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  self-adjoint operator, strong commutativity, strong anticommutativity, decomposition theorem, spectrum, spectral reflection symmetry, 
Heisenberg operator, canonical commutation relation,  Dirac operator 
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\begin{document}
\baselineskip=20pt

\title{Operator Theory on a Commutation Relation}
\author{Asao Arai\thanks{This work is supported
by the Grant-In-Aid 17340032 for scientific research from
the JSPS.}\\
Department of Mathematics, Hokkaido University\\
Sapporo 060-0810\\
Japan\\
E-mail: arai@math.sci.hokudai.ac.jp}

\maketitle


\begin{abstract}
An operator theory is developed 
for a pair $(A,H)$ with $A$ being a symmetric operator
on a Hilbert space ${\cal H}$ and $H$  an injective 
self-adjoint  
operator on  ${\cal H}$ such that 
$D(H^2)\subset D(A)$ (for a linear
operator $T$, $D(T)$ denotes its domain) and
$\lang H^2\psi,A\phi\rang=\lang A\psi, H^2\phi\rang,
\, \psi,\phi \in D(H^2)$, where
$\lang\,\cdot\,,\,\cdot\,\rang$ is the inner product of ${\cal H}$.
One of the main results of the present paper 
is concerned with a decomposition theorem
of $\bar A$ (the closure of $A$)\,: Under suitable  conditions, 
there exists a unique pair $(A_+,A_-)$ of
 self-adjoint operators such that the following hold\,:
(i) $\bar A=A_++A_-$\,;  
(ii) $A_+$ strongly commutes with $H$\,; (iii) $A_-$ 
strongly anticommutes with $H$.
Moreover, under some additional
condition,  $A_+$ strongly anticommutes with $A_-$.
It is shown that, under suitable conditions, the spectrum
of $H$ is symmetric  with respect to the origin in $\R$. 
As an application, we define  a class of
linear operators $X$ on ${\cal H}$ for which $e^{itH}Xe^{-itH}$ 
($t\in \R$) has
an explicit representation. A class of pairs $(A,H)$
to which the general theory can be applied 
is given with $H$ being an abstract form of
Dirac type operators. 
\end{abstract} 







\noindent
{\it Keywords}: self-adjoint operator, strong commutativity, strong anticommutativity, decomposition theorem, spectrum, spectral reflection symmetry,
Heisenberg operator, canonical commutation relation,  Dirac operator 


\medskip

\noindent
Mathematics Subject Classification 2000: 47B25, 47N50


\medskip

\section{Introduction}


In this paper we develop an operator theory
for  a pair $(A,H)$ with $A$ being a  
symmetric operator  on a Hilbert space ${\cal H}$ and
$H$ an injective self-adjoint operator on ${\cal H}$
such that  $D(H^2)\subset D(A)$  (for a linear
operator $T$, $D(T)$ denotes its domain) and
\begin{equation}
\lang H^2\psi,A\phi\rang=\lang A\psi, H^2\phi\rang,
\quad \psi,\phi \in D(H^2), \label{eq1-1}
\end{equation}
where
$\lang\,\cdot\,,\,\cdot\,\rang$ is the inner product of ${\cal H}$.
If $A$ is bounded, then 
(\ref{eq1-1}) is equivalent to $AH^2\subset H^2A$,
which means the commutativity of $A$ and $H^2$
(in the present paper,  we mean by 
\lq\lq{a bounded operator on ${\cal H}$}"
a bounded linear operator  {\it everywhere defined} on ${\cal H}$,
unless otherwise stated).
Hence (\ref{eq1-1}) can be regarded as a generalized version
of this commutativity to the case where $A$ is not necessarily bounded.

We introduce a subset of linear operators on ${\cal H}$:

\begin{df}{\rm 
We say that $A$ is in the set $\{H^2\}_{\rm w}'$
if $A$ is symmetric with $D(H^2)\subset D(A)$ and (\ref{eq1-1}) holds
(in this definition, $H$ is not necessarily injective).
}
\end{df}

\begin{rem}{\rm It is easy to see  that,
for each bounded, real-valued Borel measurable function $f$ on $\R$
and $\alpha \in [0,2]$, the operator $f(H)|H|^{\alpha}$ is an element
of $\{H^2\}_{\rm w}'$. But the case $A=f(H)|H|^{\alpha}$ is trivial.
Hence, in what follows, we have in mind the case
where $A$ is not such a form. See Proposition \ref{HBC}
below.
}
\end{rem}



To outline the contents of the present paper, we first  recall
two key notions:


\begin{df}\label{def1}{\rm Let $S$ and $T$ be self-adjoint operators on
a Hilbert sapce. We denote the spectral measure of $S$ by
$E_S(\cdot)$
\begin{list}{}{}
\item{(i)} We say that
$S$ and $T$  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 $J$,  $K \subset \R$
(e.g., \cite[p.271]{RS1}, \cite{Samo}).
\item{(ii)}  We say that 
$S$  strongly anticommutes with $T$ if
$e^{itT}S\subset Se^{-itT}$ for all $t\in \R$ (\cite{Ped,Vas}).
\end{list}
}
\end{df}


 

\begin{rem}{\rm One can prove that, if $S$ strongly anticommutes with $T$, then
$e^{itS}T\subset Te^{-itS}$ for all $t\in \R$ \cite{Ped}, i.e.,
$T$ strongly anticommutes with $S$.
Hence Definition \ref{def1}(ii) is symmetric in $S$ and $T$.
}
\end{rem}


One of the main results of this paper is stated as follows. 
Let $H$ be an injective self-adjoint operator
and $A \in \{H^2\}_{\rm w}'$.  Then 
there exists a unique pair $(A_+,A_-)$ of
 self-adjoint operators such that the following hold:
(i) $\bar A=A_++A_-$, where $\bar A$ is the closure of $A$ ;  
(ii)  $A_+$ strongly commutes with $H$ ; (iii)  $A_-$ 
strongly anticommutes with $H$.
If some additional condition is satisfied,
 then $A_+$ strongly anticommutes with $A_-$.

This is a kind of decomposition theorem of operators
relative to the  self-adjoint operator $H$.
It is well known that the strong commutativity and the strong anticommutativity
of self-adjoint operators make it possible to perform
more detailed analyses  in various
contexts of operator theory (e.g.,
 \cite{Ar93a}-\cite{Ar03}, \cite{Ped,Samo,Vas}).
Thus such decomposition theorems as stated above may be very useful.


We also investigate the spectral properties of $H$.
It is shown that, under suitable conditions,  the spectrum
of $H$ is symmetric with respect to the origin in $\R$. 

As an application, we define  a class of
linear operators $X$ on ${\cal H}$ for which 
the Heisenberg operator $e^{itH}Xe^{-itH}$ of $X$
($t\in \R$) with respect to $H$ has
an explicit representation. Moreover, we present a class of pairs $(A,H)$
to which the general theory can be applied with 
$H$ being an abstract form of
Dirac type operators.

One of the motivations for this work comes from
trying to understand  general operator-theoretical  structures
underlying the so-called Zitterbewegung
in relativistic quantum mechanics (\cite{Koba}, \cite[\S 1.6]{Thaller}).
But the theory presented in this paper is very general. 
It may have many applications both in operator theory and
in quantum physics\footnote{The present author plans to
discuss, in a separate paper, an application to quantum electrodynamics in connection with Koba's  work \cite{Koba}.}.


The present paper is orginaized as follows.
In Section 2, we prove some elementary facts on the set $\{H^2\}_{\rm w}'$. At the end of this section, a purely operator-theoretical
motivation for this work is given.
In Section 3 we consider a similarity transformation 
for a linear operator $A$ given by
the correspondence :$A \mapsto A_H:=HAH^{-1}$ under the condition that $H$ is injective. Some properties of the operator $A_H$ are discussed.
Section 4 is devoted to proof of the decomposition theorem stated
above. In Section 5 we prove that, under additional conditions,
the spectrum and the point spectrum of $H$ are symmetric
with respect to the origin in $\R$.
In Section 6 we introduce a class of linear operators $X$
for which the Heisenberg operator $X(t)=e^{itH}Xe^{-itH}$ ($t\in \R$) 
has an explicit representation.
The asymptotic behavor of $X(t)$ as $|t| \to \infty$
is clarified in the sense of sesquilinear form.
In Section 7 we discuss a connection with the generalized weak Weyl relation
introduced in \cite{Ar05}. We show that, under suitable conditions,
$$
\lim_{|t| \to\infty}\lang\psi,e^{-itH}A_+\phi\rang=0
$$
for all $\psi,\phi \in {\cal H}$ 
and $H$ has  no eigenvalues.
In the last section we present a class of pairs $(A,H)$ to which
the general theory developed in the previous sections can be applied.
  




\section{Preliminaries}


We first recall a useful fact:


\begin{lem}\label{Nuss}{\rm (Nussbaum)}
Let $T$ be a self-adjoint operator on ${\cal H}$ and
$S$ be a symmetric operator on ${\cal H}$ such that
$D(T)\subset D(S)$ and
$\lang T\psi,S\phi\rang=\lang S\psi,T\phi\rang,\,
\psi,\phi \in D(T)$. Then
$S$ is essentially self-adjoint
on $D(T)$ and $\bar S$ (the closure of $S$) strongly commutes with
$T$.
\end{lem}


\noindent
{\it Proof}. See \cite[Corollary 1]{NB} or \cite[Proposition 4.1]{Ped}.
\hfill \qed

\medskip




Let $H$ be a self-adjoint opeartor on ${\cal H}$.

\begin{pr}\label{pr2-1} For all
 $A\in \{H^2\}_{\rm w}'$,
$A$ is essentially self-adjoint on $D(H^2)$
and $\bar A$ strongly commutes with $H^2$.
Moreover $\bar A\in \{H^2\}_{\rm w}'$.
\end{pr}


\noindent
{\it Proof}. The first half of this proposition
follows  from a simple application of Lemma \ref{Nuss}
to the case $T=H^2$ and $S=A$.

