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To appear on Rendiconti Lincei, Matematica e Applicazioni
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Lie-Dyson expansion, uniform convergence, oscillator modes
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\begin{document}
\baselineskip=20pt
%\date{15 February 2005}
\title[Uniform convergence of the Lie-Dyson expansion with respect to  $\hbar$]{Uniform convergence of the Lie-Dyson expansion  with respect to the Planck constant}
\author{Dario Bambusi,  Mirko Degli Esposti,  Sandro Graffi}
\address{ Dipartimento di
Matematica ``F. Enriques'', Universit\`a di Milano \newline \hphantom{Wi} Via
Saldini 50, 20133 Milano, Italy.}
\email{bambusi@mat.unimi.it}
\address{Dipartimento di Matematica,  Universit\`a di Bologna  \newline
\hphantom{Wi} Piazza di Porta S Donato 5, 40127 Bologna, Italy.}
\email{desposti@dm.unibo.it, graffi@dm.unibo.it {\tiny
(corresponding author)}}

\begin{abstract}We prove that the Lie-Dyson expansion for the Heisenberg observables  has a nonzero convergence radius in the variable $\ep t$ which does not depend on the Planck constant $\hbar$. Here  the quantum  evolution  $U_{\hbar,\ep}(t)$ is generated by the \Sc\  operator  defined by the maximal action  in $L^2(\R^n)$   of  $-\hbar^2\Delta+\Q+\ep V$;  $\Q$ is a positive definite quadratic form on $\R^n$;  the observables and $V$ belong to a suitable class of pseudodifferential operators with analytic symbols.   It is furthermore proved that, up to an error of order $\ep$, the time required for an exchange of energy between the unperturbed oscillator modes  is  exponentially long time independently of $\hbar$.
\end{abstract}
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\begsection{Introduction and statement of results}
\setcounter{equation}{0}%
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Consider in $L^2(\R^{n})$ the \Sc\ operator family
\begin{eqnarray} \label{hn}
H(\ep)=H_0+\ep V, \qquad
H_{0}=-\frac{\hbar^2}{2m}\Delta+Q.
\end{eqnarray}
Here the $Q$ is the maximal multiplication operator by a non-negative quadratic form  $\Q(x):\R^n\to\R$,  $V$ is a
(semiclassical) pseudodifferential operator of order $0$ and $\ep\in\R$.   Denote  $H^2(\R^{n})$  the usual Sobolev space,
and $\hat{H}^2(\R^{n})$ its conjugate by Fourier transform,
$$
%\label{F}
%\hat{H}^2(\R^{n}):=\{f\in L^2(\R^{n})\,|\, \hat{f}\in H^2(\R^{n})\}; \quad
\hat{f}(s)=h^{-n/2}\int_{\R^n}f(x)e^{-i\la x,s\ra/\hbar}\,dx
$$
namely,
\begin{align}
\label{F} \hat{H}^2(\R^{n}):=\{\hat{f}\,\,\,|\, \int_{\R^n}(1+\vert
s\vert ^2)^2\vert\hat{f}(s)\vert^2\,ds\,<\,+\infty \}; \quad
%\hat{f}(s)=h^{-n/2}\int_{\R^n}f(x)e^{-i\la x,s\ra/\hbar}\,dx
\end{align}
%\begin{align}
%\label{F}
%\hat{H}^2(\R^{n}):=\{f\in L^2(\R^{n})\,|\, \hat{f}\in H^2(\R^{n})\}; \quad
%\hat{f}(s)=h^{-n/2}\int_{\R^n}f(x)e^{-i\la x,s\ra/\hbar}\,dx
%\end{align}
Under the above conditions $H_0$, defined on the domain
$H^2(\R^{n})\cap \hat{H}^2(\R^{n})$ is a self-adjoint operator.
