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canonical commutation relation,  Hamiltonian, time operator, time-energy uncertainty relation, phase operator,  spectrum, regular perturbation theory.
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\begin{document}

\title{\bf Time Operators of a Hamiltonian with Purely Discrete Spectrum}\author{Asao Arai
\thanks{
Corresponding author. The work is  supported  by the Grant-in-Aid No.17340032 for
Scientific Research  from Japan Society for the Promotion of Science (JSPS).}
{\,\,}${}^{,1}$ 
and  
Yasumichi Matsuzawa${}^2$\\
Department of Mathematics, Hokkaido University \\
Sapporo 060-0810\\
Japan\\
${}^1$ E-mail: arai@math.sci.hokudai.ac.jp\\
${}^2$ E-mail: s073035@math.sci.hokudai.ac.jp }





\maketitle



\begin{abstract} We develop a mathematical theory of time operators
of a Hamiltonian  with purely discrete spectrum. The main results include
boundedness, unboundedness  and spectral properties of them.  
In addition, possible connections of a time operator of $H$ with regular perturbation theory are discussed. 
\end{abstract}





\noindent

{\bf Keywords}. canonical commutation relation,  Hamiltonian, time operator, time-energy uncertainty relation, phase operator,  spectrum, regular perturbation theory.



\medskip

\noindent
{\bf Mathematics Subject Classification (2000)}. 81Q10, 47N50.



\medskip


\section{Introduction}


This paper is concerned with mathematical theory of time operators in quantum mechanics \cite{Ar05,Ar07-1,Ar07-2,ArMa,Miya1}. There are some types of time operators which are not necessarily equivalent each other.
For the reader's convenience, we first recall the definitions of them with comments.

Let $\MSH$ be a complex Hilbert space. We denote the inner product  and the norm of $\MSH$ by
$\lang \,\cdot\,, \,\cdot\, \rang$ (antilinear in the first variable) 
and $\|\cdot\|$ respectively. For a linear
operator $A$ on a Hilbert space, 
$D(A)$ denotes the domain of $A$. 

Let $H$ be a self-adjoint operator on $\MSH$ and $T$ be a symmetric operator
on $\MSH$. 

The operator  $T$  is called a {\it time operator} of $H$ if there is a dense subspace
$\MSD$ of  $\MSH$ such that $\MSD \subset D(TH)\cap D(HT)$  and the canonical commutation relation (CCR)
\begin{equation}
[T,H]:=(TH-HT)=i\label{TH}
\end{equation}
holds on $\MSD$ (i.e., 
$[T,H]\psi=i\psi, \ \forall \psi \in \MSD$), 
where $i$ is the imaginary unit. In this case, $T$ is called a {\it canonical
conjugate} to $H$ too. 

The name \lq\lq{time operator}" for the operator $T$ comes from the quantum mechanical
context  where $H$ is taken to be  the Hamiltonian of a quantum system and 
the heuristic classical-quantum correspondence based on the structure that, in 
the classical relativistic mechanics, time is a canonical conjugate variable
to energy in each  Lorentz frame of coordinates. Note also that the dimension of $T$ is that of  time if the dimension of $H$ is that of energy in the original unit system where the right hand side of (\ref{TH}) takes the form $i\hbar$ with $\hbar$ being
the Planck constant $h$ divided by $2\pi$.
We remark, however, that this name is somewhat misleading, because, in the framework of the standard quantum mechanics, time is {\it not an observable}, but just a parameter assigning the time  when a quantum event is observed. But we follow the convention in this respect.
By the same reason as just remarked,   
$T$ is not necessarily (essentially) self-adjoint. But this does not
mean that it is \lq\lq{unphysical}" \cite{Ar05,Miya1}.



From a representation theoretic  point of view, 
the pair $(T,H)$ is a symmetric representation of the CCR with one degree of freedom.
But one should remember that,
as for this original form of representation of the CCR, the von Neumann uniqueness
theorem (\cite{vN}, \cite[Theorem VIII.14]{RS1}) does not necessarily hold. In other words, 
$(T,H)$  is not necessarily unitarily
equivalent to a direct sum of the Schr\"odinger representation of 
the CCR with one degree of freedom. Indeed, for example, 
it is obvious that, if $T$ or $H$ is bounded below or bounded above, then $(T,H)$ cannot be unitarily equivalent
to a direct sum of the Schr\"odinger representation of 
the CCR with one degree of freedom. 

A classification of pairs $(T,H)$ with $T$ being a bounded self-adjoint operator
has been done by G. Dorfmeister and J. Dorfmeister \cite{DD}.  We remark, however, that
the class  discussed in \cite{DD} does not cover
the pairs  $(T,H)$ considered in this paper,
because the paper  \cite{DD} treats only the case where $T$ is bounded and  absolutely continuous.

A weak form  of time operator is defined as follows.
We say that a symmetric operator $T$ is a {\it weak time operator} of $H$
if there is a dense subspace $\MSD_{\rm w}$ of  $\MSH$ such that
$\MSD_{\rm w}\subset D(T)\cap D(H)$ and
$$
\lang T\psi,H\phi\rang-\lang H\psi,T\phi\rang=\lang \psi, i\phi\rang, 
\quad \psi,\phi\in \MSD_{\rm w},
$$
i.e., $(T,H)$ satisfies the CCR in the sense of sesquilinear form on $\MSD_{\rm w}$.
Obviously a time operator $T$ of $H$ is a weak time operator  of $H$.
But the converse is not true (it is easy to see, however,  that, if 
$T$ is a weak time operator of $H$ and $\MSD_{\rm w}\subset D(TH)\cap D(HT)$,
then $T$ is a time operator).
An important aspect of a weak time operator $T$ of $H$ is that 
a {\it time-energy uncertainty
relation} is naturally derived \cite[Proposition 4.1]{Ar05}: for all unit vectors $\psi$ in $\MSD_{\rm w}\subset D(T)\cap D(H)$,
$$
(\Delta T)_{\psi} (\Delta H)_{\psi} \geq \frac 12,
$$
where, for a linear operator $A$ on $\MSH$ and $\phi \in D(A)$ with $\|\phi\|=1$,
$$
(\Delta A)_{\phi}:=\|(A-\lang \phi,A\phi\rang)\phi\|,
$$
called the {\it uncertainty
of $A$ in the  vector $\phi$}. 


In contrast to the weak form of time operator,
there is  a strong form. We say that $T$ is a {\it strong time operator} of $H$ if, for all $t \in \BBR$,
$e^{-itH}D(T)\subset D(T)$  and 
\begin{equation}
Te^{-itH}\psi=e^{-itH}(T+t)\psi, \quad \psi \in D(T). \label{WR}
\end{equation}
We call (\ref{WR}) the {\it weak Weyl relation} \cite{Ar05}.
From a representation theoretic point of view, we call a  pair $(T,H)$
obeying the weak Weyl relation a {\it weak Weyl representation} of the CCR.
This type of representation of the CCR was extensively studied by
Schm\"udgen \cite{Sch1,Sch2}.
It is shown that a strong time operator of $H$ is a time operator of $H$ \cite{Miya1}. 
But the converse is not true. In fact, the time operators considered in the present paper
are not strong ones.


There is a generalized version of strong time operator \cite{Ar05}. 
We say that $T$ is a {\it generalized time operator} of $H$ if, for each $t\in \BBR$, there is
a {\it bounded  self-adjoint} operator $K(t)$  on $\MSH$ with
$D(K(t))=\MSH$, $e^{-itH}D(T)\subset D(T)$ and 
a {\it generalized weak Weyl relation} (GWWR)
\begin{equation}
Te^{-itH}\psi=e^{-itH}(T+K(t))\psi\quad (\forall \psi \in D(T))
\end{equation}
holds. In this case, the bounded operator-valued function $K(t)$ of $t\in \BBR$
is called the {\it commutation factor} of the GWWR under consideration.



We now come to the subject of the present paper. In his interesting paper \cite{Ga}, 
Galapon  showed by an explicit construction that,
for every self-adjoint operator $H$ (a Hamiltonian) on an abstract  Hilbert space $\MSH$ which is {\it bounded below} and has purely discrete spectrum
with some growth condition, there is a time operator $T_1$ on $\MSH$, which is
a bounded self-adjoint operator under an additional condition
 (for the definition of $T_1$, see (\ref{GT}) below).
To be definite, we call the operator $T_1$ introduced in \cite{Ga} the {\it Galapon time
operator}.

An important point of  Galapon's work \cite{Ga} is in that
it disproved the long-standing belief or folklore  among physicists that 
there is no self-adjoint operator canonically conjugate to a Hamiltonian which is bounded below (for a historical survey, see Introduction of \cite{Ga}).

Motivated by  work of Galapon \cite{Ga},
we investigate, in this paper,  properties  of time operators of
a self-adjoint operator $H$ with purely discrete spectrum.
In Section 2, we introduce a densely defined linear operator $T$ whose restriction to a subspace yields the Galapon time operator $T_1$ and
prove basic properties of $T$ and $T_1$, in particular the closedness of $T$.
It follows that, if $T$ is bounded, then $T$ is self-adjoint  with $D(T)=\MSH$ and a time operator of $H$. We denote by $T^{\#}$ one of $T_1$, $T$ and $T^*$ (the adjoint of $T$).
In Section 3, we discuss some general properties of $T^{\#}$. Moreover the reflection symmetry of the spectrum
of $T^{\#}$ with respect to the imaginary axis is proved. Sections 4--6 are the main parts 
of this paper. In Section 4, we present a general criterion  for 
$T$ to be bounded with $D(T)=\MSH$, while, in Section 5,
we give a sufficient condition for $T$ to be unbounded. In Section 6, we present a necessary and sufficient condition
for $T$ to be Hilbert-Schmidt.
In Section 7, we show that, under some condition, the Galapon time operator is
a generalized time operator of $H$, too. We also discuss non-differentiability 
of the commutation factor $K$ in the GWWR for $(T_1,H)$. In the last section,
we consider a perturbation of $H$ by a symmetric operator and try to draw out
physical meanings of $T_1$ and $K$
in the context of regular perturbation theory.

 
 





\section{Time Operators}

In this section,  we recapitulate some basic aspects of the Galapon time operator 
in more apparent manner than in \cite{Ga}. 




Let $\MSH$ be a complex  Hilbert space  and

$H$ be a self-adjoint operator on $\MSH$ which 
has the following properties (H.1) and (H.2):

\begin{list}{}{}

\item[(H.1)] The spectrum of $H$, denoted $\sigma(H)$, is purely discrete 
with $\sigma(H)=\{E_n\}_{n=1}^{\infty}$, where 
each eigenvalue $E_n $ of $H$ is  simple and 
$$
0< E_n < E_{n+1}
$$ 
for all $n\in \BBN$ (the set of positive integers).
\item[(H.2)] $\displaystyle\sum_{n=1}^{\infty} \frac{1}{{E_n}^2} < \infty$.