Since $D(A) \subset D(\bar A)$, we have $D(H^2)\subset D(\bar A)$.
By the strong commutativity of $\bar A$ and $H^2$, we have
$e^{itH^2}\bar A\subset \bar A e^{itH^2}$ for all $t\in \R$.
Hence, for all $\psi,\phi \in D(H^2)$ and $t\in \R$, we have
$\lang e^{itH^2}\psi, \bar A\phi\rang=
\lang \bar A\psi, e^{-itH^2}\phi\rang$.
Differentiating the both sides at $t=0$, we obtain
$\lang H^2\psi, \bar A\phi\rang=
\lang \bar A\psi, H^2\phi\rang$.
Thus $\bar A\in \{H^2\}_{\rm w}'$.
\hfill \qed

\medskip

The set $\{H^2\}_{\rm w}'$ has a real linear structure:

\begin{pr}\label{pr2-2} Let $A,B\in \{H^2\}_{\rm w}'$.
Then $A+B\in \{H^2\}_{\rm w}'$ and
$\lambda A\in \{H^2\}_{\rm w}'$ for all $\lambda \in \R$.
\end{pr} 

\noindent
{\it Proof}. By definition, $D(A+B)=D(A)\cap D(B)$ and 
$D(\lambda A)=D(A), \forall \lambda \in \R$.
Hence $D(H^2)\subset D(A+B)$
and $D(H^2)\subset D(\lambda A), \forall \lambda \in \R$ . 
Then it is easy to see that
$A+B$ (resp. $\lambda A$) is symmetric and (\ref{eq1-1}) with $A$
replaced by $A+B$ (resp. $\lambda A$) holds.
\hfill \qed

\medskip


Propositions \ref{pr2-1} and  \ref{pr2-2} immediately
yield the following result:

\begin{cor}\label{cor2-3}Let $A_1,\cdots,A_n\in \{H^2\}_{\rm w}'$ ($n=1,2,\cdots$).
Then, for all $a_1,\cdots,a_n \in \R$, 
the operator $\sum_{j=1}^na_jA_j$ is 
essentially self-adjoint on $D(H^2)$ and
$\overline{\sum_{j=1}^na_jA_j}$ strongly commutes with $H^2$.
\end{cor}


The following proposition shows the structure of a subset
of $\{H^2\}_{\rm w}'$.


\begin{pr}\label{HBC}
Let $B$ (resp. $C$) be a self-adjoint operator
strongly commuting (resp. anticommuting) with $H$ and
$D(H^2)\subset D(B)\cap D(C)$.
Let $A:=B+C$. Then, $A \in \{H^2\}_{\rm w}'$.
In particular, $A$ is essentially self-adjoint on $D(H^2)$
and $\bar A$ strongly commutes with $H^2$.
\end{pr}

\noindent
{\it Proof}. The strong commutativity of $B$ with $H$
implies that of $B$ with $H^2$.
Hence, (\ref{eq1-1}) with $A$ replaced by $B$
follows from the two variable functional calculus
on the pair $(H^2,B)$. Thus $B\in \{H^2\}_{\rm w}'$.
As for the operator $C$, we have 
$e^{itC}H\subset He^{-itC}$ for all $t\in \R$.
This implies that $e^{itC}H^2\subset H^2e^{itC}$ for all
$t\in \R$. Hence $C$ strongly commutes with $H^2$.
Hence,  as in the case of $B$, $C$ is in $\{H^2\}_{\rm w}'$. 
These results and Proposition \ref{pr2-2}
imply that $A\in \{H^2\}_{\rm w}'$.
Hence Proposition \ref{pr2-1}
imply the last statement of the present proposition.
\hfill \qed

\medskip
 
 


From a purely operator-theoretical point of view,
it would be interesting to investigate 
an  inverse problem concerning  Proposition \ref{HBC}:
Under what conditions 
does a self-adjoint operator 
$A$ have a decomposition as described in Proposition \ref{HBC} ?
This is one of the  motivations
for the present work.



\section{A similarity transformation}

In what follows, we need the following
condition:

\medskip

\noindent
{\bf Hypothesis (H)}:  $H$ is injective.

\medskip

We assume  (H). Then,  
for a linear operator $A$ on ${\cal H}$, we define a linear 
operator $A_H$ by
\begin{equation}
A_H:=HAH^{-1},\label{ST(A)}
\end{equation}
which is a similarity transform of $A$ by $H$.



\begin{lem}\label{lemAH}  Assume (H) and let  $A \in \{H^2\}_{\rm w}'$. Then:
\begin{list}{}{}
\item{(i)} For all $\psi,\phi \in D(A_H)\cap D(H)$, 
$\lang \psi, A_H\phi\rang=\lang A_H\psi,\phi\rang$.
\item{(ii)} For all $\psi,\phi\in D(H^2)\cap D(A_H)$,
\begin{equation}
\lang H^2\psi,A_H\phi\rang=\lang A_H\psi, H^2\phi\rang
\label{H2}.
\end{equation}
\end{list}


\end{lem}

\noindent
{\it Proof}. (i) Let $\psi,\phi \in D(A_H)\cap D(H)$.
Then $H^{-1}\psi, H^{-1}\phi \in D(H^2)$ and
\begin{eqnarray*}
\lang \psi, A_H\phi\rang
&=&\lang H^2(H^{-1}\psi), A(H^{-1}\phi)\rang
=\lang A(H^{-1}\psi),H^2(H^{-1}\phi)\rang\\
&=&\lang A_H\psi,\phi\rang,
\end{eqnarray*}
where (\ref{eq1-1}) is used to obtain the second equality.

(ii) Let $\phi\in D(H^2)\cap D(A_H)$. Then
$H^{-1}\phi \in D(A)\subset D(\bar A)$.
Hence, by the strong commutativity of $H^2$ and $\bar A$ (Proposition \ref{pr2-1}),  
$e^{-itH^2}H^{-1}\phi \in D(\bar A)$ and $\bar Ae^{-itH^2}H^{-1}\phi
=e^{-itH^2}AH^{-1}\phi$.
Note that $e^{-itH^2}H^{-1}\phi
\in D(H^2)$ and $e^{-itH^2}\phi \in D(H^{-1})$. Hence 
$$
\bar Ae^{-itH^2}H^{-1}\phi=
AH^{-1}e^{-itH^2}\phi.
$$ 
Hence
$AH^{-1}e^{-itH^2}\phi=
e^{-itH^2}AH^{-1}\phi$.
Since $AH^{-1}\phi \in D(H)$, it follows that
$e^{-itH^2}AH^{-1}\phi$ is in $D(H)$ and
$He^{-itH^2}AH^{-1}\phi=e^{-itH^2}A_H\phi$. Hence 
$AH^{-1}e^{-itH^2}\phi$ is in $D(H)$ and
$A_He^{-itH^2}\phi
=e^{-itH^2}A_H\phi$.
Therefore, for all $\psi \in D(H^2)\cap D(A_H)$, we have
\begin{eqnarray*}
\lang e^{itH^2}\psi,A_H\phi\rang&=&\lang A_H\psi, e^{-itH^2}\phi\rang.
\end{eqnarray*}
Differentiating the both sides at $t=0$,
we obtain (\ref{H2}). \hfill \qed




\medskip

Lemma \ref{lemAH} and Lemma \ref{Nuss}
imply the following
theorem:

\begin{th}\label{th4-2} Assume (H) and let $A\in \{H^2\}_{\rm w}'$.
Suppose that $D(H^2)\subset D(A_H)$.
Then $A_H\in \{H^2\}_{\rm w}'$.
In particular, $A_H$ is essentially self-adjoint on $D(H^2)$ and
$\bar A_H$ strongly commutes with $H^2$.
\end{th}

\begin{cor}\label{cor4-3} Assume (H) and let $A\in \{H^2\}_{\rm w}'$.
Then $A_{H^{-1}}\psi=A_H\psi$ for all $\psi \in D(A_H)\cap D(A_{H^{-1}})$.
In particular, if $D(H^2)\subset D(A_H)\cap D(A_{H^{-1}})$,
then $A_{H^{-1}}\in \{H^2\}_{\rm w}'$.
\end{cor}

\noindent
{\it Proof}. By Proposition \ref{pr2-1},
$A$ is essentially self-adjoint on $D(H^2)$ and
$\bar A$ strongly commutates with $H^2$.
Hence $\bar A$ strongly commutes with $|H|$ too.
Let $H=U|H|$ be the polar decomposition of $H$. Then $U$ is unitary and self-adjoint,
since $H$ is injective. 
By the two variable
functional calculus, we have $|H|A|H|^{-1}\psi=|H|^{-1}A|H|\psi$
for all $\psi\in D(|H|A|H|^{-1})\cap D(|H|^{-1}A|H|)$.
Hence, for all $\psi \in D(A_H)\cap
D(A_{H^{-1}})$,
 $A_H\psi=U|H|A|H|^{-1}U\psi=U|H|^{-1}A|H|U\psi=A_{H^{-1}}\psi$.
\hfill \qed

\medskip


\section{Decomposition theorems}


Assume (H). Then, 
for a linear operator  $A$ on ${\cal H}$, we can define
\begin{equation}
A_{\pm,0}:=\frac 12(A\pm A_H),\label{A+-}
\end{equation}
where $A_H$ is defined by (\ref{ST(A)}).
It is easy to see that
\begin{equation}
A\supset A_{+,0}+A_{-,0}.\label{decomp1}
\end{equation}

For  a linear operator $L$ and
a subspace $D\subset D(L)$, we denote by $L|D$  the restriction of $L$ to $D$.

\begin{th}\label{th4-1} Assume (H) and let $A\in \{H^2\}_{\rm w}'$.
Suppose that $D(H^2)\subset D(A_H)$.
Then:
\begin{list}{}{}
\item{(i)} $A_{\pm,0}$ are essentially self-adjoint on $D(H^2)$ and
\begin{equation}
A_{\pm}:=\bar A_{\pm,0}
\end{equation}
strongly commute with $H^2$.
\item{(ii)} The operator $A_+ +A_-$ is essentially self-adjoint
on $D(H^2)$ and the operator equality
\begin{equation}
\bar A=\overline{A_+ + A_-} \label{decomp2}
\end{equation}
holds.
\end{list}
\end{th}

\noindent
{\it Proof}. (i) By Theorem \ref{th4-2},
$A_{H} \in \{H^2\}_{\rm w}'$. Hence, by Corollary \ref{cor2-3},
the assertions follow.

