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Spectrum; time operator;  Hamiltonian; weak Weyl relation; quantum theory.
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\begin{document}
\title{\bf Spectrum of Time Operators}
\author{ Asao Arai\thanks{
This work is supported by the Grant-in-Aid No.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 

\bigskip


\begin{abstract}
Let  $H$  be a  self-adjoint operator  on a complex Hilbert space ${\cal H}$.
A symmetric operator $T$  on ${\cal H}$ is called a time operator of $H$ if,
for all $t\in \R$, $e^{-itH}D(T)\subset D(T)$ ($D(T)$ denotes the domain of $T$) and
$Te^{-itH}\psi=e^{-itH}(T+t)\psi, \ \forall t\in \R, \forall \psi \in D(T)$. In this paper,
spectral properties of $T$  are investigated. 
The following results are obtained: (i) If $H$ is bounded below, then
$\sigma(T)$, the  spectrum of $T$, is either $\C$ (the set of complex numbers)
or $\{z\in \C| \Im z \geq 0\}$. (ii) If $H$ is bounded above, then
$\sigma(T)$ is either $\C$
or $\{z\in \C| \Im z \leq 0\}$. (iii) If $H$ is bounded, then $\sigma(T)=\C$.
The spectrum of time operators of free Hamiltonians for both  nonrelativistic  and
relativistic particles is exactly identified.
Moreover spectral analysis is made on a generalized time operator.
\end{abstract}

\medskip


\noindent
{\it Keywords}: Spectrum; time operator;  Hamiltonian; weak Weyl relation; quantum theory.

\medskip

\noindent
Mathematics Subject Classification 2000: 81Q10, 47N50

\section{Introduction}

In a paper \cite{Schmu}, Schm\"udgen 
studied a pair $(T,H)$ of a symmetric operator $T$ and
a self-adjoint operator $H$ on a complex Hilbert space
${\cal H}$ (in the notation there, $T=P, H=-Q$) such that, for all $t\in \R$, $e^{-itH}D(T)\subset D(T)$
($D(T)$ denotes the domain of $T$) and
\begin{equation}
Te^{-itH}\psi=e^{-itH}(T+t)\psi, \ \forall t\in \R, \forall \psi \in D(T).\label{WWR}
\end{equation} 
This is 
a stronger version of representation of
the canonical commutation relation (CCR) with one degree of freedom, since (\ref{WWR})
implies that
\begin{equation}
\lang T\phi, H\psi\rang-\lang H\phi, T\psi\rang =\lang \phi,i\psi\rang, \quad 
\psi,\phi \in D(T)\cap D(H),
\end{equation}
i.e., $T$ and $H$ satisfy the CCR in the sense of sesquilinear  form on $D(H)\cap D(T)$ and
hence, in particular,
$TH-HT=i$ on $D(HT)\cap D(TH)$, the CCR in the original sense.
We call (\ref{WWR}) the {\it weak Weyl relation} (WWR).


About twenty years later, Miyamoto \cite{Miya1} used the WWR to 
introduce a proper concept of time operator in quantum mechanics.
Namely a symmetric operator $T$ on ${\cal H}$ is called a {\it time operator}
of $H$ if $(T,H)$ obeys the WWR (\ref{WWR}) (in \cite{Miya1},
(\ref{WWR}) is called the {\it $T$-weak Weyl relation}).
We remark that, in this terminology,  
one has in mind the case where, in application to
quantum mechanics, $H$ is the Hamiltonian of a quantum system.
Some fundamental properties of the pair $(T,H)$ 
were investigated in \cite{Miya1}.

 
The work of Miyamoto \cite{Miya1} was extended by the present author 
in a previous paper \cite{Ar05}, where
a generalized version of the WWR (\ref{WWR}), called a {\it generalized weak Weyl relation},
is given and, in terms of it,  
a concept of {\it generalized time operator} is introduced. 
We remark that a time operator as well as a generalized one of a given self-adjoint operator
$H$ is not unique \cite[Proposition 2.6, \S 11]{Ar05}.
Physically the set of generalized time operators associated with a self-adjoint operator $H$ ( a Hamiltonian)
can be regarded as  a class of  symmetric operators 
which play a role in controlling decays (in time) of survival probabilities as well as  time-energy uncertainty
relations \cite{Ar05,Miya1}.