Hence the unitary group $\ds U_0(t):=e^{i\hbi H_0 t}$, $t\in\R$,
exists. Moreover, $V$ is relatively bounded with respect to $H_0$;
therefore the \op\ family  $H(\ep): \ep\in\R$ defined on $D(H_0)$ is
also self-adjoint so that the corresponding unitary group $\ds
U_\ep(t):=e^{i\hbi H_\ep t}$, $t\in\R$, exists. Given a continuous
quantum observable, represented by a continuous self-adjoint
operator  $G$ on $L^2(\R^{n})$ , consider the corresponding
Heisenberg observables under the free and the full evolution,
respectively: \be \label{H} G_0(t):=U_0(t)GU_0(-t);\qquad
G_\ep(t):=U_\ep(t)GU_\ep(-t); \ee Perturbation theory in $\ep$
yields a formal recurrent procedure to construct the full Heisenberg
evolution $G_\ep(t)$ in terms of the free evolution $G_0(t)$. The
result, known as  Dyson's expansion, can be briefly obtained as
follows: consider the Heisenberg equation \be \label{HE}
\dot{G_\ep}(t)=i[H_\ep,G_\ep(t)]/\hbar=i[H_0,G_\ep(t)]/\hbar+i\ep
[V,G_\ep(t)]/\hbar \ee and look for a solution under the form \be
\label{SF} G_\ep(t)=G_0(t)+\ep G_1(t)+\ep^2G_2(t)+\ldots;\quad
G_0(0)=G \ee Inserting this in (\ref{HE}) and equating the
coefficients of the same powers of $\ep$ in both sides we obtain the
recurrent equations: \be \label{RE}
\dot{G_k}(t)=i[H_0,G_k(t)]/\hbar+i [V,G_{k-1}(t)]/\hbar, \quad
k=1,2,\ldots \ee Whence
\begin{align}
\nonumber
&
{G_k}(t)=\frac{i}{\hbar}\int_0^tU_0(t-s)[V,G_{k-1}(s)]U_0(s-t)\,ds=
\\
&
\left(\frac{i}{\hbar}\right)^k\int_0^t\int_0^{t_1}\ldots\int_0^{t_{k-1}}U(t_1,\ldots,t_{k})[V,\ldots [V,G_0(t_k)]\ldots]U(-t_1,\ldots,-t_{k})\,dt_{k}\cdots\,dt_1
\\
\nonumber
&
U(t_1,\ldots,t_k):=U_0(t-t_1)\cdots U_0(t_{k-1}-t_k)
\end{align}
Since $U$ is unitary, if $V$ and $G$ are  bounded operators  we immediately get
\be
\|G_k(t)\|\leq \left(\frac{2\|V\|}{\hbar}\right)^k\|G\|\int_0^t\int_0^{t_1}\ldots\int_0^{t_{k-1}}\,dt_{k}\cdots\,dt_1=\|G\|\left(2\frac{\|V\|}{\hbar}\right)^k\frac{|t|^k}{k!}
\ee
Hence  the series (\ref{SF}) converges  for all $t$, but not uniformly with respect  to the Planck constant $\hbar$.  The uniformity is interesting because for $\hbar=0$ the classical evolution of the observables is formally recovered.

Assume indeed, as usual, $G$ to be the (Weyl) quantization of the classical observable ${\mathcal G}(x,\xi)\in {\mathcal S}(\R^{2n};\R)$:
\be
\label{WGQ}
(Gu)(x)=\frac{1}{h^n}\int_{\R^n\times\R^n}{\mathcal G}(\frac{x+y}{2},\xi)e^{i\la (x-y),\xi\ra/\hbar}u(y)\,dyd\xi, \quad u(x)\in {\mathcal S}(\R^n)
\ee
or, equivalently, that $G$ admits the following distribution kernel:
$$
{\mathcal K}_G(x,y;\hbar)=\frac{1}{h^n}\int_{\R^n}{\mathcal G}(\frac{x+y}{2},\xi)e^{i\la (x-y),\xi\ra/\hbar}\,d\xi
$$
Formula (\ref{WGQ}) defines $G$ as a (semiclassical) pseudodifferential operator of symbol ${\mathcal G}(x,\xi)$. In particular, the semiclassical  symbol of $H_\ep$ is the  classical Hamiltonian function  ${\mathcal H}_\ep:\R^{2n}\to\R$ defined as
\be
\label{CH}
{\mathcal H}(x,\xi):={\mathcal H}_0(x,\xi)+\ep{\mathcal V}(x,\xi), \quad {\mathcal H}_0(x,\xi):= \frac12 \xi^2+\Q(x)
\ee
where  ${\mathcal V}(x,\xi)$ is the Weyl symbol of $V$. In general we denote $\sigma_A(x,\xi;\hbar)$  the Weyl symbol of a continuous observable $A$. Recall now that under our assumptions the following well known results hold (see e.g.\cite{Ro}):
\begin{enumerate}
\item Let  $\widehat{\sigma}_A \in L^2(\R^{2n})$. Then $A$ is a continuous operator in $L^2(\R^n)$  and
\be
\label{L1}
\|A\|:=\|A\|_{L^2\to L^2}\leq \|\widehat{\sigma}_A \|_{L^1}
\ee
\item  Denote $\G_\ep(\hbar,t):=\G_\ep(x,\xi;\hbar,t)$ the symbol of $G_\ep(t)$. Then  $\G_\ep(\hbar,t)$ admits the semiclassical expansion:
\be
\label{SC}
\G_\ep(x,\xi;\hbar,t)=\G_\ep(x,\xi;t)+\hbar \G^{(1)}_\ep(x,\xi;\hbar,t)+\hbar^2
\G^{(2)}_\ep(x,\xi;\hbar,t)+\ldots
\ee
where (semiclassical Egorov theorem: see e.g.\cite{Ro}, IV.3)
\be
\label{EG}
\G_\ep(x,\xi;t)=\G\circ\Phi_\ep^t(x,\xi)
\ee
and $(x,\xi)\mapsto \Phi_\ep^t(x,\xi)
$ the classical Hamiltonian flow generated by $\H_\ep(x,\xi)$;
\item If the two continuous observables $F$ and $G$ have (Weyl) symbols $\F$ and $\G$ then:
\be
\label{Mo1}
\sigma^W(FG)=
\F\#\G,\qquad  \sigma^W(i[F,G]/\hbar)=\{\F,\G\}_M
\ee
Here  $\{\F,\G\}_M:=\F\#\G-\G\#\F$ is the Moyal bracket of $\F$ and $\G$, and $\F\#\G$ is the composition of the symbols $\F$ and $\G$. Explicitly,
 in  Fourier space:
 (see e.g.