(ii) By (\ref{decomp1}) and (i), we have
$A|D(H^2)\subset A_++A_-$.
By this relation and  Proposition \ref{pr2-1},
$A_+ +A_-$ is essentially self-adjoint on $D(H^2)$ and
(\ref{decomp2}) holds.
\hfill \qed

\medskip


If $A_+ +A_-$ is closed, then (\ref{decomp2})
takes the form $\bar A=A_++A_-$, which gives a decomposition of $\bar A$.
Hence a next task to be done is to investigate
conditions which ensure the closedness of $A_+ +A_-$.



\begin{lem}\label{AH-H} Assume (H) and  let $A\in \{H^2\}_{\rm w}'$.
Suppose that $D(H)\subset D(A)$.
Then, for all $\psi \in D(A_H)\cap D(H)$ and
$\phi \in D(H)$,
\begin{equation}
\lang H\psi,A\phi\rang=\lang A_H\psi,H\phi\rang. \label{CR-HA}
\end{equation}
\end{lem}

\noindent
{\it Proof}. Let $\phi \in D(H^2)$ first. Then 
$\lang H\psi, A\phi\rang
=\lang H^2(H^{-1}\psi),A\phi\rang
=\lang AH^{-1}\psi, H^2\phi\rang=\lang A_H\psi, H\phi\rang$.
Hence (\ref{CR-HA}) holds.

We next consider the case $\phi \in D(H)$.
By the condition $D(H)\subset D(A)$ and the closed graph theorem,
there exists a constant $c\geq 0$ such that
\begin{equation}
\|A\psi\|\leq c(\|H\psi\|+\|\psi\|), \quad \psi \in D(H).\label{Hbdd}
\end{equation}
Since $D(H^2)$ is  a core of $H$, for each $\phi \in D(H)$,
there exists a sequence $\{\phi_n\}_{n=1}^{\infty}$
with $\phi_n\in D(H^2)$ such that $\lim_{n\to\infty}\phi_n=\phi$
and $\lim_{n\to\infty}H\phi_n=H\phi$.
By the preceding result, we have $\lang H\psi, A\phi_n\rang=
\lang A_H\psi, H\phi_n\rang$. 
By (\ref{Hbdd}), $A\phi_n\to A\phi$ as $n\to\infty$.
Thus (\ref{CR-HA}) follows. 
\hfill \qed

\medskip





\begin{lem}\label{A1} Assume (H) and  let $A\in \{H^2\}_{\rm w}'$.
Suppose that $D(H)\subset D(A)$.
Then:
\begin{list}{}{}
\item{(i)} For all $\psi,\phi \in  D(A_H)\cap D(H)$,
\begin{equation}
\lang H\psi,A_{+,0}\phi\rang=\lang A_{+,0}\psi,H\phi\rang.
\end{equation}
\item{(ii)} For all $\psi,\phi \in  D(A_H)\cap D(H)$,
\begin{equation}
\lang H\psi,A_{-,0}\phi\rang +\lang A_{-,0}\psi,H\phi\rang=0.
\end{equation}
\end{list}
\end{lem}

\noindent
{\it Proof}. Direct computations  using (\ref{CR-HA}).
\hfill \qed

\medskip


\begin{th}\label{th4-4} Assume (H) and let $A \in \{H^2\}_{\rm w}'$. 
Suppose that $D(H) \subset  D(A_H)\cap D(A)$.
Then $A_{+,0}$ is essentially self-adjoint on $D(H)$
and $A_+$ strongly commutes with $H$.
\end{th}

\noindent
{\it Proof}. 
This follows from Lemma \ref{A1}(i) and Lemma \ref{Nuss}.
\hfill \qed

\medskip

We need the following fact:

\begin{lem}\label{Pedersen} {\rm (S. Pedersen \cite[Theorem 4.3]{Ped})} 
Let $A$ and $B$ be symmetric operators on ${\cal H}$ and ${\cal D}$ be
a dense subspace such that ${\cal D}\subset D(A)\cap D(B)$.
Suppose that $(A\pm B)|{\cal D}$ are essentially self-adjoint and
$\lang A\psi, B\phi\rang+\lang B\psi,A\phi\rang=0$ for all $\psi, \phi\in
{\cal D}$. Then $A$ and $B$ are essentially self-adjoint. Moreover, 
$\bar A$ and $\bar B$ strongly anticommute. 
\end{lem}


\begin{th}\label{st-anticom} Assume (H) and let $A \in \{H^2\}_{\rm w}'$. 
Suppose that $D(H) \subset D(A)\cap D(A_H)$.
Then $A_{-,0}$ is essentially self-adjoint 
and $A_-$ strongly anticommutes with $H$.
\end{th}

\noindent
{\it Proof}. Let $T_{\pm}:=H\pm A_{-,0}$. Then
$D(T_{\pm})=D(H)$ and $T_{\pm}$ are symmetric.
For all $\psi,\phi \in D(H^2)$, we have
$\lang T_{\pm}\psi, H^2\phi\rang=
\lang H^2\psi, T_{\pm}\phi\rang$. 
Hence, by Lemma \ref{Nuss}, $T_{\pm}$ are essentially self-adjoint
on $D(H^2)$. By
Lemma \ref{A1}(ii), we have
$\lang H\psi,A_{-,0}\phi\rang+\lang A_{-,0}\psi, 
H\phi\rang=0$ for all
$\psi,\phi \in D(H^2)$.
Hence, by Lemma \ref{Pedersen}, $A_{-,0}$ is essentially self-adjoint
and $A_-=\bar A_{-,0}$ strongly anticommutes with $H$.
\hfill \qed

\medskip


\begin{th}\label{dcpth1} Assume (H) and let $A\in \{H^2\}_{\rm w}'$
with $D(H)\subset D(A)\cap D(A_H)$.
Suppose that
there exists a constant $C>0$ such that
\begin{equation}
\|A_H\psi\|^2\leq \|A\psi\|^2+2{C^2}\|\psi\|^2,\label{RB}
\quad \psi \in D(H^2).
\end{equation}
Then $A_{\pm,0}$ are essentially self-adjoint
with $D(\bar A)=D(A_+)\cap D(A_-)$ and
the operator equality
\begin{equation}
\bar A=A_+ +A_- \label{opeq1}
\end{equation}
holds. Moreover,
for all $\psi \in D(A_+)\cap D(A_-)$
\begin{equation}
\|A_+\psi\|^2+\|A_-\psi\|^2
\leq \|\bar A\psi\|^2+C^2\|\psi\|^2. \label{Abd}
\end{equation}
\end{th}

\noindent
{\it Proof}. Let $\psi\in D(H^2)$. Then
\begin{eqnarray*}
\|(A_{+,0}+A_{-,0})\psi\|^2
&=&\|A_{+,0}\psi\|^2+\|A_{-,0}\psi\|^2
+\lang A_{+,0}\psi,A_{-,0}\psi\rang+\lang A_{-,0}\psi,
A_{+,0}\psi\rang.
\end{eqnarray*}
It is easy to see that
\begin{equation}
\lang A_{+,0}\psi,A_{-,0}\psi\rang+\lang A_{-,0}\psi,
A_{+,0}\psi\rang
=\frac 12 (\|A\psi\|^2-\|A_H\psi\|^2),\label{anticom}
\end{equation}
which, together with (\ref{RB}), implies 
(\ref{Abd}) with $\psi\in D(H^2)$.
Then, by a limiting argument using the fact that
$D(H^2)$ is a core of $\bar A$, we see that
$D(\bar A)\subset D(A_+)\cap D(A_-)$
and (\ref{Abd}) holds for all $\psi \in D(\bar A)$.
Since $\overline{A_+ +A_-}=\overline{A_{+,0}+A_{-,0}}=\bar A$,
it follows that $D(A_+)\cap D(A_-)\subset D(\bar A)$.
Thus the desired results follow. \hfill \qed

\medskip 

The decomposition (\ref{opeq1}) is unique in the following sense:

\begin{th} Assume (H) and let $A \in \{H^2\}_{\rm w}'$ with
$D(H^2)\subset D(A_H)$. Suppose that
there exists a pair $(B_+,B_-)$
of self-adjoint opeartors having the following properties:
\begin{list}{}{}
\item{(B.1)} $\bar A=B_++B_-$.
\item{(B.2)}  $D(H^2)$ is a common core of
$B_{\pm}$.
\item{(B.3)} $B_+$ strongly commutes with $H$ and
$B_-$ strongly anticommutes with $H$.
\end{list}
Then $A_+=B_+$ and $A_-=B_-$.
\end{th}