In this paper, we investigate spectral properties of  (generalized) time operators.
We first recall the definition of the spectrum of a linear operator $A$ on ${\cal H}$.
The resolvent set of $A$, denoted $\rho(A)$, is defined to be 
the set of complex numbers $z$ satisfying
the following three conditions: (i) $A-z$ is injective ; (ii) $\Ran(A-z)$, the range of $A-z$,
is dense in ${\cal H}$ ; (iii) $(A-z)^{-1}$ with $D((A-z)^{-1})=\Ran(A-z)$ is bounded.
Then the spectrum of $A$, denoted $\sigma(A)$, is defined by $\sigma(A):=\C\setminus \rho(A)$,
where $\C$ is the set of complex numbers.
It follows that, if $A$ is closable, then $\sigma(\bar A)=\sigma(A)$, where $\bar A$ is the closure
of $A$,  and
$\Ran(\bar A-z)={\cal H}$ for all $z\in \rho(\bar A)=\rho(A)$.
In particular, for all symmetric operators $S$ on ${\cal H}$,
$\sigma(S)=\sigma(\bar S)$ and $\Ran(\bar S-z)={\cal H}$ for all $z\in \rho(\bar S)=\rho(S)$.

One of the motivations for this work comes from the following fact:

\begin{The}\label{th1-1}{\rm (\cite{Miya1}, \cite[Theorem 2.8]{Ar05})} 
If $H$ is a self-adjoint operator on ${\cal H}$ and
semi-bounded (i.e., bounded below or bounded above),
then every time operator $T$ of $H$
is not essentially self-adjoint .
\end{The}

This theorem combined with a general theorem \cite[Theorem X.1]{RS2} implies that, 
in the case where $H$ is semi-bounded, the spectrum $\sigma(T)$
 of $T$ ($=\sigma(\overline{T})$) is one 
of the following three sets:
\begin{list}{}{}
\item{(i)} $\C$.
\item{(ii)} $\overline {\Pi}_+$, the closure of the upper half-plane
$\Pi_+:=\{z\in \C|\Im z >0\}$.
\item{(iii)} $\overline{\Pi}_-$, the closure of the lower half-plane
$\Pi_+:=\{z\in \C|\Im z <0\}$.
\end{list}
Then it is interesting to examine which one is realized, depending on properties of $H$.

The outline of the present paper is as follows.
In Section 2, we prove a theorem on the spectrum  of time operators
(Theorem \ref{th2-1}). In Section 3 we consider time operators on direct sums of  Hilbert spaces.
In Section 4, we identify  the spectrum of concrete time operators, 
including the Aharonov-Bohm time operator
\cite{AB} and time operators of a relativistic Schr\"odinger operator.
In Section 5, we prove a theorem similar to Theorem \ref{th2-1}
in the case where $T$ is a generalized time operator.


\section{Main Result}

In this section we prove the  following theorem:

\begin{The}\label{th2-1} Let $H$ be a self-adjoint operator on ${\cal H}$ and
$T$ be  a time operator of $H$. Then the following {\rm (i)---(iii)} hold:
\begin{list}{}{}
\item{{\rm (i)}} If $H$ is bounded below, then
$\sigma(T)$ is either $\C$
or $\overline{\Pi}_+$. 
\item{{\rm (ii)}} If $H$ is bounded above, then
$\sigma(T)$ is either $\C$
or $\overline{\Pi}_-$. 
\item{{\rm (iii)}} If $H$ is bounded, then $\sigma(T)=\C$.
\end{list}
\end{The}

\begin{rem}\label{rem2-1}{\rm The time operator $T$ has no eigenvalues , i.e., 
the point spectrum $\sigma_{\rm p}(T)$ of $T$ is an empty set \cite[Corollary 4.2]{Miya1}}.
\end{rem}

\begin{rem}{\rm In the case where $\sigma(T)=\overline{\Pi}_+$ or $\overline{\Pi}_-$,
$\overline{T}$ is maximally symmetric \cite[p.141]{RS2}.}
\end{rem} 

Throughout the rest of this section, $T$ represents a time operator of $H$.
The following lemma is a key fact to prove Theorem \ref{th2-1}.

\begin{lem}\label{lem2-1} Suppose that $H$ is bounded below. Then, for all $\beta >0$,
$e^{-\beta H}D(\overline{T})\subset D(\overline{T})$ and, for all $\psi\in D(\overline{T})$
\begin{equation}
\overline{T}e^{-\beta H}\psi=e^{-\beta H}(\overline{T}-i\beta)\psi.
\end{equation}
\end{lem}

