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\begin{document}

\begin{titlepage}

\begin{center}

{\large CZECH TECHNICAL UNIVERSITY IN PRAGUE}\\


Faculty of Nuclear Sciences and Physical Engineering\\

\vspace{30ex}

{\Large\bf V\'YZKUMN\'Y \'UKOL}\\

\vspace{10ex}

{\Large  EGOROV THEOREM FOR TIME DEPENDENT CASE }\\


\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 Prague}\\

\vspace{3ex}

{\Large 30.8. 2001}

\end{center}

\end{titlepage}



\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% INTRODUCTION (section 1) %%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
\hskip25pt
According to the Bohr's correspondance principle, the quantum evolution of an observable
is close to its classical evolution as the Planck constant $\hbar$ is small. 
In the mathematical literature this result is known as Egorov Theorem \cite{ho}, \cite{ro}.
There is a lot of semiclassical methods, which attack classical limit in quantum mechanics. Nice summary of these methods, with large overview of literature
dealing with this topic is \cite{ro2}. There is also different point of view,
WKB approximation in Feynman path integral \cite{schu}.   

The goal of this work is to prove that quantum evolution of an obbservable,
in the system with time dependent Hamiltonian, may be computed 
(up to an error of order $\hbar^2$) 	with help of the classical flow, 
if the observable, Hamiltonian and the classical flow are bounded in
suitable sense. Physical interpratention of this fact is clear. When 
the quantities of the dimension same as $\hbar$, like $E.t$, $p.q$ are much larger
then numerical value of $\hbar$ (i.e. that is possible limit $\hbar 
\rightarrow 0$), then quantum evolution of an observable
tends to the classical one. 
The theory of this type use so called Weyl-quantization, as powerfull
tool to construct quantum observables (i.e. symmetric or even 
essentially self-adjoint operators) according to the principal of
corespondance. 

Starting point of this work is the preprint \cite{br}. This work deals only 
with $C^\infty$ case, but the statement of the main theorem is improved. 
The assumptions on Hamiltonian are weaker (it is possible cubic increasment in 
infinity) and the error is estimated to order $\hbar^2$ instead of $\hbar$ 
in \cite{br}. 

The heart of this work is the time dependent case. It is schown (for 
Hamiltonian of the Schr\"odinger type) that even in this more general 
case the consequences of the semiclassical theorem are the same, if observable
depends only on coordinate nad momentum.

I would like to thank very much to 
Prof. Ing. Pavel \v{S}\v{t}ov\'{\i}\v{c}ek, DrSc, supervisor of this work
for his perfect leadership. Also i would like to thank very much to
the consultant of this work, Prof. Pierre Duclos, from
 Universit\'e de Toulon et du Var, France for intoduction in this problematics.
Finally i want to thank to Dr. Michel Vittot and Prof. Pierre Duclos for their
care during my stage in Marseille, where the main part of this work came into 
begin.  

\newpage
\section{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
\[
J =  \pmatrix{0 &\quad \Id \cr
-\Id &\quad 0 \cr}.
\]
 A generic point in $Z$ is denoted $z$ and its coordinates by $(q, p)$ where $q, p \in \R^n$. We will also use notation $\langle x \rangle=(1+|x|^2)^{1/2}$.
 \\
A classical Hamiltonian is a smooth real function $H:Z\rightarrow \R$. Our basic
example will be $H(q,p) = \frac{\Vert p\Vert^2}{2m} + V(q)$ ($m>0$)
where $\Vert p\Vert^2 = <p,p>$.
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's equations
\b\label{ham}
\frac{dq}{dt} = \frac{\partial H}{\partial p}(q,p),\;\;
\frac{dp}{dt} = -\frac{\partial H}{\partial q}(q,p).
\e
The equations (\ref{ham}) generate a flow $\Phi^t$ on the phase space $Z$, defined by \\
$\Phi^t(q(0),p(0)) = (q(t),p(t))$; $\Phi^0 = \Id$. $\Phi^t$ exists locally
by the Cauchy-Lipchitz Theorem for O.D.E. But we need more assumptions
on $H$ to define $\Phi^t$ globally on $Z$.\\
 $\Phi^t$ defines a symplectic diffeomorphism (canonical transformation) group of transformations on $Z$.
Let us consider a classical observable $A$, i.e $A$ a smooth real valued function
defined on phase space $Z$.
The time evolution of $A$ can be easily computed