\noindent
{\it Proof}. Let $S_1=A_+-B_+$ and $S_2=B_--A_-$.
Then, by (B.1) and (\ref{decomp1}),
we have  $S_1\psi=S_2\psi$ for all $\psi\in D(H^2)$.
It is easy to see that,
for all $t \in \R$,
 $e^{itH}S_1\subset S_1e^{itH}$  and
$e^{itH}S_2\subset S_2e^{-itH}$.
Hence, for all $t\in \R$, 
\begin{equation}
e^{itH}S_1\psi=S_1e^{itH}\psi=S_2e^{itH}\psi
=e^{-itH}S_2\psi=e^{-itH}S_1\psi, \label{SS}
\end{equation}
i.e., $e^{2itH}S_1\psi=S_1\psi$.
This implies that $S_1\psi \in D(H)$ and $HS_1\psi=0$.
Hence $S_1\psi=0$. In particular, we have $A_+\psi=B_+\psi$
for all $\psi \in D(H^2)$. 
Since $D(H^2)$ is a common core of $A_+$ and $B_+$,
it follows that $A_+=B_+$ (operator equality).
By this result and (\ref{SS}),
we have $S_2\psi=0$ for all $\psi \in D(H^2)$.
Then, in the same way as in the case of $(A_+,B_+)$,
we see that the operator equality $A_-=B_-$ holds.
\hfill \qed

\medskip
 

\begin{pr}\label{santicom} Let the same assumption as in Theorem  \ref{th4-1} be satisfied.
Suppose that
\begin{equation}
\|A_H\psi\|^2=\|A\psi\|^2, \quad \psi \in D(H^2). \label{AH-A}
\end{equation}
Then $A_+$ and $A_-$ strongly anticommute.
\end{pr}

\noindent
{\it Proof}. By Theorem \ref{th4-1},  $(A_+ +A_-)|D(H^2)$ is essentially self-adjoint. Theorem \ref{th4-2} implies that
$(A_+-A_-)|D(H^2)$ ($=A_H|D(H^2)$) is essentially self-adjoint.
Moreover, by (\ref{anticom}) and (\ref{AH-A}), we have
$\lang A_+\psi,A_-\phi\rang+\lang A_-\psi,A_+\phi\rang=0,
\ \psi,\phi \in D(H^2)$.
Hence, we can apply Lemma \ref{Pedersen} to conclude
that $A_+$ and $A_-$ strongly anticommute.
\hfill \qed

\medskip



\begin{cor}
Assume (H) and let $A\in \{H^2\}_{\rm w}'$
with $D(H)\subset D(A)\cap D(A_H)$.
Suppose that (\ref{AH-A}) holds.
Then $A_{\pm,0}$ are essentially self-adjoint
with $D(\bar A)=D(A_+)\cap D(A_-)$, and
$A_+$ and $A_-$ strongly anticommute. Moreover,
the operator equality
\begin{equation}
\bar A=A_+ +A_- \label{opeq2}
\end{equation}
holds and
\begin{equation}
\|\bar A\psi\|^2=\|A_+\psi\|^2+\|A_-\psi\|^2, \quad \psi \in D(\bar A).
\label{normeq}
\end{equation}
\end{cor}

\noindent
{\it Proof}. It is obvious that (\ref{AH-A})  implies (\ref{RB}).
Hence the conclusion of Theorem \ref{dcpth1} holds.
The strong anticommutativity of $A_+$ and $A_-$ follows from
Proposition \ref{santicom}.
Equation (\ref{normeq}) is just an application of a general theorem
on two strongly anticommuting  self-adjoint operators
(Lemma \ref{san} below).
\hfill \qed


\medskip

\begin{lem}\label{san}{\rm (Vasilescu\cite[Theorem 2.1]{Vas})}
If $S$ ad $T$ are strongly anticommuting self-adjoint operators on a Hilbert space, then
$S+T$ is self-adjoint and
\begin{eqnarray}
&&(S+T)^2=S^2+T^2, \label{S+T}\\
&& \|(S+T)\psi\|^2=\|S\psi\|^2+\|T\psi\|^2, \quad
\forall \psi \in D(S)\cap D(T).\label{norm}
\end{eqnarray}
\end{lem}


\medskip
\section{Spectral properties of $H$}

In this section we describe some implications
of Thereom \ref{st-anticom}.
But, before doing that, we first
prove some general facts in the theory
of strongly anticommuting  self-adjoint operators.



For a linear operator $L$ on a Hilbert space,
we denote by $\sigma(L)$ (resp. $\sigma_{\rm p}(L)$)
the spectrum (resp. the point spectrum) of $L$.

We say that a subset $\Sigma$  of $\R$ is symmetric with respect to the
origin if, for all  $\lambda \in \Sigma$, $-\lambda \in \Sigma$.


Let $S$ and $T$ be strongly anticommuting self-adjoint
operators on ${\cal H}$.
We denote by ${\rm Ran}(S)$ the range of $S$.


\begin{lem}\label{spec1} (i) $T$ is reduced by $\overline{{\rm Ran}(S)}$.

(ii) We denote by $T_S$ the reduced part of $T$ to $\overline{{\rm Ran}(S)}$. Suppose that $S\not=0$. 
Then $T_S$ is unitarily equivalent to $-T_S$.
In particular, 
$\sigma(T_S)$ (resp. $\sigma_{\rm p}(T_S)$) 
is symmetric with respect to
the origin.

\end{lem}



\noindent
{\it Proof}. (i) 
Let $S=U_S|S|$
be the polar decomposition of $S$
with $\ker U_S=\ker S$.
Then, by \cite[Theorem 3.1]{Ped},
\begin{equation}
U_ST\subset -TU_S \label{UT}
\end{equation} 
and $|S|$ strongly commutes with $T$.
The latter implies that
$T$ is reduced by $\overline{{\rm Ran}(|S|)}$ (\cite[Lemma 6.5]{Ar05}).
Since $S$  and $|S|$ are self-adjoint, we have the following 
orthogonal decompositions of ${\cal H}$:
${\cal H}=\ker S\oplus \overline{{\rm Ran}(S)}
=\ker |S|\oplus\overline{{\rm Ran}(|S|)}$.
We have $\ker S=\ker |S|$.  Hence $\overline{{\rm Ran}(|S|)}
=\overline{{\rm Ran}(S)}$.
Therefore $T$ is reduced by $\overline{{\rm Ran}(S)}$.

(ii) Since $S\not=0$, the operator $V:=U_S|\overline{{\rm Ran}(S)}$
is a unitary operator on
$\overline{{\rm Ran}(S)}$. By (\ref{UT}), we have
$VT_S V^{-1}=-T_S$.
Hence $T_S$ is unitarily equivalent to $-T_S$. 
The statement of the spectral properties
of $T_S$ follows from the unitary invariance
of the  spectrum and the point spectrum of a linear
operator.
\hfill \qed

\medskip


Let 
\begin{equation}
P_+(T):=E_T((0,\infty)),\quad P_-(T):=E_T((-\infty,0)), 
\quad P_0(T):=E_T(\{0\})
\end{equation}
and
\begin{equation}
{\cal H}_{\pm}(T):={\rm Ran}(P_{\pm}(T)).
\end{equation}
Then we have
\begin{equation}
{\cal H}={\cal H}_+(T)\oplus {\cal H}_-(T)\oplus \ker T.\label{bunkai}
\end{equation}


\begin{lem}\label{map1} Let $S$ and $T$ be as above. Then 
\begin{list}{}{}
\item{(i)}
$P_{\pm}(T)S\subset SP_{\mp}(T)$ and $P_0(T)S\subset SP_0(T)$.
\item{(ii)} $S$ maps $D(S)\cap {\cal H}_{\pm}(T)$ (resp. $\ker T$) to
${\cal H}_{\mp}(T)$ (resp. $\ker T$).
\end{list}
\end{lem}

\noindent
{\it Proof}. (i) Since $S$ and $T$ strongly anticommute,
it follows from
\cite[Proposition 1.4]{Ped}
that $E_T(J)S\subset SE_T(-J)$ for all Borel sets $J\subset \R$, where $-J:=\{-\lambda|\lambda \in J\}$.
Applying this fact  with $J=(0,\infty), (-\infty,0), \{0\}$,
we obtain the desired results.

(ii) This follows from part (ii).
\hfill \qed

\medskip


We now return to the operator theory developed in the previous 
sections.




\begin{th}\label{spec2}  Assume (H). Suppose that there exists an operator
$A \in \{H^2\}_{\rm w}'$ having the following properties:
\begin{list}{}{}
\item{(a)} $D(H) \subset D(A)\cap D(A_H)$.
\item{(b)} $A_-\not=0$.
\end{list}
Then:
\begin{list}{}{}
\item{(i)} $H$ is reduced by $\overline{{\rm Ran}(A_-)}$

We denote by $\widetilde H$ 
the reduced part of $H$ to $\overline{{\rm Ran}(A_-)}$.
\item{(ii)} $\widetilde H$ is unitarily equivalent to
$-\widetilde H$. In particular, 
$\sigma(\widetilde H)$ (resp. $\sigma_{\rm p}(\widetilde H)$) 
is symmetric with respect to
the origin.
\item{(iii)}  ${\cal H}={\cal H}_+(H)\oplus {\cal H}_-(H)$. 
\item{(iv)} $P_{\mp}(H)A_-\subset A_-P_{\pm}(H)$.
\item{(v)} $A_-$ maps $D(A_-)\cap {\cal H}_{\pm}$  to
${\cal H}_{\mp}$.
\end{list}
\end{th}




\noindent
{\it Proof}. By Theorem \ref{st-anticom},
$A_-$ is self-adjoint and strongly anticommutes with $H$.
Hence we can apply Lemmas \ref{spec1} and \ref{map1} to the case
$T=H$ and $S=A_-$ and obtain the desired results 
(the injectivity of $H$ implies that $\ker H=\{0\}$).
\hfill \qed