{\it Proof}. Apply \cite[Theorem 6.2]{Ar05}. \hfill \qed

\medskip

We denote by $T^*$ the adjoint of $T$.

\begin{lem}\label{lem2-2} Suppose that $H$ is bounded below. Then, for all $\beta >0$,
$e^{-\beta H}D(T^*)\subset D(T^*)$ and, for all $\psi\in D(T^*)$
\begin{equation}
T^*e^{-\beta H}\psi=e^{-\beta H}(T^*-i\beta)\psi.
\end{equation}
\end{lem}

{\it Proof}. Lemma \ref{lem2-1} implies that 
$e^{-\beta H}(\overline{T}-i\beta)\subset \overline{T}e^{-\beta H}$.
We have $(\overline{T})^*=T^*$. For each bounded linear operator $A$ on ${\cal H}$ with $D(A)={\cal H}$ and 
all densely defined linear operators $B$ on ${\cal H}$,  $(AB)^*=B^*A^*$.
Using these facts, one can show that
 $e^{-\beta H}T^*\subset (T^*+i\beta)e^{-\beta H}$. 
Thus the desired result follows. \hfill \qed

\medskip



\noindent
{\bf Proof of Theorem \ref{th2-1}}

 (i) By the fact on the spectrum of $T$ mentioned after Theorem \ref{th1-1}, we need only to
show that the case $\sigma(T)=\overline{\Pi}_-$ is excluded. For this purpose,
suppose that $\sigma(T)=\overline{\Pi}_-$.
Then $\Pi_+=\rho(T)=\rho(\overline{T})$.

In general, we  have for all $z\in \C\setminus\R$
the orthogonal decomposition
\begin{equation}
{\cal H}=\ker(T^*-z^*)\oplus \Ran(\overline{T}-z) \label{dcp}
\end{equation}

Applying this structure with $z=i\in \Pi_+$, we obtain 
$\ker (T^*+i)=\{0\}$.
Since $T$ is not essentially self-adjoint by Theorem \ref{th1-1}, it follows that
$\ker (T^*-i)\not=\{0\}$.
Hence  there exists a non-zero vector $\psi\in D(T^*)$ such that
$T^*\psi=i\psi$. Then, by Lemma \ref{lem2-2}, $i(1-\beta)\in\sigma_{\rm p}(T^*)$.
Since $\beta>0$ is arbitrary, we can take it to be
$1< \beta$.  Then $\gamma:=i(1-\beta)\in \Pi_-$. Taking $z=\gamma^*$ in 
(\ref{dcp}), we have the orthogonal decomposition
$$
{\cal H}=\ker(T^*-\gamma)\oplus \Ran(\overline{T}-\gamma^*).
$$ 
Hence $\Ran(\overline{T}-\gamma^*)$ is not dense in ${\cal H}$.
Therefore $\gamma^*\in \sigma(\overline{T})=\sigma(T)$, i.e., $i(\beta-1) \in \sigma(T)$.
But $i(\beta-1)\in \Pi_+$. This is a contradiction. Thus
$\sigma(T)\not=\overline{\Pi}_-$.


 (ii)  If $H$ is bounded above, then $\widehat H:=-H$ is bounded below.
It is easy to see that $\widehat T:=-T$ is a time operator of $\widehat H$.
Hence, by part (i), $\sigma(\widehat T)=\C$ or $\overline{\Pi}_+$.
On the other hand,  $\sigma(T)=\{-\lambda|\lambda \in \sigma(\widehat T)\}$,
which implies that $\sigma(T)=\C$ or $\overline{\Pi}_-$.

 (iii) This follows from  (i) and (ii). \hfill \qed

\medskip




In the next section we analyze the spectrum of nontrivial examples
of time operators. Here we present only simple examples.


\begin{ex}\label{ex2-1}{\rm 
We denote by $\hat r$ the multiplication operator on $L^2([0,\infty))$
by the variable $r\in [0,\infty)$: 
$(\hat r g)(r):=rg(r), \ {\rm a.e.}r\in [0,\infty),
g\in D(\hat r)$. The operator $\hat r$ is self-adjoint and nonnegative.

