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canonical commutation relation, Weak Weyl representation, Weyl representation,
Hamiltonian, time operator, spectrum
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\begin{document}
\title{\bf Construction of
a Weyl Representation from a Weak Weyl Representation
of the Canonical Commutation Relation}
\author{Asao Arai\thanks{
Corresponding author. The work is supported by the Grant-in-Aid No.17340032 for
Scientific Research from the 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} Some aspects of weak Weyl representations of
the canonical commutation relation (CCR) with one degree of freedom is considered
in relation to the theory of time operator in quantum mechanics.
It is shown that there exists
a general structure in which a Weyl representation of the CCR can be constructed from a weak Weyl representation of the CCR.
\end{abstract}
\noindent
{\bf Mathematics Subject Classification (2000)}. 81Q10, 47N50.
\noindent
{\bf Keywords}. canonical commutation relation, Weak Weyl representation, Weyl representation,
Hamiltonian, time operator, spectrum.
\section{Introduction}
Let $H$ be a self-adjoint operator on a complex Hilbert space $\MSH$ and
$T$ be a symmetric operator on $\MSH$. We say that the pair $(T,H)$ obeys the {\it weak Weyl
relation} if, for all $t \in \BBR$, $e^{-itH}\MSD(T)\subset \MSD(T)$ ($\MSD(T)$ denotes the domain of $T$)
and
\begin{equation}
Te^{-itH}\psi=e^{-itH}(T+t)\psi, \quad \forall \psi \in \MSD(T).
\end{equation}
From the representation theoretic point of view, the pair $(T,H)$ is called a {\it weak Weyl representation}
of the canonical commutation relation (CCR) with
one degree of freedom \cite{Ar05}. This comes from the following facts: (i)
every pair $(T,H)$ obeying the weak Weyl relation
satisfies the CCR:
$$
[T,H]\psi=i\psi, \quad \forall \psi \in \MSD(TH)\cap \MSD(HT),
$$
i.e., $(T,H)$ is a representation of the CCR with one degree of freedom,
but {\it the converse is not true} ; (ii) every weak Weyl representation
$(T,H)$ with $T$ {\it self-adjoint}
is a {\it Weyl representation} of the CCR, i.e., it satisfies the {\it Weyl relation}
\begin{equation}
e^{itT}e^{isH}=e^{-ist}e^{isH}e^{itT}, \quad s,t\in \BBR\label{WR}
\end{equation}
\cite[Proposition 2.1-(iii)]{Miya1}.
Detailed analyses of weak Weyl representations were first given by
Schm\"udgen \cite{Sch1,Sch2}(in the notation there,
$T=P, H=-Q$).
Then Miyamoto \cite{Miya1} used the weak Weyl relation
to develop a theory of time operator in quantum mechanics.
In the context where
$H$ is the Hamiltonian of a quantum system,
the operator $T$ is called a {\it time operator} with respect to (w.r.t.) $H$ \cite{Miya1}.
Motivated by this work, one of the present authors
extended the framework of the theory of time operator
to a more general one, introducing a notion of generalized
time operators \cite{Ar05}. We use the term \lq\lq{time operator}"
in the general context too.
Along this line of research, spectral analysis for time operators has been made
in \cite{Ar07}.
This paper is a continuation of the previous
work \cite{Ar05,Ar07}. The main purpose of it is to
point out that
there exists a remarkable structure in which
a Weyl representation can be constructed from a weak Weyl representation.
This structure, which may have physical significance too,
makes it possible, for example, to identify
the spectrum $\sigma(|H|)$ of $|H|$ (the absolute value of $H$) as $\sigma(|H|)=[0,\infty)$ under
some condition.
The present paper is organized as follows.
In Section 2, we state the main results.
Section 3 is devoted to proofs of them.
In the last section, we discuss some examples illustrating
the abstract general results established in Section 2.
\section{Main Results}
Before stating the main results of this paper,
we first recall some known facts.
The following fact is an interesting property
to be kept in mind on weak Weyl representations:
\begin{lem}\label{lem2-1}
Let $(T,H)$ be a weak Weyl representation of the CCR on the Hilbert space $\MSH$.