\end{list}
{\it Throughout the present paper we assume (H.1) and (H.2)}.

\begin{rem}{\rm The positivity condition $E_n>0$ for the eigenvalues of $H$
does not lose generality, because, if $H$ is bounded below, but not strictly positive,
then one needs only to consider, instead of $H$, 
$\tilde H:=H+c$ with a constant $c > -\inf \sigma(H)$, which is a strictly positive self-adjoint
operator.}
\end{rem}


Property (H.2) implies that 
\begin{equation}
E_n \to \infty \quad (n\to\infty). \label{infty}
\end{equation}
Let $e_n$ be a normalized eigenvector of $H$ belonging to  eigenvalue $E_n$:
\begin{equation}
H e_n = E_n e_n, \quad n\in \BBN.
\end{equation}
Then, by property (H.1), the set $\{e_n\}_{n=1}^{\infty}$ 
is a complete orthonormal system (C.O.N.S.) of  $\MSH$.


\begin{lem}\label{lem2-1} \
\begin{list}{}{}
\item[{\rm (i) }] For all $m\in \BBN$,
\begin{equation}
\sum_{n\not=m}^{\infty}\frac 1{(E_n-E_m)^2} <\infty.
\end{equation}
In particular, for each $m\in \BBN$,
$$
\sum_{n\not=m}^{\infty}\frac 1{E_n-E_m}e_n
$$
converges in $\MSH$.
\item[{\rm (ii)}]
For all  $n\in \BBN$ and vectors $\psi$ in $\MSH$, the infinite series 

\begin{equation}
\sum_{m \not= n}^{\infty} \frac{\langle e_m,\psi \rangle}{E_n-E_m}
\end{equation}

absolutely converges.
\end{list}

\end{lem}



{\it Proof}. (i)  By (\ref{infty}),  we have
\begin{equation}
C_m:=\sup_{n\not=m}\frac{1}{\left(1-\frac{E_m}{E_n}\right)^2} <\infty.
\end{equation}
Hence we have
$$
\sum_{n \not= m}^{\infty} \frac{1}{|E_n-E_m|^2} 
\leq C_m\sum_{n\not=m}^{\infty}\frac {1}{E_n^2} <\infty.
$$ 


(ii) By the Cauchy-Schwarz inequality,
the Parseval equality and part (i),

we have

\begin{align}
\sum_{m \not= n}^{\infty}\left|\frac{\langle e_m,\psi\rangle}{E_n-E_m}\right| 
&\le \left(\sum_{m \not= n}^{\infty} \left|\langle e_m,\psi \rangle\right|^2\right)^{\frac{1}{2}}
\left(\sum_{m \not= n}^{\infty} \left|\frac{1}{E_n-E_m}\right|^2\right)^{\frac{1}{2}} \nonumber\\
&\le \|\psi\| \left(\sum_{m\not=n}^{\infty}\left|\frac{1}{E_n-E_m}\right|^2\right)^{\frac{1}{2}} <\infty.
\label{est1}
\end{align}
\ \hfill $\Box$



\medskip


By Lemma \ref{lem2-1}-(ii), 
one can define a linear operator $T$ on $\MSH$  as follows:
 \begin{eqnarray}
 D(T) &:=& \left\{\psi\in \MSH \big| \sum_{n=1}^{\infty} \left| \sum_{m \not= n} \frac{\langle e_m,\psi\rangle}{E_n-E_m}\right|^2 < \infty \right\}, \label{T1}\\
 T\psi&:=& i\sum_{n=1}^{\infty} \left(\sum_{m \not= n}^{\infty} \frac{\langle e_m,\psi \rangle}{E_n-E_m}
\right) e_n, \qquad \psi \in D(T).\label{T2}
\end{eqnarray}
Note that
\begin{equation}
\|T\psi\|^2=
\sum_{n=1}^{\infty} \left| \sum_{m \not= n}^{\infty} 
\frac{\langle e_m,\psi\rangle}{E_n-E_m}\right|^2.
\label{Tx-norm}
\end{equation}

For a subset $\MSD\subset \MSH$, we denote by
${\rm l.i.h.}(\MSD)$ the subspace algebraically spanned by
the vectors of $\MSD$.

The subspace
\begin{equation}
{\MSD}_0:={\rm l.i.h.}(\{e_n\}_{n=1}^{\infty})
\end{equation}
is dense in $\MSH$. 


\begin{lem}\label{lem2-2} The operator $T$ is densely defined with $\MSD_0 \subset D(T)$ and
\begin{equation}
Te_k=i\sum_{n\not=k}^{\infty}\frac 1{E_n-E_k} e_n, \quad k\in \BBN.\label{Tek}
\end{equation}
\end{lem}


{\it Proof}. To prove $\MSD_0 \subset D(T)$, 
it is sufficient to show that $e_k\in D(T), k\in \BBN$.
Putting
$$
c_n(k):=\sum_{m\not=n}^{\infty}\frac{\lang e_m,e_k\rang}{E_n-E_m},
$$
we have $c_k(k)=0$ and $c_n(k)=1/(E_n-E_k)$ for $n\not=k$.
Hence, by Lemma \ref{lem2-1}-(i), we have
$$
\sum_{n=1}^{\infty}|c_n(k)|^2=\sum_{n\not=k}^{\infty}\frac 1{(E_n-E_k)^2} <\infty.
$$
Hence $e_k\in D(T)$ and (\ref{Tek}) holds.
\hfill $\Box$

\medskip

In general, it is not  clear whether or not
$T$ is a symmetric operator.
But a restriction of $T$ to a smaller subspace gives  a symmetric operator. Indeed, 
we have the following fact:

\begin{lem}\label{lem-sym}{\rm (\cite{Ga})} The operator 
\begin{equation}
T_1:=T|\MSD_0\label{GT}
\end{equation}
 (the restriction of $T$ to $\MSD_0$)  
is symmetric.
\end{lem}

{\it Proof}. It is enough to show that, for all $\psi\in \MSD_0$,
$\lang \psi, T\psi\rang$ is real. For a complex number $z\in \BBC$ (the set of complex numbers), we denote
its complex conjugate by $z^*$.
We have
$$
\langle \psi,T\psi \rangle = i\sum_{n=1}^{\infty}\lang \psi,e_n\rang \sum_{m\not=n}^{\infty}\frac{\langle e_m,\psi \rangle}{E_n-E_m}.
$$
Hence 
$$
\lang \psi,T\psi\rang^*
=i\sum_{n=1}^{\infty}\lang e_n,\psi\rang \sum_{m\not=n}^{\infty}\frac{\langle \psi,e_m \rangle}{E_m-E_n}.
$$
Since $\psi$ is in $\MSD_0$, the double sum on $m,n$ with $m\not=n$ is a sum consisting of  a finite term. Hence we can exchange the sum on $n$ and that on $m$ to obtain
$$
 \lang \psi,T\psi\rang^*
=i\sum_{m=1}^{\infty}\langle \psi,e_m \rangle\sum_{n\not=m}^{\infty} \frac{\lang e_n,\psi\rang}{E_m-E_n}=\lang \psi,T\psi\rang.
$$
Hence $\lang \psi,T\psi\rang$ is real.
\hfill $\Box$

\medskip


The operator $T_1$ defined by (\ref{GT})
is the time operator introduced  by Galapon in \cite{Ga}.
Obviously we have
\begin{equation}
T_1\subset T.
\end{equation}
Hence
\begin{equation}
T^*\subset T_1^*.
\end{equation}

\begin{rem}{\rm It is asserted in \cite{Ga} that $T_1$ is essentially self-adjoint 
without additional conditions. But, unfortunately,
we find that this is not conclusive,
because the proof of it given in \cite{Ga} (pp.2678--2679) has some gap: 
the interchange of the double sum in Equation (2.30) on p.2678
in \cite{Ga} may not be justified, at least,  by the reasoning given there.
The assertion  is  true in the case where $T_1$ becomes a  bounded operator
under an additional condition for $\{E_n\}_{n=1}^{\infty}$, as we  show below in the present paper.
But, in  the case where $T_1$ is unbounded, it seems to be very difficult to prove
or disprove the essential self-adjointness of  $T_1$. We leave this problem for future study.
}
\end{rem}

\begin{lem}\label{lem1-2} The subspace 
\begin{equation}
\MSD_{\rm c}:={\rm l.i.h.}(\{e_n-e_m \in \MSH|n,m\geq 1\}).
\end{equation} 
is dense in the Hilbert space $\MSH$. Moreover
\begin{equation}
\MSD_{\rm c}\subset \MSD_0 \subset D(T). \label{dom1}
\end{equation}
\end{lem}



{\it Proof}. Let $\psi\in \MSD_{\rm c}^{\perp}$ (the orthogonal complement of $\MSD_{\rm c}$).
Then, for all $m,n\geq 1$, 
$|\langle e_n,\psi \rangle|^2 = |\langle e_m,\psi \rangle|^2$.
By the Parseval equality, $\|\psi\|^2=\lim_{N\to \infty} N
|\langle e_m,\psi\rangle|^2$�DThis implies that $|\langle e_m,\psi\rangle|^2=0$ for all $m\geq 1$ and
$\|\psi\|=0$. Hence $\psi=0$. Thus $\MSD_{\rm c}$ is dense in $\MSH$.
Inclusion relation (\ref{dom1}) is obvious.
\hfill $\Box$


\medskip



\begin{The}\label{The1-1}{\rm (\cite{Ga})} It holds that

\begin{equation}
\MSD_{\rm c} \subset D(T_1H) \cap D(HT_1)
\end{equation}
and
\begin{equation}
[T_1,H]\psi = i\psi, \qquad \psi \in \MSD_{\rm c}.
\end{equation}
\end{The}







Theorem \ref{The1-1} shows that $T_1$ is a time operator of $H$.


\begin{rem}{\rm It is easy to see that, for all $k\in \BBN$, $T_1e_k\not\in D(H)$.
Hence $\MSD_0\not\subset D(HT_1)$. Therefore one can not consider
the commutation relation $[T_1,H]$ on $\MSD_0$. Moreover,
by direct computation, we have
\begin{equation}
\lang T_1e_k,He_{\ell}\rang-\lang He_k,T_1e_{\ell}\rang=
-i(1-\delta_{k\ell}), \quad k,\ell\in \BBN.
\end{equation}
This means that
$(T_1,H)$ does not satisfy  the CCR in the  sense of sesquilinear form on $\MSD_0$
(a weak form of the CCR), either.
These facts suggest that the pair $(T_1, H)$ is  very sensitive
to the domain on which their commutation relation is applied.
}
\end{rem}


