-\hbar^2 \Delta +V)\Phi=0
\eea
can be written in the following form (we remind the definition of  $V$
and $\tilde s$ (\ref{V,tilde s}) ):
\bea 
\label{sol}
\Phi(x)= \hbar^\frac{\alpha-4}{3(\alpha+2)}
\left( \frac{\tilde s(x)}{V(x)}\right)^\frac{1}{4}
(\Ai(\hbar^{-\frac{2}{3}} \tilde s(x) ) +\varepsilon(\hbar,x) )  ,\
\eea
where the error term $\varepsilon$
is controlled by the inequality
\bea
\label{errorh}
|\varepsilon(\hbar,x) | \leq  K  \hbar^\frac{7}{6}
|\tilde s(x)|^{-\frac{1}{4}} \exp(-\frac{2}{3}\tilde s_+(x)^\frac{3}{2} )
\eea
(Again we defined  $\tilde s_+:=\max(\tilde s,0)$).\\
For simplicity, we define new function
\bea\label{tildePhi}
\tilde\Phi(x):=\left( \frac{\tilde s(x)}{V(x)}\right)^\frac{1}{4}
(\Ai(\hbar^{-\frac{2}{3}} \tilde s(x) ) +\varepsilon(\hbar,x) ) 
\implies \Phi= \hbar^\frac{\alpha-4}{3(\alpha+2)} \tilde\Phi
\eea
The quantities we are interested in, are transformed by
\bea \label{Ctrans}
&& <\Psi,\; \Psi>=<\Phi,\; \Phi>=\hbar^\frac{2(\alpha-4)}{3(\alpha+2)}
<\tilde \Phi,\; \tilde \Phi>=E^\frac{4-\alpha}{3\alpha}
<\tilde \Phi,\; \tilde \Phi> \\
 \label{eletrans}
&&
\frac{<W\,\Psi,\,\Psi>}{<\Psi,\, \Psi>}=
\frac{<W(\frac{.}{\hbar^\beta})\,\Phi,\,\Phi>}{<\Phi,\, \Phi>}=
\frac{<W(\frac{.}{\hbar^\beta})\,\tilde \Phi,\,\tilde \Phi>}{<\tilde\Phi,\, \tilde\Phi>}
\eea
In this section we will not need the estimate of $\varepsilon'$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Normalization constant %%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\subsubsection{\underline{Normalization constant}}
\begin{theorem} \label{normtheorem}
For the function $\tilde \Phi$ given by (\ref{tildePhi}) it is true that
\be 
\int^\infty _0 |\tilde \Phi|^2
=\frac{\hbar^\frac{1}{3}}{2\pi}
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}} +\0 (\hbar^\frac{4}{9}) .\
\ee
\end{theorem}

\begin{Proof}
Let us define
\bea \label{phi}
\phi(x):=(\frac{\tilde s(x)}{V(x)})^\frac{1}{4}\Ai(\hbar^{-\frac{2}{3}}
\tilde s(x))
\eea
We would like to compute  leading term of integral
\b \label{integral}
I(\hbar):=\int^\infty _0 |\phi(x)|^2 dx .\ 
\e
This expression looks very complicated, for example for $\alpha=2$ is 
the function $\tilde s$ equal to 
\[
\tilde s(x)=-\left(\frac{3}{4} (\arcsin x +x \sqrt{1-x^2})
 \right)^\frac{2}{3}  \, , \; \rm{for}\;\; x\leq 1 .\
\]
What is important, is that the function $\tilde s$ maps the point $1$
to the origin. But the point $1$ is nothing but the turning point
(of the classical problem) of our rescaled potential $V=x^\alpha-1$.\\
Now because we are interested in the $\hbar\to 0$ regime, we conclude
that there are three imported area of very different behavior of 
$\Ai(\hbar^{-\frac{2}{3}}\tilde s)$ (we suppose that
$0<\gamma<\frac{2}{3}$).

$1)$ for $x\in [\,0,1-\hbar^\gamma]$ is the argument of Airy function 
 $\hbar^{-\frac{2}{3}}\tilde s(x)$ mapped to  the ``neighborhood''

$\;\;$ of $-\infty$,  $[-\hbar^{-\frac{2}{3}}K,
 -\hbar^{-\frac{2}{3}+\gamma}K]$. We use
asymptotic (\ref{ai-infty}), $\Ai$ is fast oscillating. 

$2)$  $x\in [1-\hbar^\gamma, 1+\hbar^\gamma ]$. This small window of
 ``regular'' regime of $\Ai$, where the function 

$\;\;$ is only bounded, is  in fact mapped to the whole line. 
But the influence to integrals 

$\;\;$ like $I(\hbar)$ is controlled by $\hbar^\gamma$ and can be suppressed.

$3)$  $x\in [1-\hbar^\gamma, \infty)$  the argument of Airy function, 
 $\hbar^{-\frac{2}{3}}\tilde s(x)$ is  mapped to the

neighborhood of $+\infty$, $[\hbar^{-\frac{2}{3}+\gamma}K, +\infty)$. We use
asymptotic (\ref{Ai+infty}), $\Ai$ is exponentially decaying.


$\\$ Because of the exponential decay of $\Ai$ in the area $3)$ and
because of the fact that we can control the influence of the ``window'' 
(area $2)$ by choosing $\gamma$, we will see that the main  
contribution of  integral (\ref{integral}) will 
come from the oscillating area $1)$.\\
This strategy will be used throughout all computations, but the use of 
exponential decay will be necessary only in the part dealing with
off-diagonal matrix elements. If not necessary we will omit
it to simplify the text.\\
The ``regular'' behavior of function $\phi$ in the neighborhood
of the turning point is the reason why we chose this type of model 
solution instead of 
usual WKB. It is well-known that usual WKB (see \cite{Olver} Chapter 6)
is diverging in the turning point. Choosing our type of model solution
we can omit this fact and following the explained strategy we obtain in fact
the usual WKB as asymptotic of $\Ai$ (see (\ref{ai-infty}), 
(\ref{Ai+infty})).\\
$\\$
To compute the leading term of normalization constant it is sufficient to
split  integral \(I(\hbar)\) into following three parts (as we said
we will omit exponential decay in the \mbox{area $3)\;$).}
\[
I(\hbar)= \int^\infty _1 + \int^1 _{1-\hbar^\gamma} 
+ \int^{1-\hbar^\gamma} _0 ,\
\]
where \(\gamma>0\) will be defined later.\\
Let us introduce an auxiliary function
\bea \label{galpha}
g_\alpha (x):=\frac{\tilde s(x)}{V(x)}.
\eea
It is easy to see that \(g_\alpha\) is bounded on \( \R ^+ \) for 
$\alpha>1$.
So we may estimate (using \(\tilde s \,{\tilde s}'^2 =V \) )
\bea
\int^\infty _1 |\phi(x)|^2 dx & = & \int^\infty _1 
(\frac{\tilde s(x)}{V(x)})^\frac{1}{2}\Ai^2(\hbar^{-\frac{2}{3}}
\tilde s(x))dx \nn \\
& = &  \int^\infty _1 
(\frac{\tilde s(x)}{V(x)})\Ai^2 \left(\hbar^{-\frac{2}{3}}
\tilde s(x)\right)
{\tilde s}'(x)dx \nn \\
&\leq & \sup_ { [1,+\infty)} g_\alpha \int^\infty _1 
\Ai^2 \left(\hbar^{-\frac{2}{3}}
\tilde s(x)\right)
{\tilde s}'(x)dx \nn \\
\label{infinity}
& = & \hbar^\frac{2}{3} \sup_ { [1,+\infty)} g_\alpha
\int^\infty _0 \Ai^2(z) dz.
\eea
Because of the boundedness of $g_\alpha $ and  Airy function \(\Ai\) 
we may easily conclude
\bea \label{by turn}
 \int^1 _{1-\hbar^\gamma} |\phi(x)|^2 dx =  \0 (\hbar^\gamma)
.\
\eea
Remaining (and dominant) is this term 
\[
\int^{1-\hbar^\gamma} _0 (\frac{\tilde s(x)}{V(x)})^\frac{1}{2}
\Ai^2(\hbar^{-\frac{2}{3} }\tilde s(x))dx.\
\]
We will follow the explained strategy to the area $1)$. An easy
computation gives
\[
\tilde s(1-\hbar^\gamma)= -(\frac{3}{2} \int^1 _{1-\hbar^\gamma}
 \sqrt{1-y^\alpha}dy) ^\frac{2}{3}
 =  \,\, -(\frac{3}{2} \sqrt\alpha  )^\frac{2}{3}\hbar
^\gamma + \0(\hbar^\frac{4\gamma}{3}).\
\]
So, as we said, the argument lies in fact in a ``neighborhood''
of \(- \infty \). As was announced, we will use asymptotics
(\ref{ai-infty}) of $\Ai$. \\
Its application yields 
\bea
&& \int^{1-\hbar^\gamma}_0 (\frac{\tilde s(x)}{V(x)})^\frac{1}{2}
 \Ai^2(\hbar^{-\frac{2}{3} }\tilde s(x))dx    \nn \\
 & &  = \frac{\hbar^\frac{1}{3}} {\pi}
\int^{1-\hbar^\gamma} _0 
\frac{\cos^2(\frac{1}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{4})} 
{\sqrt{1-x^\alpha}}
(1+\0 (\hbar^{1-\frac{3}{2}\gamma} ) )dx \nn \\
& & =
\frac{\hbar^\frac{1}{3}} {\pi}
\int^{1-\hbar^\gamma} _0 \frac{1}{\sqrt{1-x^\alpha}}
 \cos^2(\frac{1}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{4})\;dx
+  \nn
 \\ \label{airaprox}
& & \,  +\,\,  \0 (\hbar^{1-\frac{3}{2}\gamma} )\; \hbar^\frac{1}{3} 
\int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}} 
\eea
The second term is of order 
$\hbar^\frac{1}{3}\0(\hbar^{1-\frac{3}{2}\gamma})$. \\
Because the argument of $\cos^2$ is large, it causes fast oscillating.
We will use  
identity \mbox{$ \cos^2(a) =\frac{1}{2}(1+\cos(2a))$} to compute 
``average'' and the rest, which is negligible thanks to the fast oscillation.
\be
&&\int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}}
 \cos^2(\frac{1}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{4})\\
& &=\frac{1}{2}\int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}}+
\frac{1}{2}\int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}}
\cos(\frac{2}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{2})
\ee
From the first integral will come the leading term
\bea \label{leader}
\int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}} =
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}}
-\int^1 _{1-\hbar^\gamma}  \frac{dx}{\sqrt{1-x^\alpha}} 
= \int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}} 
+\0(\hbar^\frac{\gamma}{2})
\eea
To suppress the second one, we will integrate by parts
\be
&& \int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}}
\sin(\frac{2}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy ) 
 = \int^{1-\hbar^\gamma} _0 \sin( \frac{2}{\hbar} f(x)) 
\frac{f'(x)}{(f'(x))^2}dx \\
&& = - \frac{\hbar}{2} 
\left[\frac{\cos(\frac{2f}{\hbar})}{(f'^2)} \right]^{1-\hbar^\gamma} _0+
\frac{\hbar}{2} 
\int^{1-\hbar^\gamma} _0 \cos( \frac{2}{\hbar} f(x)) 
\frac{f''(x)}{(f'(x))^3}dx
\ee
We defined an auxiliary function 
\b \label{f}
f(x):= \int^1 _x \sqrt{1-y^\alpha}dy  .\
\e
The fact that
\[
\frac{f''(x)}{(f'(x))^3}=-\frac{\alpha x^{\alpha-1}}{2(1-x^\alpha)^2} ,\
\]
allows us to see 
\bea
\label{cos}
\int^{1-\hbar^\gamma} _0 \frac{dx}{\sqrt{1-x^\alpha}}
\cos(\frac{2}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{4})=
\0 (\hbar^{1-\gamma})+\0 (\hbar^{1-2\gamma})
\eea
Collecting all these informations (\ref{infinity}) - (\ref{cos}) yields
\[
I(\hbar)=\frac{\hbar^\frac{1}{3}}{2\pi}
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}} +\0 (\hbar^\frac{2}{3})+
 \0 (\hbar^\gamma)+\hbar^\frac{1}{3} (\0 (\hbar^\frac{\gamma}{2})
+\0 (\hbar^{1-2\gamma}))
 \]
Now choosing $\gamma:=\frac{4}{9}$, we finally obtain the leading term
of normalization constant
\bea
\label{normmodel}
I(\hbar)=
 \int^\infty _0 
(\frac{\tilde s(x)}{V(x)})^\frac{1}{2}\Ai^2(\hbar^{-\frac{2}{3}}
\tilde s(x))dx=
\frac{\hbar^\frac{1}{3}}{2\pi}
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}} +\0 (\hbar^\frac{4}{9})
\eea \\
To finish the proof we have to show 
that the contribution of error to 
normalization constant for the exact solution of (\ref{eqpsi}) is negligible.
\[
\int^\infty _0 |\tilde \Phi|^2= 
\int^\infty _0 |\phi|^2 +\int^\infty _0 
(\frac{\tilde s}{V})^\frac{1}{2} |\varepsilon|^2
+2\int^\infty _0 (\frac{\tilde s}{V})^\frac{1}{2}  \,\,
Re (\overline{\Ai(\hbar^{-\frac{2}{3}}\tilde s)  } \,\,  \varepsilon)
\]
Thanks to the boundedness of Airy function and 
 $g_\alpha$ (see (\ref{galpha})),  we have easily 
(with help of (\ref{errorh})) 
\bea
&& \label{inteps}
\int^\infty _0 (\frac{\tilde s}{V})^\frac{1}{2} |\varepsilon|^2 
= \0(\hbar^\frac{7}{3})\\
&& \label{intaieps}
\int^\infty _0 (\frac{\tilde s}{V})^\frac{1}{2}  \,\,
Re (\overline{\Ai(\hbar^{-\frac{2}{3}} \tilde s)  } \,\,  \varepsilon) =
\0(\hbar^\frac{7}{6})
.\
\eea
So finally, using the result (\ref{normmodel}) 
we have this expression of normalization constant for arbitrary 
$\alpha>1$
\b \label{intphitilde}
\int^\infty _0 |\tilde \Phi|^2
=\frac{\hbar^\frac{1}{3}}{2\pi}
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}} +\0 (\hbar^\frac{4}{9}) .\
\e
This is nothing but the statement of  Theorem
\ref{normtheorem} . \end{Proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% Diagonal matrix elements %%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsubsection{\underline{Diagonal matrix elements}} $\\$
In this section we will follow again  the strategy explained in the
beginning of the Proof of Theorem \ref{normtheorem} to
compute leading term of diagonal matrix elements. \\
Because the order of the leading term will be 
$\hbar^{\frac{1}{3}+\frac{2}{2+\alpha}}$, what is more then in the
case of normalization constant, we will have to assume some decay
properties of the perturbation $W$, to estimate better the rest.\\
First of all let us formulate the result of this subsection.