\medskip


\begin{cor}\label{spec3}  Assume (H). Suppose that there exists an operator
$A \in \{H^2\}_{\rm w}'$ having the following properties:
\begin{list}{}{}
\item{(a)} $D(H) \subset D(A)\cap D(A_H)$.
\item{(b)} $A_-$ is injective.
\end{list}
Then $\sigma(H)$ and $\sigma_{\rm p}(H)$ are
symmetric with respect to the origin.
\end{cor}


\noindent
{\it Proof}. Condition (b) implies that $\overline{{\rm Ran}(A_-)}=
{\cal H}$. Hence $\widetilde H=H$.
Then Theorem \ref{spec2}(ii) gives the desired result.
\hfill \qed
 

\medskip

In concluding this section, we give a sufficient condition
for $A_-$ to be injective (condition (b) in Corollary \ref{spec3}).


\begin{pr}\label{prA-} Assume (H) and let $A \in \{H^2\}_{\rm w}'$
be self-adjoint.
\begin{list}{}{}
\item{(i)} Suppose that  there exists a self-adjoint operator $P$
on ${\cal H}$
having the following properties:
\begin{list}{}{}
\item{(a)} $P$ strongly commutes with $H$.
\item{(b)} $AH+HA=2P$ on $D(AH)\cap D(HA)\cap D(P)$. 
\item{(c)} $H^{-1}$ and $PH^{-1}$ are bounded.
\item{(d)} $D(H^2)\subset D(A_H)\cap D(P)$ 
\end{list}
Then $D(A_-)=D(A)$ with
\begin{equation}
A_-=A-PH^{-1}. \label{Aform}
\end{equation}
\item{(ii)} In addition to the assumption of part (i), assume
the following (f), (g) and (h):
\begin{list}{}{}
\item{(f)}   $P$ is injective and strongly commutes with $A$.
\item{(g)} $A$ is injetive with $A^{-1}$ bounded and $D(H)\subset D(A)
\cap D(HA)$.
\item{(h)} $\{\psi \in D(AP^{-1})\,|\,\|AP^{-1}\psi\|=
\|H^{-1}\psi\|\}=\{0\}$.
\end{list}
Then $A_-$ is injective. In particular,
$\sigma(H)$ and $\sigma_{\rm p}(H)$ are symmetric
with respect to the origin.
\end{list}
\end{pr}

\noindent
{\it Proof}. (i) Let $\psi \in D(H^2)$. Then
$H^{-1}\psi \in D(AH)\cap D(HA)\cap D(P)$. Hence, by condition (b), we
have
$$
HAH^{-1}\psi=2PH^{-1}\psi-A\psi,
$$
which implies that $A_{-}\psi=A\psi-PH^{-1}\psi$.
Since $(PH^{-1})^*\supset H^{-1}P$ and
$(PH^{-1})^*$ is bounded, it follows that
$H^{-1}P$ is closable with domain equal to $D(P)$ 
and $\overline{H^{-1}P}=(PH^{-1})^*$.
By the Kato-Rellich theorem,
 $B:=A-(PH^{-1}+\overline{H^{-1}P})/2$
is self-adjoint  and every core of $A$
is a core of $B$. In particular, $D(H^2)$ is a core of $B$.
It follows from the above equality that
$A_-|D(H^2)\subset B$. Hence $A_-=B$, since $A_-$ is essentially self-adjoint on $D(H^2)$. 
Thus $A_-=A-\frac 12 (PH^{-1}+\overline{H^{-1}P})$. 
It follows from  condition (a) and the boundedness of $H^{-1}$
that $H^{-1}P\subset PH^{-1}$. This implies that $\overline{H^{-1}P}
=PH^{-1}$. Thus (\ref{Aform}) follows.

(ii) To prove the injectivity of $A_-$,
we show that $\ker A_-=\{0\}$.
Let $\psi \in \ker A_-$. Then, by (\ref{Aform}),
\begin{equation}
A\psi=PH^{-1}\psi\label{AP}.
\end{equation} 
Hence $A\psi \in D(P^{-1})$ and 
\begin{equation}
P^{-1}A\psi=H^{-1}\psi. \label{A2}
\end{equation}
By conditions (f) and (g), we have $A^{-1}P\subset PA^{-1}$,
which, together with (\ref{AP}), implies
that $\psi=PA^{-1}H^{-1}\psi$. This means that
$\psi \in D(P^{-1})$ and $P^{-1}\psi=A^{-1}H^{-1}\psi$.
Hence $P^{-1}\psi \in D(A)$ and
$AP^{-1}\psi=H^{-1}\psi$. By this equation and (\ref{A2}), we obtain
that $\psi \in D(P^{-1}A)\cap D(AP^{-1})$ and
\begin{equation}
P^{-1}A\psi=AP^{-1}\psi.\label{A3}
\end{equation}
Multiplying (\ref{A2}) by $H^{-1}$ from the left and using the strong commutativity of $H$ and $P$,
we have $P^{-1}H^{-1}A\psi=H^{-2}\psi$.
Condition (b) implies that
$H^{-1}A\psi=-AH^{-1}\psi+2H^{-1}PH^{-1}\psi$.
Hence we obtain that $H^{-2}\psi=P^{-1}AH^{-1}\psi$.
By (\ref{A2}), the right hand side is equal to $P^{-1}AP^{-1}A\psi$.
By this fact and (\ref{A3}), we obtain that $\|H^{-1}\psi\|^2=
\|AP^{-1}\psi\|^2$. 
Hence, by condition (h), $\psi=0$.
 Thus $\ker A_-=\{0\}$. 
The last assertion follows from Corollary \ref{spec3}
(note that $D(H)\subset D(A_H)$,
since $H^{-1}$ is bounded and $D(H)\subset D(HA)$).
\hfill \qed

\medskip

\section{Generalized weak canonical commutation relation
and Heisenberg operators}


\subsection{A general theory}

Let $H$ be a self-adjoint operator on ${\cal H}$.
For a linear operator $X$ on ${\cal H}$, we define a linear operator
$X(t)$ ($t\in \R$) 
by
\begin{equation}
X(t):=e^{itH}Xe^{-itH}, 
\end{equation}
which is called the Heisenberg operator
of $X$ with respect to $H$.

\begin{rem}{\rm In applications to quantum mechanics, we have in mind
the case  where 
$H$ is the Hamiltonian of a quantum system and $X$ is an observable.
In this case,  $X(t)$ is interpreted as the time development
of $X$.
From this point of view too, it is important to
study properties of $X(t)$.
}
\end{rem}


In this section we present a class of linear operators $X$
such that $X(t)$ has an explicit representation.


\begin{df}{\rm  Let $C$ be a linear operator on
${\cal H}$. We say that a 
linear operator $X$ on ${\cal H}$ is in the set
${\sf K}(H,C)$ if $D(X)$ is dense and
\begin{equation}
\lang X^*\psi, H\phi\rang-\lang H\psi,X\phi\rang=
\lang\psi, iC\phi\rang,
\quad \psi,\phi \in D(X)\cap D(X^*)\cap D(C)\cap D(H).\label{X-H}
\end{equation}
}
\end{df}






\begin{rem}{\rm 
In a terminology introduced in the previous paper \cite{Ar05},
(\ref{X-H}) means that 
the triple $(X,H,C)$ obeys  the generalized weak canonical commutation
relation if $X$ is symmetric and $C$ is a bounded  self-adjoint operator on ${\cal H}$ (\cite[\S 4]{Ar05}).
}
\end{rem}

We need a lemma.

\begin{lem} Assume (H).
Then, for all $t\in \R$, the operator
\begin{equation}
G_H(t):=\frac 1{2i}H^{-1}(e^{2itH}-1)\label{GH1}
\end{equation}
is bounded and
\begin{equation}
\|G_H(t)\| \leq |t|.\label{GHB}
\end{equation}
Moreover, $G_H(t)$ is strongly differentiable in $t$ and
\begin{equation}
\mbox{{\rm s-}}\frac {dG_H(t)}{dt}=e^{2itH}, \label{GH2}
\end{equation}
where $\mbox{{\rm s-}}dG_H(t)/dt$ means the strong derivative
of $G_H(t)$.
 \end{lem}

\noindent
{\it Proof}. By the functional calculus and the
easily proven estimate
$$
|e^{2it\lambda}-1|^2 \leq  4t^2\lambda^2, \quad \lambda \in \R,
$$
we see that $D(G_H(t))={\cal H}$ and
$\|G_H(t)\psi\|\leq |t|\|\psi\|,\ \psi\in {\cal H}$.
This proves the first half of the the lemma.
Equation (\ref{GH2}) follows from  the functional
calculus.  
\hfill \qed

\medskip




The following theorem
defines a class of
linear operators $X$,  whose Heisenberg operator $X(t)$ 
with respect to $H$ has an explicit representation. 


\begin{th}\label{Hei1}
Assume (H). Let $C_{\pm}$ be self-adjoint operators
such that $C_+$ strongly commutes with $H$ and $C_-$ strongly anticommutes
with $H$, and $C:=C_+ +C_-$.
Let $X \in  {\sf K}(H,C)$.  Suppose that there exists a dense subspace $D$
such that 
\begin{list}{}{}
\item{(i)} $D\subset D(X)\cap D(X^*) \cap D(H)\cap D(C)$;
\item{(ii)}  
$e^{-itH}D\subset  D(X)\cap D(X^*)\cap D(H)\cap D(C)$ for all $t\in \R$; 
\item{(iii)} for all $\psi \in D$, $Xe^{-itH}\psi$ is strongly continuous
in $t\in \R$.
\end{list}
Then, for all $t\in \R$
\begin{equation}
X(t)=X+C_+t+G_H(t)C_-\label{Xt}
\end{equation}
on $D$.
\end{th}