\b\label{cl.prog}
\frac{d}{dt}A(\Phi^t(z)) = \{H, A\}(\Phi^t(z)),\;\; z=(q,p)
\e
where $\{H, A\}$ is the Poisson bracket defined by
\[
\{H, A\} = \partial_qH\cdot\partial_pA - \partial_pH\cdot\partial_qA.\label{poisson}
\]
Here we have used the notation $\partial_q = \frac{\partial}{\partial q}$.
Now let us assume that $H, A$ are quantizable. That means that we can associate to them
the quantum observables $\hat{H}$ and $\hat{A}$\\
 i.e self-adjoint operators in
$L^2(X)$. By solving formally the Schr\"odinger equation~:
$i\hbar\partial_t\psi_t = \hat{H}\psi_t$, we can define the one parameter group
of unitary operators \\
 $U(t) = \exp\left(-\frac{it}{\hbar}\hat{H}\right)$.
The quantum time evolution of $\hat{A}$
is then given by 
\[
\hat{A}(t) = U(-t)\hat{A}U(t)
\]
which satisfies
the Heisenberg-von Neumann equation

\b\label{hevn}
\frac{d\hat{A}(t)}{dt} = \frac{i}{\hbar}[\hat{H}, \hat{A}],
\e
where $[A, B] = AB - BA$ is the commutator of $A, B$.



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



\newpage
\section{Weyl quantization}

\begin{definition}
$A\in {\cal O}(m)$, $m\in\R$, 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 $A\in {\cal O}(m)$. We will define  $\hat{A}$ as operator from 
${\cal S}(Z) $ to itself, called $\hbar$-Weyl quantization of $A$,  by the following formula,
with $\psi\in{\cal S}(X)$,
\ba \label{weyl}
{\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.
\ea
\end{definition}
\vskip5pt
{\bf Example:}
Let us focus for example to observable $A:=x.p$. There are two 
possibilities how to compute $\hat{A}$.

\baa
&&1. \hskip10pt p\exp(\frac{i}{\hbar}(x-y)p)=-\frac{\hbar}{i}
\partial_y\exp(\frac{i}{\hbar}(x-y)p) \rightarrow \hat{A}=PQ \\
&&2. \hskip10pt p\exp(\frac{i}{\hbar}(x-y)p)=\frac{\hbar}{i}
\partial_x\exp(\frac{i}{\hbar}(x-y)p) \rightarrow \hat{A}=\frac{1}{2}(PQ+QP) 
\eaa
$P$ and $Q$ are Weyl quantizations of $p$ and $q$. The difference
between these two possibilities is equal to $\frac{1}{2}[P,Q]$,
of course is connected with usual ordering
problems in quantum mechanics and is proportional to $\hbar$.
\\
For $A\in {\cal S}(Z)$ we will compute, (with help of integral kernel) inverse prescription of Weyl symbol if operator is given as integral operator.

\[
({\hat A} \psi)(x)= (2\pi \hbar)^{-n}\int\!\!\int_Z A\left(\frac{x+y}{2}, p\right)
{\rm e}^{i\hbar^{-1}\langle x-y, p\rangle} \psi(y)dydp=
\int_{\R^n} {\cal K}_A (x,y) \psi (y) dy
\]

\[
{\cal K}_A (x,y):= (2\pi \hbar)^{-n}\int_{\R^n} A\left(\frac{x+y}{2}, p\right)
{\rm e}^{i\hbar^{-1}\langle x-y, p\rangle} dp
\]
After inverse Fourier transform.
\b\label{symbol}
A(x,\xi)=\int_{\R^n}{\cal K}_A \left(x+\frac{v}{2},x-\frac{v}{2}\right)
{\rm e}^{-\frac{i}{\hbar} v{\cdot}\xi}dv
\e
\\
The class $ {\cal O}(m)$ has nice properties. For example, 
if  $A\in {\cal O}(m)$ is real-valued function, then $\hat{A}$ is symmetric
operator in $L^2(X)$, with domain ${\cal S}(X)$. Further there exists a
product formula. For proof one can see \cite{ro}, \cite {fo} or \cite{ho}.