Suppose that
$H$ is semi-bounded (i.e., bounded below or bounded above).
\begin{list}{}{}
\item[{\rm (i)}] {\rm (\cite[Theorem 2.8]{Ar05})}
Then $T$ is not self-adjoint.
\item[{\rm (ii)}] {\rm (\cite[Theorem 2.1-(i)]{Ar07})}
If $H$ is bounded below, then the spectrum $\sigma(T)$ of $T$ is either $\BBC$ (the set of complex numbers) or
$\overline \Pi_+:=\{z\in \BBC| \Im z\geq 0\}$, the closed upper half-plane of $\BBC$.
\item[{\rm (iii)}] {\rm (\cite[Theorem 2.1-(ii)]{Ar07})}
If $H$ is bounded above, then $\sigma(T)$ is either $\BBC$ or
$\overline \Pi_-:=\{z\in \BBC| \Im z\leq 0\}$, the closed lower half-plane of $\BBC$.
\item[{\rm (iv)}] {\rm (\cite[Theorem 2.1-(iii)]{Ar07})}
If $H$ is bounded, then $\sigma(T)=\BBC$.
\end{list}
\end{lem}
For a closable linear operator $C$ on a Hilbert space, we denote its closure by $\overline C$.
Let $(T,H)$ be a weak Weyl representation of the CCR on the Hilbert space $\MSH$.
Then it is easy to see that $(\overline T,H)$ also is a weak Weyl representation of the CCR.
Hence, without loss of generality, one can assume that $T$ is closed.
In what follows we take this assumption, unless otherwise stated.
\begin{lem}{\rm (\cite[Proposition 2.1]{Miya1})}
There exists a dense subspace $\MSD\subset \MSH$ such that
$\MSD \subset \MSD(TH)\cap \MSD(HT)$ and
$H\MSD \subset \MSD$.
\end{lem}
This proposition implies that $\MSD(TH)\cap \MSD(HT)$ is dense
in $\MSH$.
Hence one can consider the adjoint $(HT)^*$ (resp. $(TH)^*$ )
of $HT$ (resp. $TH$) with $(HT)^*\supset TH$ (resp. $(TH)^*\supset HT$), which implies that $\MSD((HT)^*)$
(resp. $\MSD((TH)^*)$) is dense in $\MSH$.
Hence $HT$ and $TH$ are closable and
\begin{equation}
(TH)^*\supset \overline{HT}, \quad (HT)^*\supset \overline{TH}.
\end{equation}
Therefore the linear operator
\begin{eqnarray}
D&:=&\frac 12 (TH+\overline{HT})
\end{eqnarray}
on $\MSH$ is a symmetric operator.
The following result also is known:
\begin{lem}\label{abc}{\rm (\cite[Corollary 2]{Sch1}, \cite[Theorem 4.4]{Miya1})}
The self-adjoint operator $H$ is purely absolutely continuous.
\end{lem}
This result implies, in particular, that
the point spectrum $\sigma_{\rm p}(H)$
(the set of all the eigenvalues of $H$) is empty. In particular,
$0$ is not an eigenvalue of $H$.
Hence, via the functional calculus, we can define
a self-adjoint operator
\begin{equation}
L:=\log |H|.
\end{equation}
The first of the main results of this paper is as follows:
\begin{The}\label{th2-1} For all $t\in \BBR$, $e^{-itL}\MSD(D)=\MSD(D)$ and
\begin{equation}
De^{-itL}=e^{-itL}(D+t). \label{D}
\end{equation}
\end{The}
This theorem tells us that the pair $(D, L)$ is a weak Weyl representation
of the CCR.
Therefore Theorem \ref{th2-1} reveals
a general structure in which
every weak Weyl representation produces another weak Weyl representation.
Theorem \ref{th2-1} yields some corollaries.
\begin{cor}\label{cor2-0} Suppose that $|H|\geq c$ or $|H| \leq c$ for some constant $c>0$.