\cite{Fo},\S $3.4$):
\begin{align}
\label{comp} & \widehat{\F\#\G}(s)=\frac{1}{\hbar}\int_{\R^{2n}}
\widehat{\F}(s^1) \widehat{\G}(s-s^1)e^{\pi i\hbar(s-s^1)\wedge
s^1/{2}}\,ds^1
\\
\label{tWisted}
&
\{\F,\G\}_M^{\wedge}(s)=\frac{2}{\hbar}\int_{\R^{2n}}\widehat{\F}(s^1)
\widehat{\G}(s-s^1)
\sin{\left[{\hp}(s-s^1)\wedge s^1/{2}\right]}\,ds^1,
\end{align}
where, given to vectors $s=(u,v)$ and $s^1=(u^1,v^1)$,
$s\wedge s^1:=\la u,v_1\ra-\la v,u_1\ra$.
\item If either $\F$ or $\G$ is quadratic, then $\{\F,\G\}_M=\{\F,\G\}$.
\end{enumerate}
Then the Heisenberg equation (\ref{HE}), written in terms of symbols, becomes
\be
\label{HEM}
\dot{\G}_\ep(t,\hbar)=\{\H,\G_\ep(t,\hbar)\}_M=\{\H_0,\G_\ep(t,\hbar)\}+\ep
\{\V,\G_\ep(t,\hbar)\}_M
\ee
because $\H_0$ is a quadratic form in $(x,\xi)$.  Hence we can immediately  write  the recurrent  equations for the symbols  $\G_k(x,\xi;\hbar,t)$ of $G_k(t)$ :
\be
\label{RES}
\dot{\G}_k(\cdot;\hbar,t) =\{\H_0,{\G}_k(\cdot;\hbar,t) \}+\{\V,{\G}_{k-1}(\cdot;\hbar,t) \}_M
\ee
The solution is
\begin{align}
\nonumber
&
{\G_k}(x,\xi;\hbar,t)=\int_0^t\{\V,\G_{k-1}(\cdot;t)\}_M\circ \Phi_0^{t-s}(x,\xi)\,ds=
\\
\label{SOL}
&
\int_0^t\int_0^{t_1}\ldots\int_0^{t_{k-1}}\{\V,\ldots \{\V,\G_0(\cdot;t_k)\}_M\circ \Phi_0^{t_k-t_{k-1}}(x,\xi)\ldots\}_M\circ \Phi_0^{t-t_1}(x,\xi)\,dt_{k}\cdots\,dt_1
\\
\nonumber
&
\G_0(x,\xi;\hbar,t)=\G_0(x,\xi;t):=\G\circ\Phi^t_0(x,\xi)
\end{align}
whose solution is $\G_\ep(x,\xi;t)$ defined by (\ref{EG}) above.
 Hence for $\hbar=0$ we recover  the Lie expansion around the flow $\Phi_0^t(x,\xi)$ of $\H_0$. We  set:
 \be
\label{SFL}
\G_\ep(x,\xi;t)=\G_0(x,\xi;t)+\ep \G_1(x,\xi;t)+\ep^2\G_2(x,\xi;t)+\ldots
\ee
where $ \G_0(x,\xi;t):=\G\circ \Phi_0^t(x,\xi)$.
The recurrent equations now read
\be
\label{REC}
\dot{\G_k}(\cdot;t)=\{\H_0,\G_k(\cdot;t)\}+ \{\V,   \G_{k-1}(\cdot;t)\}, \quad k=1,2,\ldots
\ee
and formula (\ref{SOL}) then yields:
\begin{align}
\nonumber
&
{\G_k}(\cdot;t)=\int_0^t\{\V,\G_{k-1}(\cdot;t)\}\circ \Phi_0^{t-s}(x,\xi)\,ds=
\\
&
\int_0^t\int_0^{t_1}\ldots\int_0^{t_{k-1}}\{\V,\ldots \{\V,\G_0(\cdot;t_k)\}\circ \Phi_0^{t_k-t_{k-1}}(x,\xi)\}\ldots\}\circ \Phi_0^{t-t_1}(x,\xi)\,dt_{k}\cdots\,dt_1
\end{align}
\vskip 3pt\noindent
We are now in position to formulate our uniform convergence statement.
\begin{theorem}
\label{unif}
Let:
  $\ds \A_\sigma:=\{f\in L^2(\R^{2n})\,|\,\|\hat{f}\|_\sigma <\infty\}, \|\hat{f}\|_\sigma:=
\int_{\R^{2n}}|\hat{f}(s)|e^{\sigma|s|}\,ds <\infty\}$. Let
$\G\in\A_\sigma$, $\V\in \A_\sigma$ for some $\sigma >0$. Then there
exists $\Gamma(\sigma)>0$ independent of $\hbar$ such that \be
\label{stimaunif} \|\G_k(\hbar;t)\|_{\sigma/2}\leq
\Gamma(\sigma)^k|t|^k \ee
\end{theorem}
\noindent
{\bf Remarks}
\begin{enumerate}
\item Note that the norm $\| \hat{f}\|_{\sigma}$ is equivalent
to the norm
$$
\|\hat{f}\|_{\sigma}^{\H_0}:=\int_{\R^{2n}}|\hat{f}(s)|e^{\sigma\sqrt{\H_0(s)}}\,ds.
$$
\item  By (\ref{L1}), the estimate (\ref{stimaunif}) shows that the Lie-Dyson expansion converges (in $B(L^2(\R^n)$) uniformly with respect to $\hbar$ for $\ds |\ep t|< \Gamma(\sigma)^{-1}$.
