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Heisenberg operator; invariant domain ; Heisenberg equation of motion;  
Schr\"odinger operator; Dirac operator
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\begin{document}
\title{\bf Heisenberg Operators, 
 Invariant Domains and Heisenberg Equations of Motion}
\author{Asao Arai \\
Department of Mathematics \\
Hokkaido University \\
Sapporo 060-0810\\
Japan \\
E-mail:arai@math.sci.hokudai.ac.jp
}

\maketitle

\begin{abstract}
An abstract operator theory is developed
on  operators of the form 
$A_H(t):=e^{itH}Ae^{-itH}, \, t\in \R$, with
$H$ a self-adjoint operator and $A$ a 
linear operator on a Hilbert space (in the context of quantum mechanics,
$A_H(t)$ is called the Heisenberg
operator of $A$ with respect to $H$). 
The following aspects are discussed:
(i) integral equations for   $A_H(t)$ for a general class of $A$ ; 
(ii) a sufficient condition for $D(A)$, the domain of $A$, to be left invariant
by $e^{-itH}$ for all $t \in \R$ ;
(iii) a mathematically rigorous formulation of the Heisenberg equation of motion
in quantum mechanics and 
 the uniqueness of its solutions ; (iv) invariant domains in the case where $H$ is 
an abstract version
of Schr\"odinger and Dirac operators ; (v) applications to Schr\"odinger 
operators with matrix-valued potentials and Dirac operators.


\end{abstract}


\noindent
{\it Keywords}: Heisenberg operator; invariant domain ; Heisenberg equation of motion; \\ 
Schr\"odinger operator; Dirac operator

\section{Introduction}


Let ${\cal H}$ be a complex Hilbert space and
$H$ be a self-adjoint operator on ${\cal H}$.
Then, for a linear operator $A$ on ${\cal H}$ and $t \in \R$, 
one can define a linear operator by
\begin{equation}
A_H(t):=e^{itH}Ae^{-itH}.
\end{equation}
In the context of quantum mechanics, $A_H(t)$ is called
the {\it Heisenberg operator}
of $A$ with respect to $H$.
If $H$ is the Hamiltonian of a quantum system ${\sf S}$
(an operator for the total energy of ${\sf S}$)
and $A$ is a physical quantity of ${\sf S}$, then
$A_H(t)$ describes 
the time development of $A$ with $t$ being the time. 
Therefore, in quantum mechanics,  it is very important
to study Heisenberg operators. It seems, however,
that there have been few  mathematically rigorous studies of Heisenberg operators
in an {\it abstract} or a {\it general form}, although  there are ones
in concrete models (e.g., \cite{CFKS,Thaller}).
One of the reasons for this  may come from difficulties related to
treatments for domains of relevant operators in the case where
$H$ and $A$ are unbounded. 
In this paper, overcoming these difficulties, 
we present some operator theoretical analyses
on the operator $A_H(t)$.
The point of the paper is to establish a  general theory 
for  Heisenberg operators with mathematical rigor and to put a basis  for applications. 

The present paper is organized as follows. 
In Section 2 we derive  integral equations for $A_H(t)$ in the case where
$A$ is bounded. This is a preliminary version to the case where $A$ is unbounded.
In Section 3 some integral equations for $A_H(t)$ are derived in the case
where $A$ is a densely defined closed linear operator.
In particular,  we show that, under some conditions, the domain $D(A)$ of $A$  is an invariant
domain of $e^{-itH}$ for all $t \in \R$.
Section 4 is concerned with
the so-called Heisenberg equation of motion with respect to $H$.
In this case too, some difficulty arises   
in treating it  with mathematical rigor if $H$ is unbounded.
But we present a mathematically rigorous form 
for the Heisenberg equation of motion 
and discuss the uniqueness of its solutions.
In Section 5, using a Weyl representation of the canonical commutation relations
with a finite degree of freedom, we introduce an abstract version of
Schr\"odinger operators with matrix-valued potentials
and Dirac operators, and prove a theorem on invariant domains
and an integral equation for a Heisenberg operator.
The last section  is devoted to applications 
of the methods or the results established in the preceding sections to Schr\"odinger operators with matrix-valued potentials,  including
relativistic ones, 
and a class of abstract Dirac operators. As for the Schr\"odinger operators,
we derive a mathematically rigorous version of  the \lq\lq{quantum Newton equation}" of
motion for the position operator. On the other hand, 
considerations of the  abstract Dirac operators
clarify a general mathematical structure behind the usual Dirac operator
acting in $L^2(\R^3;\C^4)$ (\cite[Chapter 4]{Thaller}).
Applications to models in quantum field theory  will be discussed in a separate paper.
 



\section{Integral equations for $A_H(t)$ in the case where $A$ is bounded}



Let $H$ and $A$ be as in Introduction.
By the definition of the domain of operator
products, 
we have
\begin{equation}
D(A_H(t))=\{\psi \in {\cal H}|
e^{-itH}\psi \in D(A)\}.
\end{equation}
To treat properly domain problems arising in the analysis of  $A_H(t)$ in the case where
$H$ and $A$ are unbounded, we use some approximation 
methods. For this purpose  we introduce a class of functions:


\begin{df}{\rm Let $f$ be a real-valued 
 Borel measurable  function  on $\R$. 
We say that $f$ is in the class ${\cal F}_{+,1}(\R)$
if it is  continuous  on  $[0,\infty)$
and
\begin{eqnarray}
&&f(0)=1, \label{f0}\\
&& \sup_{\lambda \geq 0}\lambda |f(\lambda)| < \infty. \label{f1}
\end{eqnarray}
}
\end{df}

It is easy to see that 
\begin{equation}
\|f\|_{\infty}:=\sup_{\lambda \geq 0}|f(\lambda)| < \infty,\quad \forall f \in {\cal F}_{+,1}(\R).
\label{ff}
\end{equation}


We denote by $\lang \,\cdot\,, \,\cdot\,\rang$ and $\|\cdot\|$ the 
inner product and the norm of ${\cal H}$ respectively.
The Banach space of all the everywhere defined
bounded linear operators
on ${\cal H}$ is denoted ${\sf B}({\cal H})$. The operator norm
of $B\in {\sf B}({\cal H})$ is written  $\|B\|$. 


Let $S$ be a  densely defined closed
linear operator on ${\cal H}$. Then, by von Neumann's theorem,  $S^*S$ is a nonnegative self-adjoint
operator (e.g., \cite[Theorem X.25]{RS2}).
Hence, for $\varepsilon > 0$ and $f \in {\cal F}_{+,1}(\R)$,
we can define
an operator $S_{\varepsilon}$ by
\begin{equation}
S_{\varepsilon}^f:=Sf(\varepsilon |S|),
\label{reg}
\end{equation} 
where $|S|:=(S^*S)^{1/2}$. 


\begin{lem}\label{lem2-0} Let $S$ be as above. 
Then :
\begin{list}{}{}
\item{{\rm (i)}} $S_{\varepsilon}^f\in {\sf B}({\cal H})$ and
\begin{equation}
\lim_{\varepsilon \to 0}S_{\varepsilon}^f\psi=S\psi,
\end{equation}
for all $\psi \in D(S)$. 
\item{{\rm (ii)}} If  $S$ is self-adjoint, then
$S_{\varepsilon}^f$ is self-adjoint and 
\begin{eqnarray}
&&\mbox{{\rm s-}}\lim_{\varepsilon \to 0}(S_{\varepsilon}^f-z)^{-1}
=(S-z)^{-1}, \quad \forall z\in \C\setminus \R,\label{resolv}\\
&&\mbox{{\rm s-}}\lim_{\varepsilon \to 0}e^{itS_{\varepsilon}^f}=e^{itS}, \quad \forall t\in \R,
\label{exp}
\end{eqnarray}
where $\mbox{{\rm s-}}\lim$ means strong limit.
\end{list}
\end{lem}

{\it Proof}. (i) 
By the polar decomposition, we have
$S=U|S|$ with $U$ a partial isometry (e.g., 
\cite[p.334]{kato}, \cite[Theorem VIII.32]{RS1}).
It follows from  the functional calculus and (\ref{f1})
that
$|S|f(\varepsilon |S|)$ is boudned 
with $D(|S|f(\varepsilon |S|))={\cal H}$.
Hence
$S_{\varepsilon}^f\in {\sf B}({\cal H})$. 
For all $\psi \in D(S)=D(|S|)$, we have
\begin{equation}
\|S_{\varepsilon}^f\psi-S\psi\|=
\|\,|S|f(\varepsilon |S|)\psi-|S|\psi\|. \label{S1}
\end{equation}
Using  the continuity of $f$ on $[0,\infty)$ together
with (\ref{f0}), (\ref{ff}) and the Lebesgue dominated convergence 
theorem,
one can easily 
show that the right hand side of (\ref{S1}) converges to $0$ as
$\varepsilon \to 0$.

(ii)  The self-adjointness of $S_{\varepsilon}^f$ follows from
that  $S$ is self-adjoint and $S_{\varepsilon}^f=g(S)$ with
$g(\lambda)=\lambda f(\varepsilon |\lambda|), \,\lambda \in \R$.
 Formulas (\ref{resolv})
and (\ref{exp}) respectively follow from a simple  application of
a general convergence theorem (e.g., \cite[Theorem VIII.25(a), Theorem VIII.21]{RS1}). \hfill \qed

\bigskip



For  two linear operators $A$ and $B$, we set
\begin{equation}
[A,B]:=AB-BA
\end{equation}
with $D([A,B]):=D(AB)\cap D(BA)$.



The following lemma is easily proven.


\begin{lem}\label{lem2-1} For all $B\in {\sf B}({\cal H})$, 
$B_H(s)=e^{isH}Be^{-isH}$ is in ${\sf B}({\cal H})$ and  strongly continuous in $s$.
\end{lem}

By this lemma, we can define  for all $B\in {\sf B}({\cal H})$ and $\psi \in {\cal H}$ 
the strong Riemann integral
$\int_0^te^{isH}Be^{-isH}\psi ds \in {\cal H}, \, t\in \R$.