Let $p_0$ be an operator on $L^2([0,\infty))$ defined as follows:
\begin{eqnarray}
&&D(p_0):=C_0^{\infty}((0,\infty)), \\
&& (p_0g)(r):=-ig'(r), \quad g\in D(p_0),
\end{eqnarray}
where, for an open set $\Omega\subset \R^n$ ($n\in \N$), $C_0^{\infty}(\Omega)$ denotes
the set of infinitely differentiable functions on $\Omega$ with compact support in $\Omega$.
Then it is easy to see that $-p_0$ is a time operator of $\hat r$ and that
$$
\sigma(-p_0)=\overline{\Pi}_+.
$$
Hence this is an example which illustrates one of the case of Theorem \ref{th2-1}-(i).
}
\end{ex}


\begin{ex}\label{ex2-2}{\rm Let $L>0$ and $V_L:=(-L/2,L/2)\subset \R$.
We denote by $\hat x_L$ the multiplication operator on $L^2(V_L)$ by the variable
$x\in V_L$. Then $\hat x_L$ is a bounded self-adjoint operator.
We define an operator $p_L$ as follows:
\begin{eqnarray*}
&& D(p_L):=C_0^{\infty}(V_L),\\
&& p_Lf:=-if', \quad f\in D(p_L).
\end{eqnarray*}
Then it is easy to see that $-p_L$ is a time operator of $\hat x_L$ and
$$
\sigma(-p_L)=\C.
$$
Hence this is an example which illustrates Theorem \ref{th2-1}-(iii).
It should be remarked that
$p_L$ has uncountably many self-adjoint extensions \cite[pp.257--259]{RS1}.
}
\end{ex}


\section{Time Operators on Direct Sum Hilbert Spaces}

In applications, time operators on direct sum Hilbert spaces may be useful.
We briefly discuss this aspect here.
Let $H_j$  ($j=1,2$) 
be a self-adjoint operator on a complex Hilbert space ${\cal H}_j$ which has a time operator 
$T_j$. Let
\begin{equation}
{\cal H}:={\cal H}_1\oplus {\cal H}_2.
\end{equation}
Then
\begin{equation}
T:=T_1\oplus T_2
\end{equation}
is a time operator of $H_1\oplus H_2$ \cite[Proposition 2.14]{Ar05}.


\begin{The}\label{th2-2} Let $H_j$, $T_j$ and $T$ be as  above. Then:
\begin{list}{}{}
\item{{\rm (i)}} If $H_1$ is bounded below  and $H_2$ is bounded  above,  
then $\sigma(T)=\C$.
\item{{\rm (ii)}} If one of $H_1$ and $H_2$ is bounded, then
$\sigma(T)=\C$.
\end{list}
\end{The}

{\it Proof}. (i) By Theorem \ref{th2-1}, 
$\sigma(T_1)=\C$ or $\overline{\Pi}_+$, and $\sigma(T_2)=\C$ or $\overline{\Pi}_-$.
By a general theorem, we have $\sigma(T)=\sigma(T_1)\cup \sigma(T_2)$.
Hence, in each case, we have $\sigma(T)=\C$.

(ii) In this case, we can apply Theorem \ref{th2-1}-(iii) to conclude that
one of $\sigma(T_1)$ and $\sigma(T_2)$ is equal to $\C$.
Thus the desired result follows.
\hfill \qed

\medskip


\begin{rem}{\rm In each  case of Theorem \ref{th2-2}-(i) and (ii),
$H_1\oplus H_2$ can be unbounded both above and below.
}
\end{rem}


\begin{ex}{\rm Let
$$
{\cal H}_L:=L^2([0,\infty))\oplus L^2(V_L),
$$
$\hat r, p_0$ be as in Example \ref{ex2-1} and $\hat x_L, p_L$ be as in Example \ref{ex2-2}.
Then  $H_L:=\hat r\oplus \hat x_L$  on ${\cal H}_L$ is self-adjoint and
bounded below (but unbounded above).
Moreover $T_L:=(-p_0)\oplus (-p_L)$ is a time operator of $H_L$ and
$\sigma(T_L)=\C$. Thus this example shows that
the spectrum of a time operator of a self-adjoint operator which
is bounded below, but unbounded above, can be equal to $\C$.
}
\end{ex}

\section{Examples}


\subsection{Time operators of the free Hamiltonian of a nonrelativistic particle }

Let $\Delta$ be the $n$-dimensional generalized Laplacian acting in $L^2(\R_x^n)$ ($n\in \N$),
where $\R_x^n:=\{x=(x_1,\cdots,x_n)|x_j\in \R,j=1,\cdots,n\}$, and
\begin{equation}
H_0:=-\frac {\Delta}{2m}
\end{equation}
with a constant $m>0$. In the context of quantum mechanics, $H_0$ represents  the free Hamiltonian
of a  nonrelativistic  particle with mass $m$ in the $n$-dimensional space $\R_x^n$.
It is well known that  $H_0$ is a nonnegative self-adjoint operator.
We denote by $\hat x_j$ the multiplication operator on $L^2(\R_x^n)$ 
by the $j$-th variable $x_j\in \R_x^n$ and 
set
\begin{equation}
\hat p_j:=-iD_j
\end{equation}
with $D_j$ being
the generalized partial differential operator in the variable $x_j$ on $L^2(\R_x^n)$.
It is easy to see that $\hat x_j$ and $\hat p_j$ are injective.