\begin{theorem} \label{3.5}
Suppose that the perturbation $W$ satisfies decay properties
(\ref{asumW}).\\
Then for ``rescaled diagonal matrix elements'' holds true that
\bea \label{W tilde Phi}
<W(\frac{.}{\hbar^\beta}) \, \tilde \Phi, \,\tilde \Phi  > 
= \frac{\hbar^{\frac{1}{3}+\beta }}{2\pi } \left( 
\int^\infty _0 W +
\0\left(\hbar^{\min\{\beta(r-1),\,\frac{\alpha}{2+\alpha}\}} 
 \right) \right)
\eea
\end{theorem}


$\\$
Under assumptions (\ref{asumW}) the main contribution of 
(\ref{W tilde Phi}) comes from the neighborhood of the origin. $\\$
To make the proof more easy to follow we split the computation into
several parts.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%% Tail estimate %%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\underline{Tail estimate.} $\\$
In this part we will estimate the contribution of tail ($a<1$)
\be
\int^\infty _a W(\frac{x}{\hbar^\beta}) |\tilde\Phi(x)|^2 dx &\leq&
\sup\{{W(\frac{x}{\hbar^\beta}), \,\, x>a}\}  \int^\infty _a |\tilde\Phi|^2 \\
&\leq& C_\hbar ^{-2} \sup\{{W(x),\, x>\frac{a}{\hbar^\beta} }\} \\
&\leq&  C_\hbar ^{-2} \sup\{{ <x>^{-r},  x>\frac{a}{\hbar^\beta}} \} 
\| <x>^rW\|_\infty \\
&=& C_\hbar ^{-2} \frac{\hbar^{\beta r}}{ (a^2+\hbar^{2\beta})^\frac{r}{2}  } 
\| <x>^rW\|_\infty
\ee
We define normalization constant 
\[
C^{-2} _\hbar :=\int^\infty _0 |\tilde\Phi|^2
\]
So the result is
\bea
\label{tail}
\int^\infty _a W(\frac{x}{\hbar^\beta}) |\tilde\Phi(x)|^2 dx
\leq
\frac{ C_\hbar ^{-2}}{a^r}  \hbar^{\beta r} \| <x>^rW\|_\infty
\eea

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%% Error contribution%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\\$
\underline{Error contribution.} $\\$
We know that $L^2(\R_+)$  solution of (\ref{eqpsi}) can be expressed in the
form (\ref{sol}), with error $\varepsilon(\hbar,x)$ controlled by 
(\ref{errorh}).\\
 Taking into account the definition (\ref{tildePhi})
we will first of all  use estimate (\ref{tail}) to get
\be
<W(\frac{.}{\hbar^\beta}) \, \tilde\Phi, \,\, \tilde\Phi  > &=&
\int^a _0 W(\frac{x}{\hbar^\beta}) |\tilde\Phi(x)|^2 dx +
\int^\infty _a W(\frac{x}{\hbar^\beta}) |\tilde\Phi(x)|^2 dx \nn
 \\
&=&  \int^a _0 W(\frac{x}{\hbar^\beta}) |\tilde\Phi(x)|^2 dx +
\0(\hbar^{\beta r})C_\hbar^{-2}
\ee
Now we will show that the influence of error to the quantity
$<W(\frac{.}{\hbar^\beta}) \, \tilde\Phi, \,\, \tilde\Phi )>$ is of order 
$\hbar^\frac{13}{6}$ ($\phi$ is defined by (\ref{phi})).
\be
\int^a _0 W(\frac{.}{\hbar^\beta}) |\tilde\Phi|^2 &=&
\int^a _0 W(\frac{.}{\hbar^\beta})  |\phi|^2 +
\int^a _0 W(\frac{.}{\hbar^\beta})
 (\frac{\tilde s}{V})^\frac{1}{2} |\varepsilon|^2 \\
&&+2\int^a _0 W(\frac{.}{\hbar^\beta})   (\frac{\tilde s}{V})^\frac{1}{2}  \,\,
Re (\overline{\Ai(\hbar^{-\frac{2}{3}}\tilde s)  } \,\,  \varepsilon)
\ee
Similarly to (\ref{inteps}), using the boundedness of $W$ 
\be
\int^a _0 W(\frac{.}{\hbar^\beta}) 
(\frac{\tilde s}{V})^\frac{1}{2} |\varepsilon|^2 = \0(\hbar^\frac{7}{3})
.\
\ee
For the third integral the power $\frac{7}{6}$ of $\hbar$ in 
(\ref{intaieps}) is not 
sufficient. We will do the estimate more carefully
\be
&&\left |
 \int^a _0    W(\frac{.}{\hbar^\beta})
 (\frac{\tilde s}{V})^\frac{1}{2}  \,\,
Re (\overline{\Ai(\hbar^{-\frac{2}{3}}\tilde s)  } \,\,  \varepsilon)  \right|
\leq 
K \hbar^\frac{7}{6} \int^a _0 |\tilde s(x)|^{ -\frac{1}{4} }
|\Ai(\hbar^{- \frac{2}{3} }\tilde  s(x))|dx \\
\,\,\,\, & &=
K \hbar^\frac{7}{6} \hbar^\frac{1}{6}
 \int^a _0 |\tilde s(x)|^{ -\frac{1}{2} }
\left |
\cos(\frac{1}{\hbar}\int^1 _x \sqrt{-V} - \frac{\pi}{4})
\right| 
dx \,\,\, (1+\0(\hbar))  =\0(\hbar^\frac{4}{3}).\ 
\ee
So we obtain
\bea
\label{ercontrib}
\int^a _0 W(\frac{.}{\hbar^\beta}) |\tilde\Phi(x)|^2 =
\int^a _0 W(\frac{.}{\hbar^\beta})  |\phi|^2 +\0(\hbar^\frac{4}{3})
\eea

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% Leading term %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\underline {The leading term of  
$<W(\frac{.}{\hbar^\beta}) \, \tilde\Phi, \,\, \tilde\Phi ) >$.}
$\\$
Now we will concentrate on the most important term, from which the leading 
term will result. We will use the same tricks like in (\ref{airaprox}) -
(\ref{cos}).
\bea
&&\int^a _0  W(\frac{x}{\hbar^\beta})
 (\frac{\tilde s(x)}{V(x)})^\frac{1}{2}
\Ai^2 (\hbar^{-\frac{2}{3}}\tilde  s(x))dx  \nn  \\
& & =  
\frac{\hbar^\frac{1}{3}} {\pi}
\int^a _0 W(\frac{x}{\hbar^\beta}) 
\frac{\cos^2(\frac{1}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{4})} 
{\sqrt{1-x^\alpha}}
(1+\0 (\hbar) )dx   \nn \\
&&=\frac{\hbar^\frac{1}{3}} {2\pi} \int^a _0 W(\frac{x}{\hbar^\beta})
\frac{dx}{\sqrt{1-x^\alpha}}  \nn
\\ \label{rozklad}
&& +  \frac{\hbar^\frac{1}{3}} {2\pi}
\int^a _0 W(\frac{x}{\hbar^\beta}) 
\frac{dx}{\sqrt{1-x^\alpha}}
\sin(\frac{2}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy )
+\0(\hbar^\frac{4}{3}) .\
\eea
The first integral extends the window into whole line
\bea
&& \int^a _0 W(\frac{x}{\hbar^\beta}) \frac{dx}{\sqrt{1-x^\alpha}} =
\hbar^\beta \int^\infty  _0 W -\int^\infty_a
W(\frac{x}{\hbar^\beta})dx +\int^a_0
W(\frac{x}{\hbar^\beta}) \left( \frac{1}{\sqrt{1-x^\alpha}}-1 \right)dx
 \nn \\
&&= \hbar^\beta \int^\infty  _0 W -K\|W<x>^r\|_\infty\;
\hbar^{\beta\,r} \nn \\
&& \hskip10pt +K\|W<x>^u\|_\infty 
\int^a_0
  \frac{x^\alpha}{\left(1+(\hbar^{-\beta }x)^2\right)^\frac{u}{2} } 
\frac{dx}{(1+\sqrt{1-x^\alpha} )\sqrt{1-x^\alpha} }
 \nn \\
\label{intW}
&&= \hbar^\beta \int^\infty  _0 W(x) dx +
\0\left(\hbar^{\beta\,r)}\right)+\0\left(\hbar^{\beta \,u}\right)=
\0\left(\hbar^{\beta \,u}\right)  .\
\eea
We chose 
\b \label{u}
u:=\min \{r,1+\frac{\alpha}{2} \} .\
\e 
We will integrate the second term in (\ref{rozklad}) by parts
\bea
&& \int^a _0 W(\frac{x}{\hbar^\beta})
\frac{dx}{\sqrt{1-x^\alpha}}
\sin(\frac{2}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy ) 
 = \int^a _0  W(\frac{x}{\hbar^\beta})
\sin( \frac{2}{\hbar} f(x)) 
\frac{f'(x)}{(f'(x))^2}dx  \nn \\
&& = - \frac{\hbar}{2} 
\left[W(\frac{.}{\hbar^\beta})
\frac{\cos(\frac{2f}{\hbar})}{(f'^2)} \right]^a _0+ 
\hbar 
\int^a _0 \cos( \frac{2}{\hbar} f(x)) 
\frac{f''(x)}{(f'(x))^3} W(\frac{x}{\hbar^\beta})dx  \nn \\
\label{neco}
&& \hskip15pt  - \hbar^{-\beta} \,\frac{\hbar}{2} 
\int^a _0 \cos( \frac{2}{\hbar} f(x))  \frac{W'(\frac{x}{\hbar^\beta})}
{(f'(x))^2}dx
\eea
The auxiliary function $f$ is again defined by (\ref{f}).\\
We estimate roughly, (using (\ref{asumW}))
\[
\left| 
\hbar^{-\beta}  
\int^a _0 \cos( \frac{2}{\hbar} f(x))  \frac{W'(\frac{x}{\hbar^\beta})}
{(f'(x))^2}dx
\right| \leq K \hbar^{-\beta}  
\int^a _0 \left| W'(\frac{x}{\hbar^\beta}) \right| dx=
K \int^{a\hbar^{-\beta}}        _0 \left| W'(x) \right|dx =K
\]
Now because all functions in (\ref{neco}) are bounded on $[0,a]$ we have
\bea
\label{2term}
\int^a _0 W(\frac{x}{\hbar^\beta})
\frac{dx}{\sqrt{1-x^\alpha}}
\cos(\frac{2}{\hbar}\int^1 _x \sqrt{1-y^\alpha}dy - \frac{\pi}{4}) =\0(\hbar)
\eea
Putting (\ref{rozklad}), (\ref{intW}), (\ref{u}) and (\ref{2term})
together yields
\bea
\label{leadW}
&& \int^a _0  W(\frac{x}{\hbar^\beta}) (\frac{\tilde s(x)}{V(x)})^\frac{1}{2}
\Ai^2(\hbar^\frac{-2}{3}\tilde s(x))dx \nn \\
&& = \frac{\hbar^{\frac{1}{3}}}{2\pi } \left( 
\hbar^\beta \int^\infty _0 W +
\0\left(\hbar^{\beta \,u}\right)
 \right) 
+\0(\hbar^\frac{4}{3}) \nn \\
&&= \frac{ \hbar^{\beta+\frac{1}{3}}}{2\pi} \left( \int^\infty _0 W +
\0\left(\hbar^{\min \{ \beta(r-1),\,
\frac{\alpha}{\alpha+2}\} }\right) \right)
\eea


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% Proof of theorem3.5 %%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\underline {Proof of Theorem \ref{3.5}.}
$\\$
\begin{Proof}
If we collect all results (\ref{intphitilde}), (\ref{tail}), (\ref{ercontrib}) and
(\ref{leadW}),
we obtain what we wanted to prove
\be
<W(\frac{.}{\hbar^\beta}) \, \tilde\Phi, \,\, \tilde\Phi  > &=&
\int^a _0 W(\frac{.}{\hbar^\beta}) \,|\tilde\Phi|^2 +
\int^\infty _a W(\frac{.}{\hbar^\beta}) \, |\tilde\Phi|^2 \\
&=& \int^a _0 W(\frac{.}{\hbar^\beta})  |\phi|^2 +\0(\hbar^\frac{4}{3})
+C^{-2} _\hbar \0(\hbar^{\beta r})
\\
&=& \frac{\hbar^{\frac{1}{3}+\beta }}{2\pi } \left( 
\int^\infty _0 W +\0\left(\hbar^{\min\{\beta(r-1),\,\frac{\alpha}{2+\alpha}\}} 
\right)  \right) 
\ee
\end{Proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%% Proof of theorem 3.1 %%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsubsection{\underline{Proof of Theorem \ref{3.1}}}
$\\$
Now if we collect proven statements of Theorems 
\ref{normtheorem} and \ref{3.5} 
and use (\ref{Ctrans}), (\ref{eletrans}), the relationship (\ref{scaling}) 
between $E$ and $\hbar$, 
we may conclude that Theorem \ref{3.1} is
\mbox{proven. \QED}


%\newpage

%\input offdiag

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%% offdiag %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%  Offdiagonal elements %%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	