 

\begin{lem}\label{HB} Assume that
$H$ and $A$ are in ${\sf B}({\cal H})$. Then,
for all $t\in \R$ and $\psi\in {\cal H}$,
\begin{equation}
A_H(t)\psi=A\psi+\int_0^tie^{isH}[H,A]e^{-isH}\psi \,ds. \label{BAH}
\end{equation}
\end{lem}


{\it Proof}. Let $\psi \in {\cal H}$�D
Then, under the present assumption, $A_H(t)\psi$ is strongly differentiable
in $t$ with
$$
\frac{d}{dt}A_H(t)\psi=
ie^{itH}[H,A]e^{-itH}.
$$
Integrating this equation, we obtain (\ref{BAH}). \hfill \qed

\medskip






The main result of this section is
as follows:

\begin{The}\label{Heis1} 
Let  $A\in {\sf B}({\cal H})$.
Then, for all $\psi,\phi \in D(H)$,
\begin{equation}
\lang \phi, A_H(t)\psi\rang=
\lang\phi,A\psi\rang+i\int_0^t\left\{
\lang H\phi,A_H(s)\psi\rang
-\lang \phi,A_H(s)H\psi\rang\right\}ds. \label{AH}
\end{equation}
\end{The}


{\it Proof}.
Applying Lemma  \ref{HB} with $H$ replaced by
$H_{\varepsilon}^f$ ($f \in {\cal F}_{+,1}(\R)$), we have
\begin{eqnarray}
&&\lang \phi, 
e^{itH_{\varepsilon}^f}Ae^{-itH_{\varepsilon}^f}\psi\rang \nonumber\\
&&=
\lang\phi,A\psi\rang \nonumber\\
&& +i\int_0^t\left\{
\lang H_{\varepsilon}^fe^{-isH_{\varepsilon}^f}\phi,A
e^{-isH_{\varepsilon}^f}\psi\rang
-\lang A^*e^{-isH_{\varepsilon}^f}\phi,H_{\varepsilon}^f
e^{-isH_{\varepsilon}^f}\psi\rang\right\}ds \label{int1}
\end{eqnarray}
for all $\psi, \phi \in {\cal H}$ and $t\in \R$.
Applying Lemma \ref{lem2-0} with $S=H$, we have  
for all $\psi \in {\cal H}$ and $\eta \in D(H)$,
\begin{eqnarray*}
&&\mbox{s-}\lim_{\varepsilon \to 0}
e^{itH_{\varepsilon}^f}Ae^{-itH_{\varepsilon}^f}\psi=A_H(t)\psi, \quad
\mbox{s-}\lim_{\varepsilon\to 0}Ae^{-isH_{\varepsilon}^f}\psi=
Ae^{-isH}\psi,\\
&&
\mbox{s-}\lim_{\varepsilon\to 0}A^*e^{-isH_{\varepsilon}^f}
\psi=A^*e^{-isH}
\psi,
\quad
\mbox{s-}\lim_{\varepsilon \to 0}
H_{\varepsilon}^fe^{-isH_{\varepsilon}^f}\eta=He^{-isH}\eta.
\end{eqnarray*}
Moreover, we have for all $\psi,\phi \in D(H)$
\begin{eqnarray*}
&&\left|
\lang H_{\varepsilon}^fe^{-isH_{\varepsilon}^f}\phi,A
e^{-isH_{\varepsilon}}\psi\rang\right|
\leq \|H\phi\|\|A\|\|\psi\| \|f\|_{\infty}, \\
&&\left|\lang A^*e^{-isH_{\varepsilon}^f}\phi,H_{\varepsilon}^f
e^{-isH_{\varepsilon}^f}\psi\rang\right|
\leq \|A^*\| \|\phi\| \|H\psi\|\|f\|_{\infty}.
\end{eqnarray*}
Hence we can apply the Lebesgue dominated convergence
theorem to the integral on the right hand side of (\ref{int1})
to obtain (\ref{AH}).
\hfill \qed

\medskip

Formula (\ref{AH}) may be viewed as
an integral equation for $A_H(t)$ in a weak sense.


\begin{cor}
Let $A\in {\sf B}({\cal H})$.
Then, for all $\psi\in D(H)$ and $t\in \R$,
$\int_0^tA_H(s)\psi \,ds \in D(H)$ and
\begin{equation}
A_H(t)\psi=
A\psi+iH\left(\int_0^tA_H(s)\psi\,ds\right)-i\int_0^tA_H(s)H\psi\,ds. 
\end{equation}
\end{cor}

{\it Proof}. Let $\eta(t):=\int_0^tA_H(s)\psi\,ds$ and $\chi(t):=\int_0^tA_H(s)H\psi\,ds$. 
Then, by (\ref{AH}), we have for all $\phi \in D(H)$
$\lang H\phi,i\eta(t)\rang
=\lang \phi, A_H(t)\psi-A\psi+i\chi(t)\rang$.
This implies that $i\eta(t)\in D(H)$ and
$H(i\eta(t))=A_H(t)\psi-A\psi+i\chi(t)$�DThus
the desired result follows.
\hfill \qed

\medskip

\section{Integral equations for $A_H(t)$ with $A$ in a general  class and
the asymptotic behavior of $Ae^{-itH}$ as $|t|\to\infty$}


In this section, 
we  consider the  case where $A$ is not necessarily bounded.
We define a subspace $D_{H,A}$ by
\begin{equation}
D_{H,A}:=\{\psi \in {\cal H}| e^{itH}\psi \in D(A),  
\forall t \in \R\}
\end{equation}
Obviously 
$$
D_{H,A}\subset \cap_{t \in \R}D(A_H(t)) \subset  D(A).
$$



Let $A$ be a densely defined closed linear operator on ${\cal H}$.
Then, for every $\varepsilon > 0$, $t\in \R$  and 
$f \in {\cal F}_{+,1}(\R)$, we can define a sesquilinear form 
$q_{\varepsilon,t}^f:D(H) \times D(H) \to \C$ by
\begin{eqnarray}
q_{\varepsilon,t}^f(\phi,\psi)&:=&
i\int_0^t
\left\{
\lang He^{-isH}\phi,A_{\varepsilon}^fe^{-isH}\psi\rang
-\lang \left(A_{\varepsilon}^f\right)^*
e^{-isH}\phi,He^{-isH}\psi\rang\right\}ds,\nonumber\\
&& \phi,\psi\in D(H),
\end{eqnarray}
where $A_{\varepsilon}^f$ is defined by (\ref{reg})
with $S=A$.


\begin{The}\label{qt} 
Let 
$A$ be a densely defined closed linear operator on ${\cal H}$ and
$f \in {\cal F}_{+,1}(\R)$.
Then, for all $t\in \R$, $\phi\in D(H)$ and $\psi \in D_{H,A}$,
\begin{equation}
q_t(\phi,\psi):=
\lim_{\varepsilon \to 0}
q_{\varepsilon,t}^f(\phi,\psi)
\end{equation}
exists independently of $f$ and
\begin{equation}
\lang \phi, A_H(t)\psi\rang=
\lang\phi,A\psi\rang \nonumber\\
+q_{t}(\phi,\psi). \label{AH1}
\end{equation}
\end{The}

{\it Proof}. 
By (\ref{AH}), we have
\begin{equation}
\lang \phi, 
e^{itH}
A_{\varepsilon}^fe^{-itH}\psi\rang-
\lang\phi,A_{\varepsilon}^f\psi\rang=
q_{\varepsilon,t}^f(\psi,\phi) \label{AH2}
\end{equation}
for all $\phi,\psi \in D(H)$ and $t \in \R$. Let $\eta
\in D_{H,A}$. 
Then, by Lemma \ref{lem2-0},
we have
$$
\mbox{s-}\lim_{\varepsilon\to 0}
A_{\varepsilon}^fe^{i\tau H}\eta=
Ae^{i\tau H}\eta, \quad \forall \tau \in \R.
$$
Hence the left hand side of (\ref{AH2}) converges to
$\lang \phi, 
e^{itH}
Ae^{-itH}\psi\rang-
\lang\phi,A\psi\rang$ as $\varepsilon \to 0$. 
Thus the desired assertion follows.
\hfill \qed

\medskip


In some cases, the sesquilinear form $q_t$ may have more explicit
representations.


\begin{The}\label{qt2} Let $A$ be a densely defined closed linear operator on ${\cal H}$.
Suppose that there exists a subspace ${\cal D}
\subset D(H)\cap D_{H,A}\cap D_{H,A^*}$
such that 
\begin{eqnarray}
&&\sup_{0\leq |s|\leq T}\|A^{\#}e^{-isH}\phi\|<\infty,\quad
\forall T>0, \forall \phi \in {\cal D}, \label{sup}
\end{eqnarray}
where $A^{\#}$ denotes either $A$ or $A^*$.
Then,
for all $\phi,\psi \in {\cal D}$ and $t\in \R$,
\begin{equation}
\lang \phi, A_H(t)\psi\rang=
\lang\phi,A\psi\rang+i\int_0^t
\left\{\lang He^{-isH}\phi, Ae^{-isH}\psi\rang
-\lang A^*e^{-isH}\phi, He^{-isH}\psi\rang \right\}ds. \label{AH3}
\end{equation}
\end{The}

{\it Proof}.  Under the present assumption, we have for all $\psi,\phi \in {\cal D}$ and $s\in \R$
$$
\lim_{\varepsilon\to 0}A_{\varepsilon}^fe^{-isH}\psi=
Ae^{-isH}\psi,\quad
\lim_{\varepsilon\to 0}(A_{\varepsilon}^f)^*e^{-isH}\phi=
A^*e^{-isH}\phi.
$$
Let $C_{T,1}(\psi):=\sup_{0\leq |s|\leq T}\|Ae^{-isH}\psi\|$ and
$C_{T,2}(\phi):=\sup_{0\leq |s|\leq T}\|A^*e^{-isH}\phi\|$.
Then, for all $|s|\leq T$, 
\begin{eqnarray*}
&&|\lang He^{-isH}\phi,A_{\varepsilon}^fe^{-isH}\psi\rang|
\leq \|H\phi\|\|Ae^{-isH}\psi\|\|f\|_{\infty}\leq 
C_{T,1}(\psi)\|H\phi\|\|f\|_{\infty},\\
&&\lang(A_{\varepsilon}^f)^*e^{-isH}\phi, He^{-isH}\psi\rang|
\leq \|H\psi\|\|A^*e^{-isH}\phi\|\|f\|_{\infty} \leq C_{T,2}(\phi)\|H\psi\|\|f\|_{\infty}.
\end{eqnarray*}
Hence, by the Lebesgue dominated convergence theorem,
we have for $|t|\leq T$
$$
\lim_{\varepsilon \to 0}q_{\varepsilon,t}^f(\phi,\psi)=i\int_0^t\left\{
\lang He^{-isH}\phi, Ae^{-isH}\psi\rang
-\lang A^*e^{-isH}\phi, He^{-isH}\psi\rang \right\}ds.
$$
This fact and Theorem \ref{qt} yield (\ref{AH3}). \hfill \qed

\medskip


Formula  (\ref{AH3}) can be written as
\begin{equation}
\lang \phi, A_H(t)\psi\rang=
\lang\phi,A\psi\rang+i\int_0^t
\left\{\lang H\phi, A_H(s)\psi\rang
-\lang A_H(s)^*\phi, H\psi\rang \right\}ds, \quad \phi,\psi \in {\cal D}.\label{AH3p}
\end{equation}
This shows that $A_H(t)$ is a solution of the integral equation
\begin{equation}
\lang \phi, X(t)\psi\rang=
\lang\phi,A\psi\rang+i\int_0^t
\left\{\lang H\phi, X(s)\psi\rang
-\lang X(s)^*\phi, H\psi\rang \right\}ds, \quad \phi,\psi \in {\cal D}, \label{AH3p2}
\end{equation}
for  an operator-valued function $:t\mapsto X(t)$ with $X(0)=A$.