In concluding this section we discuss shortly non-uniqueness of time operators of $H$.
We introduce a set of symmetric operators associated with $H$:
\begin{eqnarray}
\{H\}_{\MSD_c}'&:=&\{S|\mbox{$S$ is a symmetric operator on $\MSH$ such that }\nonumber\\
&& \qquad \mbox{$\MSD_c\subset D(SH)\cap D(HS)$ and
$SH\psi=HS\psi,\forall \psi\in \MSD_c$}\},\label{Hcom}
\end{eqnarray}
which may be viewed as a commutant of $\{H\}$ in a restricted sense.
It is easy to see that, for all real-valued continuous  function $f$ on $\BBR$,
the operator $f(H)$ defined via the functional calculus  is in $\{H\}_{\MSD_c}'$.

\begin{prop}\label{class} For all $S\in \{H\}_{\MSD_c}'$,
$\MSD_c\subset D((T_1+S)H)\cap D(H(T_1+S))$ and
\begin{equation}
[T_1+S,H]\psi=i\psi, \quad \psi\in \MSD_c.
\end{equation}
\end{prop}

{\it Proof}. A direct computation using Theorem \ref{The1-1} and (\ref{Hcom}).
\hfill $\Box$


\medskip

\begin{prop} Let $T_2$ be a time operator of $H$ such that
$\MSD_{\rm c}\subset D(T_2H)\cap D(HT_2)$ and
$$
[T_2,H]\psi=i\psi, \quad \forall \psi\in \MSD_{\rm c}.
$$
Then $T_2=T_1+S$ with some $S\in \{H\}_{\MSD_c}'$.
\end{prop}

{\it Proof}. We need only to show that $S:=T_2-T_1$ is in $\{H\}_{\MSD_c}'$.
But this is obvious. 
\hfill $\Box$




\medskip



\section{General Properties}


\subsection{Closedness of $T$ and symmetricity of $T^*$}


\begin{lem}\label{lem-T*} $\MSD_0\subset D(T^*)$ and
$T^*|\MSD_0=T_1$, i.e., $T_1\subset T^*$.
\end{lem}

{\it Proof}. It is enough to show that, for all $k\in \BBN$, $e_k\in D(T^*)$ and
$T^*e_k=Te_k(=T_1e_k)$.
It is easy to see that, for all $\psi\in D(T)$,
\begin{equation}
\lang e_k,T\psi\rang=i\sum_{m\not=k}^{\infty}\frac {\lang e_m,\psi\rang}{E_k-E_m}.\label{ekT}
\end{equation}
By Lemma \ref{lem2-2}, the right hand side is equal to
$\lang Te_k, \psi\rang$. Hence $e_k\in D(T^*)$ and $T^*e_k=Te_k$.
\hfill $\Box$

\medskip



\begin{prop}\label{prop-T} The operator $T$ is closed and
\begin{equation}
T^*\subset T.\label{T*T}
\end{equation}
In particular, if $T$ is bounded, then $T$ is self-adjoint with $D(T)=\MSH$. 
\end{prop}

{\it Proof}. Let $\psi_k\in D(T),\, k \in \BBN$ and $\psi_k\to \psi\in \MSH, T\psi_k \to \phi\in \MSH$
 as $k\to\infty$.
Then $\sup_{k\geq 1}\|T\psi_k\|<\infty$. Hence,  by (\ref{Tx-norm}), there exists a constant $C>0$ independent of $k\in \BBN$ such that
$$
\sum_{n=1}^{\infty}\left|\sum_{n\not=m}\frac{\lang e_m,\psi_k\rang}{E_n-E_m}\right|^2\leq C.
$$
By (\ref{est1}), we have
\begin{equation}
\lim_{k\to\infty}\sum_{n\not=m}^{\infty}\frac{\lang e_m,\psi_k\rang}{E_n-E_m}=
\sum_{n\not=m}^{\infty}\frac{\lang e_m,\psi\rang}{E_n-E_m}.\label{lim}
\end{equation} 
Hence it follows that
$$
\sum_{n=1}^{\infty}\left|\sum_{n\not=m}^{\infty}\frac{\lang e_m,\psi\rang}{E_n-E_m}\right|^2\leq C.
$$
Therefore $\psi\in D(T)$. By (\ref{ekT}) and (\ref{lim}), we have for all $\ell\in \BBN$
$$
\lim_{k\to\infty}\lang e_{\ell}, T\psi_k\rang=\lang e_{\ell}, T\psi\rang.
$$
Hence $\lang e_{\ell},\phi\rang=\lang e_{\ell}, T\psi\rang, \, \ell\in \BBN$, implying
$\phi=T\psi$. Thus $T$ is closed.

To prove (\ref{T*T}), let $\psi\in D(T^*)$. Putting $\eta=T^*\psi$, we have
$\lang \eta,\chi\rang=\lang \psi, T\chi\rang$ for all $\chi\in D(T)$.
Taking $\chi=e_k$ ($k\in \BBN$), we have
\begin{equation}
\lang \eta, e_k\rang=i\sum_{n\not=k}^{\infty}\frac{\lang \psi,e_n\rang}{E_n-E_k},\label{eta}
\end{equation}
which implies that
$$
\sum_{k=1}^{\infty}\left|\sum_{n\not=k}^{\infty}\frac{\lang \psi,e_n\rang}{E_n-E_k}\right|^2
=\|\eta\|^2 <\infty.
$$
Hence $\psi\in D(T)$. 
Then, by (\ref{ekT}), the right hand side of (\ref{eta}) is equal to $\lang T\psi,e_k\rang$.
Hence  $\eta= T\psi$. Thus (\ref{T*T}) holds.

Let $T$ be bounded. Then, by the denseness of $D(T)$ and the closedness of $T$, $D(T)=\MSH$. Hence $D(T^*)=\MSH$. Thus, by (\ref{T*T}), $T^*=T$, i.e., $T$ is self-adjoint.
\hfill $\Box$

\medskip


\begin{cor}\label{Sym} The operator $T^*$ is symmetric.
\end{cor}

{\it Proof}. By Lemma \ref{lem-T*}, $T^*$ is densely defined. 
Hence, by Proposition \ref{prop-T}, $T^*\subset T=(T^*)^*$.
Thus $T^*$ is symmetric.  
\hfill $\Box$


\medskip

Thus we have
\begin{equation}
T_1\subset T^*\subset T.\label{incl}
\end{equation}

Corollary \ref{Sym} shows that $T^*$ also is a time operator of $H$.

For a closable operator $A$ on a Hilbert space, we denote its closure by
$\bar A$.

\begin{prop} $\bar T_1=T^*$.
\end{prop}

{\it Proof}. Note that $\bar{T_1}=T^*$ if and only if ${T_1}^*=T$.
By (\ref{incl}), we have $\bar{T_1} \subset T^*$. Hence
$T \subset {T_1}^*$.
Thus it is enough to show that $D({T_1}^*) \subset D(T)$.
For all $\psi \in D({T_1}^*)$, we have
$$
\lang{T_1}^*\psi,e_l\rang=\lang\psi,{T_1}e_l\rang=i\sum_{n\not=l}
\frac{\lang\psi,e_n\rang}{E_n-E_l}.$$
Hence we obtain
$$\infty>\|{T_1}^*\psi\|^2=\sum_{l=1}^{\infty}|\lang{T_1}^*\psi,e_l\rang|^2
=\sum_{l=1}^{\infty}\left|\sum_{n\not=l}
\frac{\lang\psi,e_n\rang}{E_n-E_l}\right|^2,$$
implying that $\psi \in D(T)$. Thus $D({T_1}^*) \subset D(T)$.\hfill $\Box$

\medskip








\subsection{Absence of invariant dense domains for $T$}

We first note the following general fact:
\begin{prop}\label{1-3} Let $Q$ be a bounded self-adjoint operator on $\MSH$ and $P$ be a self-adjoint operator on $\MSH$. Suppose that
there is a dense subspace $\MSD$ in $\MSH$ such that 
the following {\rm (i)--(iii)} hold:
\begin{list}{}{}
\item[{\rm (i)}] $Q\MSD\subset \MSD \subset D(P)$. 
\item[{\rm (ii)}]  $\MSD$ is a core of $P$.
\item[{\rm (iii)}]  The pair $(Q,P)$ obeys the CCR on $\MSD$ :
$[Q,P]\psi=i\psi, \forall \psi \in \MSD$.
\end{list}
Then $\sigma(P)=\BBR$.
\end{prop}
{\it Proof}. Since $Q$ is a bounded self-adjoint operator, we have for all $t\in \BBR$
$$
e^{itQ}=\sum_{k=0}^{\infty}\frac{(itQ)^k}{k!}
$$
in operator norm.
Conditions (i) and  (iii) imply that, for all $k\in \BBN$ and $\psi\in \MSD$
$$
Q^kP\psi-PQ^k\psi=ikQ^{k-1}\psi.
$$
Hence, for all $t\in \BBR$ and vectors $\psi$ in $\MSD$, we have 
\begin{eqnarray*}
e^{itQ}P\psi&=&P\psi+\sum_{k=1}^{\infty}\frac{(it)^k}{k!}Q^kP\psi \\
&=& P\psi+\sum_{k=1}^{\infty}\frac{(it)^k}{k!}(PQ^k+ikQ^{k-1})\psi \\
&=&P\psi+\sum_{k=1}^{\infty}\left\{P\frac{(itQ)^k}{k!}\psi-t\frac{(itQ)^{k-1}}{(k-1)!}\psi\right\}.
\end{eqnarray*}
It follows from  the  closedness of $P$ that $e^{itQ}\psi$ is in $D(P)$ and 
\begin{equation}
Pe^{itQ}\psi=e^{itQ}P\psi+te^{itQ}\psi.
\end{equation}
By condition (ii), this equality extends to all $\psi \in D(P)$ with
$e^{itQ}\psi \in D(P), \forall t\in \BBR,\forall \psi\in D(P)$.
Hence  the operator equality
$e^{-itQ}Pe^{itQ}=P+t$ follows. Thus $\sigma(P)=\sigma(P+t)$ for all $t\in \BBR$.
This implies that $\sigma(P)=\BBR$. 
\hfill $\Box$



\medskip




\begin{The} If $T$ is bounded (hence self-adjoint by Proposition  \ref{prop-T}),  then there is no dense subspace $\MSD$ in $\MSH$ such that 
the following {\rm (i)--(iii)} hold:
\begin{list}{}{}
\item[{\rm (i)}] $T\MSD\subset \MSD \subset D(H)$. 
\item[{\rm (ii)}]  $\MSD$ is a core of $H$.
\item[{\rm (iii)}]  The pair $(T,H)$ obeys the CCR on $\MSD$,
\end{list}
\end{The}


{\it Proof}. Suppose that there were such a dense subspace $\MSD$ as stated above.
Then  we can apply Proposition \ref{1-3}
with $(Q,P)=(T,H)$ to conclude that $\sigma(H)=\BBR$. But this is a contradiction.
\hfill $\Box$

\medskip

\begin{rem}{\rm A special case of this theorem was established in \cite[Theorem 9.5]{DD}.}
\end{rem}


\subsection{Reflection symmetry of the spectrum of $T_1, T^*$ and $T$}

We first recall the definition of the spectrum of a general linear operator
(not necessarily closed).
For a linear operator $A$ on a Hilbert space $\MSK$,
the resolvent set of $A$, denoted  $\rho(A)$, is defined by
\begin{eqnarray*}
\rho(A)&:=&\{z\in \BBC| \mbox{$\Ran(A-z)$ (the range of $A-z$) is dense in
$\MSK$}\\
&& \qquad \mbox{and $A-z$ is injective with $(A-z)^{-1}$ bounded}\}.
\end{eqnarray*}
Then the set
$$
\sigma(A):=\BBC\setminus \rho(A)
$$
is called the spectrum of $A$.