\subsection{Offdiagonal elements}	
$\\$
For understanding the behavior of all elements in matrix our precise analysis
of diagonal is not much important because the diagonal is
only ``a set of zero measure''.\\
There are several main directions in our matrix, the first one is parallel
to the diagonal, the second one is normal to the diagonal and the
third one is supposed to go through columns with a fixed row.\\
Our guess is that the behavior of elements close to the diagonal is more
or less the same as on the diagonal. We will formulate this result
with the help of Cauchy-Schwartz inequality. This estimate will be valid
for all elements, but  elements far from the diagonal decay much
faster, so we will use other, direct method to improve the decay
estimate for elements far from the diagonal.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Elements close to the diagonal %%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
\subsubsection{\underline{Elements close to the diagonal}}
$\\$
By Cauchy-Schwartz inequality we mean
\be
|F(a,b)|^2\leq q_F(a)\cdot q_F(b) ,\
\ee
where $F(.\,,.)$ is a sesquilinear form and $q_F(a):=F(a,a)$ is its
corresponding quadratic form which is supposed to be positive.\\
If we apply this inequality to the  result of diagonal case we obtain
the estimate
\begin{theorem}
Let $\alpha>1$, $\Psi_n$, $\Psi_l$ are two $L^2(\R_+)$ solutions of 
(\ref{rice}) with corresponding Energies $E_n$ and $E_l$ which depend
on natural numbers $n, l$ according to the prescription (\ref{E(n)}).\\
Suppose moreover that a function $W$ satisfies the decay properties
 (\ref{asumW}).\\
Then all matrix elements can be estimated by
\be
W_{n,\,l}:=\frac{<W\,\Psi_n,\,\Psi_l>}{\sqrt{<\Psi_n,\, \Psi_n>}\,
\sqrt{<\Psi_l,\, \Psi_l>} }\leq K 
(n\,l)^{-\frac{1}{\alpha+2}}
\ee
\end{theorem} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%% Elements far from to the diagonal %%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\subsubsection{\underline{Elements far from to the diagonal}}
$\\$
We will investigate elements for which 
\be
 n>\delta\,l \; , \hskip10pt \delta>0 .\
\ee
directly, by estimating  integral $<W\,\Psi_n,\,\Psi_l>$.\\ 
Because of the presence of different Energies $E_n$ and $E_l$ is our
scaling trick not possible, but we lose by doing it just a bit of comfort.\\
Similarly to the part dealing with diagonal elements we will divide the
area of $\int_{\R_+} W \overline{\Psi}_n\Psi_l$ into two parts, a window
and a tail. By choosing a boundary of them between the turning points of
functions $\Psi_n$ , $\Psi_l$ we get $\0(E_n ^{-\infty})$ estimate
of the tail! This is caused by exponential decay of the function $\Psi_l$
above its turning point. Again the only mathematical tool we will use is
integration by parts. \\
Let us choose (without loss of generality) the boundary between 
the window and the tail by
\[
a_n:=(\frac{E_n}{2})^\frac{1}{\alpha}
\]
and consider such $n$ and $l$ for which 
\bea \label{gap}
E_n > 4 E_l .\
\eea
Of course, instead of $4$ we can choose
$\delta^\frac{2\alpha}{\alpha+2} $.
Again we will first of all formulate the result.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%% Result %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\underline {Result.}

\begin{theorem} \label{offele}
Consider the case $n>\delta\, l$ for some  positive number $\delta$. 
$\alpha>1.$ \\
Assume that the perturbation $W$ satisfies (\ref{asumW})
and in addition that $ W''$  is defined and bounded  almost everywhere
on $\R_+$ .\\
Then  matrix elements $W_{n,\,l}$ can be estimated from above by
\be
W_{n,\,l}:=\frac{\int_{\R_+} W\, \overline{\Psi}_n \Psi_l}
{\sqrt{\int_{\R_+} |\Psi_n| ^2  }
\sqrt{\int_{\R_+} |\Psi_l| ^2  }}= 
\0\left( n^{-\frac{\alpha+1}{\alpha+2}}
l^{\frac{\alpha-1}{\alpha+2 }}  \right)=
\0\left( (n\,l)^{-\frac{1}{\alpha+2}} \;
(\frac{l}{n} )^{\frac{\alpha}{\alpha+2 }} \right) 
\ee
where $\Psi_n$ is the Dirichlet $L^2(\R_+)$ solution of equation
\be
(-\Delta +x^\alpha  )\Psi_n(x)=E_n \Psi_n(x) \; .\
\ee
So the relationship between the energy $E_n$ and  the natural number  $n$ is
given by the Bohr-Sommerfeld quantization condition (\ref{E(n)}).
\end{theorem}
$\\$
The proof of this theorem will be again given in the following text
and finished in the very last part of this subsection.\\
First of all we show that the main contribution comes from  window.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% Tail estimate %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\underline {Tail estimate.} $\\$
Set 
\be
\phi_n (x):= (\frac{s_n(x)}{x^\alpha -E_n} )^{\frac{1}{4} }
\Ai (s_n(x)).
\ee
In this part we will show that the contribution of the tail 
\be
\int^\infty _{a_n} W(x) \phi_n(x) \phi_l(x) dx
\ee
is  negligible.
\\
Using (\ref{Ai+infty}) we obtain 
\bea
&& \int^\infty _{a_n} W(x) \left (\frac{s_l(x)s_n(x) }
{(x^\alpha -E_l)(x^\alpha -E_n)  }  \right)^{\frac{1}{4} }
\Ai (s_l(x))\Ai (s_n(x))dx   \nn \\
&& \;\; = \int^\infty _{a_n} W(x) \left (\frac{s_n(x) }
{x^\alpha -E_n  }  \right)^{\frac{1}{4} } 
\frac{\Ai (s_n(x)) }{2 \sqrt\pi \sqrt[4]{|E_l-x^\alpha| } }
\exp\left(- \int^x _{E_l^\frac{1}{\alpha}}  \sqrt{y^\alpha-E_l} \; dy
\right) dx  \nn \\
&& \hskip40pt \times \left(1+ \0( E_n ^{-\frac{\alpha+2}{2\alpha}}) 
\right)  \nn \\
\label{tailn} && \;\; =\0( E_n ^{-\infty}) 
\eea
because we can estimate with the help of (\ref{gap})
\bea
\label{s_l(a_n)}
\int^{a_n} _{E_l^\frac{1}{\alpha}}  \sqrt{y^\alpha-E_l} \; dy=
E_l^{\frac {\alpha+2}{2\alpha}}
\int^{(\frac{E_n}{2E_l})^\frac{1}{\alpha} } _1 \sqrt{y^\alpha-1} \;dy
\geq K  E_n ^{-\frac{\alpha+2}{2\alpha}} 
\eea


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%% Error estimate %%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%	

$\\$
\underline {Error estimate.}
$\\$ 
 The results (\ref{error}), (\ref{error'}) 
will be often used in this part. The technique we
will apply  will be similar like in the tail estimate paragraph. \\
We shall treat the integral
\be
\int^\infty _0 W(x)  \left( \frac{s_l(x)s_n(x)}{(x^\alpha -E_l)(x^\alpha -E_n)}
\right) ^\frac{1}{4}
\biggl( \bar{\varepsilon_l}(x) \Ai(s_n(x)) + \varepsilon_n (x) \Ai(s_l(x)) +
\bar{\varepsilon _l} (x) \varepsilon_n(x)
\biggr) dx
\ee
The last term is the smallest one
\bea
\label{eps^2}
&&  \left| 
\int^\infty _0 W(x)  \left( \frac{s_l(x)s_n(x)}{(x^\alpha -E_l)(x^\alpha -E_n)}
\right) ^\frac{1}{4} \bar{\varepsilon _l} (x) \varepsilon_n(x) dx
\right| \\ \nn
&& \;\;  \leq K (E_l E_n )^{-\frac{\alpha+2}{2\alpha}} 
\int^\infty _0 |W(x)| \frac{ e^{-\frac{2}{3}s_{l+} (x)^\frac{3}{2}  }
e^{-\frac{2}{3}s_{n+} (x)^\frac{3}{2}  }  } 
{ \sqrt[4] {|x^\alpha-E_l| |x^\alpha-E_n| } } \, dx 
\\ \nn
&& \;\; \leq K (E_l E_n )^{-\frac{\alpha+2}{2\alpha} } 
\int^{2a_l} _0 | W(x) | \frac{dx} 
{ \sqrt[4] {|x^\alpha-E_l| |x^\alpha-E_n| }}   
+ \0(E_n ^{-\frac{3}{4} -\frac{1}{\alpha}} E_l ^{-\infty})
\\ \nn
&& \;\;\leq K \| W \|_\infty (E_l E_n )^{-\frac{\alpha+2}{2\alpha} } 
E_n ^{-\frac{1}{4}} \int^{2a_l} _0 \frac{dx}{ \sqrt[4] {|x^\alpha-E_l| }}
+ \0(E_n ^{-\frac{3}{4} -\frac{1}{\alpha}} E_l ^{-\infty})
\\ \nn
&& \;\; \leq K \| W \|_\infty E_n^{-\frac{3}{4}-\frac{1}{\alpha} }
E_l^{-\frac{1}{2}-\frac{1}{\alpha} -\frac{1}{4}+\frac{1}{\alpha} }
 \int^{2} _0 \frac{dx}{ \sqrt[4] {|x^\alpha-1| }}
+ \0(E_n ^{-\frac{3}{4} -\frac{1}{\alpha}} E_l ^{-\infty})
\\ \nn
&& \;\; = \0 \left( (E_n E_l)^{-\frac{3}{4}}E_n ^ { -\frac{1}{\alpha} } 
\right)
\eea	
The $\0(E^{-\infty}) $ results are obtained using exponential decay of
$\varepsilon$ (see (\ref{error})) and $\Ai$ (see (\ref{Ai+infty}))
and  with the help of (\ref{s_l(a_n)}). 
The second integral can be similarly estimated by
\bea
&& \left| \int^\infty _0 W(x)  \left( \frac{s_l(x)s_n(x)}{(x^\alpha -E_l)(x^\alpha -E_n)}
\right) ^\frac{1}{4}  \varepsilon_n (x) \Ai(s_l(x)) dx  \right|  \nn
\\
&& \;\; \leq  K  E_n ^{-\frac{\alpha+2}{2\alpha} }
\int^{a_n} _0  |W(x)|  \left( \frac{s_l(x)}{(x^\alpha -E_l)|x^\alpha -E_n|}
\right) ^\frac{1}{4} |\Ai (s_l(x))|dx +
 \0( E_n ^{-\infty}) \nn
\\
&& \;\; \leq  K E_n^{-\frac{3}{4}-\frac{1}{\alpha} }
\left(
 \int^{a_l} _0 \frac{|W(x)|dx}{ \sqrt[4] {|x^\alpha-E_l| }}
+\| \Ai \|_\infty 
 \sup_{[a_l,  \,  2a_l]} |W(x)| \int^{2a_l} _{a_l} 
 \left( \frac{s_l(x)}{x^\alpha -E_l}\right) ^\frac{1}{4} dx 
  +\0(E_l^{-\infty})\right)   \nn \\
&& \hskip20pt +  \0( E_n ^{-\infty}) \nn
\\
\label{trick}
&& \;\; = E_n^{-\frac{3}{4}-\frac{1}{\alpha} }  \left(
 \0 \left(E_l^{-\frac{1}{4} }
\right)
+\0 \left(E_l^{ -\frac{r}{\alpha}-\frac{1}{6}+\frac{7}{6\alpha} }
\right)
\right) + \0( E_n ^{-\infty}) \; ,\
\eea
where we used the fact that
\[
 \frac{s_l(x)}{x^\alpha -E_l} = E_l^{-\frac{2}{3}+\frac{2}{3\alpha} } 
u(\frac{x}{E_l ^\frac{1}{\alpha} })
\]
for a continuous  function  $u$. 
\\
To obtain sufficient estimate of the first integral, we will have to
integrate by parts,  using again estimates (\ref{error}),
(\ref{error'})  and the relation $s(x){s'}^2(x)=x^\alpha-E$.  
\bea
&&  \int^\infty _0 W(x)  \left( \frac{s_l(x)s_n(x)}
{(x^\alpha -E_l)(x^\alpha -E_n)} \right) ^\frac{1}{4} 
 \varepsilon_l (x) \Ai(s_n(x)) dx    \nn
\\
&& \;\; = -\int^{a_n} _0 W(x)  \left( \frac{s_l(x)}
{(x^\alpha -E_l)(E_n-x^\alpha)^3} \right) ^\frac{1}{4} 
 \varepsilon_l (x) \cos(f(x)) f'(x)dx +\0(E_n ^{-\infty})
\nn  \\
&& \;\; = -\left[ \frac{W(x)\sin(f(x)) \varepsilon_l(x) }{\sqrt{ s'_l(x)}
\sqrt[4]{E_n-x^\alpha} ^3 }\right]^{a_n} _0 
+ \int^{a_n} _0 
\frac{W'(x)\sin(f(x)) \varepsilon_l(x) }{\sqrt{ s'_l(x)}
\sqrt[4]{E_n-x^\alpha} ^3 } dx \nn \\
&& \;\;\;\;\;\; +  \int^{a_n} _0 
\frac{W(x)\sin(f(x)) \varepsilon_l(x) }{\sqrt{ s'_l(x)}
\sqrt[4]{E_n-x^\alpha} ^3 } \left(
\frac{\varepsilon_l'(x)}{\varepsilon_l(x)} - \frac{s''_l(x)}{2s'_l(x)} 
+\frac{3\alpha x^{\alpha-1}}{4(E_n-x^\alpha)} 
\right) dx + \0(E_n ^{-\infty})  \nn 
\eea
The auxiliary function $f$ is now defined by
\be
f(x):=  \int ^{E_n ^\frac{1}{\alpha} } 
_x  \sqrt{E_n -y^\alpha}dy -\frac{\pi}{4} \;,\, \;\;
f'(x)=- \sqrt{E_n -x^\alpha}
\ee
Let us investigate all terms 
\bea
\label{boundterm}
-\left[ \frac{W(x)\sin(f(x)) \varepsilon_l(x) }{\sqrt{ s'_l(x)}
\sqrt[4]{E_n-x^\alpha} ^3 }\right]^{a_n} _0 =
 \frac{W(0)\sin(f(0)) \varepsilon_l(0) }{\sqrt{ s'_l(0)}
E_n ^{\frac{3}{4} }} +  \0(E_n ^{-\infty}) =
\0(E_n ^{-\frac{3}{4} } E_l ^{-\frac{3}{4}
-\frac{1}{\alpha}} ) \; ,\
\eea
where we used again (\ref{error}).\\
The second term is estimated by (like in (\ref{eps^2}))
\bea
\label{holder}
&& \left| \int^{a_n} _0 
\frac{W'(x)\sin(f(x)) \varepsilon_l(x) }{\sqrt{ s'_l(x)}
\sqrt[4]{E_n-x^\alpha} ^3 } dx \right| \leq 
 E_n ^{-\frac{3}{4}} E_l^{-\frac{1}{2}-\frac{1}{\alpha}}
\left( K \int^{2a_l} _0 \frac{|W'(x)|dx}{ \sqrt[4] {|E_l-x^\alpha|}}
+\0(E_l ^{-\infty})
\right) \nn \\
&&= \0\left( E_n ^{-\frac{3}{4}} E_l ^{-\frac{3}{4}-\frac{1}{2\alpha} }
   \right)
\eea
Because of the H\"older inequality we may estimate
\be
\int^{2a}_0 |W'(x)g(\frac{x}{a})|dx \leq 
\sqrt{\| W' \|_\infty\| W' \|_{L^1(\R_+)}} \sqrt{a\int^2_0|g(x)|^2 dx} 
\ee
To treat the third integral it is useful to remark that (again with the
help of $s(x) {s'}^2(x)=x^\alpha-E$)
\bea
%\label{s}
\nn
s(x)=E^{\frac{1}{3}+\frac{2}{3\alpha} } \;
\tilde{s}'(\frac{x}{E^\frac{1}{\alpha} }) \\
\label{s'}
s'(x)=E^{\frac{1}{3}-\frac{1}{3\alpha} }\;
\tilde{s}'(\frac{x}{E^\frac{1}{\alpha} }) \\
%\label{s''}
\nn
s''(x)=\frac{V'-{s'}^3}{2ss'}=E^{\frac{1}{3}-\frac{4}{3\alpha} }\;
\tilde{s}''(\frac{x}{E^\frac{1}{\alpha} }) 
\eea
So using the same tricks and (\ref{error'}) we conclude again
\bea
&& \int^{a_n} _0 
\frac{W(x)\sin(f(x)) \varepsilon_l(x) }{\sqrt{ s'_l(x)}
\sqrt[4]{E_n-x^\alpha} ^3 } \left(
\frac{\varepsilon_l'(x)}{\varepsilon_l(x)} - \frac{s''_l(x)}{2s'_l(x)} 
+\frac{3\alpha x^{\alpha-1}}{4(E_n-x^\alpha)} 
\right) dx  \nn \\
\label{el ai n}
&& \;\;=  E_n ^{-\frac{3}{4} } \left(
\0(E_l^{-\frac{1}{4}-\frac{1}{\alpha } } ) +
\0(E_l^{-\frac{3}{4}-\frac{3}{2\alpha } } ) +
\0(E_n E_l^{\frac{1}{4}-\frac{3}{2\alpha } } )
\right)= \0(  E_n ^{-\frac{3}{4} } E_l^{-\frac{1}{4}-\frac{1}{\alpha } } )
\eea
If we collect these pieces of information (\ref{boundterm}), (\ref{holder}), 
 (\ref{el ai n}) we obtain
\bea
&&  \int^\infty _0 W(x)  \left( \frac{s_l(x)s_n(x)}
{(x^\alpha -E_l)(x^\alpha -E_n)} \right) ^\frac{1}{4} 
 \varepsilon_l (x) \Ai(s_n(x)) dx  \nn \\
&& \;\; = \0(E_n ^{-\frac{3}{4} } E_l ^{-\frac{3}{4}
-\frac{1}{\alpha}} )+\0(E_n ^{-\frac{3}{4}}
 E_l^{-\frac{3}{4}-\frac{1}{2\alpha} } )+ \0(E_n ^{-\frac{3}{4} }
 E_l ^{-\frac{1}{4}-\frac{1}{\alpha} } )  \nn
\\
\label{first}
&& \;\; = \0(E_n ^{-\frac{3}{4} } 
 E_l^{-\frac{1}{4}-\frac{1}{2\alpha} } ) 
\eea
Putting (\ref{eps^2}), (\ref{trick}),  (\ref{first})
together we see that under the assumptions $W''\in L^\infty(\R_+)$, 
(\ref{asumW}) imposed on the
perturbation $W$ the contribution of the error may be estimated by
\bea
&& \int^\infty _0 W(x) 
\left( \frac{s_l(x)s_n(x)}{(x^\alpha -E_l)(x^\alpha -E_n)}
\right) ^\frac{1}{4}
\biggl( \bar{\varepsilon_l}(x) \Ai(s_n(x)) + \varepsilon_n (x) \Ai(s_l(x)) +
\bar{\varepsilon _l} (x) \varepsilon_n(x)
\biggr) dx  \nn \\
\label{Wer}
&& \;\; = \0 \left(E_n^{-\frac{3}{4}-\frac{1}{\alpha} }
 E_l^{ -\frac{r}{\alpha}-\frac{1}{6}+\frac{7}{6\alpha} }
\right)  +\0\left(E_n ^{-\frac{3}{4}}
 E_l^{-\frac{1}{4}-\frac{1}{2\alpha} } \right)
\eea