\begin{cor}\label{cor3-3} Let 
$A$ be a densely defined closed linear operator on ${\cal H}$.
Suppose that, for some $\alpha\in [1,\infty)$, $A$ and $A^*$ are
$|H|^{\alpha}$-bounded, i.e.,  
$D(|H|^{\alpha})\subset D(A)\cap D(A^*)$ and
there exist constants $a_j\geq 0, b_j\geq 0$ ($j=1,2$) 
such that
\begin{equation}
\|A\psi\|\leq a_1\||H|^{\alpha}\psi\|+b_1\|\psi\|, \quad
\|A^*\psi\|\leq a_2\||H|^{\alpha}\psi\|+b_2\|\psi\|, \quad \forall \psi \in D(|H|^{\alpha}).
\label{relbd}
\end{equation}
Then,
for all $\phi,\psi \in D(|H|^{\alpha})$ and $t\in \R$,
\begin{equation}
\lang \phi, A_H(t)\psi\rang=
\lang\phi,A\psi\rang+i\int_0^t
\left\{\lang He^{-isH}\phi, Ae^{-isH}\psi\rang
-\lang A^*e^{-isH}\phi, He^{-isH}\psi\rang \right\}ds.\label{intform1}
\end{equation}
\end{cor}

{\it Proof}. By the present assumption, we have
$D(|H|^{\alpha})\subset D(H)\cap D_{H,A}\cap D_{H,A^*}$. Moreover
condition (\ref{relbd}) implies that, for all $\psi \in D(|H|^{\alpha})$ and $|s|\leq T, s\in \R$
($T>0$)
$$
\|Ae^{-isH}\psi\|\leq a_1\||H|^{\alpha}\psi\|+b_1\|\psi\|, \quad
\|A^*e^{-isH}\psi\|\leq a_2\||H|^{\alpha}\psi\|+b_2\|\psi\|.
$$
Hence the assumption of Theorem \ref{qt2} is satisfied with ${\cal D}=D(|H|^{\alpha})$.
Thus the desired result follows.
\hfill \qed

\medskip




We next consider conditions
under which  the integrand of the integral on the right hand side
of  (\ref{AH3}) can be written in a simpler form.

\begin{df}{\rm Let $S$ and $T$ be densely defined linear
operators on a Hilbert space ${\cal H}$ and 
${\cal D}$ be  a  subspace of ${\cal H}$ such that
${\cal D}\subset D(S)\cap D(T)\cap D(S^*)\cap D(T^*)$.
Suppose that, there exists a linear operator $W$ with $D(W)=
{\cal D}$ such that, for all $\psi, \phi \in {\cal D}$
$$
\lang S^*\psi, T\phi\rang-\lang T^*\psi, S\phi\rang=\lang \psi, W\phi
\rang.
$$
Then we say that the pair $(S,T)$ has a {\it weak commutator} $W$
on ${\cal D}$ and write
$$
W=[S,T]_{\rm w}^{\cal D}.
$$
}
\end{df}

\begin{rem}{\rm Since ${\cal D}$ is dense, the weak commutator 
$W$ (if it exists) is uniquely determined.}
\end{rem}


For a subset ${\cal S}$ of ${\cal H}$, we denote by
${\cal L}({\cal S})$ the subspace algebraically spanned by
the vectors in ${\cal S}$.


\begin{The}\label{main-th}
Let  $A$ be a densely defined closed linear operator on ${\cal H}$.
Suppose that there exists a dense subspace ${\cal D}
\subset D(H)\cap D_{H,A}\cap D_{H,A^*}$
such that the pair $(H,A)$ has a weak commutator on the subspace
\begin{equation}
{\cal E}:={\cal L}(\{e^{-isH}\psi|\psi \in {\cal D}, s\in \R\}),
\end{equation}
(note that ${\cal E}$ is dense in ${\cal H}$) and
(\ref{sup}) holds .
Assume that the mapping:
$\R\ni s \mapsto 
[H,A]_{\rm w}^{\cal E}e^{-isH}\psi$ is strongly continuous
for all $\psi \in {\cal D}$. Then 
\begin{equation}
A_H(t)\psi=
A\psi+i\int_0^t
e^{isH}[H,A]_{\rm w}^{\cal E}e^{-isH}\psi ds, \quad t \in \R,\ \psi \in
{\cal D}. \label{AH4}
\end{equation}

Moreover, if 
\begin{equation}
M_t(\psi):=\sup_{0\leq s \leq t}\|
[H,A]_{\rm w}^{\cal E}e^{-isH}\psi\| < \infty, \quad \forall t \geq 0,\forall \psi \in {\cal D}
\label{Mt}
\end{equation}
in addition, then
\begin{equation}
\|A_H(t)\psi\| \leq 
\|A\psi\|+ M_{|t|}(\psi)|t|, \quad  \forall t\in \R,\ \forall \psi \in
{\cal D}.  \label{AH5}
\end{equation}
\end{The}

{\it Proof}. Under the present assumption, for all $\psi \in {\cal D}$,
the mapping $:s \mapsto e^{isH}[H,A]_{\rm w}^{\cal E}$  
$e^{-isH}\psi$
 ($\psi \in {\cal D}$) is strongly continuous. Hence we can define
a linear  operator by
$$
K(t)\psi:=i\int_0^te^{isH}[H,A]_{\rm w}^{\cal E}e^{-isH}\psi ds, \quad \psi \in {\cal D}
$$
in the sense of strong integral. Then, by Theorem \ref{qt2},  we have
$$
\lang \phi, A_H(t)\psi\rang=\lang \phi, A\psi\rang
+\lang \phi, K(t)\psi\rang, \quad \phi,\psi \in {\cal D}.
$$
Since ${\cal D}$ is dense by the present assumption,
(\ref{AH4}) follows.

Let (\ref{Mt}) be satisfied. Then, for all $\psi \in {\cal D}$,
\begin{eqnarray*}
\|K(t)\psi\| &\leq& 
\int_{0}^{|t|}\|[H,A]_{\rm w}^{\cal E}e^{-isH}\psi\|ds \leq  M_{|t|}(\psi)|t|.
\end{eqnarray*}
Hence (\ref{AH5}) follows. 

\hfill \qed

\bigskip

Theorem \ref{main-th} implies the following theorem:


\begin{The}\label{main-th2}
Let  $A$ be a densely defined closed linear operator on ${\cal H}$.
Suppose that there exists a dense subspace ${\cal D}
\subset D(H)\cap D_{H,A}\cap D_{H,A^*}$
having the following properties {\rm (i)}--{\rm (iv)}:
\begin{list}{}{}
\item{{\rm (i)}} The pair $(H,A)$ has a weak commutator on the subspace
${\cal E}$.
\item{{\rm (ii)}} (\ref{sup}) holds.
\item{{\rm (iii)}} ${\cal D}$ is a core of $A$.
\item{{\rm (iv)}} $[H,A]_{\rm w}^{\cal E}$ is bounded.
\end{list}
Then
\begin{equation}
e^{-itH}D(A)=D(A), \quad \forall t\in \R
\end{equation}
and
\begin{equation}
\|A_H(t)\psi\| \leq 
\|A\psi\|+ \|[H,A]_{\rm w}^{\cal E}\| \|\psi\|\, |t|, \quad  \forall t\in \R,\ \forall \psi \in
D(A).  \label{AH6}
\end{equation}
\end{The}


{\it Proof}. By condition (iv), for all $\psi \in {\cal D}$, 
$[H,A]_{\rm w}^{\cal E}e^{-isH}\psi$ is strongly continuous in $s$
and $M_{|t|}(\psi)\leq C \|\psi\|,  \forall t \in \R$ with $C:=\|[H,A]_{\rm w}^{\cal E}\|$.
Hence we can apply Theorem 
\ref{main-th} to the present case. In particular, (\ref{AH5}) implies that
\begin{equation}
\|Ae^{-itH}\psi\|
\leq \|A\psi\| +C\|\psi\||t|, \quad \psi\in {\cal D}.\label{AH7}
\end{equation}
Since ${\cal D}$ is a core of $A$ by condition (iii), a limiting argument
using this inequality shows that, for all $\psi\in D(A)$,
$e^{-itH}\psi \in D(A)$ and
(\ref{AH7}) extends to all $\psi \in D(A)$.
Thus the desired reslut follows. \hfill \qed

\medskip

Theorem \ref{main-th2} tells us the following:
{\it Under the assumption of Theorem \ref{main-th2},
$D(A)$ is an invariant domain of $e^{-itH}$ for all $t\in \R$}.
We remark that this type of theorem
has been established
in \cite[Lemma 2]{GG} under a different assumption. 


Asymptotic behaviors of
$Ae^{-tH}\psi$ ($\psi \in D(A)$) as $|t|\to\infty$
can be derived from Theorem \ref{main-th2}:

\begin{cor}  Under the same assumption
as in Theorem \ref{main-th2},
\begin{equation}
\mbox{{\rm w-}}\lim_{|t|\to\infty}\frac {Ae^{-itH}}{t}\psi=0, \quad \forall \psi\in D(A),
\label{wlim}
\end{equation}
where $\mbox{{\rm w-}}\lim$ means weak limit. 

Moreover, for all compact operators $C$ on ${\cal H}$,
\begin{equation}
\lim_{|t|\to\infty}\frac{CAe^{-itH}}{t}\psi=0, \quad\forall \psi \in D(A).
\label{slim}
\end{equation}
\end{cor}

{\it Proof}. The idea of proof is same as that
of the proof of \cite[Corollary 8.7]{Thaller}.
By (\ref{AH6}), we have for all $\psi\in  D(A)$
$$
\frac{\|Ae^{-itH}\psi\|}{|t|}\leq c_0, \quad |t|\geq 1
$$
with $c_0:=\|A\psi\|+\|[H,A]_{\rm w}^{\cal E}\|\|\psi\|$.
Let $\phi \in {\cal H}$. Then, for every $\varepsilon>0$,
there exists a vector $\phi_{\varepsilon}\in {\cal D}$ such that
$\|\phi_{\varepsilon}-\phi\|<\varepsilon$�D
We have
\begin{eqnarray*}
\left|\lang \phi,\frac {Ae^{-itH}\psi}{t}\rang\right|
&\leq& \varepsilon c_0
+\frac {1}{|t|}(\|A^*\phi_{\varepsilon}\|\cdot \|\psi\|), \quad |t|\geq 1.
\end{eqnarray*}
Hence 
$$
\limsup_{|t|\to\infty}\left|\lang \phi,\frac{Ae^{-itH}\psi}{t}\rang\right|
\leq \varepsilon c_0.
$$
Since $\varepsilon>0$ is arbitrary,  (\ref{wlim}) follows. 
Formula (\ref{slim}) follows from the well-known fact
that a compact operator
maps a weakly convergent sequence to a strongly convergent one.
\hfill \qed

\medskip


\section{Generalized Heisenberg equations of motion}

Let $H$ be a self-adjoint operator on a Hilbert space ${\cal H}$ as before 
and ${\cal D}$ be a dense subspace of
${\cal H}$.
Then the Heisenberg equation of motion for an operator-valued function
$:\R\ni t\mapsto X(t)$ ($X(t)$ is a linear operator on ${\cal H}$) with respect to $(H,{\cal D})$
is defined by
\begin{equation}
\frac{dX(t)\psi}{dt}=i[H,X(t)]\psi, \quad \psi \in {\cal D}\label{HEQ}
\end{equation}
with the condtion that 
${\cal D}\subset \cap_{t\in \R}
D(HX(t))\cap D(X(t)H)$ and $X(t)\psi$ is strongly differentiable in $t$.