We denote by $T^{\#}$ any of $T_1,T^*$ and $T$.

\begin{prop}\label{1-4}
The spectrum $\sigma(T^{\#})$ of $T^{\#}$ is reflection symmetric with respect to  the imaginary axis, i.e.,
if $z \in \sigma(T^{\#})$, then $-z^* \in \sigma(T^{\#})$.
In particular, if $T$ is self-adjoint, then 
$\sigma(T)$ is reflection symmetric with respect to the origin of the real axis.
\end{prop}



{\it Proof}. We define a conjugation $J$ on $\MSH$ by
\begin{equation}
J\psi:=\sum_{n=1}^{\infty}\langle \psi,e_n \rangle e_n, \quad \psi\in \MSH. \label{J}
\end{equation}
It is easy to see that operator equality $JT^{\#}J=-T^{\#}$ holds ($JD(T^{\#})=D(T^{\#})$). 
Hence, for all $z\in \BBC$, we have $J(T^{\#}-z)J=-(T^{\#}+z^*)$. This implies that,
if $z \in \rho(T^{\#})$, then $-z^*\in \rho(T^{\#})$.
Thus the same holds for the spectrum $\sigma(T^{\#})=\BBC\setminus \rho(T^{\#})$.
\hfill $\Box$



\medskip








\section{Boundedness of $T$}

In this section we present a general criterion for the operator $T$ to be bounded.
For mathematical generality and for later use, we consider a  more general class of
operators than that of $T$. 
Let $b:=\{b_{nm}\}_{n,m=1}^{\infty}$ be a double sequence 
of complex numbers such that
\begin{equation}
\|b\|_{\infty}:=\sup_{n,m\geq 1}|b_{nm}|<\infty.
\end{equation}
Then, in the same way as in Lemma \ref{lem2-1}-(ii), for all $\psi \in \MSH$,
the infinite series
$$
\sum_{m\not=n}^{\infty}\frac {b_{nm}}{E_n-E_m} \lang e_m,\psi\rang
$$
absolutely converges. Hence one can
define a linear operator $T_b$ on $\MSH$ as follows:
\begin{eqnarray}
&& D(T_b):=\left\{\psi \in \MSH\big|
\sum_{n=1}^{\infty}\left|\sum_{m\not=n}^{\infty}\frac {b_{nm}}{E_n-E_m}\lang e_m,\psi\rang\right|^2<\infty
\right\}, \\
&& T_b\psi:=i\sum_{n=1}^{\infty}\left(\sum_{m\not=n}^{\infty}\frac {b_{nm}}{E_n-E_m}\lang e_m,\psi
\rang\right)e_n, \quad \psi \in D(T_b).
\end{eqnarray}

Obviously $T=T_b$ with $b$ satisfying $b_{nm}=1$ for all $n,m\in \BBN$.
In the same way as in the  case of $T$, one can prove the following fact:

\begin{lem}\label{Tb-cl} The operator $T_b$ is closed.
\end{lem}


The following lemma  is probably well known (but, for the completeness, we give a proof):

\begin{lem}\label{B-form} Let $A$ be a densely defined linear operator on a  Hilbert space $\MSK$.
Suppose that there exist a dense subspace $\MSD$ in $\MSK$ and a constant $C>0$  such that
$\MSD\subset D(A)$ and
$$
|\lang \psi, A\psi\rang|\leq C\|\psi\|^2, \quad \psi\in \MSD.
$$
Then $A$ is bounded with $\|\bar A\|\leq 2C$, where $\bar A$ is the closure of $A$.

If $A$ is symmetric in addition, then $\|\bar A\|\leq C$.
\end{lem}

{\it Proof}. Let $\psi,\phi\in \MSD$. Then, by the polarization identity
\begin{eqnarray*}
\lang \psi,A\phi\rang&=&\frac {1}{4}
(\lang \psi+\phi, A(\psi+\phi)\rang -\lang \psi-\phi,A(\psi-\phi)\rang \\
&& +i\lang \psi-i\phi,A(\psi-i\phi)\rang-i\lang \psi+i\phi,A(\psi+i\phi)\rang),
\end{eqnarray*}
we have 
$$
|\lang \psi,A\phi\rang| \leq \frac {C}{4}
(\|\psi+\phi\|^2+\|\psi-\phi\|^2+\|\psi-i\phi\|^2+\|\psi+i\phi\|^2) =C(\|\psi\|^2+\|\phi\|^2).
$$
Replacing $\psi\not=0$ by $\|\phi\|\psi/\|\psi\|$  we have
$$
|\lang \psi,A\phi\rang| \leq 2C\|\psi\|\|\phi\|.
$$ 
For $\psi=0$, this inequality trivially holds. Since $\MSD$ is dense, it follows from the
Riesz representation theorem that $\|A\phi\|\leq 2C\|\phi\|, \, \phi\in \MSD$.
Thus the first half of the lemma  follows.

Let $A$ be symmetric. Then, $\lang \psi,A\psi\rang \in \BBR$ for all $\psi\in D(A)$.
Hecne
$$
|\Re \lang \psi,A\phi\rang|
= \frac {1}{4}
\left|\lang \psi+\phi, A(\psi+\phi)\rang -\lang \psi-\phi,A(\psi-\phi)\rang
\right| \leq \frac {C}{2} (\|\psi\|^2+\|\phi\|^2), \quad \psi\in \MSD.
$$
We write $\lang \psi,A\phi\rang=|\lang \psi,A\phi\rang|e^{i\theta}$ with $\theta\in \BBR$.
Then $|\lang \psi,A\phi\rang|=\lang e^{i\theta}\psi, A\phi\rang$.
Hence 
\begin{eqnarray*}
|\lang \psi,A\phi\rang|&=&\Re \lang e^{i\theta}\psi,A\phi\rang
\leq \frac {C}{2}(\|e^{i\theta}\psi\|^2+\|\phi\|^2) \\
&=&\frac {C}{2}(\|\psi\|^2+\|\phi\|^2).
\end{eqnarray*}
Thus, in the same manner as above, we can obtain $|\lang \psi,A\phi\rang| \leq C\|\psi\| \|\phi\|,
\ \psi,\phi\in \MSD$.
\hfill $\Box$

\medskip



\begin{lem}\label{lem-sum0} For all  $s>1$ and $n\geq 2$,
\begin{equation}
\sum_{m=1}^{n-1}\frac 1{n^s-m^s}\leq \frac{\log n}{n^{s-1}}+\frac{1}{s(n-1)^{s-1}}.\label{cs}
\end{equation}
\end{lem}

{\it Proof}. It easy to see that
\begin{eqnarray}
\sum_{m=1}^{n-1}\frac 1{n^s-m^s} &\leq& \int_1^{n-1}\frac 1{n^s-x^s}dx+\frac 1{n^s-(n-1)^s}. 
\label{diff}
\end{eqnarray}
By the change of variable $x=ny$, we have
\begin{eqnarray}
\int_1^{n-1}\frac 1{n^s-x^s}dx&=&\frac 1{n^{s-1}}\int_{1/n}^{(n-1)/n}
\frac 1{1-y^s}dy \\
&\leq& \frac 1{n^{s-1}}\int_0^{(n-1)/n}\frac 1{1-y}dy \\
&\leq & \frac {\log n}{n^{s-1}}. \label{int1}
\end{eqnarray}
We have
$$
\frac 1{n^s-(n-1)^s}=\frac 1{(n-1)^s} \cdot \frac 1{ \left(\frac n{n-1}\right)^s-1}.
$$
By the well known inequality
\begin{equation}
x^s-1\geq s(x-1), \quad x>0, s\geq 1 \label{xs},
\end{equation}
 we obtain 
\begin{equation}
\frac 1{n^s-(n-1)^s}
\leq \frac 1{s(n-1)^{s-1}}.\label{ns-ineq}
\end{equation}
Thus (\ref{cs}) holds.\hfill $\Box$

\medskip

\begin{lem}\label{lem-sum} Let $s>1$. Then
\begin{equation}
\sup_{n\geq 2}\left(\sum_{m=1}^{n-1}\frac 1{n^s-m^s}\right) <\infty \label{cs1}
\end{equation}
and
 \begin{equation}
\sup_{m\geq 1}\left(\sum_{n=m+1}^{\infty}\frac 1{n^s-m^s}\right) <\infty. \label{ds}
\end{equation}
\end{lem}

{\it Proof}. Property (\ref{cs1}) follows from
 Lemma \ref{lem-sum0}.