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%  window %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\\$
\underline {Window.}
$\\$
We will compute the contribution of
\be
&& \int^{a_n} _0 W(x) \phi_n(x) \phi_l(x) dx= \\
\,\, &&  \int^{a_n} _0 W(x)
\left (\frac{s_l(x) }
{( E_n -x^\alpha )(x^\alpha -E_l)  }  \right)^{\frac{1}{4} }
 \cos \left( \int ^{E_n ^\frac{1}{\alpha} } 
_x  \sqrt{E_n -y^\alpha}dy -\frac{\pi}{4}
\right) \frac{dx}{\sqrt\pi}   
 \left(1+\0(E_n ^{-\frac{\alpha+2}{2\alpha}} )  \right) 
\ee
The main part is proportional to
\bea
&&  \int^{a_n} _0 W(x)
\left (\frac{s_l(x) }
{( E_n -x^\alpha )(x^\alpha -E_l)  }  \right)^{\frac{1}{4} }
 \cos \left( \int ^{E_n ^\frac{1}{\alpha} } 
_x  \sqrt{E_n -y^\alpha}dy -\frac{\pi}{4}
\right) dx  \nn \\
&& \;\; =
-\left[ \frac{W(x)\sin(f(x))\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3  } 
 \left( \frac{s_l(x) }
{x^\alpha -E_l  }  \right)^{\frac{1}{4} } 
 \right]^{a_n} _0 +
 \int^{a_n} _0 \frac{W'(x)\sin(f(x))\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) } } dx \nn \\
&& \;\;\;\; + 
 \int^{a_n} _0 W(x)\sin(f(x)) \left(
\frac{\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) } } 
\right) ' dx  \nn \\ 
\nn
%\label{mainexp}
&& \;\; =
-\left[ \frac{W(x)\sin(f(x))\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3  } 
 \left( \frac{s_l(x) }
{x^\alpha -E_l  }  \right)^{\frac{1}{4} } 
 \right]^{a_n} _0 +
 \left[ \frac{W'(x)\cos(f(x))\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^5
\sqrt{s_l'(x) } } \right]^{a_n} _0 
 \\ \nn
&& \;\;\;\; + 
 \int^{a_n} _0 W(x)\sin(f(x)) \left(
\frac{\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) } } 
\right) ' dx  \\ \label{mainexp}
&& \;\;\;\;
 -  \int^{a_n} _0 \cos(f(x))\left( \frac{W'(x)\Ai(s_l(x))}
{\sqrt[4]{E_n-x^\alpha} ^5
\sqrt{s_l'(x) } } \right)' dx ,\
\eea
where $f$ is again  defined by
\[
f(x):=  \int ^{E_n ^\frac{1}{\alpha} } 
_x  \sqrt{E_n -y^\alpha}dy -\frac{\pi}{4} \;,\, \;\;
f'(x)=- \sqrt{E_n -x^\alpha}
\]
Let us investigate all terms individually. \\
The first two of them are very small
\bea 
\label{firstboy}
&& \left[ \frac{W(x)\sin(f(x))\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3  } 
 \left( \frac{s_l(x) }
{x^\alpha -E_l  }  \right)^{\frac{1}{4} } 
 \right]^{a_n} _0= \0(E_n ^{-\frac{3}{4} } E_l ^{-\frac{3}{4}
-\frac{1}{\alpha}} ) \\
\label{secondboy}
&& \left[ \frac{W'(x)\cos(f(x))\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^5
\sqrt{s_l'(x) } } \right]^{a_n} _0 =  \0(E_n ^{-\frac{7}{4} -\frac{1}{\alpha} }
 E_l ^{-\frac{3}{4}-\frac{1}{\alpha}} )
\eea
for the same reason like (\ref{boundterm}), namely with the help of 
(\ref{ai-infty}), (\ref{bohr}), (\ref{s'}). \\
To investigate the third term
\bea
\label{blb}
 \int^{a_n} _0 W(x)\sin(f(x)) \left(
\frac{\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) } } 
\right) ' dx
\eea
of (\ref{mainexp}) we firstly compute
\be
 \left( 
\frac{\Ai(s_l(x)) }{\sqrt[4]{E_n-x^\alpha} ^k
\sqrt{s_l'(x) }} \right) '= 
 \left(
\frac{\Ai(s_l(x))} {\sqrt[4]{E_n-x^\alpha} ^k  \sqrt{s_l'(x) } } 
\right)
 \left[
\frac{\Ai'(s_l(x)) }{\Ai(s_l(x)) } s_l'(x)- \frac{s_l''(x)}{2s_l'(x)}
+\frac{k\alpha x^{\alpha-1} } {4(E_n-x^\alpha) }
\right]
\ee
Let us estimate the third expression (using similar tricks like in 
(\ref{trick}))
\bea 
&&   \left| \int^{a_n} _0 W(x)\sin(f(x)) 
\frac{\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) } } 
 \frac{ x^{\alpha-1} } {E_n-x^\alpha } dx \right|   \nn
\\ \nn
&& \;\; \leq
E_n ^{-\frac{7}{4} } \left(
\int^{a_l} _0  \frac{|W(x)|\; x^{\alpha-1} }
{\sqrt[4]{E_l-x^\alpha} } dx + K\sup_{[a_l,  \,  2a_l]} 
|W(x)| \int^{2a_l} _{a_l}  x^{\alpha-1}
 \left( \frac{s_l(x)}{x^\alpha -E_l}   
\right) ^\frac{1}{4} dx  
  +   \0(E_l ^{-\infty}) \right) 
\\
\label{guy1}
&& \;\; = E_n ^{-\frac{7}{4} }   \left(
 \0 (E_l^{\frac{3}{4} -\frac{1}{\alpha} } )
+\0 (E_l^{ \frac{5\alpha+1-6r}{6\alpha} })
\right) 
\eea
The second part of the integral above is
\bea
&& \left|  \int^{a_n} _0 \frac{W(x)\sin(f(x))\Ai(s_l(x))s_l ''(x) }
{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) }^3 } dx  \right|  \leq
E_n ^{-\frac{3}{4}} \left(
K \int^{2a_l} _0 \frac{|W(x) s''_l(x)|}{\sqrt{s_l'(x) }^3}dx 
+\0(E_l^{-\infty})  \right) \nn \\
 \label{guy2}
&& =\0\left( E_n ^{-\frac{3}{4}}
 E_l ^{-\frac{1}{6}-\frac{1}{3\alpha}} \right) 
\eea
\\
The first part of the third term from (\ref{mainexp}) can be
estimated with the help of (\ref{aibound})
\bea
\label{guy3}
 \int^{a_n} _0 \frac{W(x)\sin(f(x))\Ai'(s_l(x)) \sqrt{s_l'(x) } }
{\sqrt[4]{E_n-x^\alpha}^3  } dx =
\0\left( E_n ^{-\frac{3}{4} }
 E_l ^{\frac{1}{4} } \right) 
\eea
The results of (\ref{guy1}), (\ref{guy2}), (\ref{guy3}) allow us to
say that the third integral (\ref{blb}) in (\ref{mainexp})
\bea
\label{thirdboy}
 \int^{a_n} _0 W(x)\sin(f(x)) \left(
\frac{\Ai(s_l(x))}{\sqrt[4]{E_n-x^\alpha} ^3
\sqrt{s_l'(x) } } 
\right) ' dx = \0\left( E_n ^{-\frac{3}{4} }
 E_l ^{\frac{1}{4} } \right) 
\eea
The very last term in (\ref{mainexp}) is
\bea
\label{lastboy}
\int^{a_n} _0 \cos(f(x))\left( \frac{W'(x)\Ai(s_l(x))}
{\sqrt[4]{E_n-x^\alpha} ^5
\sqrt{s_l'(x) } } \right)' dx =  \0\left( E_n ^{-\frac{7}{4} }
 E_l ^{\frac{1}{4} } \right) 
\eea
by the same tricks like in the previous case (\ref{blb}).\\ \\
The main result of this subsection is obtained by collecting
(\ref{firstboy}), (\ref{secondboy}), (\ref{thirdboy}) and
(\ref{lastboy}).
\be
&& \int^{a_n} _0 W(x) \phi_n(x) \phi_l(x) dx= \\
&& = \0(E_n ^{-\frac{3}{4} } E_l ^{-\frac{3}{4}
-\frac{1}{\alpha}} ) +
 \0(E_n ^{-\frac{7}{4} -\frac{1}{\alpha} }
 E_l ^{-\frac{3}{4}-\frac{1}{\alpha}} ) \\
&& \;\; +
\0\left( E_n ^{-\frac{3}{4} }
 E_l ^{\frac{1}{4} } \right) +
 \0\left( E_n ^{-\frac{7}{4} }
 E_l ^{\frac{1}{4} } \right) ,\
\ee
which yields
\bea
\label{window}
\int^{a_n} _0 W(x) \phi_n(x) \phi_l(x) dx=
\0\left( E_n ^{-\frac{3}{4} }
 E_l ^{\frac{1}{4} } \right)
\eea