In the case where $H$ and $X(t)$ are both in ${\sf B}({\cal H})$,
this definition has no problems and it is easy to see that
a ${\sf B}({\cal H})$-valued solution $X(t)$ of (\ref{HEQ}) 
with the initial condition $X(0)=A\in {\sf B}({\cal H})$  and ${\cal D}={\cal H}$ 
is unique and given by $X(t)=e^{itH}Ae^{-itH}$. 
But, in the case where
at least one of $H$ and $X(t)$ is unbounded,
the treatment of (\ref{HEQ}) becomes somewhat difficult because
of the domain problems and the regularity problem (the non-triviality of strong differentiability
of $X(t)$).
To overcome this difficulty, we propose a weak version of
(\ref{HEQ}). 

\begin{df}\label{defH}{\rm Let ${\cal D}$ be a dense subspace of ${\cal H}$.
We say that an operator-valued function $X(t)$ with $D(X(t))$ dense for all $t\in \R$
obeys a {\it generalized Heisenberg equation of motion with respect to $(H,{\cal D})$}
if ${\cal D}\subset \cap_{t\in \R}D(X(t))\cap D(X(t)^*)\cap D(H)$ and, for all $\phi,\psi\in {\cal D}$, $\lang \phi,X(t)\psi\rang$ is
differentiable in $t\in \R$ with
\begin{equation}
 \frac {d}{dt}\lang \phi,X(t)\psi\rang
=i\left(\lang H\phi, X(t)\psi\rang-\lang X(t)^*\phi, H\psi\rang\right), \quad t\in \R.
\label{HEQ2}
\end{equation}
}
\end{df}


\begin{pr}\label{sol1} Let $A$ and ${\cal D}$ be as in Theorem \ref{qt2}.
Then $A_H(t)$ is a solution of (\ref{HEQ2}) with $X(0)=A$.
\end{pr}

{\it Proof}. This follows from differentiating  (\ref{AH3p}) in $t$.
\hfill \qed

\medskip


We next consider the problem on the uniqueness of solutions of (\ref{HEQ2}).
We introduce a class of operator-valued functions.

\begin{df}{\rm Let ${\cal D}$ be a dense subspace of ${\cal H}$ and
\begin{equation}
{\cal F}_{\cal D}:=\{\psi| e^{itH}\psi\in {\cal D}, \forall t\in \R\}.
\end{equation}
We say that an operator-valued function $X(\cdot)$ with $D(X(t))$ dense for all $t\in \R$
is in the set ${\sf X}_{\cal D}$ if the following (X.1) and (X.2) hold:
\begin{list}{}{}
\item{(X.1)}
${\cal D}\subset \cap_{t\in \R}D(X(t))\cap D(X(t)^*)\cap D(H)$.
\item{(X.2)} For all $\psi\in{\cal F}_{\cal D}$ and $t\in \R$, $X(t+s)e^{itH}\psi$ and
$X(s)^*e^{isH}\psi$  are strongly continuous in $s\in \R$
or $X(s)e^{isH}\psi$ and
$X(t+s)^*e^{itH}\psi$  are strongly continuous in $s\in \R$.
\end{list}
}
\end{df}


\begin{The}\label{unique} Suppose that ${\cal F}_{\cal D}$ is dense in ${\cal H}$.
Let $X_0$ be a densely defined 
linear operator on ${\cal H}$ such that ${\cal D}\subset D(X_0)\cap D(X_0^*)$. 
Then the solution  $X(\cdot)$ of (\ref{HEQ2})
such that $X(\cdot)\in {\sf X}_{\cal D}$ and $X(0)=X_0$
is unique on ${\cal D}$.
\end{The}

{\it Proof}.  It is sufficient to prove that, if  $X(0)|{\cal D}=0$, then
$X(t)|{\cal D}=0, \,\forall t\in \R$.
Let $Y(t):=e^{-itH}X(t)e^{itH}, \ t\in \R$ and
$$
L_{\varepsilon}:=\frac{e^{i\varepsilon H}-1}{\varepsilon}, \quad\varepsilon\in \R\setminus\{0\}. 
$$
We first consider the case where the first condition in (X.2) is satisfied.
We have for all $\phi,\psi \in {\cal F}_{\cal D}$
\begin{eqnarray*}
\lang\phi, \frac{Y(t+\varepsilon)-Y(t)}{\varepsilon}\psi\rang
&=& \lang X(t+\varepsilon)^*e^{i(t+\varepsilon)H}\phi, L_{\varepsilon}e^{itH}\psi\rang\\
&&+\lang e^{itH}L_{\varepsilon}\phi, X(t+\varepsilon)e^{itH}\psi\rang \\
&&+\lang e^{itH}\phi, \frac{X(t+\varepsilon)-X(t)}{\varepsilon}
e^{itH}\psi\rang.
\end{eqnarray*}
This implies that
$\lang \phi, Y(t)\psi\rang$ is differentiable in $t$ with
\begin{eqnarray*}
\frac{d}{dt}\lang \phi, Y(t)\psi\rang&=&
i\lang X(t)^*e^{itH}\phi, He^{itH}\psi\rang  -i\lang e^{itH}H\phi, X(t)e^{itH}\psi\rang \\
&& +\frac {d}{ds}\lang e^{itH}\phi,X(s)e^{itH}\psi\rang|_{s=t}.
\end{eqnarray*}
By  (\ref{HEQ2}), the right hand side is equal to zero.
Hence $\lang\phi, Y(t)\psi\rang=\lang \phi, Y(0)\psi\rang=0$.
Since ${\cal F}_{\cal D}$ is dense, it follows that $Y(t)\psi=0$ for all $\psi\in{\cal F}_{\cal D}$.
Hence it follows from  the denseness of ${\cal F}_{\cal D}$ that
 $X(t)^*\phi=0$ for all $\phi \in {\cal D}$.
This implies that $X(t)|{\cal D}=0$.

In the case where the second condition in (X.2) holds, we need only to rewrite
$\lang \phi, \varepsilon^{-1}[(Y(t+\varepsilon)-Y(t))]\psi\rang$ as
\begin{eqnarray*}
\lang\phi, \frac{Y(t+\varepsilon)-Y(t)}{\varepsilon}\psi\rang
&=& \lang X(t+\varepsilon)^*e^{itH}\phi, L_{\varepsilon}e^{itH}\psi\rang\\
&&+\lang e^{itH}L_{\varepsilon}\phi, X(t+\varepsilon)e^{i(t+\varepsilon)H}\psi\rang \\
&&+\lang e^{itH}\phi, \frac{X(t+\varepsilon)-X(t)}{\varepsilon}
e^{itH}\psi\rang
\end{eqnarray*}
and proceed in the same way as in the preceding case.
\hfill \qed

\medskip




\begin{cor}\label{uniq-cor} Let $A$ be a densely defined closed linear operator on ${\cal H}$.
Suppose that there exists a dense subspace ${\cal D}
\subset D(H)\cap D_{H,A}\cap D_{H,A^*}$
such that the following {\rm (i)} and {\rm (ii)} hold:
\begin{list}{}{}
\item{{\rm (i)}} For all $\psi \in {\cal D}$,  $Ae^{itH}\psi$ and $A^*e^{itH}$
are strongly continuous in $t$.
\item{{\rm (ii)}} ${\cal F}_{\cal D}$ is dense in ${\cal H}$.
\end{list}
Then
$X(t)=A_H(t)$ is the unique solution of (\ref{HEQ2})
such that $X(\cdot)\in {\sf X}_{\cal D}$ and $X(0)|{\cal D}=A|{\cal D}$.
\end{cor}


{\it Proof}. Condition (i) implies  (\ref{sup}) in Theorem \ref{qt2}.
Hence, by Proposition \ref{sol1}, $X(t)=A_H(t)$ is a solution of (\ref{HEQ2})
with $X(0)|{\cal D}=A|{\cal D}$ and ${\cal D}\subset 
\cap_{t\in \R}D(A_H(t))\cap D(A_H(t)^*)\cap D(H)$.
For all  $\psi\in {\cal F}_{\cal D}$,
we have $A_H(t+s)e^{itH}\psi=e^{i(t+s)H}Ae^{-isH}\psi$
and $A_H(s)^*e^{isH}\psi=
e^{isH}A^*\psi$.
Hence $A_H(t+s)e^{itH}\psi$ and $A_H(s)^*e^{isH}\psi$ are
strongly continuous in $s$. 
Therefore $A_H \in {\sf X}_{\cal D}$.
Thus, by Theorem \ref{unique}, the uniqueness of $A_H(\cdot)$  follows.
\hfill \qed

\medskip
 

\begin{rem}{\rm  In applications, 
Corollary \ref{uniq-cor} can be used as follows:
Suppose that one finds an operator $X(t)$  satisfying (\ref{HEQ2})
with $X(\cdot)\in {\sf X}_{\cal D}$ and $X(0)|{\cal D}=A|{\cal D}$ and that
$X(t)$ has an explicit representation.
Then, by Corollary \ref{uniq-cor}, $A_H(t)|{\cal D}=X(t)|{\cal D}$, 
which gives an explicit representation for  $A_{H}(t)|{\cal D}$.
Elementary examples in quantum physics  for which this method works
are as follows: (i)  quantum particles in $\R^n$ with potentials of
quadratic polynomials, including  harmonic oscillators ; (ii) quasi free quantum
field models. But we do not discuss them here.
}
\end{rem}
 
  
\section{Weyl Representations of canonical commutation relations and
invariant domains}


In this section we show that there is a general mathematical structure
for existence of invariant domains of the strongly continuous one-parameter 
unitary group generated by
an abstract self-adjoint operator which
is made from the Weyl representation  of canonical commutation relations (CCR) with
a finite degree of freedom. 





Let ${\cal H}$  be a complex  Hilbert space and
$\{Q_j, P_j|j=1,\cdots,n\}$ be a set of self-adjoint operators on ${\cal H}$ obeying
the following conditions:
\begin{list}{}{}
\item{(i)} For all $s,t\in \R$ and $j,k=1,\cdots,n$
$$
e^{itQ_j}e^{isQ_k}=e^{isQ_k}e^{itQ_j}, \quad
e^{itP_j}e^{isP_k}=e^{isP_k}e^{itP_j}.
$$
\item{(ii)} 
For all $j,k=1,\cdots,n $ and $s,t\in \R$
\begin{equation}
e^{itQ_j}e^{isP_k}=e^{-its\delta_{jk}}e^{isP_k}e^{itQ_j}.\quad \label{WW}
\end{equation} 
\end{list}
The set $\{Q_j, P_j|j=1,\cdots,n\}$ is called
a {\it Weyl representation of
the CCR with $n$ degrees of freedom}. It is well known that,
if ${\cal H}$ is separable, then 
$\{Q_j,P_j|j=1,\cdots,n\}$ is unitarily equivalent to a direct sum
of the Schr\"odinger representation of the CCR with $n$ degrees of freedom \cite{VN}.
But  we work with  $\{Q_j, P_j|j=1,\cdots,n\}$, since
it is suitable for analysis
of general mathematical structures that the Weyl representation of  CCR has.


Here we recall the definition of strong commutativity 
on self-adjoint operators. For a self-adjoint operator $S$, we denote by
$E_S(\cdot)$ its spectral measure.

\begin{df}{\rm Two self-adjoint operators  $S$ and $T$  on a Hilbert space are
said to {\it strongly commute} if their spectral measures commute, i.e.,
$E_S(J)E_T(K)=E_T(K)E_S(J)$ for all Borel sets $J,K \subset \R$. In this case,
we say also that $S$ strongly commutes with $T$.