We introduce
\begin{equation}
\Omega_j:=\{k=(k_1,\cdots,k_n)\in \R_k^n| k_j\not=0\}, \quad j=1,\cdots,n.
\end{equation}
For  a real-valued, Borel measurable  function $G$ on $\R_k^n$ which is continuous 
on $\Omega_j$, we define a linear operator
on $L^2(\R_x^n)$ by
\begin{equation}
G(\hat p):={\cal F}^{-1}G{\cal F},\label{G}
\end{equation}
where ${\cal F}:L^2(\R_x^n)\to L^2(\R_k^n)$ is the Fourier transform:
\begin{equation}
({\cal F}f)(k):=\frac {1}{(2\pi)^{n/2}}\int_{\R^n_x}f(x)e^{-ikx}dx, \quad 
f\in L^2(\R_x^n),\,k=(k_1,\cdots,k_n)
\in \R_k^n,
\end{equation}
in the $L^2$-sense,  
$\hat p:=(\hat p_1,\cdots, \hat p_n)$ and $G$ on the right hand side of (\ref{G}) represents the multiplication operator on
$L^2(\R_k^n)$  by the function $G$.
Since the Lebesgue measure of  the set
$\R_k^n\setminus\Omega_j$ is zero, it follows that $G(\hat p)$ is self-adjoint.
 

For each $j=1,\cdots,n$,  one can define a linear  operator on $L^2(\R_x^n)$ by
\begin{equation}
T_{j}(G):=\frac m{2}
\left(\hat x_j\hat p_j^{-1}+\hat p_j^{-1}\hat x_j\right)+G(\hat p)
\end{equation}
with domain
\begin{equation}
D(T_{j}(G)):={\cal F}^{-1}C_0^{\infty}(\Omega_j),
\end{equation}
where $C_0^{\infty}(\Omega_j)$  denotes the
set of infinitely many differentiable functions on $\Omega_j$
with compact support in $\Omega_j$. 
It is easy to see that $T_j(G)$ is a symmetric operator on $L^2(\R_x^n)$.

\begin{lem} The operator $T_j(G)$ is a time operator of $H_0$.
\end{lem}


{\it Proof}. We write 
$$
T_j(G)=T_j+G(\hat p)
$$
with
\begin{equation}
T_j:=\frac m{2}
\left(\hat x_j\hat p_j^{-1}+\hat p_j^{-1}\hat x_j\right).
\end{equation}
The operator $T_j$ is  a time operator of $H_0$
(\cite{Miya1}, \cite[\S 10]{Ar05}).
By using the Fourier transform, one can show that 
 $e^{-itH_0}G(\hat p)\subset G(\hat p)e^{-itH_0}$ for all $t\in \R$.  Hence, by applying
\cite[Proposition 2.6]{Ar05}, $T_j(G)$ is a time operator of $H_0$.
\hfill \qed

\medskip


\begin{rem}{\rm 
The operator $T_j$ is called the {\it Aharonov-Bohm time operator}
\cite{AB}. Hence $T_j(G)$ is a perturbed Aharonov-Bohm  time operator.
}
\end{rem}

As for the spectrum of $T_j(G)$, we have the following theorem:


\begin{The}\label{th3-1} $\sigma(T_j(G))=\overline{\Pi}_+,\ j=1,\cdots,n$.
\end{The}  