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%% Proof of theorem 1.8 %%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$\\$
\underline {Proof of the Theorem \ref{offele}. }
$\\$
Now we are ready to state upper bound of non-diagonal matrix
elements.\\
\begin{Proof}
Taking (\ref{tailn}), (\ref{Wer}) and (\ref{window}) together
we  get
\be &&
\int_{\R_+} W \Psi_n \Psi_l= \0( E_n ^{-\infty}) +
\0 \left(E_n^{-\frac{3}{4}-\frac{1}{\alpha} }
 E_l^{ -\frac{r}{\alpha}-\frac{1}{6}+\frac{7}{6\alpha} }
\right)  +\0\left(E_n ^{-\frac{3}{4}}
 E_l^{-\frac{1}{4}-\frac{1}{2\alpha} } \right)+
\0\left( E_n ^{-\frac{3}{4} }
 E_l ^{\frac{1}{4} } \right) \\
&& = \0\left( E_n ^{-\frac{3}{4} }
 E_l ^{\frac{1}{4} } \right)
\ee
because $E_n>4E_l$.\\
We use the knowledge about normalization constant (\ref{C_E})
\be
\int_{\R_+} \Psi_n ^2= \frac{1}{2\pi}E_n ^{-\frac{1}{2}+\frac{1}{\alpha} }
\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}}+ 
\0\left( E_n^{\frac{16-10\alpha}{18\alpha}} \right)
\ee
to get 
\be
W_{n,\;l}:=\frac{\int_{\R_+} W \Psi_n \Psi_l}{\sqrt{\int_{\R_+} \Psi_n ^2  }
\sqrt{\int_{\R_+} \Psi_l ^2  }}=
\0\left( E_n^{-\frac{1}{2}-\frac{1}{2\alpha}} 
 E_l^{\frac{1}{2}-\frac{1}{2\alpha}} 
\right ) ,\
\ee
what is just the statement of Theorem \ref{offele}.
\end{Proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%% Summary %%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Summary}
$\\$
If we sum up all results we can state the main theorem about
asymptotic behavior of matrix elements.

\begin{theorem} \label{sum}
Let  $\Psi_n$ is Dirichlet $L^2(\R_+)$ solution of the equation
\be
(-\Delta +x^\alpha  )\Psi_n(x)=E_n \Psi_n(x) \; ,\
\ee
where $\alpha>1$.\\
Dirichlet boundary condition implies that there is a 
relationship between energy $E_n$ and a natural number  $n$ 
given by the Bohr-Sommerfeld quantization condition 
\be
E_n= \left(\frac{\pi}{\int^1_0  \sqrt{1-x^\alpha} dx } (n-\frac{1}{4})
\right)^\frac{2\alpha}{\alpha+2}
 + \0\left((\frac{1}{n})^\frac{2\alpha}{\alpha+2}\right).
\ee 
Assume that a perturbation $W$ satisfies
the following decay properties \\
\mbox{($ <x>=\sqrt{1+x^2}$).}
\be
W, W' \in L^1(\R^+)   ,\
<x>^rW \in L^\infty(\R^+) \, ,\ \,\,\, \rm{for\,\, some} \,\,
\, r>1 
\ee   
and in addition that $ W''$  is defined and bounded  almost everywhere
on $\R_+$ .\\
Then for matrix elements $W_{n,\,l}$ it holds true that
\be
W_{n,\,n}:=\frac{<W\,\Psi_n,\,\Psi_n>}{<\Psi_n,\, \Psi_n>}=
\left(\frac{\pi (n-\frac{1}{4})}{\int^1_0  \sqrt{1-x^\alpha} dx }
\right)^{-\frac{2}{\alpha+2}}
\left(\int^1 _0 \frac{dx}{\sqrt{1-x^\alpha}}\right)^{-1}
\left( \int_{\R_+}W + \0\left(n^{-t } \right) 
\right) ,\
\ee
(where $t:=\min\{\frac{1}{9}, \frac{2(r-1)}{\alpha+2}\} >0$)
for the diagonal elements. Other elements are estimated from above by
\be
W_{n,\,l}:=\frac{<W\,\Psi_n,\,\Psi_l>}{\sqrt{<\Psi_n,\, \Psi_n>}\,
\sqrt{<\Psi_l,\, \Psi_l>} }\leq K 
(n\,l)^{-\frac{1}{\alpha+2}} ,\
\ee
or if moreover  $n>\delta\,l$,  for some $\delta>0$, by a better estimate
\bea \label{offdiag}
W_{n,\,l}\leq
K (n\,l)^{-\frac{1}{\alpha+2}} \;
(\frac{l}{n} )^{\frac{\alpha}{\alpha+2 }} .\
\eea
\end{theorem}





%\input egor

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%% egor %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newpage
\section{Egorov theorem for time dependent case} \label{Egorov}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% Introduction %%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Introduction}
$\\$
This section is dealing with Egorov theorem, mathematically rigorous
tool for approximation of quantum time evolution with help of the
classical one. It is proven that the difference between exact quantum
evolution and approximation is proportional to $\hbar^2$ in the sense
of the statement of Theorem \ref{main}.\\
Physical interpretation of such result is clear,  when 
observables of given physical system with the same dimension as
$\hbar$
like action are much larger
then numerical value of $\hbar$ (i.e. is possible limit $\hbar 
\rightarrow 0$), then quantum evolution of an observable
tends to the classical one. 
The theory of this type uses so called Weyl-quantization, as powerful
tool to construct quantum observables (i.e. symmetric or even 
essentially self-adjoint operators) according to the principal of
correspondence. \\
The result of this section, formulated in Theorem \ref{main}, is
improvement of \cite{Lev}. It is again based on \cite{br}, but the
generalization to time dependent systems is led directly using the
trick (\ref{basic}). This is idea of Joachim Asch, I would like to
thank him for it.\\
The statement of Theorem \ref{main} is improved with respect to
\cite{Lev} in several ways. The first is arbitrary dimension, then
general time dependent Hamiltonian despite the Schr\"odinger one is
included. The assumptions on the growth of $H(\tau)$ are also weaken.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%% Notation %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Notation and conventions} 
$\\$
Let us denote by $X = \R^n$ the configuration 
space of a classical mechanical system
with $n$ degrees of freedom. The corresponding phase 
space $Z$ is identified with $\R^{2n}$ equipped  with the 
symplectic form $\sigma$  defined  by
\b
\sigma(z;z^\prime) = \langle Jz, z^\prime \rangle.
\label{sympform}
\e
where  $\langle \;,\; \rangle$ is the Euclidean scalar product and  
$J$ is the $2n\times 2n$ matrix
\[
J =  \begin{pmatrix} 
0 &\quad \Id \cr
-\Id &\quad 0 \cr
\end{pmatrix}.
\]
A generic point in $Z$ is denoted $z$ and its coordinates by 
$(q, p)$ where $q, p \in \R^n$. 
We will again use notation $\langle x \rangle=(1+|x|^2)^{1/2}$.
 \\
A classical time dependent Hamiltonian is a time dependent real
 function, defined on the phase space $H:[s,t]\times Z\rightarrow \R\;$. 
Basic example is $H(\tau, q,p) = \frac{\Vert p\Vert^2}{2m} +
V(\tau,q)$, ($m>0$)
where $\Vert p\Vert^2 = <p,p>$ and $V$ is time dependent potential.\\
In what follows we put emphasis on the case $X = \R^n$.\\
The motion of the classical system is determined by the system of
 Hamilton equations
\b\label{ham}
\frac{dq}{dt} = \frac{\partial H(t)}{\partial p}(q,p),\;\;
\frac{dp}{dt} = -\frac{\partial H(t)}{\partial q}(q,p).
\e
We will suppose that the equations (\ref{ham}) generate a 
flow $\Phi(t,s)$ on the phase space $Z$, defined by 
\[
\Phi(t,s,q,p) = (q(t,s),p(t,s))\; , \;\Phi(s,s) = \Id.
\] 
We define Poisson bracket by
\[
\{A, B\} = \partial_qA\cdot\partial_pB - \partial_pA\cdot\partial_qB.
\label{poisson}
\]
Here we have used the notation $\partial_q = \frac{\partial}{\partial q}$.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%% weyl quantization%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\subsection{Weyl quantization}
$\\$
First of all we define a class of classical observables, which
are quantizable by Weyl quantization, introduced after.
\begin{definition}
$A\in \0(m)$, $m\in\R$, if and only if $Z\stackrel{A}{\rightarrow}\C$
is $C^\infty$ in $Z$ and for every multi-index $\gamma\in\N^{2n}$ there exists $C>0$ such that
$$
\vert \partial^\gamma_z A(z)\vert
\leq C\langle z\rangle^m,\;\forall z\in Z.
$$
\end{definition}
\begin{definition}
Let $A\in \0(m)$. We will define  $\hat{A}$ as operator from 
${\mathcal S}(X) $ to itself, called $\hbar$-Weyl quantization of $A$,  by the following formula,
with $\psi\in{\mathcal S}(X)$,
\be 
%\label{weyl}
{\hat A}\psi(x) = (2\pi \hbar)^{-n}\int_X \left(\int_X A\left
(\frac{x+y}{2}, p\right)
{\rm e}^{i\hbar^{-1}\langle x-y, p\rangle} \psi(y)dy\right)dp.
\ee
\end{definition} 
$ \\$
{\bf Example:}
Let us focus for example to observable $A:=x.p$. To compute $\hat{A}$ we will
use 
\be
\hskip10pt p\exp(\frac{i}{\hbar}(x-y)p)=-\frac{\hbar}{i}
\partial_y\exp(\frac{i}{\hbar}(x-y)p) 
\ee
and after integration by parts and using inverse Fourier transform  we get
\\
$ \hat{A}=\frac{1}{2} (\hat{p}\hat{q}+ \hat{q}\hat{p} )$.
\\
For $A\in {\mathcal S}(Z)$ we will compute, 
(with the help of integral kernel) inverse 
prescription of Weyl symbol if operator is given as an integral operator.
\be
&& ({\hat A} \psi)(x)= (2\pi \hbar)^{-n}\int\!\!\int_Z A\left(\frac{x+y}{2}, p\right)
{\rm e}^{i\hbar^{-1}\langle x-y, p\rangle} \psi(y)dydp=
\int_{\R^n} {\mathcal K}_A (x,y) \psi (y) dy\\
&& {\mathcal K}_A (x,y):= (2\pi \hbar)^{-n}\int_{\R^n} A\left(\frac{x+y}{2}, p\right)
{\rm e}^{i\hbar^{-1}\langle x-y, p\rangle} dp
\ee
After inverse Fourier transform.
\b\label{symbol}
A(x,\xi)=\int_{\R^n}{\mathcal K}_A \left(x+\frac{v}{2},x-\frac{v}{2}\right)
{\rm e}^{-\frac{i}{\hbar} v{\cdot}\xi}dv
\e
\\
The class $ \0(m)$ has nice properties. For example, 
if  $A\in \0(m)$ is a real-valued function, then $\hat{A}$ is symmetric
operator in $L^2(X)$, with domain ${\mathcal S}(X)$. Further there exists a
product formula. For the proof one can see \cite{ro}, \cite {fo} or \cite{ho}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% Product formula %%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Product formula}
\begin{theorem}Let $A\in \0(m)$,  $B\in \0(p)$, then there exists  $C\in \0(m+p)$, such that 
\[ 
 \hat{A}\cdot\hat{B}=\hat{C} 
\]
And there exists following asymptotic expansion of Weyl symbol of this operator 
\[
C(q,p) =
\exp\left(\frac{i\hbar}{2}\sigma(D_q,D_p;D_{q^\prime},D_{p^\prime})\right)A(q,p)
B(q^\prime,p^\prime)\vert_{(q,p)=(q^\prime,p^\prime)},
\]
where $\sigma$ is the symplectic bilinear form (\ref{sympform}) and $D=i^{-1}\nabla$.
By expanding the exponential term, we get
\[
C(q,p) =
\sum_{j\geq 0}\frac{\hbar^j}{j!}(\frac{i}{2}\sigma(D_q,D_p;D_{q^\prime},D_{p^\prime}))^j
A(q,p)B(q^\prime,p^\prime)\vert_{(q,p)=(q^\prime,p^\prime)}.
\]
So that $C(q,p)$ is a formal power series in $\hbar$ with coefficients given by
\b
\label{AB}
C_j(q,p) = \frac{1}{2^j}\sum_{\vert\alpha+\beta\vert=j}
\frac{(-1)^{\vert \beta\vert}}{\alpha!\beta!}
(D^\beta_q\partial^\alpha_p A).( D^\alpha_q\partial^\beta_p B)(q,p).
\e
\end{theorem}
\vskip8pt
\begin{Proof}
We will show this for $A,B\in {\mathcal S}(Z) $
\[
\left(\hat{A}\cdot\hat{B}\psi\right) (x)
= \int_{\R^n} {\mathcal K}_{A\cdot B} (x,y) \psi (y) dy
\]
where
\[
{\mathcal K}_{A\cdot B} (x,y)=
\frac{1}{(2\pi \hbar)^{2n}} 
  \int_{\R^{3n}} A\left(\frac{x+z}{2}, p\right)B\left(\frac{z+y}{2}, \eta 
\right)
{\rm e}^{-\frac{i}{\hbar}\langle x-z, p\rangle +\langle z-y, \eta \rangle
} dzdpd\eta
\]
Let's define 
\[
\hat{C}:=\hat{A} \cdot \hat{B}
\]
then Weyl symbol of $\hat{C}$ can be computed with help of formula 
(\ref{symbol}).
\be
C(x,\xi) &=& \int_{\R^n}{\mathcal K}_{A\cdot B} 
\left(x+\frac{v}{2},x-\frac{v}{2}
\right)
{\rm e}^{-\frac{i}{\hbar} v{\cdot}\xi}dv\;  \\
&=&\frac{1}{(2\pi \hbar)^{2n}}
  \int_{\R^{4n}} A\left(\frac{x+z-\frac{v}{2}}{2}, p\right)
B\left(\frac{z+y-\frac{v}{2}}{2}, \eta \right) \times \\
&& \times{\rm e}^{-\frac{i}{\hbar}\left (\langle x+\frac{v}{2}-z, p\rangle +
\langle z-x +\frac{v}{2}, \eta \rangle -\langle v, \xi \rangle
 \right)} dzdpd\eta dv
\ee
We apply the change of variables
\[
s:=\frac{1}{2}(x+z)+ \frac {1}{4}v, \hskip10pt 
t:=\frac{1}{2}(x+z)- \frac {1}{4}v
\Rightarrow z=s+t-x, \hskip10pt v=2(s-t), \hskip10pt dzdv=4^ndsdt  
\] 
to get
\[
C(x, \xi) =\frac{1}{(\pi \hbar)^{2n}} 
  \int_{\R^{4n}} A\left(s, p\right)
B\left(t, \eta \right)
{\rm e}^{-\frac{2i}{\hbar}\left(\langle x-s, \xi -\eta \rangle -
\langle x -t, \xi -p \rangle \right)} dsdpdtd\eta 
\]
Now using Taylor formula
\be
&&A(s,p)B(t,\eta) = \sum_{|\alpha+\beta|<N} \partial_x ^{\alpha} A(x,p)
\hskip5pt
\partial_x ^{\beta} B(x,\eta) 
\frac{(s-x)^{\alpha}}{\alpha !} \frac{(t-x)^{\beta}}{\beta !} \;   \\
&& \hskip10pt +  \sum_ {|\alpha+\beta|=N} {\int_0 ^1 (1-\sigma)^N
\partial_x ^{\alpha} A(x+\sigma(s-x),p)
\hskip5pt
\partial_x ^{\beta} B(x+\sigma(t-x),\eta) 
\frac{(s-x)^{\alpha}}{\alpha !} \frac{(t-x)^{\beta}}{\beta !} d \sigma} 
\ee
we obtain
\be
 C(x, \xi) &=& \frac {1}{(\pi \hbar )^{2n}}
 \sum_{|\alpha+\beta|<N} \int_{\R ^{4n}} \partial_x ^{\alpha} A(x,p)
\hskip5pt
\partial_x ^{\beta} B(x,\eta) 
\frac{(s-x)^{\alpha}}{\alpha !} \frac{(t-x)^{\beta}}{\beta !} \times \\
&& \times {\rm e}^{\frac{2i}{\hbar} \left(\langle x-s, \xi -\eta \rangle -
\langle x -t, \xi -p \rangle  \right)} dsdpdtd\eta 
+   R_{N-1} (A,B;z;\hbar),
\ee
where the remainder is defined by:
\b
\label{remainder}
C(z) - \sum_{0\leq j \leq N}\hbar^jC_j(z) =: R_N(A,B;z;\hbar).
\e
\[
z=(x,\xi)\in Z  
\]
Using
\[
(s-x)^\alpha (t-x)^\beta 
{\rm e}^{\frac{2i}{\hbar} \left( \langle x-s, \xi -\eta \rangle -
\langle x -t, \xi -p \rangle  \right)} =
\left(\frac{\hbar}{2i} \right)^{|\alpha|+|\beta|}(-1)^{|\beta|}
\hskip5pt
\partial_\eta ^\alpha \partial_p ^\beta
\hskip5pt
{\rm e}^{\frac{2i}{\hbar} \left(
\langle x-s, \xi -\eta \rangle -\langle x -t, \xi -p \rangle  \right)}  
\]
and integrating by parts we get
\be
&& C(x, \xi) = \\
&&\hskip10pt  \frac {1}{(\pi \hbar )^{2n}} 
 \sum_{|\alpha+\beta|<N} 
\left(\frac{\hbar}{2i} \right)^{|\alpha|+|\beta|}(-1)^{|\alpha|}
\int_{\R ^{4n}}
\frac{ \partial_p ^\beta  \partial_x ^{\alpha} A(x,p) }{\alpha !} 
\frac{\partial _\eta ^\alpha \partial_x ^{\beta} B(x,\eta)}{\beta !} \times \\
&& \hskip10pt\times \hskip5pt
{\rm e}^{\frac{2i}{\hbar} \left( \langle x-s, \xi -\eta \rangle -
\langle x -t, \xi -p \rangle \right) } dsdpdtd\eta 
\hskip5pt +  R_{N-1} (A,B;z;\hbar).
\ee
Finally 
\be
 C(x, \xi) &=&  \sum_{|\alpha+\beta|<N} 
\left(\frac{\hbar}{2i} \right)^{|\alpha|+|\beta|}(-1)^{|\alpha|}
\frac{\partial_\xi ^\beta  \partial_x ^{\alpha} A(x,\xi)}{\alpha !} 
\frac{\partial _\xi ^\alpha \partial_x ^{\beta} B(x,\xi)}{\beta !}+  \\
&+&  R_{N-1} (A,B;z;\hbar).
\ee
Remainder (\ref{remainder}) can be also expressed as
\be
&& R_{N-1} (A,B;z;\hbar):= \\
&& \hskip15pt 
\frac {1}{(\pi \hbar )^{2n}} 
\sum_ {|\alpha+\beta|=N} \int_0 ^1 (1-\sigma)^N
\partial_x ^{\alpha} A(x+\sigma(s-x),p)
\hskip5pt
\partial_x ^{\beta} B(x+\sigma(t-x),\eta) \times \\
&& \hskip15pt
 \times
\frac{(s-x)^{\alpha}}{\alpha !} \frac{(t-x)^{\beta}}{\beta !}  
{\rm e}^{\frac{2i}{\hbar} \left( \langle x-s, \xi -\eta \rangle -
\langle x -t, \xi -p \rangle  \right) } d\sigma dsdpdtd\eta 
\ee
After the change of variables
\[
\sigma (s-x)=: s^\prime ,  \hskip5pt \sigma (t-x)=: t^\prime ,  
\hskip5pt  \eta - \xi =: \eta^\prime , \hskip5pt p-\xi =: p^\prime
\]
and using the same trick with integrating by parts (and in what follows 
omitting primes) we finally get
\be
&&R_{N-1} (A,B;z;\hbar):= \\
&&\hskip10pt 
\left(\frac{\hbar}{2i} \right)^N
\int_0 ^1 \frac {(1-\sigma)^N}{(\pi \hbar \sigma)^{2n}} 
\sum_ {|\alpha+\beta|=N} \frac{(-1)^{\alpha}}{\alpha !\beta !} 
\partial_p ^\beta  \partial_x ^{\alpha} A(x+s,p+\xi )
\hskip5pt
\partial _\eta ^\alpha \partial_x ^{\beta} B(x+t,\eta+\xi ) \times 
\\
&&\hskip10pt
 \times
\exp \left(\frac{2i}{\hbar \sigma}\left( \langle s,\eta \rangle -
\langle t, p \rangle\right)  \right) d\sigma dsdpdtd\eta  \\
&&\hskip10pt =
\left(\frac{\hbar}{2i} \right)^N
\int_0 ^1  \frac{(1-\tau)^N}{(\pi \hbar \tau)^{2n}}
\int_{Z\times Z} \exp(-\frac{2i}{\tau\hbar}\sigma(u,v))\sigma^{N}(D_u,D_v)
A(u+z)B(v+z)dudvd\tau.
\ee  
\end{Proof}
Now we estimate this remainder and his derivatives.