A set $\{S_j\}_{j=1}^n$  ($n\in \N$)   of self-adjoint operators 
on a Hilbert space is said to be {\it strongly commuting}
if, for each pair $(j,k)$ with $j\not=k$ ($j,k=1,\cdots,n$), $S_j$ and $S_k$ strongly commute.
}
\end{df}


\begin{lem}\label{lem5-1} Let $Q_j$ and $P_j$ be as above. Then:
\begin{list}{}{}
\item{{\rm (i)}} $\{Q_j\}_{j=1}^n$ is  strongly commuting.
\item{{\rm (ii)}} $\{P_j\}_{j=1}^n$ is  strongly commuting.
\item{{\rm (iii)}}
For all $j,k=1,\cdots,n$ with $j\not=k$,
$P_j$ and $Q_k$  strongly commute.
\item{{\rm (iv)}} For all $t\in \R$ and $k,j=1,\cdots,n$
\begin{equation}
e^{itQ_k}D(P_j)=D(P_j)\label{WW1}
\end{equation}
and 
\begin{equation}
e^{itQ_k}P_je^{-itQ_k}=P_j-\delta_{jk}t.\label{WW2}
\end{equation}
\end{list}
\end{lem}

{\it Proof}. The facts of (i) and (ii) follow from a general theorem
(e.g.,\cite[Theorem VIII.13]{RS1}).

Formula (\ref{WW}) with $j\not=k$ gives
$e^{itQ_j}e^{isP_k}=e^{isP_k}e^{itQ_j}$ for all $s,t\in \R$. Hence, by a general theorem 
(\cite[Theorem VIII.13]{RS1}) again, the desired result follows. 

To prove (iii), let $\psi \in D(P_j)$. 
Then, by (\ref{WW}), $e^{itQ_k}e^{ist\delta_{jk}}e^{isP_j}\psi=e^{isP_j}e^{itQ_k}\psi$
for all $s,t\in \R$. The vector-valued function $e^{isP_j}\psi$ of $s$ is
strongly differentiable in $s$. Hence the left hand side is strongly differentiable
in $s$. Therefore the right hand side is strongly differentiable in $s$.
Thus $e^{itQ_k}\psi\in D(P_j)$ and
\begin{equation}
e^{itQ_k}t\delta_{jk}\psi+e^{itQ_k}P_j\psi=
P_je^{itQ_k}\psi.\label{WW3}
\end{equation}
Hence $e^{itQ_k}D(P_j)\subset D(P_j)$.
This implies (\ref{WW1}). Then (\ref{WW2}) follows from (\ref{WW3}).
\hfill \qed

\medskip




By the strong commutativity of $Q:=\{Q_j\}_{j=1}^n$ [Lemma \ref{lem5-1})-(i)], there exists
a unique $n$-dimensional spectral measure $\hat E_Q$, called the joint spectral measure
of $Q$,  such that,
for all Borel sets $B_j$ in $\R$ ($j=1,\cdots,n$)
$$
\hat E_Q(B_1\times \cdots \times B_n)=E_{Q_1}(B_1)\cdots E_{Q_n}(B_n).
$$
Then one can develop the functional calculus with respect to $\hat E_Q$. Namely,
for each Borel measurable function $F$ on $\R^n$, we can define a
linear operator $F(Q)$ on ${\cal H}$ by
\begin{equation}
F(Q):=\int_{\R^n}F(\lambda)d\hat E_Q(\lambda),
\end{equation}
i.e.,
\begin{eqnarray*}
&& D(F(Q))=\left\{\psi \in {\cal H}\big| \int_{\R^n}|F(\lambda)|^2d\|\hat E_{Q}(\lambda)\psi\|^2<\infty\right\},\\
&& \lang \phi, F(Q)\psi\rang=\int_{\R^n}F(\lambda)d\lang \phi, \hat E_Q(\lambda)\psi\rang, \ \phi \in {\cal H}, \psi \in D(F(Q)).
\end{eqnarray*}
If $F$ is real-valued such that $\hat E_Q(\{\lambda \in \R^n|\,
|F(\lambda)|=+\infty\})=0$, then $F(Q)$ is self-adjoint.
Similarly $P:=\{P_j\}_{j=1}^n$ has the joint spectral measure
$\hat E_P$. In what follows, we define, using $\hat E_Q$ and $\hat E_P$,
an abstract  version of quantum mechanical Hamiltonians,
including the usual  Schr\"odinger type  operators with matrix-valued potentials 
and  Dirac type ones.





For $\ell=1,\cdots,L$ ($L\in \N$), let $T_{\ell}:\R^n \to \R; \R^n\ni \lambda \mapsto T_{\ell}
(\lambda) \in\R$ be  a continuously differentiable function 
satisfying the following conditions (T.1) and (T.2):
\begin{list}{}{}
\item{(T.1)} For each $
j=1,\cdots,n$ and $\ell=1,\cdots,L$, 
there exist positive constants $a_{\ell j}, b_{\ell j}$ such that, for 
all sufficiently small $|t|$ ($t\in \R$),
$$
|T_{\ell}(\lambda+t\e_j)|^2
\leq a_{\ell j}|T_{\ell}(\lambda)|^2+b_{\ell j}, \ \forall \lambda \in \R^n,
$$
where $\e_j:=(0,\cdots,0,\stackrel{\mbox{\footnotesize $j$-th}}{1},0,\cdots,0)\in 
\R^n$ ($j=1,\cdots,n$).
\item{(T.2)} For $j=1,\cdots,n$ and $\ell=1,\cdots,L$,  
there exist positive constants $c_{\ell j},d_{\ell j}$ such that, for 
all sufficiently small $|t|$,
$$
|\partial_jT_{\ell}(\lambda+t\e_j)|^2\leq c_{\ell j}|T_{\ell}(\lambda)|^2+d_{\ell j}, 
\forall \lambda \in \R^n,
$$
where $(\partial_jT_{\ell})(\lambda):=\partial T_{\ell}(\lambda)/\partial \lambda_j$.
\end{list}

Let ${\cal K}$ be a Hilbert space and
$\{A_{\ell},B_{\ell}|\ell=1,\cdots,L\}$ be a set of bounded self-adjoint
operators on ${\cal K}$ ($A_{\ell},B_{\ell}\in {\sf B}({\cal K})$).
Let $V_{\ell}$  ($\ell=1,\cdots,L$) be a real-valued Borel measurable function on $\R^n$
such that $\hat E_Q(\{\lambda \in \R^n|
\,|V_{\ell}(\lambda)|=+\infty\})=0$.
Then $T_{\ell}(P)$ and $V_{\ell}(Q)$ are self-adjoint.
We define an linear operator $H$ acting in ${\cal K}\otimes {\cal H}$ as follows:
\begin{equation}
H:=\sum_{\ell=1}^L\left[A_{\ell}\otimes T_{\ell}(P)+B_{\ell}\otimes V_{\ell}(Q)\right].
\end{equation}
We set
\begin{equation}
H_0(P):=\sum_{\ell=1}^LA_{\ell}\otimes T_{\ell}(P), \quad
H_{\rm I}:=\sum_{\ell=1}^LB_{\ell}\otimes V_{\ell}(Q).
\end{equation}
Hence
\begin{equation}
H=H_0(P)+H_{\rm I}.
\end{equation}
We assume the following:

\bigskip

\noindent
{\bf Hypothesis (H)}
\begin{list}{}{}
\item{(H.1)} Each $A_{\ell}$ has a bounded inverse $A_{\ell}^{-1}
\in {\sf B}({\cal H})$.
\item{(H.2)}
The operator $H$ is self-adjoint.
\end{list}

\medskip

The main result of this section is the following theorem:


\begin{The}\label{mainth1} Under Hypothesis  (H), the following {\rm (i)} and 
{\rm (ii)} hold:
\begin{list}{}{}
\item{{\rm (i)}}
For all $t\in \R$ and $j=1,\cdots,n$,
\begin{equation}
e^{-itH}D(I\otimes Q_j)\cap D(H)=D(I\otimes Q_j)\cap D(H),\label{invdom}
\end{equation} 
where $I$ denotes identity.
Moreover, for all $\psi \in D(I\otimes Q_j)\cap D(H)$,
\begin{equation}
e^{itH}(I\otimes Q_j)e^{-itH}\psi
=I\otimes Q_j\psi
+\sum_{\ell=1}^L\int_0^te^{i\tau H}
A_{\ell}\otimes (\partial_jT_{\ell})(P)e^{-i\tau H}\psi d\tau,\label{HQ}
\end{equation}
where the integral on the right hand side is taken in the strong sense.
\item{{\rm (ii)}} If each $\partial_jT_{\ell}$ ($j=1,\cdots,n, \ell=1,\cdots,L$)
is bounded and 
 $D(I\otimes  Q_j)\cap D(H)$ is a core for $I \otimes Q_j$ in addition, then
\begin{equation}
e^{-itH}D(I\otimes Q_j)=D(I\otimes Q_j)\label{invdom1}, \quad \forall t\in \R
\end{equation} 
and (\ref{HQ}) holds  for all $\psi \in D(I\otimes Q_j)$.
\end{list}
\end{The}


To prove this theorem, we need some lemmas.
It follows from  (H.2) and the closed graph theorem that there exists a constant $C>0$
such that
\begin{equation}
\|A_{\ell}\otimes T_{\ell}(P)\psi\|\leq
C(\|H\psi\|+\|\psi\|), \quad \forall \psi \in D(H), \ell=1,\cdots,L.
\end{equation}
Condition (H.1) implies that
$$
\|I\otimes T_{\ell}(P)\psi\|
\leq \|A_{\ell}^{-1}\| \|A_{\ell}\otimes T_{\ell}(P)\psi\|, \quad
\forall \psi \in D(A_{\ell}\otimes T_{\ell}(P)), \ell=1,\cdots,L.
$$
Hence
\begin{equation}
\|I\otimes T_{\ell}(P)\psi\|\leq
\|A_{\ell}^{-1}\|C(\|H\psi\|+\|\psi\|), \quad \forall \psi \in D(H), \ell=1,\cdots,L.
\label{H-bd}
\end{equation}



\begin{lem}\label{unit} Let $U$ be a unitary operator from a Hilbert space ${\cal H}_1$ to a
Hilbert space ${\cal H}_2$ and $R_j$ ($j=1,\cdots, N, \, N\in \N $) be
a linear operator from ${\cal H}_1$ to ${\cal H}_2$ with domain $D(R_j)\subset {\cal H}_1$.
Then the operator equality
$$
U\left(\sum_{j=1}^NR_j\right)U^{-1}=\sum_{j=1}^NUR_jU^{-1}
$$
holds.
\end{lem}


{\it Proof}. We need only to show that $D\left( U\left(\sum_{j=1}^NR_j\right)U^{-1}\right)
=\cap_{j=1}^ND(UR_jU^{-1})$. But this is straightforward.
\hfill \qed

\medskip
 

\begin{lem}\label{lem5-2} For all $t\in \R$, $e^{it(I\otimes Q_j)}H_{\rm I}e^{-it(I\otimes Q_j)}
=H_{\rm I}$ (operator equality).
\end{lem}