To prove (\ref{ds}),  we write
$$
\sum_{n=m+1}^{\infty}
\frac 1{n^s-m^s}
\leq \int_{m+1}^{\infty}\frac 1{x^s-m^s}dx +\frac 1{(m+1)^s-m^s}.
$$
We fix  a constant $R > 2 (\geq (m+1)/m)$.  
By the change of variable $x=my$, we have
\begin{eqnarray*}
\int_{m+1}^{\infty}\frac 1{x^s-m^s}dx 
&=&\frac 1{m^{s-1}}\left(
\int_{(m+1)/m}^R
\frac 1{y^s-1}ds +C_R\right),
\end{eqnarray*}
where
$$
C_R:=\int_R^{\infty}\frac 1{y^s-1}ds <\infty.
$$
Using (\ref{xs}) we have
\begin{eqnarray*}
\int_{(m+1)/m}^R\frac 1{y^s-1}dy
 &\leq&  
\int_{(m+1)/m}^R\frac 1{s(y-1)}dy \\
&=& \frac 1{s} \left(\log (R-1)+\log m\right).
\end{eqnarray*}
Hence
$$
\int_{m+1}^{\infty}\frac 1{x^s-m^s}dx 
\leq \frac {\log m}{sm^{s-1}}
+\frac {\log (R-1)}{sm^{s-1}}
+\frac 1{m^{s-1}}C_R.
$$
Thus (\ref{ds}) follows. \hfill $\Box$

\medskip


Let
\begin{eqnarray}
&&c_H(n):=\sum_{m=1}^{n-1}\frac{E_n}{(E_n-E_m)E_m}, \quad n\geq 2,\\
&&d_H(m):=\sum_{n=m+1}^{\infty}\frac{E_m}{(E_n-E_m)E_n}, \quad m\geq 1.
\end{eqnarray}
Since $c_H(n)$ and $d_H(m)$ are positive (recall that $E_n>0, \forall n\in \BBN$), 
one can define constants
\begin{eqnarray}
c_H&:=& \sup_{n\geq 2}c_H(n), \\
d_H&:=&\sup_{m\geq 1}d_H(m),
\end{eqnarray}
which are finite or  infinite.

\begin{The}\label{B-thm} Suppose that there exist constants $\alpha>1$, $C>0$ and $a>0$ such that
\begin{equation}
E_n-E_m \geq C(n^{\alpha}-m^{\alpha}), \quad n>m > a.\label{E-bound}
\end{equation}
Then $T_b$ is a bounded operator with $D(T_b)=\MSH$ and
\begin{equation}
\|T_b\|\leq 4\|b\|_{\infty} \sqrt{c_Hd_H}.\label{Tb1}
\end{equation}
Moreover, if $b_{nm}^*=b_{mn}$ for all $m,n\in \BBN$, then
$T_b$ is a bounded self-adjoint operator with $D(T_b)=\MSH$ and
\begin{equation}
\|T_b\|\leq 2\|b\|_{\infty} \sqrt{c_Hd_H}.\label{Tb2}
\end{equation}
In particular, $T$ is a bounded self-adjoint operator with $D(T)=\MSH$.
\end{The}


{\it Proof}. By Lemma \ref{B-form}, it is enough to show that $c_H$ and $d_H$ are finite and 
\begin{equation}
|\lang \psi, T_b\psi\rang |\leq 2\|b\|_{\infty}
\sqrt{c_Hd_H}\|\psi\|^2, \quad \psi\in \MSD_0.\label{bound}
\end{equation}
Then $T_b$ is bounded with (\ref{Tb1}). Since  $T_b$ is densely defined and closed, it follows that $D(T_b)=\MSH$.
As in the case of $T$, one can show that, if
$b_{nm}^*=b_{mn}$ for all $m,n\in \BBN$, then $T_b|\MSD_0$ is symmetric and
hence $T_b$ is a bouned self-adjoint operator with $D(T_b)=\MSH$ and (\ref{Tb2}) holds. Therefore the desired result follows.

To prove (\ref{bound}),
we first note that, for $\psi \in \MSD_0$,
$$
\lang \psi,T_b\psi\rang=i\sum_{m,n=1, m\not=n}^{\infty}
\frac {b_{nm}}{E_n-E_m}\lang \psi,e_n\rang \lang e_m,\psi\rang.
$$
Hence
$$
|\lang \psi,T_b\psi\rang|\leq 2\|b\|_{\infty}A(\psi),
$$
where
\begin{eqnarray*}
&&A(\psi):=\sum_{n>m\geq 1}\frac {|\lang e_m,\psi\rang| |\lang \psi,e_n\rang|}{E_n-E_m}.
\end{eqnarray*}
Inserting $1=\sqrt{E_m/E_n}\cdot \sqrt{E_n/E_m}$ into the summand on the right hand side and
using the Cauchy-Schwarz inequality,  we have
$$
A(\psi)^2 \leq B(\psi)C(\psi)
$$
with
\begin{eqnarray*}
&&B(\psi)=\sum_{n>m\geq 1}\frac{|\lang e_n,\psi\rang|^2}{E_n-E_m}\cdot\frac{E_n}{E_m}, \\
&&C(\psi)=\sum_{n>m\geq 1}\frac{|\lang e_m,\psi\rang|^2}{E_n-E_m}\cdot\frac{E_m}{E_n}.
\end{eqnarray*}
One can rewrite  and estimate $B(\psi)$  as follows:
\begin{eqnarray*}
B(\psi)&=& \sum_{n=2}^{\infty}|\lang e_n,\psi\rang|^2
c_H(n) \\
&\leq & \|\psi\|^2c_H.
\end{eqnarray*}
Similarly we have
\begin{equation}
C(\psi)\leq \|\psi\|^2d_H.
\end{equation}
Hence
\begin{equation}
|\lang \psi,T_b\psi\rang|\leq 2\|b\|_{\infty}\sqrt{c_Hd_H} \|\psi\|^2.
\end{equation}
Therefore we need only to 
prove that $c_H$ and $d_H$ are finite.

We can write
$$
c_H(n)=\sum_{m=1}^{n-1}\frac 1{E_n-E_m} +\sum_{m=1}^{n-1}\frac 1{E_m}.
$$
By  assumption (\ref{E-bound}), we have
\begin{equation}
\frac 1{E_n-E_m} \leq \frac 1{C(n^{\alpha}-m^{\alpha})}, \quad n >m > a.
\end{equation}
Since we have
$$
\sum_{n=1}^{\infty}\frac 1{n^{\alpha}}<\infty,
$$
it follows that
$$
\varepsilon_1:=\sum_{m=1}^{\infty}\frac 1{E_m}<\infty.
$$
Thus
$$
c_H(n) \leq \sum_{m=1}^{n-1}\frac 1{E_n-E_m} +\varepsilon_1.
$$
Let $n_0\geq 2$ be a natural number such that $n_0> a$.
Then, for all $n>n_0$
$$
\sum_{m=1}^{n-1}\frac 1{E_n-E_m}
\leq \sum_{m=1}^{n_0-1} \frac 1{E_n-E_m}+\frac 1{C}\sum_{m=n_0}^{n-1}\frac {1}{n^{\alpha}-m^{\alpha}}.
$$
By (\ref{cs}), the right hand side is uniformly bounded in $n$.
Thus we have $c_H<\infty$.

To prove $d_H<\infty$, we write for $m > a$
\begin{eqnarray*}
d_H(m)&=&\sum_{n=m+1}^{\infty}\frac{1}{(E_n-E_m)} -\sum_{n=m+1}^{\infty} \frac 1{E_n} \\
&\leq& \sum_{n=m+1}^{\infty}\frac{1}{(E_n-E_m)} \\
&\leq & \frac 1{C}\sum_{n=m+1}^{\infty}\frac 1{n^{\alpha}-m^{\alpha}}.
\end{eqnarray*}
Hence, by (\ref{ds}) in Lemma \ref{lem-sum}, we have
$$
\sup_{m> a}d_H(m)<\infty.
$$
Thus it follows that $d_H<\infty$.
\hfill $\Box$


\medskip


\medskip




\begin{ex}\label{ex-B1}{\rm Let $\lambda >0,\alpha >1$ and $P(x)$ be a real polynomial of $x\in \BBR$
with degree $p <\alpha$. Then it is easy to see that
the sequense $\{E_n\}_{n=1}^{\infty}$ defined by
$$
E_n:=\lambda n^{\alpha}+P(n)
$$
satisfies the assumptions (H.1), (H.2) and (\ref{E-bound}).
Thus, by Theorem \ref{B-thm}, in the present example,
$T_b$ is bounded.
}
\end{ex}




We remark that Theorem \ref{B-thm} does not cover the case $E_n=\lambda n+\mu$ with constants $\lambda >0$ and $\mu \in \BBR$.
For this case, we have the following theorem:

\begin{The}\label{GB2} 
Suppose that there exist constants $\lambda >0, \mu\in \BBR$ and $a >0$ such that
\begin{equation}
E_n=\lambda n+\mu, \quad n >a.
\end{equation}
Then $T$ is a bounded self-adjoint operator with $D(T)=\MSH$.
\end{The}


{\it Proof}. Let $k_0$ be the greatest integer such that $k_0\leq a$.
Let $a_n:=\lang e_n,\psi\rang$ ($\psi\in \MSH$). Then, by the Parseval equality, we have
$\sum_{n=1}^{\infty}|a_n|^2=\|\psi\|^2$. 
Let $\psi\in \MSD_0$. Then we can write:
$$
\lang \psi, T\psi\rang =S_I+S_2+S_3+S_4,
$$
where
\begin{eqnarray*}
S_1&:=i&\sum_{n=1}^{k_0}\sum_{m\not=n}^{k_0}\frac{a_n^*a_m}{E_n-E_m},\\
S_2&:=i&\sum_{n=1}^{k_0}\sum_{m\geq k_0+1}^{\infty}\frac{a_n^*a_m}{E_n-E_m},\\
S_3&:=i&\sum_{n\geq k_0+1}^{\infty}\sum_{m=1}^{k_0}\frac{a_n^*a_m}{E_n-E_m},\\
S_4&:=&i\frac 1{\lambda}\sum_{n=k_0+1}^{\infty}\sum_{m\not=n, m\geq k_0+1}^{\infty}
\frac {a_n^*a_m}{n-m}.
\end{eqnarray*}
By the Schwarz inequality, we have
$$
|S_j|\leq C_j\|\psi\|^2,\quad j=1,2,3,
$$ 
where $C_j>0$ is a constant.
To estimate $|S_4|$, we use the following well known inequality \cite[Theorem 294]{HLP}:
$$
\left|\sum_{n,m=1,n\not=m}^{\infty}\frac {x_ny_m}{n-m}\right|
\leq \pi \sqrt{\sum_{n=1}^{\infty}x_n^2}\sqrt{\sum_{m=1}^{\infty}y_m^2}
$$
for all real sequences $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$.
Hence
$$
|S_4|\leq \pi \|\psi\|^2.
$$
Therefore it follows that $|\lang\psi,T\psi\rang| \leq {\rm const.} \|\psi\|^2$.
Thus $T$ is bounded.
\hfill $\Box$

\medskip






\begin{ex}{\rm A physically interesting case  is the case where $E_n=\omega (n+\frac 12), 
\quad n\in \{0\}\cup\BBN$ with a constant $\omega >0$. In this case, by Theorem \ref{GB2}, $T$  is a bounded self-adjoint operator with $D(T)=\MSH$ and
takes the form
$$
T\psi=\frac {i}{\omega}\sum_{n=1}^{\infty}
\left(\sum_{m\not=n}^{\infty}\frac {\lang e_m,\psi\rang}{n-m}\right)e_n, \quad \psi\in\MSH.
$$
Moreover, one can prove that
$$
\sigma(T)=[-\pi/\omega,\pi/\omega]
$$
(\cite[Theorem 2.1]{Ar07-2}).