\begin{theorem} \label{estimR}
Suppose that $\lambda>1$.\\
For every $m\in \N$, $m> 2n$, for
every $s > 4n$ there exists a constant $\rho_{n,m,s}$ such that
for every $N\geq 1$, for every multi-index $\gamma$, 
there exists $K_{n,N,|\gamma|}>0$, such that
the following estimate holds, for every  $A, B \in 
{\mathcal S}(Z)$,  $z\in Z$,
\bea\label{remest1}
&& \vert \partial_z^\gamma\left(C(z) - 
\sum_{0\leq j \leq N}\hbar^jC_j(z)\right)\vert
\leq \hbar^{N+1}\rho_{n,m,s}K_{n,N,\vert \gamma\vert} \times \nonumber\\
&& \times \sup_{(*)}\left[(1+u^2+v^2)^{(s-m)/2}
\vert\partial_u^{(\alpha,\beta)+\mu}A(u+z)\vert
\vert\partial_v^{(\beta,\alpha)+\nu}B(v+z)\vert\right]
\eea
where $\displaystyle{\sup_{(*)}}$ means that the supremum holds
under the conditions\\ $u, v \in Z$, $\vert \mu\vert +\vert \nu\vert\leq m+\vert\gamma \vert$,
$\vert\alpha\vert+\vert\beta\vert = N+1$ ($\mu, \nu\in\N^{2n}$, $\alpha, \beta\in\N^n$).
\end{theorem} $\\$ {\bf Proof :}~
We shall use the following lemma to estimate $R_N(A,B;z,\hbar) $.
\begin{lemma}\label{foi}
Let us consider $F\in{\mathcal S}(Z\times Z)$ and the integral ($\lambda>1$)
\be
I(\lambda) = \lambda^{2n}\int_{Z\times Z}\exp[-i\lambda \sigma(u, v)]F(u,v)dudv.
\ee
Then for every real number $s>4n$ and every integer $m> 2n$ there exists\\ 
$\kappa(n,s,m) >0$ depending only on $n, s, m$ (but independent of $F$)
such that \\ the following  estimate holds
\be
\vert I(\lambda)\vert \leq \kappa(n,s,m)
\sup_{u, v\in Z, \atop \vert \mu\vert+\vert \nu\vert\leq m}
(1+u^2+v^2 )^{(s-m)/2}
\vert \partial^\mu_u\partial^\nu_v F(u,v)\vert.
\ee
\end{lemma} $\\$
{\bf Proof :}  Let us introduce a cut-off $\chi_0$, $C^\infty$ on $\R$,
$\chi_0(x) = 1$ for $\vert x\vert\leq 1/2$ and
$\chi_0(x) = 0$ for $\vert x\vert\>\geq1$.\\
 We split $I(\lambda)$ into three pieces
\be
I_0(\lambda) = \lambda^{2n}\int_{Z\times Z}\exp[-i\lambda \sigma(u, v)]
\chi_0(u^2+v^2) \chi_0(\lambda(u^2+v^2))F(u,v)dudv, \\
I_1(\lambda) = \lambda^{2n}\int_{Z\times Z}\exp[-i\lambda \sigma(u, v)]
\left(1-\chi_0(\lambda(u^2+v^2)) \right)
\chi_0(u^2+v^2)F(u,v)dudv,\\
I_2(\lambda) = \lambda^{2n}\int_{Z\times Z}\exp[-i\lambda \sigma(u, v)]
(1-\chi_0(u^2+v^2))F(u,v)dudv.
\ee
\\
For $I_0(\lambda)$, we easily have
\be
\vert I_0(\lambda)\vert \leq \omega_{4n}\sup_{u^2+v^2\leq 1} \vert F(u,v)\vert,
\ee
where $\omega_{4n}$ is the volume of the unit ball in $Z^2$.
For $I_1(\lambda)$ and $I_2(\lambda)$, we integrate by parts with the differential
operator
\be
L = \frac{i}{u^2+v^2}\left(Ju\frac{\partial}{\partial v}-Jv\frac{\partial}{\partial u}\right),
\ee
where $J$ is the matrix associated to the symplectic form ($\sigma(u,v)=<Ju,v>$).
It holds true that
\[ 
L\left( \exp[-i\lambda \sigma(u, v)] \right)=\lambda 
\exp[-i\lambda \sigma(u, v)].
\]
Performing $2n$ integrations by parts, we can see that it exists a constant $c_n$
such that
\be
\vert I_1(\lambda)\vert \leq c_n
\sup_{u^2+v^2\leq 1\atop \vert \mu\vert+\vert \nu\vert\leq 4n }
\vert\partial^\mu_u \partial^\nu_vF(u,v)\vert.
\ee
Similarly, performing $m$ integrations by parts,
\be
\vert I_2(\lambda)\vert  &\leq& 
\vert\int_{Z\times Z}\exp[-i\sigma(u, v)]
\left(1-\chi_0 (\frac{1}{\lambda}(u^2+v^2)\right) 
F(\frac{u}{\sqrt\lambda},\frac{v}{\sqrt\lambda})\hskip5pt dudv\vert \\
&\leq& c_m\int_{Z \times Z}(\frac{|u|+|v|}{u^2+v^2})^m
\vert\partial^\mu_u \partial^\nu_v 
\left((1-\chi_0 (\frac{1}{\lambda}(u^2+v^2))
F(\frac{u}{\sqrt\lambda},\frac{v}{\sqrt\lambda})\right)\vert dudv. 
\ee
Using the existence of integral
\[
\int_{Z \times Z}(\frac{|u|+|v|}{u^2+v^2})^m 
(1+u^2+v^2)^{\frac{m-s}{2}}dudv
\]
we get for a constant $c(n,s,m)$,
\be
\vert I_2(\lambda)\vert \leq c(n,s,m)
\sup_{u,v \in Z\atop \vert \mu\vert+\vert \nu\vert\leq m}
(1+u^2+v^2 )^{(s-m)/2}\vert\partial^\mu_u \partial^\nu_vF(u,v)\vert.
\ee
\QED
Now we can complete the proof of the theorem by using Lemma \ref{foi}, the Leibniz formula
and the following elementary estimate, using in $Z$
the coordinates $u=(x,\xi)$, $v=(y,\eta)$,
\be
\vert \sigma^N(\partial_x,\partial_\xi; \partial_y,\partial_\eta))A(x,\xi)
B(y,\eta)\vert \leq
(2n)^N\sup_{\vert \alpha\vert+\vert \beta\vert=N}
\vert\partial_x^\alpha\partial_\xi^\beta A(x,\xi)\partial_y^\beta\partial_\eta^\alpha B(y,\eta)\vert.
\ee
\QED
\begin{remark}
We can easily extend the estimate (\ref{remest1}) for observables $A, B$ with polynomial
growth at infinity, by choosing $m$ large enough to get a finite r.h.s.
Let us assume that $A\in\0(\mu_A)$, $B\in\0(\mu_B)$,
where $\mu_A, \mu_B \in\R$. Then we can apply (\ref{remest1}) to
$A_\varepsilon(u) = {\rm e}^{-\varepsilon u^2}A(u)$ and
$B_\varepsilon(v) = {\rm e}^{-\varepsilon v^2}B(v)$ for $\varepsilon >0$ and
pass to the limit $\varepsilon \rightarrow 0$ with $m-s\geq \mu_A+\mu_B$.
\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%   main theorem %%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\subsection{Main theorem}