{\it Proof}. For all $\psi \in {\cal K}\hat \otimes D(V_{\ell}(Q))$
($\hat \otimes$ denotes algebraic tensor product),
$e^{it(I\otimes Q_j)}B_{\ell}\otimes V_{\ell}(Q)e^{-it(I\otimes Q_j)}\psi
=B_{\ell}\otimes V_{\ell}(Q)\psi$ ($\ell=1,\cdots,L$).
Hence $B_{\ell}\otimes V_{\ell}(Q)\subset e^{it(I\otimes Q_j)}B_{\ell}\otimes V_{\ell}(Q)e^{-it(I\otimes Q_j)}$. Since the both sides are self-adjoint, we have
$B_{\ell}\otimes V_{\ell}(Q)=e^{it(I\otimes Q_j)}B_{\ell}\otimes V_{\ell}(Q)e^{-it(I\otimes Q_j)}$.
By this relation and Lemma \ref{unit}, we obtain  the desired result. \hfill \qed

\medskip

\begin{lem}\label{lem5-3} For all $t\in \R$, $e^{it(I\otimes Q_j)}H_0(P)e^{-it(I\otimes Q_j)}
=H_0(P-t\e_j)$ (operator equality).
\end{lem}

{\it Proof}. By (\ref{WW2}) and the functional calculus, we have
$e^{itQ_j}T_{\ell}(P)e^{-itQ_j}=T_{\ell}(P-t\e_j)$, which implies that
$e^{it(I\otimes Q_j)}A_{\ell}\otimes T_{\ell}(P)e^{-it(I\otimes Q_j)}=A_{\ell}\otimes T_{\ell}(P-t\e_j)$. 
This relation and Lemma \ref{unit} give the desired result.
\hfill \qed

\medskip



\begin{lem}\label{lem5-4} There exists a constant $\delta>0$ such that,
for all $|t|<\delta$,
\begin{equation}
D(I\otimes T_{\ell}(P-t\e_j))=D(I\otimes T_{\ell}(P)), \quad j=1,\cdots,n, \ell=1,\cdots,L,
\end{equation}
and 
\begin{eqnarray}
\|[I\otimes T_{\ell}(P-t\e_j)-I\otimes T_{\ell}(P)]\psi\|
&\leq&  K|t|(\|H\psi\|+\|\psi\|), \quad \psi\in D(H),\nonumber\\
&& \quad j=1,\cdots,n, \ell=1,\cdots,L,
\label{H-bd1}
\end{eqnarray}
where $K>0$ is a constant independent of $t$.
\end{lem}

{\it Proof}. The first half of the lemma  follows from
condition (T.1) and the functional calculus.
By the mean value theorem and condition (T.2), we have for all sufficiently small $|t|$
$$
|T_{\ell}(\lambda-t\e_j)-T_{\ell}(\lambda)|^2
\leq C_1^2|t|^2(|T_{\ell}(\lambda)|^2+1),
$$
where $C_1>0$ is a constant. Hence
$$
\|[T_{\ell}(P-t\e_j)-T_{\ell}(P)]\phi\|
\leq C_1|t|(\|T_{\ell}(P)\psi\|+\|\phi\|), \quad \phi
\in D(T_{\ell}(P)).
$$
By this estimate and (\ref{H-bd}), we obtain (\ref{H-bd1}).
\hfill \qed

\medskip




For $j=1,\cdots,n$ and $t\in \R$, we define
\begin{equation}
H_j(t):=H_0(P-t\e_j)+H_{\rm I}.
\end{equation}

\begin{lem}\label{lem5-5} For $j=1,\cdots,n$ and all sufficiently small $|t|$,
$H_j(t)$ is self-adjoint with $D(H_j(t))=D(H)$ and
\begin{equation}
e^{it(I\otimes Q_j)}He^{-it(I\otimes Q_j)}=H_j(t).\label{unitrans}
\end{equation}
In particular, for all $t\in \R$, $e^{-it(I\otimes Q_j)}D(H)=D(H)$.
\end{lem}

{\it Proof}. 
We can write
$$
H_j(t)=H+\sum_{\ell=1}^LW_{j\ell}(t),
$$
where $W_{j\ell}(t):=A_{\ell}\otimes T_{\ell}(P-t\e_j)-A_{\ell}\otimes T_{\ell}(P)$.
Let $\psi \in D(H)$. Then, by Lemma \ref{lem5-4}, we obtain
$$
\|W_{j\ell}\psi\|\leq C_2|t|(\|H\psi\|+\|\psi\|),
$$
where $C_2>0$ is a constant independent of $t$.
Hence, by the Kato-Rellich theorem, for all $|t|$ such that $LC_2|t|<1$, 
$H_j(t)$ is self-adjoint with $D(H_j(t))=D(H)$.
Formula (\ref{unitrans}) follows from
Lemmas \ref{lem5-2} and \ref{lem5-3}.

The result of the preceding paragraph implies that, for all sufficiently small $|t|$,
$$
e^{-it(I\otimes Q_j)}D(H)=D(H).
$$
But, by using the group property of $e^{-it(I\otimes Q_j)}$
in $t$, one can show that this  holds for all $t\in \R$.
\hfill \qed

\medskip


\begin{lem}\label{Du1} Let $A$ and $B$ be self-adjoint operators on ${\cal H}$ such that $A+B$ is self-adjoint and $B$ is $A$-bounded. 
Then, for all $\psi \in D(A)$ and $t\in \R$,
\begin{equation}
e^{-it(A+B)}\psi=e^{-itA}\psi-i\int_0^te^{-i(t-s)(A+B)} Be^{-isA}\psi ds,
\label{dform1}
\end{equation}
where the integral on the right hand side is taken in the sense of strong Riemann  integral.
\end{lem}


{\it Proof}. This lemma is probably well known. But, for completeness,
we give a proof. We first consider the case where both $A$ and $B$ are in
${\sf B}({\cal H})$. Let $\phi(t)
=e^{-itA}\psi-i\int_0^te^{-i(t-s)(A+B)} Be^{-isA}\psi ds$. 
Then it is easy to show that $\phi(t)$ is strongly differentiable
in $t$ and 
$$
i\frac {d}{dt}\phi(t)=(A+B)\phi(t).
$$
Hence $\phi(t)=e^{-i(A+B)t}\phi(0)=e^{-i(A+B)t}\psi$. Therefore
(\ref{dform1}) holds.

We next consider the case where $A$ is not necessarily bounded and $B\in
{\sf B}({\cal H})$. Let $A_{\varepsilon}^f$
be  the operator $S_{\varepsilon}^f$ defined by (\ref{reg}) 
with $S$ replaced by $A$. Then, by the preceding result, we have
\begin{equation}
e^{-it(A_{\varepsilon}^f+B)}\psi=e^{-itA_{\varepsilon}^f}\psi-
i\int_0^te^{-i(t-s)(A_{\varepsilon}^f+B)} Be^{-isA_{\varepsilon}^f}\psi ds.
\label{dform2}
\end{equation}
For all $\phi \in D(A)$, $\lim_{\varepsilon \to 0}(A_{\varepsilon}+B)\phi=
(A+B)\phi$.
Hence, by a general convergence theorem, 
$\lim_{\varepsilon\to 0}e^{-it(A_{\varepsilon}^f+B)}\psi=
e^{-it(A+B)}\psi$. 
Moreover
\begin{eqnarray*}
&& \|\int_0^te^{-i(t-s)(A_{\varepsilon}^f+B)}Be^{-isA_{\varepsilon}^f}\psi ds
-\int_0^te^{-i(t-s)(A+B)} Be^{-isA}\psi ds\| \\
&& \leq \int_0^{|t|}
\|B\|\|e^{-isA_{\varepsilon}^f}\psi-e^{-isA}\psi\|ds
+\int_{0}^{|t|}\|(e^{-i(t-s)(A_{\varepsilon}^f+B)}-e^{-i(t-s)(A+B)})Be^{-isA}\psi\|
ds.
\end{eqnarray*}
Hence, using the Lebesgue dominated convergence theorem,
we see that the right hand side converges to $0$ as $\varepsilon \to 0$.
Thus, taking the limit $\varepsilon\to 0$
in (\ref{dform2}), we obtain (\ref{dform1}).
 

Finally we consider the case where both $A$ and $B$ satisfy the assumption
of the present lemma. By the result of the preceding paragraph, we have
\begin{equation}
e^{-it(A+B_{\varepsilon}^f)}\psi=e^{-itA}\psi-i\int_0^te^{-i(t-s)(A+B_{\varepsilon}^f)} B_{\varepsilon}^fe^{-isA}\psi ds.
\label{dform3}
\end{equation}
Since $B$ is $A$-bounded, it follows that $D(A)\subset D(B)$ and $e^{-isA}\phi \in D(A)\subset D(B)$ for all $\phi \in D(A)$ and $s\in \R$. Hence 
$$
\lim_{\varepsilon\to 0}B_{\varepsilon}^fe^{-isA}\psi=Be^{-isA}\psi.
$$
Using the $A$-boundedness of $B$, one can show that $e^{-i(t-s)(A+B)}
Be^{-isA}\psi$ is strongly continuous in $s\in \R$. 
Hence, in the same way as in the preceding paragraph, we can show that
taking the limit $\varepsilon \to 0$ in (\ref{dform3}) yields (\ref{dform1}).
\hfill \qed

\medskip

\bigskip


\noindent
{\bf Proof of Theorem \ref{mainth1}}

Throughout the proof, we set $\hat Q_j:=I\otimes Q_j$. 