\item  It is well known that, given $\ep >0$,  the Lie expansion yields a good approximation to the perturbed dynamics only up to a time $t$ such that $\ep t$ is small. The above statement shows that this assertion holds uniformly  with respect to $\hbar$.
\end{enumerate}
For the energy of the individual oscillator modes the  quantum normal form constructed in \cite{BGP1} under diophantine conditions on the frequencies  allows us to obtain a stability result valid for  a time scale exponentially long in the perturbation strength $\ep$, uniformly with respect to $\hbar$, up to an error of order $\ep$ also independent of $\hbar$.  Assume without loss of generality
\be
\label{osc}
\Phi_0^t(x,\xi):=\left\{\begin{array}{l} \ds x_k(t)=x_k\cos{\om_k t}+\frac{\xi_k}{\om_k}\sin{\om_k t} \\
\ds
\xi_k(t)=-\om_kx_k\sin{\om_k t}+\xi_k\cos{\om_k t}
\end{array}\right.\quad k=1,\ldots,n
\ee
and denote
\be
z:=(x,\xi);\quad \Phi_0^\theta(z):=\{\Phi_0^t(x,\xi):\om_kt=\theta_k, k=1,\ldots,n\}.
\ee
Given $g\in L^1(\R^{2n})$, set:
\be
\label{fc}
\tilde{g}_k(z):=\frac1{(2\pi)^n}\int_{\T^n}g(\Phi_0^\theta(z))e^{-i\la k,\theta\ra}\,d\theta, \quad k\in\Z^n
\ee
Hence, if $g\in C^1\cap L^1$, one has pointwise
$$
g(\Phi_0^\theta(z))=\sum_{k\in\Z^n}\,\tilde{g}_k(z)e^{i\la k,\theta\ra}
$$
Whence, for $\theta=0$:
\be
\label{FF}
g(z)=\sum_{k\in\Z^n}\,\tilde{g}_k(z).
\ee
Given $\rho>0$, $\sigma>0$,  define the norm
\be
\label{nars}
\|g\|_{\rho,\sigma}:=\sum_{k\in\Z^n}\,e^{{\rho}|k|}\|{\tilde{g}}_k(s)\|_\sigma.
\ee
Remark that
\be
\label{dis}
\|g\|_\sigma \leq \|g\|_{\rho,\sigma}
\ee
and denote
\be
\label{ars}
\A_{\rho,\sigma}:=\{g:\R^{2n}\to\C\,|\,\|g\|_{\rho,\sigma}<+\infty\}
\ee
The following result is proven in \cite{BGP1}, Proposition 3.1.
\begin{proposition}
\label{prop}
Let the frequencies $\om:=\om_k:  k=1,\ldots,n$ fulfill the diophantine condition:
\be
\label{Dio}
|\la\om,\nu\ra|\geq \gamma |\nu|^{-\tau},\quad \forall\,\nu\in \Z^n\setminus \{0\}
\ee
for some $\tau>n-1$ and some $\gamma >0$.  Let $\V\in \A_{\rho,\sigma}$ for some $\rho>0$, $\sigma >0$. Then  there exists $\ep^\ast >0$ and, for $|\ep|<\ep^\ast $, a family $T(\ep)$ of unitary maps in $L^2$ such that
\be
\label{fn}
S(\ep):=T(\ep)H_\ep T(\ep)^\ast =H_0+\ep Z(\ep)+R(\ep).
\ee
Here the following properties hold:
\par\noindent
(1)
 Let ${\mathcal J}_k, k=1,\ldots,n$ be the the energy of the $k-$ oscillator mode:
$$
{\mathcal J}_k:=\frac12(\xi_k^2+\om^2_kx^2_k); \qquad \H_0(x,\xi)=\sum_{k=1}^n\J_k(x,\xi)
$$
and  $J_k$ be the corresponding $k$-th quantized oscillator mode
$$
J_k=-\frac{\hbar^2}{2}\frac{d^2}{dx_k^2}+\om^2_kx^2_k;\quad H_0=\sum_{k=1}^nJ_k
$$
Then $Z$ depends only on $(J_1,\ldots,J_n)$ and $\hbar$ or, equivalently, ${\mathcal Z}$ depends only on  $(\J_1,\ldots,\J_n)$ and $\hbar$, so that $[H_0,Z(\ep)]=0$.