{\it Proof}. By Theorem \ref{th2-1}-(i) and (\ref{dcp}), we need only to show that $\ker (T_j(G)^*-i)
=\{0\}$. 
Let $f\in\ker (T_j(G)^*-i)$. Then $T_j(G)^*f=if$. This implies the equation
$$
D_{k_j}\hat f(k)=\left(\frac{1}{2k_j}+\frac{k_j}{m}+
\frac {i}{m} k_jG(k)\right)\hat f(k)
$$
in the sense of distribution on $\Omega_j$, where $\hat f:={\cal F}f$ and $D_{k_j}$ is
the generalized  partial differential operator in the variable $k_j$.
Hence
$$
\hat f(k)=c(k_1,\cdots, k_{j-1},k_{j+1},\cdots, k_n)
\sqrt{|k_j|}e^{k_j^2/(2m)}e^{iG_j(k)/m}, \quad k\in \Omega_j
$$
where $c(k_1,\cdots, k_{j-1},k_{j+1},\cdots, k_n)\in \C$ is 
independent of $k_j$ and  $G_j$ is a differentiable function
on $\Omega_j$ such that $\partial G_j(k)/\partial k_j=k_jG(k), \ k\in \Omega_j$. 
Since $\hat f$ is in $L^2(\R_k^n)$, it follows that $c(k_1,\cdots, k_{j-1},k_{j+1},\cdots, k_n)=0$ (a.e.).
Hence  $f=0$.
Thus $\ker(T_j(G)^*-i)=\{0\}$.  \hfill \qed

\medskip


\subsection{Time operators of the  free Hamiltonian of a relativistic particle }

A Hamiltonian of a free relativistic  particle with mass $m\geq 0$ moving in $\R_x^n$ is given by
\begin{equation}
H_{\rm rel}:=\sqrt{-\Delta+m^2}
\end{equation} 
acting in $L^2(\R_x^n)$. It is shown that
the operator
\begin{equation}
T_j^{\rm rel}(G):=\sqrt{-\Delta+m^2}\,\hat p_j^{-1}\hat x_j+\hat x_j\sqrt{-\Delta+m^2}\,\hat p_j^{-1}+G(\hat p)
\end{equation}
with $D(T_j^{\rm rel}(G)):={\cal F}^{-1}C_0^{\infty}(\Omega_j)$  is a time operator
of $H_{\rm rel}$ \cite[Example 11.4]{Ar05}.



\begin{The}\label{th3-2} $\sigma(T_j^{\rm rel}(G))=\overline{\Pi}_+,\ j=1,\cdots,n$.
\end{The}  



{\it Proof}. As in Theorem \ref{th3-1}, we need only to show that $\ker (T_j^{\rm rel}(G)^*-i)
=\{0\}$. 
Let $f\in\ker (T_j^{\rm rel}(G)^*-i)$ and $\omega(k):=\sqrt{k^2+m^2}, \, k\in \R_k^n$. Then 
$$
D_{k_j}\hat f(k)=\frac{1}{2}\left(\frac{k_j}{\omega(k)}-\frac{k_j}{\omega(k)}
\left(\frac{\partial}{\partial k_j}\frac{\omega(k)}{k_j}\right)
-\frac{k_jG(k)}{\omega(k)i}\right)\hat f(k)
$$
in the sense of distribution on $\Omega_j$.
Hence
$$
\hat f(k)=c(k_1,\cdots, k_{j-1},k_{j+1},\cdots, k_n)
\sqrt{\frac{|k_j|}{\omega(k)}}e^{\omega(k)/2}e^{iF_j(k)/2}, \quad k\in \Omega_j
$$
where $c(k_1,\cdots, k_{j-1},k_{j+1},\cdots, k_n)\in \C$ is 
independent of $k_j$ and  $F_j$ is a differentiable function
on $\Omega_j$ such that $\partial F_j(k)/\partial k_j=k_jG(k)/\omega(k), \ k\in \Omega_j$. 
Since $\hat f$ is in $L^2(\R_k^n)$, it follows that $c(k_1,\cdots, k_{j-1},k_{j+1},\cdots, k_n)=0$ (a.e.).
Hence  $f=0$.
Thus $\ker(T_j^{\rm rel}(G)^*-i)=\{0\}$.  \hfill \qed

\medskip

\section{A Class of Generalized Time Operators}

In this section we consider spectral properties of  a class of generalized time operators.
Let $H$ be a self-adjoint operator on a complex Hilbert space ${\cal H}$
and $T$ be a symmetric operator on ${\cal H}$.

We call the operator $T$ a {\it generalized time operator} of $H$ 
if $e^{-itH}D(T)\subset D(T)$ for all $t\in \R$ and
there exists a bounded self-adjoint operator $C\not=0$ on ${\cal H}$
with $D(C)={\cal H}$ such that
\begin{equation}
Te^{-itH}\psi=e^{-itH}(T+tC)\psi, \quad \psi\in D(T).\label{GWWR}
\end{equation}
We call  $C$ the {\it noncommutative factor} for $(H,T)$.