\newpage

\begin{theorem}Let $A\in {\cal O}(m)$,  $B\in {\cal O}(p)$, then there exists  $C\in {\cal O}(m+p)$, such that 
\[ 
 \hat{A}\cdot\hat{B}=\hat{C} 
\]
And there exists following asympotic 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 serie 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

\b \label{remainder}
C(z) - \sum_{0\leq j \leq N}\hbar^jC_j(z) =: R_N(A,B;z;\hbar).
\e

\end{theorem}
\vskip8pt
We will show this for $A,B\in {\cal S}(Z) $
\[
\left(\hat{A}\cdot\hat{B}\psi\right) (x)
= \int_{\R^n} {\cal K}_{A\cdot B} (x,y) \psi (y) dy
\]
where
\[
{\cal K}_{A\cdot B} (x,y)=
\frac{1}{(2\pi \hbar)^{2n}} 
  \int_{\R^{3n}} A\left(\frac{x+z}{2}, p\right)B\left(\frac{z+y}{2}, \eta 
\right)
{\rm e}^{-\frac{i}{\hbar}\langle x-z, p\rangle +\langle z-y, \eta \rangle
} dzdpd\eta
\]
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}).
\baa
C(x,\xi) &=& \int_{\R^n}{\cal 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 -v \xi \right)} dzdpd\eta dv
\eaa
We proceed 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

\baa
&&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} 
\eaa
we obtain
\baa
 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).
\eaa

\[
z=(x,\xi)\in Z  
\]
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
\baa
&& 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).
\eaa
Finally 
\baa
 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).
\eaa
Remainder (\ref{remainder}) can be also expressed as
\baa
&& 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 
\eaa
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
\baa
&&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+1}(D_u,D_v)
A(u+z)B(v+z)dudvd\tau.
\eaa  
Now we estimate this remainder and his derivatives.
\begin{theorem} \label{estimR}
 There exists $K_n>0$ and for every $m\in \N$, $m\geq 4n$, for
every $s > 4n$ there exists a constant $\rho_{n,m,s}$ such that
for every $A, B \in {\cal S}(Z)$, for every $N\geq 1$, for every multi-index $\gamma$,
the following estimate holds, for every $z\in Z$,
\ba\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]
\ea
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 $F\in{\cal S}(Z\times Z)$ and the integral
\baa
I(\lambda) = \lambda^{2n}\int_{Z\times Z}\exp[-i\lambda \sigma(u, v)]F(u,v)dudv.
\eaa
Then for every real number $s>4n$ and every integer $m\geq 4n$ there exists\\ 
$\kappa(n,s,m) >0$ depending only on $n, s, m$ (but independent of $F$)
such that \\ the following  estimate holds
\baa
\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.
\eaa
\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
\baa
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.
\eaa
\\
For $I_0(\lambda)$, we easily have
\baa
\vert I_0(\lambda)\vert \leq \omega_{4n}\sup_{u^2+v^2\leq 1} \vert F(u,v)\vert,
\eaa
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
\baa
L = \frac{i}{u^2+v^2}\left(Ju\frac{\partial}{\partial v}-Jv\frac{\partial}{\partial u}\right),
\eaa
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 $4n$ integrations by parts, we can see that it exists a constant $c_n$
such that
\baa
\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.
\eaa
Similarly, performing $m$ integrations by parts,

\baa
\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& \int_{Z \times Z}\vert\frac{u+v}{u^2+v^2}\vert^m
 \sup_{\vert \mu\vert+\vert \nu\vert\leq m} 
\vert\partial^\mu_u \partial^\nu_v
 \left(1-\chi_0 (\frac{1}{\lambda}(u^2+v^2)\right)
F(\frac{u}{\sqrt\lambda},\frac{v}{\sqrt\lambda})\vert dudv. 
\eaa
Using existence of integral
\[
\int_{Z \times Z}\vert\frac{u+v}{u^2+v^2}\vert^m 
(1+u^2+v^2)^{\frac{m-s}{2}}dudv
\]
we get for a constant $c(n,s,m)$,
\baa
\vert I_2(\lambda)\vert \leq c_n
\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.
\eaa
\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 $u=(x,\xi)$, $v=(y,\eta)$,
\baa
\vert \sigma^N(\partial_x,\partial_\xi; \partial_y,\partial_\eta))F(x,\xi)G(y,\eta)\vert \leq
\nonumber\\
(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.
\eaa
\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 $A\in{\cal O}(\mu_A)$, $B\in{\cal O}(\mu_B)$,
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 %%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\newpage 
\section{Main theorem}