\par\noindent
(2) $S(\ep)$, $T(\ep)$, $Z(\ep)$ and $R(\ep)$ are semiclassical
pseudodifferential operators in $L^2$. In terms of their symbols
$\Sigma_\ep(x,\xi;\hbar)$, ${\mathcal T}_\ep(x,\xi;\hbar)$,
${\mathcal Z}_\ep(x,\xi;\hbar)$, ${\mathcal R}_\ep(x,\xi;\hbar)$,
(\ref{fn}) becomes \be \label{fn1}
\Sigma_\ep(x,\xi;\hbar)=\H_0(x,\xi)+\ep {\mathcal
Z}_\ep(x,\xi;\hbar)+{\mathcal R}_\ep(x,\xi;\hbar),\;\;
\{\H_0,{\mathcal Z}_\ep\}_M=0. \ee
\par\noindent
(3) There are  $C_1(\rho,\sigma,\tau,\ep^\ast)>0, C_1(\rho,\sigma,\tau,\ep^\ast)$ independent of $\hbar$ such that
\be
\label{resto}
\|Z(\ep)\|\leq \|{\mathcal Z}\|_{\rho/2,\sigma/2} \leq C_1;\quad
\|R(\ep)\|\leq  \|{\mathcal R}\|_{\rho/2,\sigma/2} \leq C_1\exp{[-C_2{\ep}^{-1/(\tau+2)}]}
\ee
\end{proposition}
\noindent
{\bf Remark} Formula (\ref{fn})   represents the quantum normal form of $H_\ep$ with a remainder of order $\ds \exp{[-C_2{\ep}^{-1/(\tau+2)}]}$ uniform with respect to $\hbar$. For $\hbar=0$ it reduces to the corresponding classical Birkhoff normal form with remainder.  Then we have:
\par\noindent
\begin{theorem}
\label{modi}
Let $\V\in\A_{\rho,\sigma}$ for some $\rho>0$, $\sigma>0$. Let  $J$ denote any one of the operators $J_k, k=1,\ldots,n$.  Then there are $D_1(\rho,\sigma,\tau,\ep^\ast)>0$,  $D_2(\rho,\sigma,\tau,\ep^\ast)>0$ independent of $\hbar$  such that, for any $0<d<\rho/2$:
$$
\|[J_\ep(t)-J]\|\leq \|[J_\ep(t)-J]\|_{\rho/2-d,\sigma/2} \leq D_1\exp{[-C_2{\ep}^{-1/(\tau+2)}]}|t|+D_2|\ep|,\quad |\ep| <\ep^\ast.
$$
\end{theorem}
\noindent
{\bf Remark.}  Let $\psi_\nu(\hbar): \nu\in\Z^n$ be the eigenvector corresponding to the simple eigenvalue $\ds E_\nu(\hbar):=\sum_{k=1}^n\,\om_k(\nu_k+\frac12)\hbar$.
Then Theorem 1.2 entails:
\begin{align}
\label{FPU1}
&
|\la\psi_\nu(\hbar),J_\ep(t)\psi_\nu(\hbar)\ra-(\om\nu+1/2)\hbar|\leq D_1\exp{[-C_2{\ep}^{-1/(\tau+2)}]}|t|+D_2|\ep|,\quad |\ep| <\ep^\ast
\end{align}
At the classical limit $\nu_k\to\infty$, $\hbar\to 0$, $\nu_k\hbar\to A_k$, $A_k$ the action of the $k-$th mode,  we have (see \cite{DEGH}) $\la\psi_\nu(\hbar),J_\ep(t)\psi_\nu(\hbar)\ra\to \J_k\circ \Phi_\ep^t(x,\xi)$, $(\om\nu_k+1/2)\hbar\to \J_k(x,\xi)$. Therefore (\ref{FPU1}) entails that the absence of energy exchange between the different energy modes for an exponentially long time (see e.g. \cite{BGG}, \cite{BGP2}, \cite{BP}) is stable under the process of quantization, up to an error of order $\ep$ independent of time.
\vfill\eject
\section{Proof of the statements}
\noindent
To  prove the estimate (\ref{stimaunif})  we work in the Fourier representation. First we prove the invariance of $\A_\sigma$ under the action of the  flow $\Phi_o^t(x,\xi)$:
%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}
\label{InW}
Let $\G(x,\xi)\in A_\sigma$. Then  $\G\circ\Phi_o^t(x,\xi)\in A_\sigma$ $\forall t\in\R$.
\end{lemma}
\noindent {\bf Proof.}  Let $z:=(x,\xi)\in\R^{2n}$.  Since
$\H_0(x,\xi)$ is a positive-definite quadratic form we can assume
without loss of generality that the map $z\mapsto
\Phi_0^t(z):\R^{2n}\leftrightarrow \R^{2n}$ leaves the
$\|\hat{f}\|_{\sigma}^{\H_0}$ invariant for all $t\in\R$. Now:
$$
(\G\circ \Phi_0^t)^{\wedge}(s)=\int_{\R^{2n}}(\G\circ \Phi_0^t)(z)e^{-i\la s,z\ra}\,dz=
\int_{\R^{2n}}\G(z)e^{-i\la s,\Phi_0^{-t}(z)\ra}\,dz =\hat{\G}(\Phi_0^t(s)).
$$
Therefore $\| \hat{\G}(\Phi_0^t(s))\|_\sigma^{\H_0}= \|
\hat{\G}(s)\|_\sigma^{\H_0}\;\forall t\in\R$. Since
$\|\hat{f}\|_\sigma$ and $\|\hat{f}\|_\sigma^{\H_0}$ are equivalent,
this proves the Lemma.