\noindent
{\it Proof}.  For each vector 
$\eta\in {\cal H}$, we set
$\eta(t):=e^{-itH}\eta, \ t\in \R$. Let $\psi,\phi\in D$ and
$f(t):=\lang\psi,X(t)\phi\rang, \,t\in \R$.
Then, it is easy to see that $f$ is differentiable with
$$
\frac{df(t)}{dt}=i\{\lang H\psi(t),X\phi(t)\rang-\lang X^*\psi(t),H\phi(t)
\rang\}.
$$ 
Since $\psi(t)$ and $\phi(t)$ are in $D(H)\cap D(C)\cap D(X)\cap D(X^*)$
by the present assumption, we can compute, using (\ref{X-H}),
 the right hand side
of the above equation  to obtain
$df(t)/dt=\lang \psi, e^{itH}Ce^{-itH}\phi\rang$.
Note that $e^{-itH}\phi \in D(C)$ due to the present assumption. Hence
$e^{itH}Ce^{-itH}\phi
=e^{itH}C_+e^{-itH}\phi+e^{itH}C_-e^{-itH}\phi$.
The strong commutativity of $C_+$ and $H$ implies that
$e^{itH}C_+e^{-itH}=C_+$, while the strong anticommutativity
of $C_-$ with $H$ implies that
$e^{itH}C_-e^{-itH}=e^{2itH}C_-$.
Therefore
$$
\frac {df(t)}{dt}=\lang \psi, C_+\phi\rang+
\lang \psi, e^{2itH}C_-\phi\rang.
$$
Integrating this equation and using (\ref{GH2}), we obtain
$$
f(t)=\lang \psi, X\phi\rang
+\lang\psi, C_+t\phi\rang
+\lang\psi, G_H(t)C_-\phi\rang.
$$
Hence (\ref{Xt}) on $D$ follows.
\hfill \qed

\medskip


\begin{cor}\label{Hei2} Let the same assumption as in Theorem \ref{Hei1}
be satisfied. Suppose that $X$ is closed, $D$ is a core of
$X$ and  there exist constants $a_{\pm}\geq 0$ and $b_{\pm}\geq 0$
such that 
\begin{equation}
\|C_{\pm}\psi\| \leq a_{\pm}\|X\psi\|+b_{\pm}\|\psi\|,
\quad \psi\in D. \label{XBDD}
\end{equation}
Then, for all $t\in \R$, $e^{-itH}D(X)= D(X)$ and
(\ref{Xt}) holds on $D(X)$. Moreover
\begin{equation}
\|X(t)\psi\| \leq \|X\psi\|+(\|C_+\psi\|+\|C_-\psi\|)|t|, \quad
\psi \in D(X), t\in \R. \label{X2}
\end{equation}
\end{cor}

\noindent
{\it Proof}. 
By (\ref{Xt}) and (\ref{GHB}), we have for all $\psi\in D$
\begin{equation} 
\|Xe^{-itH}\psi\|
\leq \|X\psi\|
+(\|C_+\psi\|+\|C_-\psi\|)|t|. \label{X3}
\end{equation}
Using (\ref{XBDD}), we obtain
$$
\|Xe^{-itH}\psi\|
\leq \left(1+(a_++a_-)|t|\right)\|X\psi\|+(b_++b_-)|t|\,\|\psi\|.
$$
By this estimate and the condition that $D$ is a core of $X$, 
we can show by a simple limiting argument that
$e^{-itH}D(X) \subset D(X)$. Since $t\in \R$ is arbitrary and
$\{e^{-itH}\}_{t\in \R}$ is a unitary group, it follows that
$e^{-itH}D(X)=D(X)$ for all $t\in \R$.
Then, by a limiting argument again,
we can extend (\ref{Xt}) to an equation on $D(X)$.
Similarly (\ref{X3}) extends to all $\psi \in D(X)$ giving
(\ref{X2}).
\hfill \qed

\medskip


In the context of quantum mechanics, 
the quantity $\lang \psi, T\phi\rang$
with $T$ a linear operator on a Hilbert space ${\cal H}$ 
and $\psi\in {\cal H}$, $\phi \in D(T)$ is called a matrix
element of $T$ and has  physical importance.

The following result gives an explicit  formula
for each element in a subset of  matrix elements of $X(t)$.


\begin{cor}\label{Hei3} Let the same assumption as in Theorem
\ref{Hei1} be satisfied. Let
$\psi,\phi \in {\cal H}_+(H)\cap D$ or
$\psi,\phi \in {\cal H}_-(H)\cap D$. Then
\begin{equation}
\lang \psi,X(t)\phi\rang
=\lang \psi, X\phi\rang+
\lang \psi, C_+\phi\rang t.\label{Xinv}
\end{equation}
\end{cor}

\noindent
{\it Proof}. By the functional calculus, 
$G_H(t)^*{\cal H}_{\pm}(H)\subset {\cal H}_{\pm}(H)$.
This fact and Lemma  \ref{map1}(ii) with
$S=C_-$ and $T=H$  imply that
$\lang \psi, G_H(t)C_-\phi\rang=0$. Hence
(\ref{Xinv}) follows.
\hfill \qed

\medskip

The following theorem
describes the asymptotic behavior
of matrix elements of $X(t)$  as $t\to\pm\infty$. 

We denote by ${\cal H}_{\rm ac}(H)$ the subspace of absolute continuity
with respect to  $H$.

\begin{th}\label{Hei4} Let the same assumption as in Theorem
\ref{Hei1} be satified. Let 
$\psi$ $\in$  $D(H^{-1})$ $\cap$  $D$  and $\phi\in {\cal H}_{\rm ac}(H)\cap D$.
Then
\begin{equation}
\lang \psi, X(t)\phi\rang=
\lang \psi,X\phi\rang -
\lang H^{-1}\psi, \frac 1{2i}C_-\phi\rang+
\lang \psi, C_+\phi\rang t +o\left(\frac {1}{t}\right) \quad
(t\to\pm\infty).
\end{equation}
\end{th}

\noindent
{\it Proof}.
Let 
\begin{equation}
\delta(t):=\frac 1{2i}\lang H^{-1}\psi, e^{2itH}C_-\phi\rang.
\end{equation}
Then we have by (\ref{Xt})
$$
\lang \psi, X(t)\phi\rang=
\lang \psi,X\phi\rang-
\lang H^{-1}\psi,\frac 1{2i}C_-\phi\rang+
\lang \psi, C_+\phi\rang t +\delta(t).
$$
By the strong anticommutativity of $C_-$ and $H$,
we have $E_H(J)C_- \subset C_-E_H(-J)$ for all Borel sets
$J\subset \R$. This implies that, if $\phi$ is in ${\cal H}_{\rm ac}(H)
\cap D(C_-)$,
then $C_-\phi \in {\cal H}_{\rm ac}(H)$.
Hence $\lim_{t\to\pm\infty}\delta(t)=0$.
\hfill \qed

\medskip


\subsection{The case $C=A_+ +A_-$ with
$A\in \{H^2\}_{\rm w}'$}


We now apply the results of the precdeding subsection
to the operator theory
concerning the pair $(A,H)$ with
$A \in \{H^2\}_{\rm w}'$.


\begin{th}\label{Hei5}
Assume (H) and let $A\in \{H^2\}_{\rm w}'$ with
$D(H)\subset D(A_H)\cap D(A)$ (hence Theorems \ref{th4-4} and
\ref{st-anticom} hold). 
Let $X \in  {\sf K}(H,A_++A_-)$.  Suppose that there exists a dense subspace  $D_0$
such that 
\begin{list}{}{}
\item{(i)} $D_0\subset D(X)\cap D(X^*) \cap D(H)$;
\item{(ii)}  
$e^{-itH}D_0\subset  D(X)\cap D(X^*)\cap D(H)$ for all $t\in \R$; 
\item{(iii)} for all $\psi \in D_0$, $Xe^{-itH}\psi$ is strongly continuous
in $t\in \R$.
\end{list}
Then, for all $t\in \R$,
\begin{equation}
X(t)=X+A_+t+G_H(t)A_-\label{form1}
\end{equation}
on $D_0$.
\end{th}

\noindent
{\it Proof}. We can apply Theorem \ref{Hei1} 
with $C_{\pm}=A_{\pm}$ to obtain the desired result.
\hfill \qed

\medskip



\begin{cor}\label{Hei6} Let the same assumption as in Theorem \ref{Hei5}
be satisfied. Suppose that $X$ is closed, $D_0$ is a core of
$X$ and  there exist constants $a_{\pm}\geq 0$ and $b_{\pm}\geq 0$
such that 
\begin{equation}
\|A_{\pm}\psi\| \leq a_{\pm}\|X\psi\|+b_{\pm}\|\psi\|,
\quad \psi\in D_0. 
\end{equation}
Then, for all $t\in \R$, $e^{-itH}D(X)= D(X)$ and
(\ref{form1}) holds on $D(X)$. Moreover
\begin{equation}
\|X(t)\psi\| \leq \|X\psi\|+(\|A_+\psi\|+\|A_-\psi\|)|t|, \quad
\psi \in D(X), t\in \R. 
\end{equation}
\end{cor}

\noindent
{\it Proof}. A simple application of Corollary \ref{Hei2}.
\hfill \qed

\medskip

 

\begin{cor}\label{Hei7} Let the same assumption as in Theorem
\ref{Hei5} be satisfied. Let
$\psi,\phi \in {\cal H}_+(H)\cap D_0$ or
$\psi,\phi \in {\cal H}_-(H)\cap D_0$. Then
\begin{equation}
\lang \psi,X(t)\phi\rang
=\lang \psi, X\phi\rang+
\lang \psi, A_+\phi\rang t.
\end{equation}
\end{cor}