Let $\hat N:=\omega^{-1}H-1/2$ and $\hat \theta :=\omega T$.
Then it follows that
\begin{eqnarray}
&& \sigma(\hat N)=\{0\}\cup \BBN, \quad \sigma(\hat\theta)=[-\pi,\pi],\label{phase1}\\
&& [\hat\theta, \hat N]\psi=i\psi, \quad \psi\in \MSD_{\rm c}.\label{phase2}
\end{eqnarray}
 
As is well known, in the context of quantum mechanics, 
the sequence $\{\omega(n+\frac 12)\}_{n=1}^{\infty}$
appears  as the spectrum of the one-dimensional
quantum harmonic oscillator Hamiltonian with mass $m>0$
$$
H_{\rm os}:=\frac {p^2}{2m}+\frac 12 m\omega^2 q^2 
$$
in the Schr\"odinger representation $(q,p)$ of the CCR, where 
$p:=-iD$ with $D$ being the generalized partial differential operator on $L^2(\BBR)$
and $q$ is the multiplication operator by the variable $x\in \BBR$.
In this context, the operator $\hat N$ is called the number operator and,
in view of (\ref{phase1}) and (\ref{phase2}), the operator $\hat \theta$
is interpreted as a {\it  phase operator} \cite{DD}.
}
\end{ex}



\section{Unboundedness of $T$}

As for the unboundedness of $T$, we have the following fact:

\begin{The}\label{UB-thm}
If $\{E_n\}_{n=1}^{\infty}$ satisfies
\begin{equation}
\inf_{n \in \mathbb{N}}(E_{n+1}-E_n)=0, \label{BD2}
\end{equation}
then $T$ is unbounded. 
\end{The}



{\it Proof}. By (\ref{BD2}), there exists a subsequence $\{E_{p_k}\}_{k=1}^{\infty}$ of $\{E_p\}_{p=1}^{\infty}$ such that 
\begin{equation}
\lim_{k\to\infty}(E_{p_k+1}-E_{p_k})=0.
\end{equation}
Hence we have
\begin{eqnarray*}
\|Te_{p_k}\|^2 &=& \sum_{n=1}^{\infty} \left| \sum_{m\not=n}^{\infty} \frac{\lang e_m,e_{p_k} \rang}{E_n-E_m}\right|^2
= \sum_{n\not=p_k}^{\infty} \left| \frac{1}{E_n-E_{p_k}} \right|^2
\\
&\geq&  \left| \frac{1}{E_{p_k+1}-E_{p_k}} \right|^2
\to \infty \ (k\to\infty).
\end{eqnarray*}
Thus $T$ is unbounded. 
\hfill $\Box$

\medskip

\begin{ex}{\rm Let 
$$
E_n=\lambda n^{\alpha}+\mu
$$
with constants $\lambda>0$, $\alpha >1/2$ and $\mu \in \BBR$. 
Then $\{E_n\}_{n=1}^{\infty}$ satisfies the assumptions (H.1) and (H.2).
As we have already seen, $T$ is bounded if $\alpha \geq 1$

Let $1/2 <\alpha <1$. Then one easily sees that
$$
\lim_{n\to\infty}(E_{n+1}-E_n)=0.
$$
Hence $\inf_{n\in \BBN}(E_{n+1}-E_n)=0$.
Therefore, in this case, $T$ is unbounded.
Thus $T$ is bounded if and only if $\alpha \geq 1$.
}
\end{ex}




\section{Hilbert-Schmidtness of $T$}

In this section we investigate 
Hilbert-Schmidtness  of the  operator $T$.




\begin{prop}\label{HS}
The operator $T$ is Hilbert-Schmidt if and only if
\begin{equation}
\displaystyle\sum_{n=1}^{\infty}\sum_{m\not=n}^{\infty}\frac{1}{(E_n-E_m)^2}<\infty.\label{SS}
\end{equation} 
In that case, $T$ is  self-adjoint  with
\begin{equation}
\|T\|_2^2=\sum_{n=1}^{\infty}\sum_{m\not=n}^{\infty}\frac{1}{(E_n-E_m)^2}, \label{HSnorm}
\end{equation}
where $\|\cdot\|_2$ denotes Hilbert-Schmidt norm.
In particular, there exist a C.O.N.S. $\{f_n\}_{n=1}^{\infty}$ of $\MSH$ and
real numbers $t_n, \, n\in \BBN$
such that $Tf_n=t_nf_n$ and $t_n \to 0$ ($n\to\infty$).
\end{prop}



{\it Proof}. Suppose that $T$ is Hilbert-Schmidt. Then
$\sum_{n=1}^{\infty}\|Te_n\|^2<\infty$.
On the other hand, we have
\begin{equation}
\sum_{n=1}^{\infty}\|Te_n\|^2
=\sum_{n=1}^{\infty}\sum_{m\not=n}^{\infty} \frac 1{(E_n-E_m)^2} \label{HSnorm1}
\end{equation}
Hence (\ref{SS}) follows with (\ref{HSnorm}).

Conversely, (\ref{SS}) holds.  Hence, by (\ref{HSnorm1}),
$\sum_{n=1}^{\infty}\|Te_n\|^2<\infty$. Therefore $T$ is Hilbert-Schmidt.
The last statement follows from the Hilbert-Schmidt theorem
(e.g., \cite[Theorem VI.16]{RS1}).
\hfill $\Box$



\medskip

\begin{rem}{\rm In  Proposition \ref{HS},
the number $t_n\not=0$ is an eigenvalue of $T$ with a finite multiplicity.
Since $T$ is self-adjoint in the present case, it may be an observable in the
context of  quantum mechanics.  
If this is the case, 
then Proposition \ref{HS} shows that
the observable described by $T$  (\lq\lq{time}" in any sense ?) is quantized (discretized) 
in the quantum system whose Hamiltonian is  $H$ with eigenvalues $\{E_n\}_{n=1}^{\infty}$
satisfying (\ref{SS}). 
}
\end{rem}

The next theorem gives a class of $H$ such that  $T$ is  Hilbert-Schmidt:

\begin{The}\label{HB} Suppose that (\ref{E-bound}) in Theorem \ref{B-thm}
holds with $\alpha>3/2$.
Then $T$ is  Hilbert-Schmidt and self-adjoint.
\end{The}


{\it Proof}. Since $1/(E_n-E_m)^2$ is symmetric in $n$ and $m$,  it is sufficient to show that
$\sum_{n>m\geq 1}1/(E_n-E_m)^2<\infty$.
By the present assumption, we need only to show that
$$
\Sigma:=\sum_{n>m\geq 1}\frac{1}{(n^{\alpha}-m^{\alpha})^2}<\infty
$$
for all $\alpha>3/2$. We have
\begin{eqnarray*}
\Sigma&=&\sum_{n=2}^{\infty}\sum_{m=1}^{n-1}\frac 1{(n^{\alpha}-m^{\alpha})^2} \\
&\leq & \sum_{n=2}^{\infty}\frac 1{n^{\alpha}-(n-1)^{\alpha}}
\cdot \sum_{m=1}^{n-1}\frac 1{(n^{\alpha}-m^{\alpha})}.
\end{eqnarray*}
Using  (\ref{cs}) and (\ref{ns-ineq}), we obtain
$$
\Sigma\leq \sum_{n=2}^{\infty}
\frac 1{\alpha(n-1)^{\alpha-1}}\frac {\log n}{n^{\alpha-1}}
+\sum_{n=2}^{\infty}
\frac 1{\alpha^2(n-1)^{2(\alpha-1)}}.
$$
Each infinite series on the right hand side converges for all $\alpha >3/2$.
Thus the desired result follows.
\hfill $\Box$

\medskip




\section{The Galapon Time Operator  as a Generalized Time Operator} 


It is shown that  every  self-adjoint
operator which has a strong time operator
is absolutely continuous \cite{Miya1}.
Hence the Galapon time operator $T_1$ is not a strong time operator of $H$.
But it may be a generalized time operator of $H$.
In this section we investigate this aspect.


\subsection{An operator-valued function on $\BBR$}

In the same way as in Lemma \ref{lem2-1}-(ii), one can show that,
for all $\psi \in \MSH$, $n\in \BBN$ and all $t\in \BBR$, 
the infinite series
$$
\sum_{m\not=n}^{\infty}\frac {e^{it(E_n-E_m)}-1}{E_n-E_m}\lang e_m,\psi\rang
$$
absolutely converges. Hence,
for each $t\in \BBR$, one can  define a linear operator $K(t)$ as follows:
\begin{eqnarray}
&& D(K(t)):=\left\{\psi \in \MSH
|\sum_{n=1}^{\infty}
\left|\sum_{m\not=n}^{\infty}\frac {e^{it(E_n-E_m)}-1}{E_n-E_m}\lang e_m,\psi\rang\right|^2<\infty\right\},\\
&& K(t)\psi:=i\sum_{n=1}^{\infty}\left(
\sum_{m\not=n}^{\infty}\frac {e^{it(E_n-E_m)}-1}{E_n-E_m}\lang e_m,\psi\rang\right)e_n, \quad
\psi \in D(K(t)).
\end{eqnarray}
It is easy to see that, for all $t\in \BBR$,  
\begin{equation}
\MSD_0 \subset D(K(t))
\end{equation}
and
\begin{equation}
K(t)e_k=i\sum_{n\not=k}\frac{e^{it(E_n-E_k)}-1}{E_n-E_k}e_n, \quad k\in \BBN. \label{Kek}
\end{equation}
The correspondence $K: \BBR \ni t\mapsto K(t)$ gives an
operator-valued function on $\BBR$.
In the notation in Section 4, $K(t)$ is the operator $T_b$ with
$b_{nm}=e^{it(E_n-E_m)}-1, \ n,m\in \BBN$.


\begin{rem}{\rm Equation (\ref{Kek}) shows that
$K(t)\not=tI|\MSD_0$. Hence $T$ cannot be a strong time operator of $H$, 
as already remarked based on the general theory of strong time operators.}
\end{rem} 

\begin{prop}\label{K-closed} For all $t\in \BBR$, $K(t)$ is a densely defined closed operator. 
\end{prop}

{\it Proof}. Similar to the proof of Proposition \ref{prop-T}.
\hfill $\Box$

\medskip


\begin{prop}\label{K-sym} For all $t\in \BBR$, $K(t)|\MSD_0$ is symmetric.
\end{prop}

{\it Proof}. Similar to the proof of Lemma \ref{lem-sym}. 
\hfill $\Box$

\medskip



\begin{The}\label{GWWR} For all $\psi \in D(T_1)(=\MSD_0)$ and $t\in \BBR$,
$e^{-itH}\psi \in D(T_1)$ and
\begin{equation}
T_1e^{-itH}\psi=e^{-itH}(T_1+K(t))\psi.\label{GWWR1}
\end{equation}
\end{The}

{\it Proof}. We need only to prove the statement in the case  $\psi=e_k$ ($\forall k\in \BBN$).
Since $e^{-itH}e_k=e^{-itE_k}e_k$, it follows that $e^{-itH}e_k \in D(T_1)$ with
$$
T_1e^{-itH}e_k=e^{-itE_k}\sum_{n\not=k}^{\infty}\frac i{E_n-E_k}e_n.
$$
We have
$$
e^{-itH}T_1e_k=i\sum_{n\not=k}^{\infty}\frac {e^{-itE_n}}{E_n-E_k}e_n.
$$
It follows from these equations that
$$
T_1e^{-itH}e_k-e^{-itH}T_1e_k=e^{-itH}K(t)e_k.
$$
Thus the desired result follows. 
\hfill $\Box$