\begin{theorem}\label{main}
Let us consider a Hamiltonian H and an observable A satisfying
\ba\label{asham}
\vert\partial_z^\gamma H(z)\vert \leq C_{\gamma},\; &{\rm for}&\;
\vert\gamma\vert \geq 3 \nonumber \\
\vert\partial_z^\gamma A(z)\vert\leq C_\gamma, \;&{\rm for}&\; 
\vert\gamma\vert \geq 1  \\
\vert\partial_z^\gamma \Phi^\tau(z)\vert\leq K_\gamma(t), \;
&{\rm for}& \; \vert \gamma\vert \geq 1,  \; \tau \in [0,t] \nonumber
\ea
If moreover $\hat{H}, \hat{A}$ are essentially self-adjoint operators 
in $L^2(X)$, with core ${\cal S}(X)$, 
then the quantum evolution \hbox {$U(t) =\exp(-\frac{it}{\hbar}\hat{H})$}
is well defined for all $t\in\R$.
And the quantum evolution of observable $\hat{A}$ can be approximated by the 
operator with the Weyl symbol
\b
\label{A0}
A_0(t,z)= A(\Phi^t(z))
\e
in the following sense:
\[
\Vert \hat{A}(t) - \widehat{A_0(t)}  \Vert_{L^2} \leq 
 t\hbar^2 K_{n,m,s}(t) 
\]


\end{theorem}
{\bf Proof}: 
By Taylor formula
\[
H \in {\cal O}(4) \;, \;\; A\in {\cal O}(2)
\]
For any $\psi \in \cal{S} (X)$ it holds true that 
\[
\frac{d}{dt}\widehat{A_0} \psi= \widehat{(\frac{d}{dt}{A_0})}\psi
\]  
Under assumptions propagator exists and conserves $\cal{S} (X)$, because Hamiltonian does. 
On Schwartz space holds equation
\ba \label{basic}
U(-t)\hat{A}U(t) - \widehat{A_0}(t) =
\int_0^t U(-s)\left(\frac{i}{\hbar}
[\hat{H}, \widehat{A_0}(t-s)] - \widehat{\{{H}, A_0\}\Phi^{t-s}}\right)U(s)ds 
\ea
where (\ref{cl.prog}), (\ref{hevn}), conservation of Poisson bracket and Hamiltonian along the classical flow were used.
\hskip5pt
With help of (\ref{AB}) we compute that the Weyl symbol of 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
\ba \label{R2}
&& R_2(H,A;z,\hbar)= \\
&& \left(\frac{\hbar}{2i} \right)^3
\int_0 ^1  \frac{(1-\tau)^N}{(\pi \hbar \tau)^{2n}}
\int_{Z\times Z} \exp(-\frac{2i}{\tau\hbar}\sigma(u,v))\sigma^{3}(D_u,D_v)
A(u+z)B(v+z)dudvd\tau. \nonumber
\ea
Using (\ref{asham}) we deduce 
\baa
&& K_{n,m,s}(t) := \\   
&&\sup_{(*)}\left[(1+u^2+v^2)^{(s-m)/2}
\vert\partial_u^{(\alpha,\beta)+\mu}H(u+z) \vert 
\vert\partial_v^{(\beta,\alpha)+\nu} A(\Phi^{\tau}(u+z)
\vert\right]< \infty \nonumber
\eaa
where $\displaystyle{\sup_{(*)}}$ means that the supremum holds
under the conditions\\ $u, v, z \in Z, \tau\in [0,t]
$, $\vert \mu\vert +\vert \nu\vert\leq m+
[n/2]+4$,
$\vert\alpha\vert+\vert\beta\vert =3$ ($\mu, \nu\in\N^{2n}$, $\alpha, \beta\in\N^n$). 
There is also estimate on derivatives of composition of functions
\[
\vert \partial^\alpha(f\circ g) \vert\leq K_{n,\vert\alpha\vert}
\sup_{\beta, \gamma  \neq 0  \atop \vert\beta\vert,\vert\gamma\vert\leq
\vert \alpha \vert}
\vert \partial^\beta f \vert\vert \partial^\gamma g \vert
\]
But there is a Calderon- Vaillancourt Theorem with improvement by A. 
Boulkhemair \cite{bo} , to estimate the norm of Weyl-quantized operator 
\b\label{cv}
\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.
\e
\\
Aplying $L^2$ norm to (\ref{basic}), combining (\ref{cv}) with 
the statement of the {\bf Theorem} \ref{estimR},
we finish the proof.  
\QED