\par
Our second preliminary result is an estimate on the Moyal brackets worked out in \cite{BGP1}, and reproduced here for convenience of the reader.
\begin{lemma} Let $g\in\A_\sigma$, $g^\prime\in \A_\sigma$. Then:
 \be
\label{M1}
\|\{g,g^{\prime}\}_M\|_{\sigma-\delta-\eta} \leq \frac{1}{e^2\eta(\delta+\eta)}\|g\|_\sigma\,\|g^{\prime}\|_{\sigma-\delta},
\quad 0<\eta<\sigma-\delta
\ee
\end{lemma}
\noindent {\bf Proof} Since  $(s-s^1)\wedge s^1=s\wedge s^1$, and
$|s\wedge s^1|\leq |s|\cdot|s^1|$, by definition of ${\mathcal
A}_\sigma$-norm and (\ref{tWisted}) we get:
\begin{align*}
&
\|(\{g,g^{\prime}\}_M)\|_{\sigma-\delta}=
\\
&
\frac{2}{\hbar}\int_{\R^{2n}}e^{(\sigma-\delta)|s|}\,ds \int_{\R^{2n}}|\hat{g}(s^1)\hat{g}^\prime(s-s^1)\sin{(\hbar(s-s^1)\wedge s^1)/2}|\,ds^1
\\
&
\leq \frac{2}{\hbar}\int_{\R^{2n}}\,ds \int_{\R^{2n}}e^{(\sigma-\delta)(|s|+|s^1|)}|\hat{g}(s)\hat{g}^\prime(s^1)\sin{(\hbar(s \wedge s^1)/2}|\,ds^1
\\
&
\leq \int_{\R^{2n}}e^{(\sigma-\delta)|s|}|\hat{g}(s)|\,ds \int_{\R^{2n}}e^{(\sigma-\delta)|s^1|}|\hat{g}^\prime(s^1)s \wedge s^1|\,ds^1
\\
&
\leq \int_{\R^{2n}}e^{(\sigma-\delta)|s|}|\hat{g}(s)||s|\,ds \int_{\R^{2n}}e^{(\sigma-\delta)|s^1|}|\hat{g}^\prime(s^1)s| |s^1||\,ds^1
\end{align*}
whence the assertion because  $\ds xe^{-\delta x}\leq \frac1{e\delta}$, $\forall\,x>0$, $\forall\delta>0$.
\begin{corollary} Under the above assumptions on $g$ and $g^{\prime}$:
 \be
\label{M2}
\|\{g,g^{\prime}\}_M\circ\Phi^t_0(x,\xi)\|_{\sigma-\delta-\eta} \leq \frac{1}{e^2\eta(\delta+\eta)}\|g\|_\sigma\,\|g^{\prime}\|_{\sigma-\delta},
\quad 0<\eta<\sigma-\delta,\;\forall\,t\in\R
\ee
\end{corollary}
\noindent
{\bf Proof} By Lemma \ref{InW}  both  $g\circ\Phi^t_0(x,\xi)$ and $g^{\prime}\circ\Phi^t_0(x,\xi)$ belong to $\A_\sigma$ $\forall\,t\in\R$  whenever $g$ and $g^{\prime}$ do.
Then the assertion follows from (\ref{M1}), and this proves the Corollary.
\newline
Iterating $k$ times exactly the same arguments we get:
\begin{corollary}
\label{M3}
Let $g\in\A_\sigma$, $\V\in \A_\sigma$, $0<k\eta<\sigma-\delta$. Then:
 \begin{align}
\label{M1b} &
\|\{\V,\{\ldots,\{\V,g\}_M\circ\Phi^t_0(x,\xi)\ldots,\}_M\circ\Phi^t_0(x,\xi)\|_{\sigma-\delta-k\eta}
\\
\nonumber
&
 \leq \frac{1}{e^2\eta^k(\delta+\eta)^k}\|\V\|^k_\sigma\,\|g\|_{\sigma-\delta}, \;\quad \forall\,t\in\R
\end{align}
\end{corollary}
\noindent
{\bf Proof of Theorem \ref{unif}.}  Consider formula (\ref{SOL}).  By Corollary \ref{M3} above we have:
\begin{align}
\nonumber
&
\|\{\V,\ldots \{\V,\G_0(\cdot;t_k)\}_M\circ \Phi_0^{t_k-t_{k-1}}(x,\xi)\ldots\}_M\circ \Phi_0^{t-t_1}(x,\xi)\|_{\sigma-\delta-k\eta}
\\
\nonumber
&
\leq \frac{1}{e^2\eta^k(\delta+\eta)^k}\|\V\|^k_\sigma\,\|\G_0\|_{\sigma-\delta}, \;\quad \forall\,t\in\R
\end{align}
Now fix $0<\delta <\sigma/2$ and set:
$$
\eta=\frac{\delta}{2k}
$$
Therefore:
\begin{align}
\nonumber
&
\|\{\V,\ldots \{\V,\G_0(\cdot;t_k)\}_M\circ \Phi_0^{t_k-t_{k-1}}(x,\xi)\ldots\}_M\circ \Phi_0^{t-t_1}(x,\xi)\|_{\sigma-2\delta}
\\
&
\leq e^{-2}2^k\delta^{2k}(1+\frac1{2k})^kk^k\|\V\|^k_\sigma\,\|\G_0\|_{\sigma-\delta}\leq H\Gamma(\sigma)^k k!