\medskip

\begin{cor}\label{GTOP} Suppose that, for all $t\in \BBR$, $K(t)$ is bounded.
Then $T_1$ is a generalized  time operator of $H$ with commutation factor $K$.
\end{cor}

{\it Proof}. This follows from
Theorem \ref{GWWR}, Proposition \ref{K-closed} and Proposition \ref{K-sym}.
\hfill $\Box$


\medskip


In view of Corollary \ref{GTOP}, we need to investigate conditions for $K(t)$ to be bounded.





\begin{prop}\label{K1} Suppose that (\ref{E-bound}) holds with $\alpha >1$. 
Then, for all $t\in \BBR$,  $K(t)$ is a bounded self-adjoint operator with $D(K(t))=\MSH$.
\end{prop}

{\it Proof}. This follows from an application of Theorem \ref{B-thm}
to the case where $b_{nm}=e^{it(E_n-E_m)}-1, \ n,m\in \BBN$.
\hfill $\Box$

\medskip




\begin{prop} Suppose that (\ref{SS}) holds. Then, for all $t\in \BBR$,
 $K(t)$ is Hilbert-Schmidt and self-adjoint with
\begin{equation}
\|K(t)\|_2^2=\sum_{k=1}^{\infty}\sum_{n\not=k}^{\infty}
\left|\frac{e^{it(E_n-E_k)}-1}{E_n-E_k}\right|^2.
\end{equation} 
\end{prop}

{\it Proof}. Similar to the proof of Proposition \ref{HS}.
\hfill $\Box$

\medskip





\subsection{Non-differentiability of $K$}

From the view-point of the theory of generalized time operators \cite{Ar05}, it is
interesting to examine differentiability of the operator-valued function $K$.

\begin{prop}\label{non-dif} For all $k\in \BBN$, 
the $\MSH$-valued function $: \BBR \ni t \mapsto K(t)e_k$  is not
strongly differentiable on $\BBR$.
\end{prop}

{\it Proof}. We first show that $K(t)e_k$ is not strongly differentiable at $t=0$.
Since $K(0)e_k=0$, we have for all $t\in \BBR\setminus \{0\}$ and $N>k$
\begin{eqnarray*}
\left\|\frac {K(t)e_k-K(0)e_k}t\right\|^2&=&
\sum_{n\not=k}^{\infty}\frac{|e^{it(E_n-E_k)}-1|^2}{t^2|E_n-E_k|^2}  \\
&\geq& \sum_{n\not=k}^{N+1}\frac{|e^{it(E_n-E_k)}-1|^2}{t^2|E_n-E_k|^2}.
\end{eqnarray*}
Hence
$$
\liminf_{t\to 0}\left\|\frac {K(t)e_k-K(0)e_k}t\right\|^2
\geq \sum_{n\not=k}^{N+1}1=N.
$$
Since $N>k$ is arbitrary, it follows that
$$
\lim_{t\to 0}\left\|\frac {K(t)e_k-K(0)e_k}t\right\|^2=+\infty.
$$
This implies that $K(t)e_k$ is not strongly differentiable at $t=0$.

We next show that $K(t)e_k$ is not strongly differentiable at each $t\not=0$.
By (\ref{GWWR1}), we have for all $s\in \BBR\setminus \{0\}$
$$
\frac{K(t+s)e_k-K(t)e_k}{s}
=e^{it(H-E_k)}\frac{K(s)e_k}{s}.
$$
Hence
$$
\left\|\frac{K(t+s)e_k-K(t)e_k}{s}\right\|
=\left\|\frac{K(s)e_k}{s}\right\|.
$$
By the preceding result, the right hand side diverges to $+\infty$ as $s\to 0$.
Therefore $K(t)e_k$ is not strongly differentiable at $t$.
\hfill $\Box$

\medskip


\begin{rem}{\rm We have
\begin{equation}
\lang e_{\ell}, K(t)e_k\rang=\left\{
\begin{array}{ll}
i\frac{e^{it(E_{\ell}-E_k)}-1}{E_{\ell}-E_k} \ &; \ \ell\not=k \\
0 \ & ; \ \ell=k
\end{array}
\right.\label{K-mat}
\end{equation}
Hence, for all $k,\ell \in \BBN$, 
$\lang e_{\ell}, K(t)e_k\rang$ is differentiable in $t\in \BBR$ and
\begin{equation}
\frac {d}{dt}\lang e_{\ell}, K(t)e_k\rang=(\delta_{\ell k}-1)e^{it(E_{\ell}-E_k)}.
\end{equation}
}
\end{rem}


Proposition \ref{non-dif}
tells us  some singularity of $K(t)$ acting on $\MSD_0$. But, as shown in the next
proposition,   $K(t)$ restricted to $\MSD_c$ is strongly differentiable  at $t=0$.

\begin{prop}\label{prop-K0} For all $\psi \in \MSD_c$,
the $\MSH$-valued function $K(t)\psi$ is strongly differentiable at $t=0$
with
\begin{equation}
\frac d{dt}K(t)\psi\bigg|_{t=0}=\psi.\label{K0}
\end{equation}
\end{prop}

{\it Proof}. We need only to prove the statement for $\psi$ of the form
$\psi=e_k-e_{\ell}$ ($k,\ell \in \BBN,k\not=\ell$). For all $t\in \BBR\setminus \{0\}$, we have
$$
\frac {K(t)(e_k-e_{\ell})}{t}
=A(t)+B(t),
$$
where
\begin{eqnarray*}
&& A(t):=i\frac{e^{it(E_{\ell}-E_k)}-1}{t(E_{\ell}-E_k)}e_{\ell}
-i\frac{e^{it(E_{k}-E_{\ell})}-1}{t(E_{k}-E_{\ell})}e_k, \\
&& B(t):=i\sum_{n\not=k,\ell}^{\infty}
\left(\frac{e^{it(E_n-E_k)}-1}{t(E_n-E_k)}
-\frac{e^{it(E_n-E_{\ell})}-1}{t(E_n-E_{\ell})}\right)e_n.
\end{eqnarray*}
It is easy to see that
$$
\lim_{t\to 0}A(t)=e_k-e_{\ell}.
$$
As for $B(t)$, we have
\begin{eqnarray*}
\|B(t)\|^2&=&\sum_{n\not=k,\ell}^{\infty}|F_n(t)|^2,
\end{eqnarray*}
where
$$
F_n(t):=\frac{e^{it(E_n-E_k)}-1}{t(E_n-E_k)}
-\frac{e^{it(E_n-E_{\ell})}-1}{t(E_n-E_{\ell})}.
$$
It is easy to see that
$$
\lim_{t\to 0}F_n(t)=0.
$$
Moreover, one can show that 
$$
|F_n(t)| \leq \frac{C}{|E_n-E_k|}, \quad n\not=k,\ell,
$$
where $C>0$ is a constant independent of $n$ and $t$.
Since $\sum_{n\not=k}^{\infty}1/|E_n-E_k|^2 <\infty$, one can apply the dominated convergence theorem
to conclude that $\lim_{t\to 0}\|B(t)\|^2=0$. Thus
$K(t)(e_k-e_{\ell})$ is strongly differentiable at $t=0$ 
and (\ref{K0}) with $\psi=e_k-e_{\ell}$ holds.
\hfill $\Box$

\medskip



\begin{prop} For all $k,\ell \in \BBN$ with $k\not=\ell$,
the $\MSH$-valued function $K(t)(e_k-e_{\ell})$ is not strongly differentiable at $t\not\in \{2{\pi}n/(E_k-E_{\ell})|n\in \BBZ\}$.
\end{prop}


{\it Proof}. Let $t\not=2{\pi}n/(E_k-E_{\ell})$ ($n\in \BBZ$) and $s\in \BBR\setminus\{0\}$.
Then, by (\ref{GWWR1}), we have
$$
\frac{(K(t+s)-K(t))(e_k-e_{\ell})}{s}
=e^{itH}\frac {K(s)}{s}e^{-itH}(e_k-e_{\ell}).
$$
Hence
$$
\left\|\frac{(K(t+s)-K(t))(e_k-e_{\ell})}{s}\right\|=\|u(s)\|
$$
with
$$
u(s):=
\frac {K(s)}{s}(e^{-itE_k}e_k-e^{-itE_{\ell}}e_{\ell}).
$$
We write
$$
u(s)=u_1(s)+u_2(s)
$$
with
$$
u_1(s):=e^{-itE_k}\frac {K(s)}{s}(e_k-e_{\ell}), \quad u_2(s):=(e^{-itE_k}-e^{-itE_{\ell}})
\frac {K(s)}{s}e_{\ell}.
$$
By Proposition \ref{prop-K0}, we have $\lim_{s\to 0}u_1(s)=e^{-itE_k}(e_k-e_{\ell})$.
On the other hand,  we have from the proof of Proposition \ref{non-dif} and
the assumed condition for $t$
$$
\lim_{s\to 0}\|u_2(s)\|=+\infty.
$$
Hence $\lim_{s\to 0}\|u(s)\|=+\infty$. Thus the desired result follows.
\hfill $\Box$

\medskip
\section{Possible Connections with Regular Perturbation Theory}

We consider a perturbation of $H$ by a symmetric operator $H_I$ on $\MSH$:
\begin{equation}
H(\lambda):=H+\lambda H_I,
\end{equation}
where $\lambda \in \BBR$ is a perturbation parameter.
For simplicity, we assume that $H_I$ is $H$-bounded: $D(H)\subset D(H_I)$ and there exist
 constants $a,b\geq 0$ such that
$$
\|H_I\psi\|\leq a\|H\psi\|+b\|\psi\|, \quad \psi \in D(H).
$$
Then, by the Kato-Rellich theorem (e.g., \cite[Theorem X.12]{RS2}),
for all $\lambda \in \BBR$ satisfying
\begin{equation}
a|\lambda| < 1,\label{lam}
\end{equation}
$H(\lambda)$ is self-adjoint and bounded below.
In what follows, we assume (\ref{lam}).