The following facts are known:

\begin{The}\label{KR} Let $T$ be a generalized time operator of $H$ with noncommutative factor $C$.
\begin{list}{}{}
\item{{\rm (i)}}{\rm (\cite[Theorem 2.8]{Ar05})}
Let $H$ be semi-bounded and
\begin{equation}
CT\subset TC. \label{TC}
\end{equation}
Then $T$ is not essentially self-adjoint .
\item{{\rm (ii)}}{\rm (\cite[Corollary 5.3-(ii)]{Ar05})} If $C\geq 0$ or $C\leq 0$, then
$\sigma_{\rm p}(\overline{T}|[D(\overline{T})\cap (\ker C)^{\perp}])=\emptyset$.
\item{{\rm (iii)}}{\rm (\cite[Theorem 6.2-(ii)]{Ar05})} Let $H$ be bounded below. Then, for all $\beta>0$,
$e^{-\beta H}D(\overline{T})\subset D(\overline{T})$ and 
\begin{equation}
\overline{T}e^{-\beta H}\psi-e^{-\beta H}\overline{T}\psi=-i\beta e^{-\beta H}C\psi,
\quad \psi \in D(\overline{T}).
\end{equation}
\item{{\rm (iv)}}{\rm (\cite[Proposition 6.4, Corollary 6.6]{Ar05})}
The operators $H$ and $C$ strongly commute (i.e., their spectral measures commute)
and $H$ is reduced by $\overline{\Ran(C)}$.
\end{list}
\end{The}

In what follows,  $T$ is a generalized time operator of $H$ with  noncommutative
factor $C$ satisfying (\ref{TC}). 

\subsection{The case where $C$ has a non-zero eigenvalue}


We first consider the case  where $C$ has a non-zero eigenavlue $a \in \R\setminus\{0\}$,
i.e.,
 \begin{equation}
{\cal K}_a:=\ker (C-a)\not=\{0\}.
\end{equation}
We have the orthogonal decomposition
\begin{equation}
{\cal H}={\cal K}_a\oplus {\cal K}_a^{\perp}. \label{orth}
\end{equation}

Relation (\ref{TC}) implies that
\begin{equation}
C\overline{T}\subset \overline{T}C.
\end{equation}
Then it follows that $\overline{T}$ is reduced by ${\cal K}_a$ and hence by ${\cal K}_a^{\perp}$.
We denote the reduced part of $\overline{T}$ to ${\cal K}_a$  and ${\cal K}_a^{\perp}$ 
by $\overline{T}_a$ and $\overline{T}_a^{\perp}$ respectively. Hence we have
\begin{equation}
\overline{T}=\overline{T}_a\oplus \overline{T}_a^{\perp}
\end{equation}
relative to the orthogonal decomposition (\ref{orth}). Therefore
\begin{equation}
\sigma(T)=\sigma(\overline{T})=\sigma(\overline{T}_a)\cup \sigma(\overline{T}_a^{\perp}).
\end{equation}

By the strong commutativity of $H$ and $C$ (Theorem \ref{KR}-(iv)),
$H$ also is reduced by ${\cal K}_a$. We denote the reduced part of $H$ to ${\cal K}_a$
by $H_a$.

\begin{lem}\label{lem3-1} The operator $\overline{T}_a$ is a time operator of $H_a/a$.
\end{lem}

{\it Proof}. Let $\psi \in D(\overline{T}_a)=D(\overline{T})\cap {\cal K}_a$.
Then, for all $t\in \R$,  $e^{-itH_a}\psi=e^{-itH}\psi\in D(\overline{T})\cap {\cal K}_a
=D(\overline{T}_a)$
and, by (\ref{GWWR}), 
$$
\overline{T}_ae^{-itH_a}\psi=e^{-itH_a}(\overline{T}_a+ta)\psi.
$$
Thus the desired result follows.
\hfill \qed