\\
&
H:=\frac{1}{e^2\sqrt{2\pi}}\|\G_0\|_{\sigma-\delta}; \quad \Gamma(\sigma):=3e\sigma ^2\|\V\|_\sigma
\end{align}
Here we have used the inequalities $\delta<\sigma$, $\ds (1+\frac1{2k})< \frac{3}{2}, \,k=1,\ldots$, and the Stirling formula.  Hence, again by (\ref{SOL}):
$$
\|\G_k(x,\xi;\hbar,t)\|_{\sigma-2\delta}\leq H\Gamma(\sigma)^k k!
\int_0^t\int_0^{t_1}\ldots\int_0^{t_{k-1}}\,dt_{k}\cdots\,dt_1\leq H\Gamma(\sigma)^k |t|^k,\;k=1,2\ldots
$$
The assertion of the Theorem is now proved if we choose $\delta
<\sigma/4$. \vskip 0.3cm\noindent Let us turn now to Theorem 1.2.
We first prove an auxiliary result.
\begin{lemma}
\label{lim}
Let $\B\in\A_{\rho,\sigma}$ for some  $\rho>0, \sigma>0$, and $B$ the corresponding operator in $L^2(\R^n)$ defined by its Weyl quantization.  Then there is $K>0$ independent of $\hbar$ such that, for any $0<d<\rho$:
$$
\|[B,J]/\hbar\|\leq \|\{\B,J\}\|_{\rho-d,\sigma}\leq \frac{K}{d}\|\B\|_{\rho,\sigma}
$$
\end{lemma}
\noindent
{\bf Proof.} We have:
$$
 \frac{1}{\hbar}\|[B,J]\|\leq \|\{{\mathcal B},\J\}_M\|_{\rho,\sigma} =\|\{{\mathcal B},\J\}\|_{\rho,\sigma}
 $$
 because $\J(x_1,\xi_1)$ is quadratic.   Here we have set, without loss,
 $$
 \J=\J(x_1,\xi_1)=\frac12(\xi_1^2+\om_1^2x_1^2).
 $$
  Explictly:
 $$
 \{{\mathcal B},\J\}=\nabla_{\xi_1} {\mathcal B}\cdot \om_1x_1-\nabla_{x_1}{\mathcal B}\cdot \xi_1
 $$
 whence, in Fourier space
 $$
 \{{\mathcal B},\J\}^{\wedge}(s)=\int_{\R^{2n}}(s_2 \om_1 \partial_{s_1}-s_1 \partial_{s_2})\widehat{{\mathcal B}}(s)\,ds.
$$
To estimate $\|\{{\mathcal B},\J\}\|_{\rho,\sigma}$  we apply the arguments of \cite{BGP1}, Lemma 3.3.   Denote $\phi_0^\theta(z):=\{\phi_0^t(z):\om_1 t=\theta\}$ the flow of $\J$. Here
$\phi_0^t(z)=\Phi_0^t(x_1,\xi_1)$.
Then we can write, on account of (\ref{FF})
$$
 \{{\mathcal B},\J\}(z)=\left.\frac{d}{dt}\right|_{t=0}{\mathcal B}(\phi_0^{t}(z))=\sum_{k\in\Z^n}\om_1k_1\tilde{\mathcal B}_k(z)
$$
Therefore
$$
\| \{{\mathcal B},\J\}(z)\|_{\rho,\sigma}\leq \sum_{k\in\Z^n}|\om_1k_1|e^{|\rho k_1|}\|\tilde{\mathcal B}_k\|_\sigma
$$
Whence, as in Lemma 3.8 of \cite{BGP1}:
\begin{align*}
&
\| \{{\mathcal B},\J\}(z)\|_{\rho/2-d,\sigma/2}= \sum_{k\in\Z^n}|\om_1k_1|e^{(|\rho|/2-d)k_1|}\|\tilde{\mathcal B}_k\|_{\sigma/2} \leq
\\
&
 \leq \| {\mathcal B}\|_{\rho/2,\sigma/2}\sup_{k_1}{|k_1|e^{-d|k_1|}}=\frac{\om_1 \| {\mathcal B}\|_{\rho/2,\sigma/2}}{d}, \quad 0<d<\rho.
\end{align*}
Taking $K:=\om_1$ this proves the Lemma.