Then $D$ is not essentially self-adjoint. Moreover the following {\rm (i)}--{\rm (iii)}
hold:
\begin{list}{}{}
\item[{\rm (i)}] If $|H|\geq c$, then $\sigma(D)$ is either $\BBC$ or
$\overline \Pi_+$.
\item[{\rm (ii)}] If $|H|\leq c$, then $\sigma(D)$ is either $\BBC$ or
$\overline \Pi_-$.
\item[{\rm (iii)}] If $|H|$ is bounded with $00$, we define a function $f_{\varepsilon}$ on $\BBR$ by
$$
f_{\varepsilon}(\lambda):=e^{-it\log \sqrt{\lambda^2+\varepsilon^2}},
$$
It is easy to see that $f_{\varepsilon}$ is a continuously differentiable bounded function on $\BBR$ and $f_{\varepsilon}'$ is bounded on $\BBR$ with
\begin{eqnarray*}
&& |f_{\varepsilon}(\lambda)|=1,\quad
f_{\varepsilon}'(\lambda)=-\frac {it\lambda}{\lambda^2+\varepsilon^2}f_{\varepsilon}(\lambda).
\end{eqnarray*}
Hence we can apply Lemma \ref{lem2-2} to obtain that
\begin{equation}
Tf_{\varepsilon}(H)\psi-f_{\varepsilon}(H)T\psi=if'_{\varepsilon}(H)\psi, \quad
\psi \in \MSD(T). \label{Tf}
\end{equation}
Let $\phi\in \MSD(TH)$.
Then $H\phi \in \MSD(T)$. Hence
$$
Tf_{\varepsilon}(H)H\phi-f_{\varepsilon}(H)TH\phi=if'_{\varepsilon}(H)H\phi.
$$
It follows from the functional calculus that
\begin{eqnarray*}
&&\lim_{\varepsilon \to 0}if'_{\varepsilon}(H)H\phi=t e^{-itL}\phi, \quad \lim_{\varepsilon \to 0}f_{\varepsilon}(H)TH\phi=e^{-itL}TH\phi,\quad
\lim_{\varepsilon \to 0}f_{\varepsilon}(H)H\phi=He^{-itL}\phi.
\end{eqnarray*}
By the closedness of $T$, we conclude that
$He^{-itL}\phi \in \MSD(T)$ and
\begin{equation}
THe^{-itL}\phi=e^{-itL}(TH+t)\phi, \quad \phi \in \MSD(TH). \label{TH}
\end{equation}
For all $\eta \in \MSD(H)$ and $\psi \in \MSD(HT)$, we have by (\ref{Tf})
$$
\lang H\eta, Tf_{\varepsilon}(H)\psi\rang=
\lang \eta, f_{\varepsilon}(H)HT\psi\rang+
\lang \eta, iHf_{\varepsilon}'(H)\psi\rang.
$$
Hence $Tf_{\varepsilon}(H)\psi\in \MSD(H)$ and
$$
HTf_{\varepsilon}(H)\psi=f_{\varepsilon}(H)HT\psi+ iHf_{\varepsilon}'(H)\psi.
$$
Hence
$$
\lim_{\varepsilon \to 0}HTf_{\varepsilon}(H)\psi=
e^{-itL}HT\psi+te^{-itL}\psi.
$$
Therefore $e^{-itL}\psi \in \MSD(\overline{HT})$ and
$$
\overline{HT}e^{-itL}\psi=e^{-itL}(HT+t)\psi, \quad \psi \in \MSD(HT).
$$
By a simple limiting argument,
one can extend this result to all $\psi \in \MSD(\overline{HT})$ with
$e^{-itL}\MSD(\overline{HT})=\MSD(\overline{HT})$ and
\begin{equation}
\overline{HT}e^{-itL}\psi=e^{-itL}(\overline{HT}+t)\psi, \quad \psi \in \MSD(\overline{HT})\label{HT}
\end{equation}
It follows from (\ref{TH}) and (\ref{HT}) that
$$
e^{-itL}\MSD(TH)\cap \MSD(\overline{HT})=\MSD(TH)\cap \MSD(\overline{HT})
$$
and, for all $\psi \in \MSD(\overline{HT})\cap \MSD(TH)$,
\begin{eqnarray*}
&&De^{-itL}\psi=e^{-itL}(D+t)\psi.