(i) We can write
$$
H_j(s)=H+C(s)
$$
with $C(s):=H_0(P-s\e_j)-H_0(P)$, where $|s|$ ($s\in \R$) is sufficiently small.
Then, by Lemma \ref{lem5-5} and the functional calculus, we have for all Borel measurable function $u$ on $\R$
$$
e^{is\hat Q_j}u(H)e^{-is\hat Q_j}=u(H+C(s)).
$$
Hence, taking $u(\lambda)=e^{-it\lambda}, \lambda \in \R$, 
we obtain for all $t\in\R$ and
sufficiently small $|s|$
$$
e^{is\hat Q_j}e^{-itH}=e^{-it(H+C(s))}e^{is \hat Q_j}.
$$
Hence, for all  $\psi\in D(\hat Q_j)$ ($s\not=0$),
\begin{eqnarray}
\frac{(e^{is\hat Q_j}-1)}{s}e^{-itH}\psi
&=&e^{-it(H+C(s))}\left(\frac{(e^{is\hat Q_j}-1)\psi}{s}-i\hat Q_j\psi\right)
+e^{-it(H+C(s))}i\hat Q_j\psi \nonumber
\\
&& +\frac{e^{-it(H+C(s))}-e^{-itH}}{s}\psi.\label{bibun}
\end{eqnarray}
We have
$$
\lim_{s\to 0}
\left(\frac{(e^{is\hat Q_j}-1)\psi}{s}-i\hat Q_j\psi\right)=0.
$$
Since $e^{-it(H+C(s))}$ is unitary, it follows that
$$
\lim_{s\to 0}
e^{-it(H+C(s))}\left(\frac{(e^{is\hat Q_j}-1)\psi}{s}-i\hat Q_j\psi\right)=0.
$$
By Lemma \ref{lem5-4},  $C(s)$ is $H$-bounded and
$\lim_{s\to 0}C(s)\eta=0$ for all $\eta\in D(H)$.
Hence, by general convergence theorems \cite[VIII.21, VIII.25(a)]{RS1}, we have
$$
\lim_{s\to 0}e^{-it(H+C(s))}\phi=e^{-itH}\phi, \quad \phi\in {\cal H}.
$$
Hence 
$$
\lim_{s\to 0}e^{-it(H+C(s))}i\hat Q_j\psi=ie^{-itH}\hat Q_j\psi.
$$
By Lemma \ref{Du1}, we have
$$
\frac{e^{-it(H+C(s))}-e^{-itH}}{s}\psi=
-i\int_0^te^{-i(t-\tau)(H+C(s))}\frac{C(s)}{s}e^{-i\tau H}\psi d\tau.
$$
By the functional calculus, we can show that, for all $\phi \in D(H)$, 
$\lim_{s\to 0}C(s)\phi/s=-\sum_{\ell=1}^L
A_{\ell}\otimes (\partial_jT_{\ell})(P)\phi$ and $\|C(s)\phi/s\|
\leq a\|(H+1)\phi\|$, where $a$  is a positive constant independent of
$s$.
Hence it follows that
\begin{eqnarray*}
\lim_{s\to 0}\frac{e^{-it(H+C(s))}-e^{-itH}}{s}\phi&=&
i\sum_{\ell=1}^L\int_0^te^{-i(t-\tau)H}A_{\ell}\otimes (\partial_jT_{\ell})(P))
e^{-i\tau H}\phi d\tau, \quad \forall \phi \in D(H).
\end{eqnarray*}
Putting these results into (\ref{bibun}),
one can conclude that, for all $\psi \in D(\hat Q_j)\cap D(H)$,
 $e^{-itH}\psi \in D(\hat Q_j)\cap D(H)$ and
$$
\hat Q_je^{-itH}\psi=
e^{-itH}\hat Q_j\psi+
\sum_{\ell=1}^L\int_0^te^{-i(t-\tau)H}A_{\ell}\otimes (\partial_jT_{\ell})(P))
e^{-i\tau H}\psi d\tau.
$$
Hence (\ref{invdom}) and (\ref{HQ}) follow. 

(ii) In this case, each $A_{\ell}\otimes (\partial_jT_{\ell})(P)$ is 
in ${\sf B}({\cal K}\otimes {\cal H})$. 
By using this property and that
$D(\hat Q_j)\cap D(H)$ is a core of $\hat Q_j$, one can  extend, by a simple limiting argument,
(\ref{invdom}) to all $\psi \in D(\hat Q_j)$, at the same time, showing
that, for all $\psi \in D(\hat Q_j)$ and all $t\in \R$,  $e^{-itH}\psi\in D(\hat Q_j)$.
Thus the desired result follows.
\hfill \qed

\medskip

\begin{rem}\label{sym}{\rm The operator $H$ is symmetric in the exchange of $(A_{\ell}, T_{\ell}(P))$ and
$(B_{\ell}, V_{\ell}(Q))$ ($\ell=1,\cdots,L$). Hence Theorem \ref{mainth1} holds with $Q_j$ 
(resp. $A_{\ell}, T_{\ell}(P)$) replaced by $-P_j$ (resp. $B_{\ell}, V_{\ell}(Q)$)
under the assumption for $(A_{\ell}, T_{\ell}(P))$ ($\ell=1,\cdots,L$)
replaced by $(B_{\ell}, V_{\ell}(Q))$ ($\ell=1,\cdots,L$)(note that, in the present case,  $-P_j$ plays the role of $Q_j$ in Theorem \ref{mainth1}).
}
\end{rem}



\section{Applications to Schr\"odinger and  Dirac Operators}

In this section we apply the results in the preceding section to
Schr\"odinger and  Dirac operators with operator-valued potentials.

\subsection{Schr\"odnger operators}
 
Let $\{q_j,p_j\}_{j=1}^n$ be the Schr\"odinger representaton
of the CCR with $n$ degrees of freedom, i.e., $q_j$ is the multiplication operator
by the $j$-th coordinate variable $x_j$ in $x=(x_1,\cdots,x_n)\in \R^n$ acting in
$L^2(\R^n)$ and 
$p_j:=-iD_j$, where $D_j$ is the generalized partial differential
operator in $x_j$ acting in $L^2(\R^n)$.
It is well-known (or easy to see) that $\{q_j,p_j\}_{j=1}^n$  is a Weyl representation.
In the context of quantum mechanics, $q_j$ and $p_j$ are called the  $j$-th position operator 
and the
$j$-th momentum operator respectively. 
We consider the following case in the notation of the preceding section:
\begin{eqnarray*}
&&{\cal H}=L^2(\R^{n}), \quad Q_j=q_j,P_j=p_j,\ j=1,\cdots,n,\\
&&  A_{\ell}=I,\ \ell=1,\cdots,L; \, T_1(\lambda)=\frac{\lambda^2}{2m},\ \lambda \in \R^n, \ \, T_{\ell}=0, \ell=2,\cdots,L, 
\end{eqnarray*}
where $m>0$ is a constant.
Then, under the natural identification ${\cal K}\otimes L^2(\R^n)\cong
L^2(\R^n;{\cal K})$, $H$ takes the following form:
$$
H_{\rm S}:=-\frac {\Delta}{2m}+\sum_{\ell=1}^LB_{\ell}V_{\ell}(q),
$$
where $\Delta:=\sum_{j=1}^nD_j^2$ is the generalized Laplacian acting in $L^2(\R^n)$ and
$q=(q_1,\cdots,q_n)$. This is an $n$-dimensional Schr\"odinger
operator with a ${\sf B}({\cal K})$-valued potential. Hence $H_{\rm S}$ is a generalization of 
Schr\"odinger operators with matrix-valued potentials.
  
\begin{The} 
Suppose that $H_{\rm S}$ is self-adjoint. Then, 
for all $t\in \R$ and $j=1,\cdots,n$,
\begin{equation}
e^{-itH_{\rm S}}D(q_j)\cap D(H_{\rm S})=D(q_j)\cap D(H_{\rm S})\label{invdom2}
\end{equation} 
and, for all $\psi \in D(q_j)\cap D(H_{\rm S})$,
\begin{equation}
q_j(t)\psi
=q_j\psi
+\int_0^t
\frac{p_j(s)}{m}\psi ds,
\end{equation}
where 
$$
q_j(t):=e^{itH_{\rm S}}q_je^{-itH_{\rm S}},
\quad p_j(t):=e^{itH_{\rm S}}p_je^{-itH_{\rm S}}.
$$
In particular, $q_j(t)\psi$ is strongly differentiable in $t\in \R$ and
\begin{equation}
\frac{d}{dt} q_j(t)\psi=\frac 1{m}p_j(t)\psi.\label{q-Newton1}
\end{equation} 
\end{The}

{\it Proof}. By direct computations, one can  easily check that 
the functions $T_{\ell}$ in the present case satisfy (T.1) and (T.2) with
$\partial_jT_1(\lambda)=\lambda_j/m$ and $\partial_jT_{\ell}=0, \,\ell\geq 2$.
Also Hypothesis (H) holds. 
Thus Theorem \ref{mainth1} 
yields the desired  result. \hfill \qed

\medskip
 


As for the momentum operator $p_j$, we have the following result:


\begin{The}\label{S2} Assume the following {\rm (i)}--{\rm (iv)}:
\begin{list}{}{}
\item{{\rm (i)}} $H_{\rm S}$ is self-adjoint.
\item{{\rm (ii)}} For $\ell=1, \cdots L$, $B_{\ell}$ is bijective and $B_{\ell}^{-1}$ is bounded. 
\item{{\rm (iii)}} For each $
j=1,\cdots,n$ and $\ell=1,\cdots,L$, 
there exist positive constants $a_{\ell j}, b_{\ell j}$ such that, for 
all sufficiently small $|t|$ ($t\in \R$),
$$
|V_{\ell}(x+t\e_j)|^2
\leq a_{\ell j}|V_{\ell}(x)|^2+b_{\ell j}, \ \forall x \in \R^n.
$$
\item{{\rm (iv)}} Each $V_{\ell}$ ($\ell=1,\cdots,L$) is continuously differentiable on $\R^n$ and, for $j=1,\cdots,n$ and $\ell=1,\cdots,L$,  
there exist positive constants $c_{\ell j},d_{\ell j}$ such that, for 
all sufficiently small $|t|$,
$$
|\partial_jV_{\ell}(x+t\e_j)|^2\leq c_{\ell j}|V_{\ell}(x)|^2+d_{\ell j}, 
\forall x \in \R^n.
$$
\end{list}
Then, 
for all $t\in \R$ and $j=1,\cdots,n$,
\begin{equation}
e^{-itH_{\rm S}}D(p_j)\cap D(H_{\rm S})=D(p_j)\cap D(H_{\rm S})\label{invdom3}
\end{equation} 
and, for all $\psi \in D(p_j)\cap D(H_{\rm S})$,
\begin{equation}
p_j(t)\psi
=p_j\psi
+\int_0^tF_j(s)\psi ds,
\end{equation}
where 
$$
F_j(s):=-\sum_{\ell=1}^Le^{isH_{\rm S}}B_{\ell}\partial_jV_{\ell}(q)e^{-isH_{\rm S}}.
$$
In particular, $p_j(t)\psi$ is strongly differentiable in $t\in \R$ and
\begin{equation}
\frac {d}{dt}p_j(t)\psi=F_j(t)\psi.\label{q-Newton2}
\end{equation}
\end{The}

{\it Proof}. This follows from Remark \ref{sym} and the present assumption. \hfill \qed

\medskip


\begin{rem}{\rm The system of differential equations (\ref{q-Newton1}) and (\ref{q-Newton2}) gives a mathematically 
rigorous form of the Heisenberg equation of motion for the pairs ($q_j(t), p_j(t)$), 
$j=1,\cdots,n$ (a \lq\lq{quantum Newton equation}").
}
\end{rem}




\subsection{Relativistic Schr\"odinger operators}

Let
$$
H_{\rm rel}:=\sqrt{-\Delta+m^2}+\sum_{\ell=1}^LB_{\ell}V_{\ell}(q),
$$
a relativistic Schr\"odinger operator with a ${\sf B}({\cal K})$-valued potential.


\begin{The} Assume that $V_{\ell}\in L^2_{\rm loc}(\R^n),\, \ell=1,\cdots,L$,
i.e., $\int_{|x|\leq R}|V_{\ell}(x)|^2dx<\infty$ for all $R>0$, and that
$H_{\rm rel}$ is self-adjoint. Then, for all $t\in \R$ and $j=1,\cdots,n$,
\begin{equation}
e^{-itH_{\rm rel}}D(q_j)=D(q_j)
\end{equation} 
and, for all $\psi \in D(q_j)$,
\begin{equation}
{\widetilde q}_j(t)\psi
=q_j\psi
+\int_0^t
{{\widetilde p}_j(s)}({\widetilde p}(s)^2+m^2)^{-1}\psi ds,
\end{equation}
where 
$$
{\widetilde q}_j(t):=e^{itH_{\rm rel}}q_je^{-itH_{\rm rel}}, \quad
{\widetilde p}_j(t):=e^{itH_{\rm rel}}p_je^{-itH_{\rm rel}}.
$$
In partcular, ${\widetilde q}_j(t)\psi$ is strongly differentiable in $t\in \R$ and
\begin{equation}
\frac {d}{dt}{\widetilde q}_j(t)\psi={\widetilde p}_j(t)({\widetilde p}(t)^2+m^2)^{-1}\psi.
\end{equation}
\end{The}