\newpage



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%% time dependent case %%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section {Time dependent case}

We will consider the time dependent Hamiltonian of special type
\[
H= \frac{1}{2}p^2+V(q,t)
\]
on line.
The trick is to extend the phase space about the time and energy,
to define new parameter $\sigma$ which describes the trajectory
and to take Floquet Hamiltonian instead of original one.
\[ \label{floq}
K(q_1,q_2,p_1,p_2):= \frac{1}{2}p_1+V(q_1,q_2)+p_2,
\] 
where 
\baa
&& q_1=q  \;\; \rm{is \;the\;original \; coordinate} \\
&& p_1=p  \;\; \rm{is \;the\;original \; momentum}  \\
&& q_2=t  \;\; \rm{is \; time}\\
&& q_1=E  \;\; \rm{is \;energy.}
\eaa
New Hamilton equations of motion will be

\baa 
&&\frac{dq_1}{d\sigma} = \frac{\partial K}{\partial p_1},  \;\; 
\frac{dp_1}{d\sigma} =- \frac{\partial K}{\partial q_1} \\
&&\frac{dq_2}{d\sigma} = \frac{\partial K}{\partial p_2},  \;\; 
\frac{dp_2}{d\sigma} = -\frac{\partial K}{\partial q_2} 	
\eaa
Hence
\ba \label{he}
&&\frac{dq}{d\sigma} =  p , \;\; 
\frac{dp}{d\sigma} =- \frac{\partial V}{\partial q} \\
&&\frac{dt}{d\sigma} =1 , \;\;
\frac{dE}{d\sigma} =- \frac{\partial V}{\partial t} \nonumber
\ea
We need initial conditions
\b \label{pocpod}
q(\sigma=0)=q, \; p(\sigma=0)=p, \; t(\sigma=0)=t, \;E(\sigma=0)=E
\e
But because the right hand sides of Hamiltonian equations 
for $q, p, t$ does not depend on energy $E$ and so doesn`t the initial 
conditions, we may conclude that classical flow $\Phi^{\sigma}$ 
will be of this type
\[ \label{init}
\Phi^\sigma
 \pmatrix{q \cr
          p \cr
	  t \cr
          E \cr}
=\pmatrix{q(q,p,t,\sigma) \cr
          p(q,p,t,\sigma) \cr
	  \sigma+t \cr
          E(q,p,t,\sigma) \cr}
\]
So we see that the trajectories of $q, p, t$
are independent on energy $E$. 
This cause that we can make weaker assumptions on potential $V$, 
then in case of $\R^4$.

\begin{theorem}
Consider the time dependent Hamiltonian of type
\[
H= \frac{1}{2}p^2+V(q,t)
\]
on line. Denote the classical flow $\Phi$, given by Hamilton equations in
extended space (\ref{he}). Suppose that
\baa
\vert\partial_x^{3+\alpha} \partial_t^\beta  V(x,\tau)\vert \leq
C_{\alpha \beta}, \; &{\rm for}&  \; \tau \in [0,t] \\
\vert\partial_z^\gamma A(z)\vert\leq C_\gamma, \;&{\rm for}&\; 
\vert\gamma\vert \geq 1  \\
\vert\partial_z^\gamma \Phi^\tau(z)\vert\leq K_\gamma(t), \;
&{\rm for}& \; \vert \gamma\vert \geq 1,  \; \tau \in [0,t] 
\eaa
\end{theorem}
{\bf Proof:}
The situation is almost the same as in the {\bf Theorem} \ref{main},
except the expression (\ref{R2}), which may by substituted by the following
better estimate
\baa 
&& R_2(K,A;z,\hbar)= \\
&& \left(\frac{-\hbar}{3!2i} \right)^3
\int_0 ^1  \frac{(1-y)^N}{(\pi \hbar y)^{2n}}
\int_{\R ^8} \exp(-\frac{2i}{y\hbar}(ec+fd-ga-hb)) \times \\
&& \times
\partial^3 _x V(x+a,t+b)\partial^3 _p A\circ \Phi^\tau(x+e,t+f,p+g)dadbdc
dddedfdg.
\eaa
End of the proof follows the proof of {\bf Theorem} \ref{main} .
\QED