\end{eqnarray*}
Thus the desired results follow.
\medskip
\subsection{Proof of Corollary \ref{cor2-0}}
Under the assumption that $|H|\geq c>0$ (resp. $|H|\leq c$),
we have $ L\geq \log c$ (resp. $L\leq \log c$).
Hence $L$ is semi-bounded.
Therefore, by Lemma \ref{lem2-1}-(i),
$D$ is not essentially self-adjoint.
(i) If $|H|\geq c$, then $L$ is bounded below as shown above. Hence, by Lemma \ref{lem2-1}-(ii),
$\sigma(D)$ is either $\BBC$ or
$\overline \Pi_+$.
(ii) If $|H|\leq c$, then $L$ is bounded above as already seen.
Hence, by Lemma \ref{lem2-1}-(ii), $\sigma(D)$ is either $\BBC$ or
$\overline \Pi_-$.
(iii) Under the present assumption, $L$ is bounded.
Hence, by Lemma \ref{lem2-1}-(iii), $\sigma(D)=\BBC$.
\medskip
\subsection{Proof of Corollary \ref{cor2-1}}
By (\ref{D}), we have $
e^{itL}\overline De^{-itL}=\overline D+t$.
Since $e^{itL}$ is unitary with $(e^{itL})^{-1}=e^{-itL}$, it follows that
$$
e^{itL}e^{is \overline D}e^{-itL}=e^{is(\overline D +t)}
=e^{is \overline D}
e^{ist}.
$$
Hence (\ref{W1}) holds.
\medskip
\subsection{Proof of Corollary \ref{cor2-2}}
By Corollary \ref{cor2-1}, $(\overline D,L)$ is a Weyl representation
of the CCR. Since $\mathscr{H}$ is separable by the present assumption,
the first half of Corollary \ref{cor2-2}
follows from the von Neumann uniqueness theorem (\cite{vN}, \cite[Theorem VIII.14]{RS1}).
Hence we have
\begin{eqnarray*}
&&\sigma (L)=\sigma(p)=\BBR, \quad \sigma_{\rm p}(L)=\sigma_{\rm p}(p)=\emptyset,\\
&&\sigma (\overline D)=\sigma(q)=\BBR, \quad \sigma_{\rm p}(\overline D)=\sigma_{\rm p}(q)=\emptyset.
\end{eqnarray*}
By the spectral mapping theorem, we have (\ref{specH}).
\hfill \qed
\medskip
\section{Examples}
\subsection{Quantum mechanics on the half-line $(0,\infty)$}
Let $\BBR_+:=\{\lambda |\lambda > 0\}=(0,\infty)$@
and define the operators $H$ and $T$ on $L^2(\BBR_+)$ as
follows:
\begin{equation}
H:=\hat \lambda_+ :=M_{\lambda}, \quad T:=-\hat p_+:=i\frac d{d\lambda}, \ \MSD(\hat p_+):=C_0^{\infty}(\BBR_+).
\end{equation}
Then it is easy to see that
$\hat \lambda_+$ is self-adjoint, $-\hat p_+$ is symmetric and
$(-\hat p_+,\hat \lambda_+)$ is a weak Weyl representation of the CCR.
As already pointed out in \cite[Example 2.1]{Ar07}, one has
\begin{equation}
\sigma(\hat \lambda_+)=[0,\infty), \quad \sigma(-\hat p_+)
=\overline{\Pi}_+.
\end{equation}
In the present case we have
\begin{eqnarray*}
&&L=\log |H|= B:=\log \hat \lambda_+, \\
&& D=A:=-\frac 12 (\hat p_+\hat \lambda_+ + \overline{\hat\lambda_+\hat p_+}).