\subsection{Eigenvalues of $H(\lambda)$}

We fix $n\in \BBN$ arbitrarily.
By a general theorem in regular perturbation theory (e.g., \cite[Theorem XII.9]{RS4}),
there exists a constant $c_n>0$ such that, for all $|\lambda| <c_n$,
$H$ has a unique, isolated  non-degenerate eigenvalue $E_n(\lambda)$  near $E_n$.
Moreover, $E_n(\lambda)$ is analytic in $\lambda$  with Taylor expansion
\begin{equation}
E_n(\lambda)=E_n+E_n^{(1)}\lambda +E_n^{(2)}\lambda^2+\cdots,
\end{equation}
where
\begin{equation}
E_n^{(1)}:=\lang e_n, H_Ie_n\rang,\quad
E_n^{(2)}:=\sum_{m\not=n}^{\infty}\frac{|\lang e_n, H_Ie_m\rang|^2}
{E_n-E_m}.
\end{equation}

As an eigenvector of $H(\lambda)$ with eigenvalue $E_n(\lambda)$,
one can take a vector $\psi_n(\lambda)$ analytic in $\lambda$ with Taylor expansion
\begin{equation}
\psi_n(\lambda)=e_n+e_n^{(1)}\lambda +\cdots,
\end{equation}
where
\begin{equation}
e_n^{(1)}:=\sum_{m\not=n}^{\infty}\frac{\lang e_m,H_Ie_n\rang}{E_n-E_m} e_m.
\end{equation}

By Lemma \ref{lem2-2}, we have 
$$
\lang e_n, Te_m\rang=\left\{
\begin{array}{ll}
\displaystyle{\frac{i}{E_n-E_m}} \ & \ ; n\not=m\\
0 \ & \ ; n=m
\end{array}
\right.
$$
Hence $E_n^{(2)}$ can be written
\begin{equation}
E_n^{(2)}=(-i)
\sum_{m=1}^{\infty}|\lang e_n, H_Ie_m\rang|^2\lang e_n,Te_m\rang.\label{E2}
\end{equation}
To rewrite the right hand side only in terms of $e_n$ and  linear operators
on $\MSH$, we  note that
$$
\sum_{m=1}^{\infty}|\lang e_n,H_Ie_m\rang|^2=\|H_Ie_n\|^2 <\infty
$$
by the Parseval equality. Hence 
$$
\sum_{m=1}^{\infty}|\lang e_n,H_Ie_m\rang|^4 <\infty.
$$
Therefore the infinite series
\begin{equation}
f_n:=\sum_{m=1}^{\infty}|\lang e_n,H_Ie_m\rang|^2e_m 
\end{equation}
strongly converges and defines a vector in $\MSH$.
Thus we can define a linear operator $V$ on $\MSH$ as follows:
\begin{eqnarray}
&& D(V):=\MSD_0, \\
&& V\psi:=-i\sum_{n=1}^{\infty}\lang e_n,\psi\rang f_n, \quad \psi \in \MSD_0\label{V2}
\end{eqnarray}
where the right hand side of (\ref{V2}) is a  sum over a finite term.
It is easy to see that $V$ is a symmetric operator.

\begin{prop}
For all $n\in \BBN$,
\begin{equation}
E_n^{(2)}=\lang Te_n,Ve_n\rang.
\end{equation}
\end{prop}

{\it Proof}. We have $Ve_n=-if_n$.  Hence $\lang Te_n,Ve_n\rang
=-i\lang Te_n, f_n\rang$, which is equal to the right hand side of (\ref{E2}). 
\hfill $\Box$

\medskip


This proposition suggests some role of the time operator $T_1=T|\MSD_0$
in the perturbation expansions of the eigenvalues of $H$.



As for the first order term $e_n^{(1)}\lambda $ of the eigenvector $\psi_n(\lambda)$, we have
\begin{equation}
e_n^{(1)}=(-i)\sum_{m=1}^{\infty}\lang e_m,H_Ie_n\rang\lang e_n, Te_m\rang e_m.
\end{equation}




\subsection{Transition probability amplitudes}

In the context of quantum mechanics where $H(\lambda)$ is the Hamiltonian of a quantum system, 
the complex number
$\lang \phi, e^{-itH(\lambda)}\psi\rang$ with
unit vectors $\phi,\psi \in \MSH$ is called the
{\it transition probability amplitude} for the probability such  that
the state of the quantum system at time $t$ is found in the state $\phi$
under the condition that the state of the quantum system at time zero is  $\psi$.

\begin{lem} Let $\phi, \psi \in D(H)$. Then, for all $t\in \BBR$,
\begin{equation}
\lang  \phi, e^{-itH(\lambda)}\psi\rang
=\lang \phi, e^{-itH}\psi\rang-i\lambda\int_0^t\lang e^{i(t-s)H}\phi, H_Ie^{-isH}\psi\rang ds
+O(\lambda^2). \label{tp1}
\end{equation}
\end{lem}

{\it Proof}. By a simple application of a general formula for
the unitary  group generated by a self-adjoint operator
(\cite[Lemma 5.9]{Ar07-3}), we have
\begin{equation}
e^{-itH(\lambda)}\psi=
e^{-itH}\psi
-i\lambda \int_0^te^{-i(t-s)H(\lambda)}H_Ie^{-isH}\psi ds, \label{exp}
\end{equation}
where the integral is taken in the strong sense.
Hence
\begin{eqnarray*}
\lang \phi, e^{-itH(\lambda)}\psi\rang
&=&
\lang \phi, e^{-itH}\psi\rang 
-i\lambda \int_0^t\lang e^{i(t-s)H(\lambda)}\phi, H_Ie^{-isH}\psi\rang ds \\
&=&
\lang \phi, e^{-itH}\psi\rang 
-i\lambda \int_0^t\lang e^{i(t-s)H}\phi, H_Ie^{-isH}\psi\rang ds  +R(\lambda),
\end{eqnarray*}
where
$$
R(\lambda):=-i\lambda\int_0^t\lang \left(e^{i(t-s)H(\lambda)}-e^{i(t-s)H}\right)\phi, 
H_Ie^{-isH}\psi\rang ds.
$$
Using (\ref{exp}) again, we have
$$
R(\lambda)
=-\lambda^2\int_0^tds
\int_0^{-(t-s)}ds'\lang e^{i(t-s+s')H(\lambda)}H_Ie^{-is'H}\phi, H_Ie^{-isH}\psi\rang.
$$
Hence
$$
|R(\lambda)|
\leq \lambda^2
\int_0^{|t|}ds\int_0^{|t-s|}ds'
\|H_Ie^{-is'H}\phi\|
\|H_Ie^{-isH}\psi\|
$$
Therefore $R(\lambda)=O(\lambda^2)$.  
Thus (\ref{tp1}) holds.
\hfill $\Box$

\medskip

Applying  (\ref{tp1}) with $\phi=e_m$ and $\psi=e_n$ ($n\not=m$), we have
\begin{equation}
\lang  e_m, e^{-itH(\lambda)}e_n\rang
=-\lambda \frac {e^{-itE_n}-e^{-itE_m}}{E_m-E_n}\lang e_m, H_Ie_n\rang
+O(\lambda^2), \label{tp2}
\end{equation}
which, combined with  (\ref{K-mat}), gives 
\begin{equation}
\lang  e_m, e^{-itH(\lambda)}e_n\rang
=i\lambda \lang e_n, e^{-itH}K(t)e_m\rang\lang e_m, H_Ie_n\rang
+O(\lambda^2). \quad m\not=n.
\end{equation}
This suggests a physical meaning of the commutation factor $K$.

By Theorem \ref{GWWR}, one can rewrite the first term on the right hand side  in terms of
$T_1$ and $e^{-itH}$, obtaining
\begin{equation}
\lang  e_m, e^{-itH(\lambda)}e_n\rang
=i\lambda \lang e_n, [T_1,e^{-itH}]e_m\rang\lang e_m, H_Ie_n\rang
+O(\lambda^2), \quad m\not=n. 
\end{equation}
This also is suggestive on  physical meaning 
of the time operator $T_1$.



\begin{thebibliography}{99}


\bibitem{AB}Y. Aharonov and D. Bohm, 
Time in the quantum theory and the uncertainty
relation for time and energy,
{\it Phys. Rev.} {\bf 122} (1961), 1649--1658.
\bibitem{Ar05}A. Arai, Generalized  weak Weyl relation and decay of quantum dynamics,
{\it Rev. Math. Phys.} {\bf 17} (2005) , 1071--1109 .
\bibitem{Ar07-1}A. Arai, Spectrum of time operators, {\it Lett.  Math.  Phys.} {\bf 80} (2007), 211--221.
\bibitem{Ar07-2}A. Arai, Some aspects of time operators, 
to be published in the {\it Proceedings
of International Conference in QBIC} (Quantum Bio-Informatics Center), 2007,
Tokyo University of Science.
\bibitem{Ar07-3}A. Arai, Heisenberg operators, invariant domains and Heisenberg equations of motion,
{\it Rev. Math. Phys.} {\bf 19} (2007), 1045--1069.

\bibitem{ArMa}A. Arai and Y. Matsuzawa, Construction of 
a Weyl representation from a weak Weyl representation
of the canonical commutation relation, 
 mp{\_}arc 07-255, {\it Lett. Math. Phys.} (2008) DOI 10.1007/s11005-008-0220-4.

\bibitem{DD}G. Dorfmeister and J. Dorfmeister, Classification of certain pairs of operators
$(P,Q)$ satisfying $[P,Q]=-i{\rm Id}$, {\it J. Funct. Anal. } {\bf 57} (1984), 301--328.

\bibitem{Ga}E. A. Galapon, Self-adjoint time operator is the rule for discrete semi-bounded
Hamiltonians, {\it Proc. R. Soc. Lond. A} {\bf 458} (2002), 2671--2689.
 
\bibitem{HLP}G. H. Hardy, J. E. Littlewood and
G. P\'olya, {\it Inequalities} (Cambridge University Press, London, 1934).
\bibitem{Miya1}M.~Miyamoto, A generalized
Weyl relation approach to the time operator
and its connection to the survival probability, 
{\it J. Math. Phys.} {\bf 42} (2001), 1038--1052.

\bibitem{vN}J.~von~Neumann, Die Eindeutigkeit der
Schr\"odingerschen Operatoren, {\it Math. Ann.}  (1931)
{\bf 104}, 570--578.
\bibitem{RS1}M.~Reed and B.~Simon, 
{\it Methods of Modern Mathematical Physics I: Functional Analysis} 
(Academic Press, New York, 1972).
\bibitem{RS2}M.~Reed and B.~Simon, 
{\it Methods of Modern Mathematical Physics II: Fourier Analysis, Self-adjointness}
(Academic Press, New York, 1975).
\bibitem{RS4}M.~Reed and B.~Simon, 
{\it Methods of Modern Mathematical Physics IV: Analysis of Operators} 
(Academic Press, New York, 1978).
\bibitem{Sch1}K.~Schm\"udgen, On the Heisenberg
commutation relation. I, {\it J. Funct. Anal.} {\bf  50} (1983), 8--49.
\bibitem{Sch2}K.~Schm\"udgen, On the Heisenberg
commutation relation. II, {\it Publ. RIMS, Kyoto Univ.} {\bf  19} (1983), 601--671.
\end{thebibliography}
 
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