\vskip 0.3cm\noindent
{\bf Proof of Theorem \ref{modi}.} Introduce the Heisenberg observable corresponding to $J$ along the evolution generated by $S(\ep)$ defined by (\ref{fn}):
$$
J^S_\ep(t)=e^{iS(\ep)t/\hbar}J e^{-iS(\ep)t/\hbar}
$$
so that
\be
\label{DE}
\dot{J}_\ep^S(t)= \frac{i}{\hbar}e^{iS(\ep)t}[S(\ep),J]e^{-iS(\ep)t}=\frac{i}{\hbar}
e^{iS(\ep)t}[R(\ep),J]e^{-iS(\ep)t}
\ee
because $[H_0,J]=[Z(\ep),J]=0$.  Hence, upon integration of (\ref{DE}):
$$
\|J^S_\ep(t)-J\|_{\rho/2-d,\sigma/2}\leq \int_0^t\| \{{\mathcal R}(\ep),\J\}(z)\|_{\rho/2-d,\sigma/2}\,dt \leq
$$
because $\|[R(\ep),J]/i\hbar\|\leq \|\{{\mathcal R}(\ep),\J\}_M\|_{\rho/2-d,\sigma/2}\leq
\|\{{\mathcal R}(\ep),\J\}\|_{\rho/2-d,\sigma/2}$ since $\J$ is quadratic. Then,  on account of (\ref{resto}), the above Lemma yields:
\be
\label{primastima}
\|J^S_\ep(t)-J\|_{\rho/2-d,\sigma/2}\leq D\exp{[-C_2{\ep}^{-1/(\tau+2)}]}|t|, \;D:=\frac{C_1\om_1}{d}
\ee
We have now to estimate the difference
\begin{align*}
&
\|J_\ep(t)-J\|=\|T(\ep)[e^{iH_\ep t/\hbar} J_\ep e^{-iH_\ep t/\hbar}-J]T(\ep)^{-1}\|=
\\
&
\|e^{iS_\ep t/\hbar}T(\ep) J_\ep T(\ep)^{-1} e^{-iS_\ep t/\hbar}-T(\ep)JT(\ep)^{-1}\|
\end{align*}
By \cite{BGP1}, Proposition 3.2, we can write $\ds T(\ep)=e^{iW(\ep)/\hbar}$, where  $W(\ep)$ is, for $|\ep|<\ep^\ast$,  a bounded self-adjoint semiclassical pseudodifferential operator with symbol $\W_\ep(x,\xi;\hbar)$ such that
\be
\label{Wlim}
\|W(\ep)\|\leq \|\W_\ep(x,\xi;\hbar)\|_{\rho/2,\sigma/2}\leq E|\ep|
\ee
for some $E>0$ independent of $\hbar$.  Hence, by Lemma \ref{lim} and (\ref{dis}), we get:
$$
\|[W(\ep),J]/i\hbar\|\leq \|\{\W_{\ep},\J\}\|_{\sigma/2}\leq \|\{\W_{\ep},\J\}\|_{\rho/2-d,\sigma/2}\leq \frac{E}{d}, \;0<d<\rho/2
$$
and   Corollary 2.2 for $t=0$ yields the estimate:
\begin{align*}
&
\|[W(\ep),[\ldots [W(\ep),J]\ldots]/(i\hbar)^k\|\leq \|\{\W_\ep,\{\ldots \{\W_\ep,J\}_M\ldots\}_M\|{\rho/2-d,\sigma/2}\leq
\\
&
\leq  \frac{1}{e^2\eta^k(\delta+\eta)^k}\|\W_\ep\|_{\rho/2-d,\sigma/2}^k \|\{\W_{\ep},\J\}\|_{\rho/2-d,\sigma/2-\delta},\quad 0<k\eta <\sigma/2-\delta
\end{align*}
As in the proof of Theorem 1.1, fix $0<\delta <\sigma/4$ and set:
$$
\eta:=\frac{\delta}{4k}
$$
Then
\begin{align}
\label{stima2}
&
\|[W(\ep),[\ldots [W(\ep),J]\ldots]/(i\hbar)^k\| \leq L\Theta(\sigma)^kk!
\\
&
L:=\frac{1}{e^2\sqrt{2\pi}d}, \quad \Theta(\sigma):=5e\sigma^2 E|\ep|
\end{align}
Therefore  the commutator expansion
$$
T(\ep)JT(\ep)^{-1}=J+\sum_{k=1}^\infty\frac{[W(\ep),[\ldots [W(\ep),J]\ldots]}{(i\hbar)^kk!}
$$
has a nonzero convergence radius,  uniformly with respect to $\hbar$. We can therefore write
$$
T(\ep)JT(\ep)^{-1}=J+\ep M(\ep)
$$
where $\|M(\ep)\|$  is bounded uniformly with respect to $\hbar$.   Hence:
\begin{align*}
&
\|e^{iS_\ep t/\hbar}T(\ep) J_\ep T(\ep)^{-1} e^{-iS_\ep t/\hbar}-T(\ep)JT(\ep)^{-1}\|=
\\
&
\|e^{iS_\ep t/\hbar} J e^{-iS_\ep t/\hbar}-J+\ep e^{iS_\ep t/\hbar} M(\ep) e^{-iS_\ep t/\hbar}-\ep M(\ep)\|
\\
&
\leq \|J_\ep(t)-J\|+|\ep| \|e^{iS_\ep t/\hbar} M(\ep) e^{-iS_\ep t/\hbar}-M(\ep)\|
\end{align*}
Setting $2\|M(\ep)\|=D_2$ this estimate concludes the proof of Theorem 1.2.
\vfill
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\end{thebibliography}

\end{document}
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