\end{eqnarray*}
\begin{lem}\label{lem4-1} The operator $A$ is essentially self-adjoint
on $C_0^{\infty}(\BBR_+)$. Moreover
\begin{equation}
(e^{-i\theta \overline A}f)(\lambda)=e^{\theta/2}
f(e^{\theta}\lambda), \quad f\in L^2(\BBR_+).\label{DF}
\end{equation}
\end{lem}
{\it Proof}. Similar to the proof of the fact stated in Remark \ref{rem2-2}.
\hfill \qed
\medskip
\begin{rem}{\rm Every unitary operator $u(\theta)$
mentioned in
Remark \ref{rem2-2} is reduced by $L^2(\BBR_+)$. We denote its reduced part by $u_+(\theta)$.
Then $\{u_+(\theta)\}_{\theta\in \BBR}$ gives
a unitary representation on $L^2(\BBR_+)$ of the dilation group
$\{e^{\theta}\}_{\theta\in \BBR}$ acting on $\BBR_+$. The lemma shows that
$-\overline A$ is the generator of this unitary group.
}
\end{rem}
By Lemma \ref{lem4-1} and Corollary \ref{cor2-2},
$(A,B)$ is unitarily equivalent to a direct sum of the Schr\"odinger
representation $(q,p)$.
It may be instructive to
find explicitly the unitary operator implementing the equivalence.
Indeed this is possible as done below (cf. \cite[Appendix B]{Ar83}).
We denote by $\MSF$ the Fourier transform on
$L^2(\BBR)$:
$$
(\MSF u)(k):=\frac 1{\sqrt{2\pi}}\int_{\BBR}e^{-ikx}u(x)dx, \quad u \in L^2(\BBR), \ {\rm a.e.}
k\in \BBR,
$$
in the $L^2$-sense. For each $f \in L^2(\BBR_+)$, we define $f_e\in L^2(\BBR)$ by
$$
f_e(k):=f(e^k)e^{k/2}, \quad {\rm a.e.}k\in \BBR,
$$
and introduce $E:L^2(\BBR_+) \to L^2(\BBR)$ by
$$
Ef:=f_e, \quad f\in L^2(\BBR_+).
$$
Then $E$ is unitary.
Using $E$ and $\MSF$,
we define an operator $M:L^2(\BBR_+)\to L^2(\BBR)$ by
\begin{equation}
M:=\MSF^{-1} E.
\end{equation}
Obviously $M$ is unitary.
Explicitly we have
$$
(Mf)(x)=\frac 1{\sqrt{2\pi}}
\int_{\BBR}f(\lambda)\lambda ^{(\frac 12 +ix)-1} d\lambda, \quad x\in \BBR,
$$
for all $ f\in L^2(\BBR_+)$ such that
$$
\int_{\BBR_+}\frac{|f(\lambda)|}{\sqrt{\lambda}}d\lambda <\infty.
$$
Hence $M$ is a Mellin transform.
It is not so difficult to show that
$$
\quad MAM^{-1}=q, \quad MBM^{-1}=p.
$$
For every constant $m>0$,
the pair $(-\hat p_+, \hat \lambda_++m)$ is obviously a weak Weyl representation.
In this case we have $\hat \lambda_+ +m\geq m >0$.
Hence, by Corollary \ref{cor2-0}, the operator
$$
A_m:=-\frac 12 (\hat p_+(\hat \lambda_++m)+\overline{(\hat \lambda +m) \hat p_+})
$$
is not essentially self-adjoint. This is an interesting phenomenon too (one can prove this fact directly by computing
the deficiency indices of $A_m$). Moreover we can show that
$\sigma(A_m)=\overline \Pi_+$ (the proof is similar to that of \cite[Theorem 4.2]
{Ar07}, but, in the present case, it is very easy).
\subsection{The free Hamiltonian of a non-relativistic quantum particle in $\BBR$ and the
Aharonov-Bohm time operator}
Let $(q,p)$ be the Schr\"odinger representation of the CCR with one degree of freedom as in
(\ref{qp}).