\begin{theorem}\label{main}
Let $[t,s]$ be a compact interval.
Let us consider time dependent Hamiltonian function  $H(\tau)\in\0(c)$ 
and an observable $A\in\0(d)$ for some $c,d$, satisfying
\bea\label{asham}
\vert\partial_z^\gamma H(\tau,z)\vert \leq C_{\gamma},\; &{\rm for}&\;
3\leq \vert\gamma\vert \leq 4n+[\frac{n}{2}]+5 
 \, ,  \; \tau \in [s,t] \nonumber \\
\vert\partial_z^\gamma A(z)\vert\leq C_\gamma, \;&{\rm for}&\; 
1\leq \vert\gamma\vert \leq 4n+[\frac{n}{2}]+5  \\
\vert\partial_z^\gamma \Phi(t,\tau,z)\vert\leq C_\gamma, \;
&{\rm for}& \; 
1\leq \vert\gamma\vert \leq 4n+[\frac{n}{2}]+5, 
 \; \tau \in [s,t] ,\ \nonumber
\eea
where the classical flow $\Phi(t,s) $ is given by classical
Hamilton equations (\ref{ham}) and is assumed to be unique. \\
We know that $\widehat{H(\tau)}, \hat{A}$ are symmetric operators  on 
Schwartz space $\forall \tau\in[s,t]$.
We assume moreover that $A$ is essentially self-adjoint operator 
in $L^2(X)$, with core ${\mathcal S}(X)$.\\
Finally we suppose that quantum evolution 
given by Schr\"odinger equation is defined  on  Schwartz space 
by the propagator $U$
\bea
\label{schrodinger}
i\hbar \;d_\tau U(\tau,s)\psi=\widehat{H(\tau)}U(\tau,s)\psi ,\
\eea
which conserves  Schwartz space and which satisfies initial condition
\bea
\label{schinit}
U(\tau,\tau)=\Id \;,\; \forall \tau\in[s,t]
\eea
valid also on Schwartz space.
Then the quantum evolution of observable $\hat{A}$ can be approximated by the 
operator with the Weyl symbol
\be
%\label{A0}
A_0(t,s,z)= A(\Phi(t,s,z))
\ee
in the following sense:
\[
\Vert \widehat{A}(t,s) - \widehat{A_0(t,s)}  \Vert_{L^2(X)} \leq 
 \hbar^2 C_{n}(t,s) 
\]
\end{theorem} 

$\\$
\begin{Proof}
For any $B(t)\in \0(m),\, \psi \in \mathcal{S} (X)$  it holds true that 
\b \label{change}
\frac{d}{dt}\widehat{B(t)} \psi= \widehat{(\frac{d}{dt}{B(t)})}\psi
\e  
On Schwartz space holds equation
\bea \label{basic}
&&U(s,t)\hat{A}U(t,s) - \widehat{A_0}(t,s) = 
\int^t_s \frac{d}{d\tau}\left(U(s,\tau)\widehat{A\circ\Phi(t,\tau)}
U(\tau,s) \right)\,{\rm d}\tau \nonumber
\\
&&\hskip20pt=\int_s^t U(s,\tau)\left(\frac{i}{\hbar}
[\widehat{H(\tau)}, \widehat{A\circ\Phi(t,\tau)}] +
\widehat{\frac{d}{d\tau}A\circ\Phi(t,\tau)}\right)U(\tau,s)
\;{\rm d}\tau \nonumber
\\
&&\hskip20pt=
\int_s^t U(s,\tau)\left(\frac{i}{\hbar}
[\widehat{H(\tau)}, \widehat{A\circ\Phi(t,\tau)}]
 -\widehat{\{ H(\tau), A\circ\Phi(t,\tau) \} } \right)U(\tau,s)\;{\rm d}\tau, 
\eea
where we used (\ref{change}), 
 Schr\"odinger equation (\ref{schrodinger}), with initial condition
(\ref{schinit}). \\
Differentiating a basic property of the propagator, 
$U(\tau,s)U(s,\tau)\psi=\psi,$ yields
\be
 d_\tau U(s,\tau)=-U(s,\tau)\biggl(d_\tau U(\tau,s) \biggr) U(s,\tau)  .\
\ee
And finally it was used
\be
\frac{d}{d\tau}A\circ\Phi(t,\tau)=-\{ H(\tau), A\circ\Phi(t,\tau) \} ,\
\ee
which can be obtained by differentiating $\Phi(t,\tau,y)=y$ with respect
to $t$ (see (\ref{ham})) and then putting $y:=\Phi(t,\tau,x)$.\\
With the help of (\ref{AB}) we compute that the Weyl symbol of the expression 
\[
\widehat{C}:=\frac{i}{\hbar} [\hat{H},\hat{A}]
\] 
is
\[
C=0+\{H,A\}+0+2\frac{i}{\hbar}R_2(H,A;z,\hbar)\; ,
\]
where
\be
% \label{R2}
&& R_2(H,A;z,\hbar)= \\
&& \left(\frac{\hbar}{2i} \right)^3
\int_0 ^1  \frac{(1-\tau)^3}{(\pi \hbar \tau)^{2n}}
\int_{Z\times Z} \exp(-\frac{2i}{\tau\hbar}\sigma(u,v))\sigma^{3}(D_u,D_v)
H(u+z)A(v+z)dudvd\tau. \nonumber
\ee
Applying $L^2$ norm to (\ref{basic}) we obtain
\[
\Vert U(s,t)\hat{A}\, U(t,s) - \widehat{A_0(t,s)}  \Vert_{L^2} \leq 
 2\frac{|t-s|}{\hbar}\Vert\widehat{R_2}(H,A;z,\hbar)   \Vert_{L^2}. 
\]
But we can apply Calderon-Vaillancourt Theorem with improvements by A. 
Boulkhemair \cite{bo}, which estimates the norm of the Weyl-quantized operator 
\be
\Vert\hat{B}\Vert_{L^2} \leq \gamma_n
 \sup_{\vert\alpha\vert, \vert\beta\vert \leq[n/2]+1\atop z\in Z}
\vert \partial^{\alpha,\beta}_z B(z)\vert.
\ee
So we need to estimate
\[
\sup_{\vert\alpha\vert, \vert\beta\vert \leq[n/2]+1\atop z\in Z}
\vert \partial^{\alpha,\beta}_z  R_2(H,A;z,\hbar) \vert.
\]
This is exactly what we have prepared in Theorem
\ref{estimR}. 
Using (\ref{asham}) we deduce that for  $s=m:=4n+1$ there exists
$ K_{n}(t,s)$, such that:
\be
 K_{n} (t,s) := 
\sup_{(*)}\left[\vert\partial_u^{(\alpha,\beta)+\mu}H(\tau,u+z) \vert 
\vert\partial_v^{(\beta,\alpha)+\nu} A\biggl(\Phi(t,\tau,u+z)\biggr)
\vert\right]< \infty \nonumber
\ee
where $\displaystyle{\sup_{(*)}}$ means that the supremum holds
under the conditions\\ $u, v, z \in Z, \tau\in [s,t]
$, $\vert \mu\vert +\vert \nu\vert\leq 5n+
[n/2]+1$,
$\vert\alpha\vert+\vert\beta\vert =3$ ($\mu, \nu\in\N^{2n}$, $\alpha, \beta\in\N^n$).\\ 
We have also used the following estimate on 
derivatives of composition of functions
\[
\vert \partial^\alpha(f\circ g) \vert\leq K_{n,\vert\alpha\vert}
\sup_{\beta, \gamma  \neq 0  \atop \vert\beta\vert,\vert\gamma\vert\leq
\vert \alpha \vert}
\vert \partial^\beta f \vert\vert \partial^\gamma g \vert 
\]
Combining all these facts we may finish the proof.  
\end{Proof}



%\input weak

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%% weak %%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newpage
\section{Weakly regular Floquet Hamiltonians with pure point spectrum}

\subsection{Introduction}
$\\$
The problem we address in this part concerns spectral analysis of so called
Floquet Hamiltonians.  The study of stability of non autonomous quantum
dynamical systems is an effective tool to understand most of quantum problems
which involve a small number of particles. When these systems are periodic the
spectral analysis of the evolution operator over one period can give a fairly
good information on this stability, see e.g. \cite{EnssVeselic}. In fact this
type of result generalizes the celebrated RAGE theorem concerned with 
time-independent systems (one can consult \cite{ReedSimon} for a
summary). As shown in \cite{Howland}
and \cite{Yajima} the spectral analysis of the evolution operator over one
period (so called monodromy operator or Floquet operator) is equivalent to the
spectral analysis of the corresponding 
Floquet Hamiltonian (also called operator of quasi-energy). 
This is also what we are aiming for in this part. More
precisely, we analyze periodic quantum systems which are weakly regular in time
and \char`\"{}space\char`\"{} in the sense of an appropriately chosen norm,
and give sufficient conditions to insure that the Floquet Hamiltonians has a
pure point spectrum.\\
Another generalization is that in the present result 
(Theorem \ref{MainTheorem})
we allow degenerate eigenvalues of the unperturbed Hamilton operator (denoted
\( H \) in what follows). The degeneracy of eigenvalues \( h_{m} \) of \( H \)
can grow arbitrarily fast with \( m \) provided the time-dependent perturbation
is sufficiently regular. To our knowledge this is a 
new feature in this context. 
Previously two conditions were usually imposed, namely bounded degeneracy and
a growing gap condition on eigenvalues \( h_{m} \), reducing this way the scope
of applications of this theory to one dimensional confined systems. Owing to
the generalization to degenerate eigenvalues we are able to consider also some
models in higher dimensions, for example the \( N \)-dimensional quantum top,
i.e., the \( N \)-dimensional version of the pulsed rotor. A short description
of this model is also given in subsection \ref{Models}.\\
Because the proof of Main Theorem \ref{MainTheorem} is quite long we
will address reader to \cite{D2}. Further generalization to unbounded
perturbations is treated in \cite{D3}. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%  Main theorem %%%%%%%%%%%%%%%%%%%%%$%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Main theorem} \label{Sec:MainTheorem}
$\\$
The central object we wish to study in this part is a self-adjoint
operator of the form \( \bK +\bV  \) acting in the Hilbert space\[
\KK =L^{2}([\, 0,T\, ],\textrm{dt})\otimes \HH \cong L^{2}([\, 0,T\,
],\HH ,\textrm{dt})\]
where \( T=2\pi /\omega  \), \( \omega  \) is a positive number
(a frequency) and \( \HH  \) is a fixed separable Hilbert space.
The operator \( \bK  \) is self-adjoint and has the form\[
\bK =-i\, \partial _{t}\otimes 1+1\otimes H\]
where the differential operator \( -i\partial _{t} \) acts in \(
L^{2}([\, 0,T\, ],\textrm{dt}) \)
and represents the self-adjoint operator characterised by periodic
boundary conditions. This means that the eigenvalues of \( -i\partial
_{t} \)
are \( k\omega  \), \( k\in \Z  \), and the corresponding normalised
eigenvectors are \( \chi _{k}(t)=T^{-1/2}\exp (ik\omega t). \) \( H \)
is a self-adjoint operator in \( \HH  \) and is supposed to have
a discrete spectrum. Finally, \( \bV  \) is a bounded Hermitian
operator
in \( \KK  \) determined by a measurable operator-valued function
\( t\mapsto V(\omega t)\in \BB (\HH ) \) such that \( \sup _{t\in \R
}\| V(t)\| <\infty  \),
\( V(t) \) is \( 2\pi  \)-periodic, and for almost all \( t\in \R  \),
\( V(t)^{\ast }=V(t) \). Naturally, \( (\bV \psi )(t)=V(\omega t)\psi
(t) \)
in \( \KK \cong L^{2}([\, 0,T\, ],\HH ,\textrm{dt}) \).\\
Let\[
\sum _{k\in \Z }k\omega \, P_{k}\]
be the spectral decomposition of \( -i\partial _{t} \) in \(
L^{2}([\, 0,T\, ],\textrm{dt}) \)
and let\[
H=\sum _{m\in \N }h_{m}\, Q_{m}\]
be the spectral decomposition of \( H \) in \( \HH  \). Thus we
can write\[
\HH =\sideset {}{^{\oplus }}\sum _{m\in \N }\HH _{m}\]
where \( \HH _{m}=\Ran Q_{m} \) are the eigenspaces. We suppose that
the multiplicities are finite,\[
M_{m}=\dim \HH _{m}<\infty ,\textrm{ }\forall m\in \N .\]
Hence the spectrum of \( \bK  \) is pure point and its spectral
decomposition
reads\be
%\label{K_spec_decomp-def}
\bK =\sum _{k\in \Z }\sum _{m\in \N }(k\omega +h_{m})\, P_{k}\otimes
Q_{m},
\ee
implying a decomposition of \( \KK  \) into a direct sum,\[
\KK =\sideset {}{^{\oplus }}\sum _{(k,m)\in \Z \times \N }\Ran
(P_{k}\otimes Q_{m})\, .\]
Here is some additional notation. Set\begin{equation}
\label{Vknm_eq_int0T}
V_{knm}=\frac{1}{T}\int ^{T}_{0}e^{-ik\omega t}\, Q_{n}V(\omega t)\,
Q_{m}\, \textrm{dt}=\frac{1}{2\pi }\int ^{2\pi }_{0}e^{-ikt}\,
Q_{n}V(t)\, Q_{m}\, \textrm{dt}\in \BB (\HH _{m},\HH _{n})\, .
\end{equation}
Further,\[
\Delta _{mn}=h_{m}-h_{n},\]
and\[
\Delta _{0}=\inf _{m\neq n}|\Delta _{mn}|.\]
Finally we set\[
\mu _{mn}=M_{m}M_{n}.\]
Now we are able to formulate our main result. Though not indicated
explicitly in the notation the operator \( \bK +\bV  \) is considered
as depending on the parameter \( \omega  \).