\noindent
{\it Proof}. A simple application of Corollary \ref{Hei3}.
\hfill \qed

\medskip

.

\begin{th}\label{Hei8} Let the same assumption as in Theorem
\ref{Hei5} be satified. Let 
$\psi$ $\in$ $D(H^{-1})$ $\cap$  $D_0$ and $\phi\in {\cal H}_{\rm ac}(H)\cap D_0$.
Then
\begin{equation}
\lang \psi, X(t)\phi\rang=
\lang \psi,X\phi\rang-
\lang H^{-1}\psi, \frac 1{2i}A_-\phi\rang+
\lang \psi, A_+\phi\rang t +o\left(\frac {1}{t}\right) \quad
(t\to\pm\infty).
\end{equation}
\end{th}

\noindent
{\it Proof}. A simple application of 
Theorem \ref{Hei5}. \hfill \qed

\medskip




\section{Connection with the generalized weak Weyl relation}

In the previous paper \cite{Ar05}, we introduced
a notion, which we call the generalized weak Weyl relation (GWWR).
This is defined as follows:
Let $T$ be a symemtric operator
on ${\cal H}$, $H$ be a self-adjoint operator on ${\cal H}$,
 and $K(t)$ ($t\in \R$) be a bounded self-adjoint
operator on ${\cal H}$  for all $t\in \R$.
We say that the triple $(T,H,K)$ obeys the GWWR
if $e^{-itH}D(T)\subset D(T)$ for all $t\in \R$ and
$$
Te^{-itH}\psi=e^{-itH}(T+K(t))\psi,\quad \forall \psi\in D(T), 
\forall t\in \R.
$$

In this section we point out that there is a connection
of the operator theory developed in the previous sections
with the theory of GWWR.


\begin{th}\label{th7-1} Suppose that the same assumption as in Theorem \ref{Hei5}
holds 
with $X$ being symmetric
 and that $A_{\pm}$ and $H^{-1}$ are bounded.
Let
\begin{equation}
K_H(t):=A_+t +G_H(t)A_-.
\end{equation}
Then $K_H(t)$ is a bounded self-adjoint operator
on ${\cal H}$ for all $t\in \R$
and $(X,H,K_H)$ obeys the GWWR.
\end{th}

\noindent
{\it Proof}. Under the present assumption,
the assumption of Corollary \ref{Hei6} is satisfied.
Hence (\ref{form1}) holds on $D(X)$.  Hecne it is sufficient to prove that
$K_H(t)$ is a bounded self-adjoint operator.
The boundedness of $K_H(T)$ is obvious, since $A_{\pm}$ are bounded
by the present assumption.
Using the fact that
$A_-e^{-2itH}=e^{2itH}A_-$ and
$A_-H^{-1}=-H^{-1}A_-$, which follow from the strong
anticommutativity of $A_-$ and $H$,
one can show that $G_H(t)A_-$ is self-adjoint.
\hfill \qed

\medskip

\begin{cor} Let the same assumption as in Theorem
\ref{th7-1} be satisfied. Then, for all $\psi,\phi \in {\cal H}$,
\begin{equation}
\lim_{t \to\pm\infty}\lang \psi, e^{-itH}A_+\phi\rang=0.
\label{infty}
\end{equation}
\end{cor}

\noindent
{\it Proof}. 
We have
$$
\mbox{{\rm s-}}\lim_{t \to\pm\infty}
\frac {K_H(t)}{t}=A_+.
$$
Hence, applying \cite[Corollary 3.3]{Ar05},
we obtain (\ref{infty}).
\hfill \qed

\medskip
 
\begin{th} Assume (H) with  $H^{-1}$ bounded on ${\cal H}$.
Suppose that there exist
a  closed symmetric operator $X$ and a bounded
self-adjoint operator $A\in \{H^2\}_{\rm w}'$ such 
that $X\in {\sf K}(H,A)$ and
$A_{\pm}$ are bounded with $\ker A_+=\{0\}$. Then
$\sigma_{\rm p}(H)=\emptyset$.
\end{th}

\noindent
{\it Proof}. The present assumption imply
that the assumption of Theorem \ref{Hei5} is satified. 
Hence Theorem \ref{th7-1} holds. Then, a  simple application
of \cite[Corollary 3.4]{Ar05} yields the desired result.
\hfill \qed

\medskip
 
\section{Applications}

Let $(A,B,L)$ be a triple of self-adjoint operators
on a Hilbert space on ${\cal H}$ obeying the following conditions:
\begin{list}{}{}
\item{(A.1)} $A$ is bounded and injective with $A^{-1}$ bounded.
\item{(A.2)} $A$ and $B$ strongly anticommute.
\item{(A.3)} $L$ strongly commutes with $A$ and $B$.
\end{list}

We define an operator $H$ by
\begin{equation}
H:=AL+B\label{AD}
\end{equation}

This operator is an abstract form of
Dirac operators. A typical example is given as follows.

\begin{ex}\label{ex8-1}{\rm Let $\alpha_j\,(j=1,2,3)$ and $\beta$
be $4\times 4$ Hermitian matrices satisfying
the anticommutation relations
\begin{eqnarray*}
&&\{\alpha_j,\alpha_k\}=2\delta_{jk}, \quad \{\alpha_j,\beta\}=0,
\quad j,k=1,2,3,\\
&& \beta^2=1,
\end{eqnarray*}
where $\{X,Y\}:=XY+YX$. Let
$m>0$ be a constant and $D_j$ be the generalized partial
differential
operator in the variable $x_j$ ($x=(x_1,x_2,x_3)\in \R^3$)
acting in $L^2(\R^3)$. Then the free Dirac operator with mass $m$
is defined by
$$
H_{\rm D}:=\sum_{j=1}^3\alpha_j(-iD_j)+m\beta
$$
acting in the Hilbert space $L^2(\R^3;\C^4)$.
It is well known that $H$ is self-adjoint \cite[Theorem 1.1]{Thaller}.
 Let $B_j=\sum_{k\not=j}\alpha_k(-iD_k)+m\beta$ ($j=1,2,3$).
Then $H_{\rm D}=\alpha_j(-iD_j)+B_j$ and $(\alpha_j,  B_j, -iD_j)$
satisfies (A.1)--(A.3) with
$A=\alpha_j, B=B_j$ and $L=-iD_j$.
The triple $(\alpha_j, B_j,-iD_j)$ 
satisfies also the conditions
assumed in the discussions below.

For other examples, see, e.g.,  \cite{Ar93a,Ar93b,Ar95b,Ar05}.
}
\end{ex}




We first prove the following fact.

\begin{th}\label{8-1} The operator $H$ is self-adjoint and
\begin{equation}
H^2=A^2L^2+B^2.\label{eqH2}
\end{equation}
Moreover
\begin{equation}
\|H\psi\|^2=\|AL\psi\|^2+\|B\psi\|^2, \quad
\psi \in D(H)=D(L)\cap D(B). \label{Hnorm}
\end{equation}
\end{th}


To prove this theorem, we need two lemmas.

\begin{lem}\label{lem8-2} 
Let $S$ be an injective,  bounded self-adjoint operator with $S^{-1}$
bounded 
and $T$ be a  self-adjoint operator strongly commuting with $S$.
Then $ST$ is self-adjoint and $ST=TS$.
\end{lem}

\noindent
{\it Proof}. Since $D(ST)=D(T)$, $ST$ is densely defined. 
By the boundedness of $S$, we have $(ST)^*=TS$.
The strong commutativity of $S$ and $T$ implies that $ST \subset TS$.
Hence $(ST)^*\supset ST$. Therefore $ST$ is symmetric.
By the strong commutativity of $S$ and $T$, there exists
a unique two dimensional spectral measure $E$ such that
$E(J\times K)=E_S(J)E_T(K)$ for all Borel sets $J,K\subset \R$
and $S=\int_{\R^2}\lambda \,dE(\lambda,\mu), T=
\int_{\R^2}\mu \,dE(\lambda,\mu)$.
The operator $F:=\int_{\R^2}\lambda \mu \,dE(\lambda,\mu)$
is self-adjoint with $D(F)=
\{\psi \in {\cal H}
|\int_{\R^2}\lambda^2\mu^2d\|E(\lambda,\mu)\psi\|^2 <\infty\}$.
It is easy to see that $ST\subset F$.
By the condition that $S^{-1}$ is bounded, the support of $E$ is included
in $[-\|S\|, -\delta]\cup[\delta,\|S\|\,]\times \R$ for some $\delta >0$. 
Using this fact, one can show that $D(F)\subset D(T)$.
Thus $ST=F$, showing that $ST$ is self-adjoint. 
\hfill \qed

\medskip

\begin{lem}\label{lem8-3} 
Let $S$ and $T$ be as in Lemma \ref{lem8-2}.
Let $C$ be a self-adjoint operator strongly anticommuting with $S$
and strongly commuting with $T$.
Then  $C$ strongly anticommutes with $ST$.
\end{lem}

\noindent
{\it Proof}. For all $t \in \R$, we have
$e^{itC}S=Se^{-itC}$ (strong anticommutativity of $C$ and $S$)
and $e^{itC}T\subset Te^{itC}$ (strong commutativity of $C$ and $T$).
Hence it follows that $e^{itC}ST\subset STe^{-itC}$.
Thus $C$ strongly anticommutes with $ST$.
\hfill \qed

\medskip


\noindent
{\bf Proof of Theorem \ref{8-1}}

We can apply Lemmas \ref{lem8-2} and \ref{lem8-3} with
$S=A, T=L$ and $C=B$ 
to conclude that $AL$ is self-adjoint strongly anticommuting
with $B$.
Hecne, by Lemma \ref{san}, $H$ is self-adjoint and
(\ref{eqH2}) holds. 
\hfill \qed