\medskip

\begin{The} Let $T$  be a generalized time operator of $H$ with noncommutative factor $C$ 
satisfying (\ref{TC}).
Then the following {\rm (i)---(v)} hold:
\begin{list}{}{}
\item{{\rm (i)}} If $H_a$ is bounded below and $a>0$, then
$\sigma(\overline{T}_a)$ is either $\C$
or $\overline{\Pi}_+$. 
\item{{\rm (ii)}} If $H_a$ is bounded above and $a<0$, then
$\sigma(\overline{T}_a)$ is either $\C$
or $\overline{\Pi}_+$. 
\item{{\rm (iii)}} If $H_a$ is bounded above and $a>0$, then
$\sigma(\overline{T}_a)$ is either $\C$
or $\overline{\Pi}_-$. 
\item{{\rm (iv)}} If $H_a$ is bounded below and $a<0$, then $\sigma(\overline{T}_a)=\C$
or $\overline{\Pi}_-$.
\item{{\rm (v)}} If $H_a$ is bounded, then $\sigma(\overline{T}_a)=\C$. 
\end{list}
\end{The}


{\it Proof}. In (i) and (ii), $H_a/a$ is bounded below. Hence, Lemma \ref{lem3-1} and
Theorem \ref{th2-1}-(i) yield the results stated in (i) and (ii).
Similarly other cases follow.
\hfill \qed

\medskip


\subsection{The case where $\Ran(C)$ is closed}

We next consider the case where $\Ran(C)$ is closed. Then ${\cal H}$ is decomposed as
\begin{equation}
{\cal H}=\ker C\oplus \Ran(C).
\end{equation}
By the closed graph theorem,
there exists a constant $M>0$ such that
\begin{equation}
\|C\psi\|\geq M\|\psi\|, \quad \psi \in (\ker C)^{\perp}=\Ran(C). \label{C1}
\end{equation}
The operators $T$ and $C$ are reduced by $\Ran(C)$. We denote by $\widetilde T$ and $\widetilde C$
the reduced part of $T$ and $C$ to $\Ran(C)$ respectively.
The operator $\widetilde T$ is symmetric and $\widetilde C$ is a bounded self-adjoint operator
on $\Ran(C)$ which
is bijective with $\widetilde C^{-1}$ bounded. 
It follows from (\ref{TC}) that
\begin{equation}
\widetilde C\widetilde T\subset \widetilde T\widetilde C.\label{TC1}
\end{equation}

\begin{lem}\label{lem-TC} The operator
\begin{equation}
T_C:=\widetilde C^{-1}\widetilde T=\widetilde T\widetilde C^{-1}|D(\widetilde T)
\end{equation}
on $\Ran(C)$ is a symmetric operator.
\end{lem}

{\it Proof}. We have $D(T_C)=D(T)\cap \Ran(C)$. Hence $D(T_C)$ is dense in $\Ran(C)$.
Relation 
(\ref{TC1}) implies that 
\begin{equation}
\widetilde C^{-1}\widetilde T\subset \widetilde T\widetilde C^{-1}.
\end{equation}
Hence $T_C^*=\widetilde T^*\widetilde C^{-1}\supset \widetilde T\widetilde C^{-1}\supset
T_C$. Thus the desired result follows.
\hfill \qed


\medskip

By Theorem \ref{KR}-(iv), $H$ is reduced by $\Ran(C)$. We denote the reduced part of $H$
to $\Ran(C)$ by $\widetilde H$.

\begin{The} The operator $T_C$ is a time operator of $\widetilde H$  and
the following {\rm (i)---(iii)} hold:
\begin{list}{}{}
\item{{\rm (i)}} If $\widetilde H$ is bounded below, then
$\sigma(T_C)$ is either $\C$
or $\overline{\Pi}_+$. 
\item{{\rm (ii)}} If $\widetilde H$ is bounded above, then
$\sigma(T_C)$ is either $\C$
or $\overline{\Pi}_-$. 
\item{{\rm (iii)}} If $\widetilde H$ is bounded, then $\sigma(T_C)=\C$.
\end{list}
\end{The}


{\it Proof} Since $D(T_C)=D(T)\cap \Ran(C)$, it follows that $e^{-it\widetilde H}D(T_C)\subset
D(T_C)$ for all $t \in\R$. 
Using (\ref{GWWR}), one can see that  $T_C$ satisfies
$$
T_Ce^{-it\widetilde H}\psi=e^{-it\widetilde H}(T_C+t)\psi, \quad \psi \in D(T_C).
$$
These facts and Lemma \ref{lem-TC} imply that $T_C$ 
is a time operator of $\widetilde H$.
Then (i)---(iii)  follow from application of  Theorem \ref{th2-1}.
\hfill \qed

\medskip

\begin{ex}{\rm A simple example is given by the case where $C\not=0$ is 
an orthogonal projection. Then $T_C=\widetilde T$. To construct such examples, see 
\cite[\S 11]{Ar05}.
}
\end{ex}

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