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%% examples %%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newpage

\section{Examples}
\vskip10pt
{\bf Example 1:} \\
Firstly, we will deal with time independent harmonic oscilator.
Hamiltonian is 
\[
H=\frac{1}{2}(p^2+q^2)
\]
Either by direct computations of the remainder (\ref{AB}), 
or by estimate (\ref{remest1}), one can see that in this case 
semiclassical approximation of quantum evolution of an 
observable (\ref{A0}) is exact solution (even any observalbe
$A \in {\cal O} (m)$ without restrictions (\ref{asham}) ).
If we observe expectation values $\langle \psi, \hat{A}\psi \rangle$
of such quantities as $\hat{q},\; \hat{p}$, which 
are linear, on wave packets 
\[
\Psi_{a,b,\; (\Delta a)^2} (x):=(2\pi(\Delta a)^2)^{-\frac{1}{4}}
\exp(-\frac{(x-a)^2}{4(\Delta a)^2 }+\frac{i}{\hbar}bx )
\] 
then thanks to relations
\baa
\langle \Psi_{a,b,\; (\Delta a)^2}, \hat{q}\;\Psi_{a,b,\;(\Delta a)^2} 
\rangle=a \\
\langle \Psi_{a,b,\; (\Delta a)^2}, \hat{p\;}\Psi_{a,b,\;(\Delta a)^2}
\rangle=b
\eaa 
expactation values follow exactly the classical evolution of these
quantities.
On the other hand if we focus on hamiltonian, which is of course 
conserving, then because of dispersion of mimimalizing wave packets:
\baa
(\Delta_{\Psi_{a,b,\; (\Delta a)^2}}\; \hat{q})^2 &=&  (\Delta a)^2 \\
(\Delta_{\Psi_{a,b,\; (\Delta a)^2}}\; \hat{p})^2 &=& 
\frac{\hbar^2}{4(\Delta a)^2} =(\Delta b)^2
\eaa
we yield
\[
\langle \Psi_{a,b,\; (\Delta a)^2}, \hat{H}\;\Psi_{a,b,\;(\Delta a)^2} 
\rangle= \frac{1}{2}(a^2+b^2)+(\Delta a)^2+\frac{\hbar^2}{4(\Delta a)^2}
 =\frac{1}{2}(a^2+b^2)+(\Delta a)^2+(\Delta b)^2
\]
Additional terms are not quantum effects, but they are
 the consequences of error during measuring.
\vskip10pt
{\bf Example 2:}
\\
Our next Hamiltonian will be
\[
H:= \frac{1}{2}p^2-q\cos t
\]
Hamilton equations in extended space (\ref{he}) are
\baa 
&&\frac{dq}{d\sigma} =  p , \;\; 
\frac{dp}{d\sigma} = \cos t \\
&&\frac{dt}{d\sigma} =1 , \;\;
\frac{dE}{d\sigma} = q\sin t
\eaa
The solution with initial conditions (\ref{pocpod}) is desribed by relations:

\baa
q(\sigma) &=& q+(p-\sin t)\sigma-\cos (t+\sigma)+\cos t  \\ 
p(\sigma) &=& p+\sin (t+\sigma)-\sin t  \\ 
t(\sigma) &=& \sigma+t \\ 
E(\sigma) &=&  E+(\cos t+q)(\cos t-\cos (t+\sigma)+
\frac{1}{4}(\cos 2(t+\sigma)-\cos 2t)+\\
&+&(p-\sin t)(\sin (t+\sigma)-\sigma\cos (t+\sigma)-\sin t)
\eaa
We will not care about observables explicitly dependent on $t,\; E$.
After Weyl quantization procedure (\ref{weyl}) in extended space, we
discover that the result would be the same (except the measuring of 
the time from point $t$, instead of $0$), if we computeted it directly as time
independent hamiltonian.

\newpage
\tableofcontents
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%% REFERENCES %%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\bibitem[Sch]{schu} L.S. Schulman, {\em Techniques and Applications of Path 
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\end{thebibliography}
\end{document}