\medskip


\begin{cor} $A\in \{H^2\}_{\rm w}'$.
\end{cor}

\noindent
{\it Proof}. Since $A$ strongly anticommutes with
$B$, it strongly commutes with $|B|$ \cite[Theorem 3.1]{Ped}.
Hence, for all $t\in \R$, we have by (\ref{eqH2})
$e^{itA}H^2=H^2e^{itA}$, which implies the strong commutativity
of $A$ and $H^2$ and that $AH^2\subset H^2A$.
Thus, as remarked in Introduction, $A \in \{H^2\}_{\rm w}'$.
\hfill \qed

\medskip

\begin{lem} The operator $L$ strongly commutes with $H$.
\end{lem}

\noindent
{\it Proof}. One can easily see that, for all $t \in \R$,
$e^{itL}H\subset He^{itL}$. This implies the
strong commutativity of $L$ and $H$.
\hfill \qed

\medskip


\begin{lem}\label{A-inv} $AD(H)\subset D(H)$. 
\end{lem}

\noindent
{\it Proof}. We have $D(H)=D(L)\cap D(B)$.
Since $A$ strongly commutes with $L$ and strongly anticommutes with $B$,
it follows that $AD(L)\subset D(L)$ and $AD(B)\subset D(B)$.
Hence $AD(H)\subset D(H)$.
\hfill \qed

\medskip
 

By Lemma \ref{A-inv}, we have
\begin{equation}
D(H)\subset D(HA)\cap D(AH)\cap D(L).
\end{equation}


\begin{lem} $AH+HA=2A^2L$ on $D(H)$.
\end{lem}

\noindent
{\it Proof}. Let $\psi \in D(H)$.
Then $AH\psi=A^2L\psi+AB\psi=A^2L\psi-BA\psi$ and
$HA\psi=A^2L\psi+BA\psi$.
Hence the desired result follows.
\hfill \qed
\medskip


In what follows, we  assume the following condition too:

\medskip

\noindent
{\bf Hypothesis (B)}:
$B^2 \geq M^2$ for some constant $M>0$.

\medskip




 
\begin{lem}\label{lemH} Assume (B). Then: 
\begin{list}{}{}
\item{(i)}  $H$ is injective and $H^{-1}$ is bounded with
$\|H^{-1}\|\leq M^{-1}$.
\item{(ii)} $LH^{-1}$ is bounded with
$\|LH^{-1}\| \leq \|A^{-1}\|$. 
\item{(iii)} $BH^{-1}$ is bounded with $\|BH^{-1}\| \leq 1$.
\item{(iv)} $D(A_H)={\cal H}$ and $A_H$ is bounded.
\end{list}
\end{lem}

\noindent
{\it Proof}. (i) By (\ref{Hnorm}), we have
$\|H\psi\| \geq M\|\psi\|, \ \psi \in D(H)$.
This implies that $H$ is injective and $H^{-1}$ is bounded with
$\|H^{-1}\| \leq M^{-1}$.

(ii) By (\ref{Hnorm}), we have $\|L\psi\| \leq \|A^{-1}\| \|H\psi\|$,
which implies that $\|LH^{-1}\phi\| \leq 
\|A^{-1}\|\|\phi\|$,   $\phi \in {\cal H}$.

(iii) Similar to part (ii).

(iv) For all $\psi \in {\cal H}$, $H^{-1}\psi\in D(H)$.
By Lemma \ref{A-inv}, $AH^{-1}\psi \in D(H)$. Hence $\psi\in D(A_H)$.
Thus $D(A_H)={\cal H}$.
Morever, by (\ref{Hnorm}), 
\begin{eqnarray*}
\|A_H\psi\|&\leq &\|ALAH^{-1}\psi\|+\|BAH^{-1}\psi\|\\  
&\leq & (\|A\|^2\|LH^{-1}\| +\|A\|\|BH^{-1}\|)
\|\psi\|,
\end{eqnarray*}
where we have used the relation $AB\subset -BA$,
which follows from the strong anticommutativity of $A$ and $B$.
Hence $A_H$ is bounded.
\hfill \qed 

\medskip

In the present example, by Lemma \ref{lemH}, $A_{\pm}$
are bounded and self-adjoint and
\begin{equation}
A=A_+ +A_-.
\end{equation}
Moreover, by the above results, we can apply
Theorems \ref{th4-4} and \ref{st-anticom} to
conclude that $A_{+}$ strongly commutes with $H$ and
$A_-$ strongly anticommutes with $H$.
Also we can apply  Proposition \ref{prA-} with $P=A^2L$ to obtain
\begin{equation}
A_-=A-A^2LH^{-1}.
\end{equation}
Hence
\begin{equation}
A_+=A^2LH^{-1}.
\end{equation}

\begin{pr} The operators $A_+$ and $A_-$ strongly anticommute.
\end{pr}

\noindent
{\it Proof}. Since $A_{\pm}$ are bounded, it is sufficient to show that
$A_-A_++A_+A_-=0$. But this is easily done.
An alternative proof is to check (\ref{AH-A}).
\hfill \qed

\medskip


\begin{th}\label{rsym} Suppose that $L$ is injective.
Then $\sigma(H)$ and $\sigma_{\rm p}(H)$ are
symmetric with respect to the origin.
\end{th}

\noindent
{\it Proof}. We show that
condition (h) in Proposition \ref{prA-} holds in
the present case ($P=A^2L$). Let $\psi \in D((A^2L)^{-1})$ and
$\|A(A^2L)^{-1}\psi\|=\|H^{-1}\psi\|$.
Then 
$$
\lang L^{-1}\psi, A^{-2}L^{-1}\psi\rang=
\lang \psi, (A^2L^2+B^2)^{-1}\psi\rang
\leq \lang \psi, (A^2L^2+M^2)^{-1}\psi\rang.
$$
Since $A$ and $L$ strongly commute, there exists a two-dimensional
spectral measure $E_{A,L}$ such that
$E_{A,L}(J\times K)=E_A(J)E_L(K)$ for all Borel sets $J,K\subset \R$
and  
$$
A=\int_{\R^2}\lambda \,dE_{A,L}(\lambda,\mu),\quad
L=\int_{\R^2}\mu \,dE_{A,L}(\lambda,\mu).
$$
By the above inequality, we have
$$
\int_{\R^2}\left( \frac 1{\lambda^2\mu^2+M^2}-\frac 1{\lambda^2\mu^2}
\right)d\|E_{A,L}(\lambda,\mu)\psi\|^2 \geq 0,
$$
which  implies that
$$
\int_{\R^2}\frac 1{\lambda^2\mu^2(\lambda^2\mu^2+M^2)}
d\|E_{A,L}(\lambda,\mu)\psi\|^2=0.
$$
Hence $\psi=0$.
Thus we can apply Proposition \ref{prA-} to
obtain the desired result.
\hfill \qed


\medskip


\begin{rem}{\rm 
Theorem \ref{rsym} is interesting
in that it clarifies a general structure
for a self-adjoint operator
$H$ of the form (\ref{AD}) to have 
the spectral reflection symmetry.
}
\end{rem}

 

 

Let $X$ be a closed symmetric operator 
satisfying the following conditions:
\begin{list}{}{}
\item{(X.1)} $X \in {\sf K}(L, I)\cap {\sf K}(B,0) $.
\item{(X.2)} $AX\subset XA$.
\item{(X.3)} There exists a core $D_0$ of $X$ such that
\begin{list}{}{}
\item{(i)} $D_0\subset D(X)\cap D(H)$.
\item{(ii)} $e^{-itH}D_0\subset D(X)\cap D(H)$ for all $t\in \R$.
\item{(iii)} for all $\psi \in D_0$,
$Xe^{-itH}\psi$ is strongly continuous in $t \in \R$.
\end{list}
\end{list}

\begin{th}\label{th-X} For all $t\in \R$, $e^{-itH}D(X)=D(X)$ and
\begin{equation}
X(t)=X+A_+t+G_H(t)A_- 
\end{equation}
on $D(X)$. Moreover
$$
\|X(t)\psi\| \leq \|X\psi\|
+(\|A_+\|+\|A_-\|)\|\psi\| |t|, \quad \psi \in D(X), \, t\in \R.
$$
\end{th}

\noindent
{\it Proof}. Using (X.1) and (X.2), we can show that
$X \in {\sf K}(H,A)$.
Then, applying Corollary \ref{Hei6}, we obtain the desired
results.
\hfill \qed

\medskip

Also we have a corollary and a theorem
following from Corollary \ref{Hei7} and Theorem
\ref{Hei8} respectively.
But we omit writing them down here.


\begin{ex}\label{ex8-2}{\rm As for Example \ref{ex8-1}, we have the decomposition
$$
\alpha_j=\alpha_{j,+}+\alpha_{j,-}
$$
with $\alpha_{j,-}=\alpha_j-(-iD_j)H_{\rm D}^{-1}$, where
$\alpha_{j,+}$ strongly commutes with $H_{\rm D}$ and
$\alpha_{j,-}$ strongly anticommutes with $H_{\rm D}$.
In the context of quantum mechanics, the operator $\alpha_{j,+}=(-iD_j)H_{\rm D}^{-1}$
is called the classical velocity operator and
the $\alpha_{j,-}$ describes the so-called {\it Zitterbewegung}
\cite[\S 1.6]{Thaller}.

The position operator $x_j$ satisfies
all the conditions imposed on $X$ in the above theory.
Hence we can apply Theorem \ref{th-X}
to obtain an explicit formula for the Heisenberg operator
$x_j(t)=e^{itH_{\rm D}}x_je^{-itH_{\rm D}}$, which conincides with
the known formula \cite[Theorem 1.3]{Thaller}.
}
\end{ex}




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