\begin{theorem}
\label{MainTheorem}Fix \( J>0 \) and set \( \Omega _{0}:=[\,
\frac{8}{9}J,\frac{9}{8}J\, ] \).
Assume that \( \Delta _{0}>0 \) and that there exists \( \sigma >0 \)
such that\[
\Delta _{\sigma }(J):=J^{\sigma }\sum _{\substack {m,n\in \N \cr
\Delta _{mn}>J/2}}\frac{\mu _{mn}}{(\Delta _{mn})^{\sigma }}<\infty
\, .\]
Then for every \( r>\sigma +\frac{1}{2} \) there exist positive
constants
(depending, as indicated, on \( \sigma  \), \( r \), \( \Delta _{0} \)
and \( J \) but independent of \( V \)), \( \epsilon _{\star
}(r,\Delta _{0},J) \)
and \( \delta _{\star }(\sigma ,r,J) \), with the property: if\[
\epsilon _{V}:=\ \sup _{n\in \N }\sum _{k\in \Z }\sum _{m\in \N }\|
V_{knm}\| \, \max \{|k|^{r},1\}<\min \left\{ \epsilon _{\star
}(r,\Delta _{0},J),\frac{|\Omega _{0}|}{\delta _{\star }(\sigma
,r,J)}\right\} \]
(here \( |\Omega _{\ast }| \) stands for the Lebesgue measure of
\( \Omega _{\ast } \)) then there exists a measurable subset \(
\Omega _{\infty }\subset \Omega _{0} \)
such that \begin{equation}
\label{measureJinfty_geq_RHS}
|\Omega _{\infty }|\geq |\Omega _{0}|-\delta _{\star }(\sigma ,r,J)\,
\epsilon _{V}
\end{equation}
and the operator \( \bK +\bV  \) has a pure point spectrum for all
\( \omega \in \Omega _{\infty } \)
\end{theorem}
\begin{rem*} $\\$
1) It is sometimes necessary to consider
potentials ${\bf V}$ which depend on the frequency $\omega$ in a more
elaborate way. Suppose that $V:\R\times\R_+\to\BB(\HH)$ is a bounded
measurable function, which is $2\pi$ periodic with respect to the
first variable and such that for almost all $t\in\R$ and
$\omega\in\R_+$, $V(t,\omega)^\star=V(t,\omega)$; then $({\bf
V}\psi)(t)=V(\omega t,\omega)\psi(t)$ defines an operator family which
is uniformly bounded on $\KK$ with respect to the variable $\omega$.
Let now
$$
\epsilon_V:=\sup_{\omega,\omega'\in\Omega_0}\sup_{n\in\N}\sum_{k\in\Z,m\in\
N}
\left(\|V_{knm}(\omega)\|+J\|\partial
V_{knm}(\omega,\omega')\|\right)\max\{|k|^r,1\}
$$
where
$$
V_{knm}(\omega):=\frac{1}{2\pi}\int_0^{2\pi}e^{-ikt}Q_nV(t,\omega)Q_mdt
$$
and where the symbol \( \partial  \) is the discrete derivative
in \( \omega  \),
$$
\partial
V_{knm}(\omega,\omega'):=\frac{V_{knm}(\omega)-V_{knm}(\omega')}{\omega-
\omega'}.
$$
It is not difficult to check that Theorem~1 applies in exactly the same
conditions.\\
2) In the course of the proof is shown even more.
Namely, for all \( \omega \in \Omega _{\infty } \) and any eigenvalue
of \( \bK +\bV  \) the corresponding eigen-projector \( P \) belongs
to the Banach algebra with the norm\[
\| P\| =\sup _{n\in \N }\sum _{k\in \Z }\sum _{m\in \N }\|
P_{knm}\| \, \max \{|k|^{r-\sigma -\frac{1}{2}},1\}.\]
This shows that \( P \) is \( (r-\sigma -1/2) \)-differentiable
as a map from \( [\, 0,T\, ] \) to the space of bounded operators
in \( \HH  \). $\\$ 
3) The constants \( \epsilon _{\star }(r,\Delta _{0},J) \) and \(
\delta _{\star }(\sigma ,r,J) \)
are in fact known quite explicitly .
\[
\epsilon _{\star }(r,\Delta _{0},J)=\frac{\min \left\{4 \Delta
_{0},\, J\right\}}{270\, e^{3}} ,\]
and\begin{eqnarray*}
\delta _{\star }(\sigma ,r,J) & = & 720\pi
\, e^{5}2^\sigma\left(
\frac{2\sigma +1}{\left( 1-e^{-\frac{2}{r}}\right) e}\right) ^{\sigma
+\frac{1}{2}}\left( \sum _{s=1}^{\infty
}s^{2}e^{-\frac{2}{r}(r-\sigma -\frac{1}{2})s}\right) \Delta _{\sigma
}(J)\\
 & = & 720\pi\left( \frac{2\sigma +1}{\left( 1-e^{-\frac{2}{r}}\right)
e}\right) ^{\sigma +\frac{1}{2}}2^\sigma e^{3+\frac{2}{r}\left(
\sigma +\frac{1}{2}\right) }\frac{1+e^{-2+\frac{2}{r}\left( \sigma
+\frac{1}{2}\right) }}{\left( 1-e^{-2+\frac{2}{r}\left( \sigma
+\frac{1}{2}\right) }\right) ^{3}}\, \Delta _{\sigma }(J)
\end{eqnarray*}
where we used the identity
\[
\sum _{s=1}^{\infty }s^{2}e^{-2xs}=\frac{\cosh (x)}{4\, \sinh
(x)^{3}}
\]
\end{rem*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%% Models %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Two models} \label{Models}
$\\$
We conclude this subsection with a brief description of two models
illustrating
the effectiveness of Theorem \ref{MainTheorem}. In the first model
we set \( \HH =L^{2}([\, 0,1\, ],\textrm{dx}) \), \( H=-\partial
_{x}^{2} \)
with Dirichlet boundary conditions, and \( V(t)=z(t)x^{2} \) where
\( z(t) \) is a sufficiently regular \( 2\pi  \)-periodic function.
As shown in \cite{Seba} the spectral analysis of this simple model
is essentially equivalent to the analysis of the so called quantum
Fermi accelerator. The particularity of the latter model is that the
underlying Hilbert space itself is time-dependent, \( \HH
_{t}=L^{2}([\, 0,a(t)\, ],\textrm{dx}) \)
where \( a(t) \) is a strictly positive periodic function. The
time-dependent
Hamiltonian is \( -\partial _{x}^{2} \) with Dirichlet boundary
conditions.
Using a convenient transformation one can pass from the Fermi
accelerator
to the former model getting the function \( z(t) \) expressed in
terms of \( a(t) \), \( a'(t) \) and \( a''(t) \). But let us return
to the analysis of our model. Eigenvalues of \( H \) are
non-degenerate,
\( h_{m}=m^{2}\pi ^{2} \) for \( m\in \N  \), with normalised
eigenfunctions
equal to \( \sqrt{2}\sin (m\pi x) \).
A straightforward
calculation gives\[
V_{knm}=z_{k}\times \left\{ \begin{array}{c}
\frac{8(-1)^{m+n}mn}{(m^{2}-n^{2})^{2}\pi ^{2}}\quad \textrm{if
}m\neq n,\\
\\
\frac{1}{3}-\frac{1}{2m^{2}\pi ^{2}}\quad \textrm{ if }m=n,
\end{array}\right. \]
where \( z_{k}=\frac{1}{2\pi }\int _{0}^{2\pi }e^{-ikt}z(t)\, dt \)
is the Fourier coefficient of \( z(t) \). Hence one derives that\[
\epsilon _{V}=\sup _{n\in \N }\left( \frac{1}{3}+\frac{2}{n^{2}\pi
^{2}}+\frac{4}{\pi ^{2}}\sum ^{n-1}_{j=1}\frac{1}{j^{2}}\right) \sum
_{k\in \Z }|z_{k}|\max \{|k|^{r},1\}=\sum _{k\in \Z }|z_{k}|\max
\{|k|^{r},1\}.\]
For any \( J>0 \), \( \Delta _{\sigma }(J) \) is finite if and only
if \( \sigma >1 \). On the other hand, to have \( \epsilon _{V} \)
finite it is sufficient that \( z(t)\in C^{s} \) where \(
s>r+1>\sigma +\frac{1}{2}+1>\frac{5}{2} \).
So \( z(t)\in C^{3} \) suffices for the theory to be applicable.
This may be compared to an older result in \cite{DuclosStovicek},
\S4.2, giving a much worse condition, namely \( z(t)\in C^{17} \).
$\\ \\$
The second model is the pulsed rotator in \( N \) dimensions. In
this case \( \HH =L^{2}(S^{N},d\mu ) \), with \( S^{N}\subset \R
^{N+1} \)
being the \( N \)-dimensional unit sphere with the standard
(rotationally
invariant) Riemann metric and the induced normalised measure \( d\mu
\),
and \( H=-\Delta _{LB} \) is the Laplace-Beltrami operator on \(
S^{N} \).
The spectrum of \( H \) is well known, \cite{Nenciu}: \( \Spec
(H)=\{h_{m}\}_{m=0}^{\infty } \),
where\[
h_{m}=m(m+N-1)\]
and the multiplicities are\[
M_{m}={m+N\choose N}-{m+N-2\choose N}.\]
The time-dependent operator \( V(t) \) in \( \HH  \) acts via
multiplication,
\( (V(t)\varphi )(x)=v(t,x)\varphi (x) \), where \( v(t,x) \) is
a real measurable bounded function on \( \R \times S^{N} \) which
is \( 2\pi  \)-periodic in the variable \( t \). Consequently, \( \KK
\cong L^{2}([\, 0,T\, ]\times S^{N},dt\, d\mu ) \)
and \( (\bV \psi )(t,x)=v(\omega t,x)\psi (t,x) \). Note that the
asymptotic behavior of the eigenvalues and the multiplicities, as
\( m\to \infty  \), is \( h_{m}\sim m^{2} \), \( M_{m}\sim
(2/(N-1)!)\, m^{N-1} \).
So \( \Delta _{\sigma }(J) \) is finite, for any \( J>0 \), if and
only if\[
\sum
_{m^{2}-n^{2}>J/2}\frac{(mn)^{N-1}}{(m^{2}-n^{2})^{\sigma
}}<\infty \, .\]
To ensure this condition we require that \( \sigma
>2(N-1)+1 \).
Let us assume that there exist \( s,u\in \Z _{+} \) such that, for
any system of local (smooth) coordinates \( (y_{1},\ldots ,y_{N}) \)
on \( S^{N} \), the derivatives \( \partial _{t}^{\, \alpha }\partial
_{y_{1}}^{\, \beta _{1}}\ldots \partial _{y_{N}}^{\, \beta
_{N}}v(t,y_{1},\ldots ,y_{N}) \)
exist and are continuous for all \( \alpha  \), \( \beta  \), \(
\alpha \leq s \)
and \( \beta _{1}+\ldots +\beta _{N}\leq u \). If \( u\geq 4 \)
then \( [\, H,[\, H,V(t)\, ] \){]} is a well defined second order
differential operator with continuous coefficient functions and the
operator \( [\, H,[\, H,V(t)\, ]](1+H)^{-1} \) is bounded. Clearly,\[
\frac{(h_{m}-h_{n})^{2}}{1+h_{m}}\, Q_{n}V(t)Q_{m}=Q_{n}[\, H,[\,
H,V(t)\, ]](1+H)^{-1}Q_{m}.\]
Using this relation one derives an estimate on \( V_{knm} \),\[
\| V_{knm}\| \leq \textrm{const}\, \frac{1+\min
\{h_{n},h_{m}\}}{|k|^{s}(h_{m}-h_{n})^{2}}\, ,\]
valid for \( k\neq 0 \) and \( m\neq n \). The number\[
\sup _{n\in \Z _{+}}\sum _{m\in \Z _{+},\, m\neq n}\frac{1+\min
\{h_{n},h_{m}\}}{(h_{m}-h_{n})^{2}}\]
is finite. To see it one can employ the asymptotics of \( h_{m} \)
and the fact that the sequence\[
a_{n}=\sum _{m\in \Z _{+},\, m\neq n}\frac{1+\min
\{n^{2},m^{2}\}}{(m^{2}-n^{2})^{2}}=\left( 1+\frac{1}{n^{2}}\right)
\frac{\pi
^{2}}{12}-\frac{3}{16n^{2}}+\frac{5}{16n^{4}}-\frac{1}{2n}\sum
^{2n-1}_{m=1}\frac{1}{m}\, ,\]
\( n=1,2,3,\ldots  \), is bounded. It follows that the norm \(
\epsilon _{V} \)
is finite if \( s>r+1>\sigma
+\frac{1}{2}+1>2(N-1)+1+\frac{3}{2}=2N+\frac{1}{2} \).
Thus the theory is applicable provided \( u\geq 4 \) and \(
s>2N+\frac{1}{2} \).
The same example has also been treated by adiabatic methods in
\cite{Nenciu}.
In that case the assumptions are weaker. It suffices that \( v(t,x) \)
be \( (N+1) \)-times differentiable in \( t \) with all derivatives
\( \partial _{t}^{\, \alpha }v(t,x) \), \( 0\leq \alpha \leq N+1 \),
uniformly bounded. However the conclusion is somewhat weaker as well.
Under this assumption \( \bK +\bV  \) has no absolutely continuous
spectrum but nothing is claimed about the singular continuous
spectrum.







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