For a constant $m>0$, we define
$$
H_0:=\frac 1{2m} p^2=-\frac 1{2m} \Delta_x,
$$
acting in $L^2(\BBR)$,
where $\Delta_x:=D_x^2$. The operator $H_0$
is called the one-dimensional free Hamiltonian
of a non-relativistic quantum particle with mass $m$.
It is well known that $H_0$ is a non-negative self-adjoint operator on $L^2(\BBR)$
and its spectrum is purely absolutely continuous with
$$
\sigma(H_0)=[0,\infty).
$$
Let
\begin{equation}
T_{\rm AB}:=\frac m{2}
\left(qp^{-1}+p^{-1}q\right)
\end{equation}
with domain
\begin{equation}
\MSD(T_{\rm AB}):=\MSF^{-1}C_0^{\infty}(\BBR\setminus\{0\}).
\end{equation}
It is easy to see that $T_{\rm AB}$ is a symmetric operator on $L^2(\BBR)$.
Moreover,
$(T_{\rm AB},H_0)$ is a weak Weyl representation
of the CCR \cite{Miya1}. The operator $T_{\rm AB}$ is called the {\it Aharonov-Bohm time
operator} \cite{AB}. By Lemma \ref{lem2-1}, $T_{\rm AB}$ is not essentially self-adjoint.
It is proven in \cite{Ar07} that
$$
\sigma(T_{\rm AB})=\overline{\Pi}_+.
$$
In the present example, we have
\begin{eqnarray*}
&&L=P_0:=\log H_0,\\
&&D=Q_0:=\frac 1{2}\left(T_{\rm AB}H_0+\overline{H_0T_{\rm AB}}\right) \supset q_0,
\end{eqnarray*}
with
$$
q_0:=\frac 12 D_{\rm S}|\MSD(T_{\rm AB}),
$$
where $D_{\rm S}$ is given by (\ref{DS}).
Therefore we can apply Corollary \ref{cor2-2} to
conclude that $(\overline Q_0,P_0)$ is unitarily equivalent to a direct sum
of the Schr\"odinger representation $(q,p)$ of the CCR.
In fact one can prove explicitly that $(\overline Q_0,P_0)$ is unitarily
equivalent to a two direct sum of the Schr\"odinger representation $(q,p)$ of the CCR.
This is essentially done in \cite[Appendix B]{Ar83}.
Hence we omit the details.
\subsection{A relativistic free Hamiltonian and a time operator of it }
A one-dimensional relativistic free Hamiltonian of a quantum particle
with mass $m\geq 0$ and without spin is given y
$$
H(m):=\sqrt{p^2+m^2}
$$
acting in $L^2(\BBR)$. A time operator w.r.t. $H(m)$ is defined by
$$
T(m):=\frac 12 \{H(m)p^{-1}q+qp^{-1}H(m)\}
$$
with domain $\MSD(T(m)):=\MSF^{-1}C_0^{\infty}(\BBR\setminus \{0\})$
(\cite[Example 11.4]{Ar05}).
In this case we have
\begin{eqnarray*}
&& L=P(m):=\log H(m),\\
&& D=Q(m):=\frac 12 (T(m)H(m)+\overline{H(m)T(m)})\supset
2(D_{\rm S}+mT_{\rm AB}).
\end{eqnarray*}
Hence the following results hold:
\begin{list}{}{}
\item[(i)] Consider the case $m=0$. Then $Q(0)$ is essentially self-adjoint.
Hence $(\overline{Q(0)}, P(0))$ is unitarily equivalent to a direct sum
of the Schr\"odinger representation $(q,p)$.
\item[(ii)] If $m>0$, then $Q(m)$ is not essentially self-adjoint.
\end{list}
In case (i), we can show, as in the case of the preceding example,
that $(\overline{Q(0)}, P(0))$ is unitarily equivalent to a two direct sum
of the Schr\"odinger representation $(q,p)$.
In case (ii), we can prove that
$\sigma(Q(m))=\overline \Pi_+$ (the proof is similar to that of
\cite[Theorem 4.3]{Ar07}).
\begin{rem}{\rm Results obtained in this section
can be extended easily to higher dimensional
versions of the examples.
}
\end{rem}
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\end{document}
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