{\it Proof}. In the present case $T_1(\lambda)=\sqrt{\lambda^2+m^2}, \,
T_{\ell}(\lambda)=0, \, \ell \geq 2, \lambda \in \R^n$.
Hence (T.1) holds. We have  
$\partial_jT_1(\lambda)=\lambda_j/\sqrt{\lambda^2+m^2}, \ 
\partial_jT_{\ell}=0, \, \ell\geq 2$.
Hence $\partial_jT_{\ell}$ is bounded. Moreover, $D(q_j)\cap D(H_{\rm rel})$
is a core for $q_j$, since ${\cal K}\hat \otimes C_0^{\infty}(\R^n)\subset
D(q_j)\cap D(H_{\rm rel})$ and $C_0^{\infty}(\R^n)$ is a core for $q_j$.
Thus  Theorem \ref{mainth1}-(ii)
yields the desired result. \hfill \qed

\medskip

\begin{The} Assume that $H_{\rm rel}$ is self-adjoint and that conditions {\rm (ii)}--{\rm (iv)}
in Theorem \ref{S2} holds. Then, 
for all $t\in \R$ and $j=1,\cdots,n$,
\begin{equation}
e^{-itH_{\rm rel}}D(p_j)\cap D(H_{\rm rel})=D(p_j)\cap D(H_{\rm rel})\label{invdom4}
\end{equation} 
and, for all $\psi \in D(p_j)\cap D(H_{\rm rel})$,
\begin{equation}
{\widetilde p}_j(t)\psi
=p_j\psi
+\int_0^tF_j(s)\psi ds.
\end{equation}
In particular, ${\widetilde p}_j(t)\psi$ is strongly differentable in $t\in \R$ and
\begin{equation}
\frac {d}{dt}{\widetilde p}_j(t)\psi=F_j(t)\psi.
\end{equation}
\end{The}

{\it Proof}. Similar to the proof of Theorem \ref{S2}. \hfill \qed

\medskip




\subsection{Abstract Dirac operators}

In this subsection we consider a class of abstract Dirac operators.
Let $\{Q_j,P_j|j=1,\cdots,n\}$ be as in Section 5 and 
$\{\Gamma_j,B|j=1,\cdots,n\}$
be a set of bounded self-adjoint operators on ${\cal K}$ satisfying 




\begin{eqnarray}
&&\{\Gamma_j,\Gamma_k\}=2\delta_{jk}, \quad \{\Gamma_j,B\}=0, \quad j,k=1,\cdots,n,  \label{Cl1}\\
&& B^2=I,\label{Cl2}
\end{eqnarray}
where $\{X,Y\}:=XY+YX$.



Let $M,V_{\ell}:\R^n \to \R$ ($\ell=1,\cdots,N$)  be  Borel measurable functions such that
$$
\hat E_P(\{\lambda\in \R^n| |M(\lambda)|=+\infty\})=0, \quad 
\hat E_Q(\{\lambda\in \R^n| |V_{\ell}(\lambda)|=+\infty\})=0.
$$
Then $M(P)$ and $V_{\ell}(Q)$ are self-adjoint. 
Let
$C_{\ell}$ be a bounded self-adjoint operator on ${\cal K}$. Then we define
\begin{equation}
H_{\rm D}:=\sum_{j=1}^n\Gamma_j\otimes P_j+B\otimes M(P)+\sum_{\ell=1}^NC_{\ell}\otimes V_{\ell}(Q)
\end{equation}
As is easily noted, $H_{\rm D}$ is an abstract version of the usual Dirac 
operator acting in $L^2(\R^3;\C^4)$ (\cite[Chapter 4]{Thaller}). Let
\begin{equation}
H_0=\sum_{j=1}^n\Gamma_j\otimes P_j+B\otimes M(P).
\end{equation}




Before proving the self-adjointness of $H_0$,  we recall the notion of strong anticommutativity on self-adjoint operators: 

\begin{df}{\rm  Let $\{S_j\}_{j=1}^n$ be  a set of self-adjoint operators on a Hilbert space.
We say that   $\{S_j\}_{j=1}^n$ is {\it strongly anticommuting} if,
 for each pair $(j,k)$ with $j\not=k$ ($j,k=1,\cdots,n$), 
$e^{itS_j}S_k\subset S_ke^{-itS_j},\forall t\in \R$.
}
\end{df}



\begin{rem}{\rm A posteriori one can show that the definition
of the strong anticommutativity is symmetric
in $S_j,\, j=1,\cdots,k$ \cite{Ped,Vas} and that $e^{itS_j}S_k=S_ke^{-itS_j},\, \forall t\in \R, 
j\not=k$. 
}
\end{rem}

The following theorem is a fundamental result in the theory of
strongly anticommuting self-adjoint operators:

\begin{The}\label{vasilescu}{\rm (Vasilescu \cite{Vas})} Let $\{S_j\}_{j=1}^n$ be  strongly anticommuting on a Hilbert space.
Then $T:=\sum_{j=1}^nS_j$ is self-adjoint and
$$
T^2=\sum_{j=1}^nS_j^2.
$$
\end{The}

We set
$$
P^2:=\sum_{j=1}^nP_j^2,
$$
which is a nonnegative self-adjoint operator.

\begin{lem}\label{D-lem} The operator $H_0$ is self-adjoint and
\begin{equation}
\|H_0\psi\|=\|I \otimes (P^2+M(P)^2)^{1/2}\psi\|, \quad \psi \in D(H_0).
\label{norm}
\end{equation}
\end{lem} 


{\it Proof}. Since $\{P_1,\cdots, P_n, M(P)\}$ is a set of strongly commuting self-adjoint operators and (\ref{Cl1}) and (\ref{Cl2}) hold, it follows from a genreral theorem \cite[Theorem 3.4]{Ar93}
that $\Gamma_j\otimes P_j$ strongly anticommutes with $\Gamma_k\otimes P_k$ ($k\not=j$) and
$B\otimes M(P)$. Hence we can apply  
Vasilescu's theorem (Theorem \ref{vasilescu}) to conclude that $H_0$ is self-adjoint and that $H_0^2=\sum_{j=1}^nI\otimes P_j^2+I\otimes M(P)^2$, implying
(\ref{norm}). \hfill \qed

\medskip

\begin{lem}\label{lemD-2} Assume the following {\rm (i)}--{\rm (ii)}:
\begin{list}{}{}
\item{{\rm (i)}} For each $\ell=1,\cdots,N$, 
$D(\sqrt{P^2})\subset D(V_{\ell}(Q))$ and there exist positive constant $\varepsilon_{\ell}$ and $b_{\ell}$ such that  
$$
\|V_{\ell}(Q)\psi\| \leq \varepsilon_{\ell}\|\sqrt{P^2}\psi\|+b_{\ell}\|\psi\|, \quad \psi
\in D(\sqrt{P^2}).
$$
\item{{\rm (ii)}} The function $M$ is bounded  on $\R^n$. 
\end{list}
Let 
\begin{equation}
\sum_{\ell=1}^N\|B_{\ell}\|\varepsilon_{\ell}<1.
\end{equation}
Then $H_{\rm D}$ is self-adjoint with $D(H_{\rm D})=D\left(\sum_{j=1}^n\Gamma_j
\otimes P_j\right)=D(I\otimes \sqrt{P^2})$.
\end{lem}

{\it Proof}. Let $W:=\sum_{\ell=1}^NB_{\ell}V_{\ell}(Q)$. Then, by condition (i), we have for all $\psi \in D(I\otimes \sqrt{P^2})=D(\sum_{j=1}^n\Gamma_j\otimes P_j)$
$$
\|W\psi\| \leq \left(\sum_{\ell=1}^N\|B_{\ell}\|\varepsilon_{\ell}\right)
\|\left(\sum_{j=1}^n\Gamma_j\otimes P_j\right)\psi\|
+\left(\sum_{\ell=1}^N\|B_{\ell}\|b_{\ell}\right)\|\psi\|,
$$
where we have used (\ref{norm}) with the case $M=0$.
Hence, by the Kato-Rellich theorem, $H_{\rm D}':=
\sum_{j=1}^n\Gamma_j\otimes P_j+W$ is self-adjoint with $D(H_{\rm D}')=
D\left(\sum_{j=1}^n\Gamma_j
\otimes P_j\right)=D(I\otimes
\sqrt{P^2})$. Since $H_{\rm D}=H_{\rm D}'+B\otimes M(P)$ and
$B\otimes M(P)$ is a bounded self-adjoint operator under the present assumption,
it follows that $H_{\rm D}$ is self-adjoint with $D(H_{\rm D})=
D(H_{\rm D}')$. \hfill \qed

\medskip



\begin{The}\label{th-Dirac} Suppose that the assumption of Lemma \ref{lemD-2}
holds  and that $M$ is continuously differentiable  on $\R^n$ 
with $\partial_j M$ being bounded on $\R^n$ 
($j=1,\cdots,n$). 
Then, for all $t\in \R$ and $j=1,\cdots,n$,
\begin{equation}
e^{-itH_{\rm D}}D(I\otimes Q_j)=D(I\otimes Q_j) 
\end{equation} 
and, for all $\psi \in D(I\otimes Q_j)$,
\begin{equation}
e^{itH_{\rm D}}(I\otimes Q_j)e^{-itH_{\rm D}}\psi
=I\otimes Q_j\psi
+\int_0^te^{isH_{\rm D}}
[\Gamma_j\otimes I+B\otimes (\partial_jM)(P) ]e^{-isH_{\rm D}}\psi ds.
\end{equation}
\end{The}


{\it Proof}. We apply Theorem \ref{mainth1} to the following case:
\begin{eqnarray*}
&&L:=\max\{n+1, N\}, \\
&& A_{\ell}=\left\{
\begin{array}{lc}
\Gamma_{\ell} &;\  \ell=1,\cdots,n, \\
B & ; \ \ell=n+1,\\
A_{\ell}=0 & ; \  \quad \ell>n+1
\end{array}
\right.,
\quad T_{\ell}(\lambda)=\left\{
\begin{array}{lc}
\lambda_{\ell}& ; \  \ell=1,\cdots,n, \\
M(\lambda) &; \, \ell=n+1,\\
T_{\ell}=0 &;  \ell>n+1
\end{array}
\right., \\
&& B_{\ell}=\left\{
\begin{array}{lc}
C_{\ell} & ; \ \ell=1,\cdots,N, \\
0 & ; \ \ell>N
\end{array}
\right..
\end{eqnarray*}
We also note that $D(I\otimes Q_j)\cap D(\sum_{j=1}^n\Gamma_j\otimes P_j)$ is  a core of $I\otimes Q_j$, since $\{Q_j,P_j|j=1,\cdots,n\}$ is a Weyl representation
of CCR. Thus we obtain the desired result from Theorem \ref{mainth1}-(ii). 
\hfill \qed

\medskip

\begin{rem}{\rm  One can also obtain a result on $I\otimes P_j$.
But we omit writing it down.
}
\end{rem}

\section*{Acknowledgment}

This work is supported  by the Grant-In-Aid 17340032 for scientific  research  from the Japan Society for the Promotion of Science (JSPS).

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