# Conjugate Operators of $1$D-harmonic Oscillator

Fumio Hiroshima111Faculty of Mathematics, Kyushu University and Noriaki
Teranishi222Faculty of Science, Department of Mathematics, Hokkaido University

###### Abstract

Conjugate operator $T$ of 1D-harmonic oscillator
$N=\frac{1}{2}(p^{2}+q^{2}-{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})$ is defined by an
operator satisfying canonical commutation relation $[N,T]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$
on some domain but not necessarily dense. The angle operator
$T_{A}=\frac{1}{2}(\arctan q^{-1}p+\arctan pq^{-1})$ and Galapon operator
$T_{G}=i\sum_{n=0}^{\infty}\left(\sum_{m\neq
n}\frac{(v_{m},\cdot)}{n-m}v_{n}\right)$ are examples of conjugate operators,
where $\\{v_{n}\\}$ denotes the set of normalized eigenvectors of $N$. Let
$\mathscr{T}$ be a subset of conjugate operators of $N$. A classification of
$\mathscr{T}$ are given as
$\mathscr{T}=\mathscr{T}_{\\{0\\}}\cup\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}\cup\mathscr{T}_{\partial\mathbb{D}}$,
and $T_{A}\in\mathscr{T}_{\\{0\\}}$ and
$T_{G}\in\mathscr{T}_{\partial\mathbb{D}}$ are shown. Here the classification
is specified by a pair of parameters
$(\omega,m)\in\mathbb{C}\times\mathbb{N}$. Finally the time evolution
$T_{\omega,m}(t)=e^{itN}T_{\omega,m}e^{-itN}$ for $T_{\omega,m}\in\mathscr{T}$
is investigated, and show that $T_{\omega,m}(t)$ is periodic.

##  1 Introduction

Let us consider conjugate operator $T$ of 1D-harmonic oscillator

$N=\frac{1}{2}(p^{2}+q^{2}-{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}).$

Here $p=-i\frac{d}{dx}$ and $q=M_{x}$ be the multiplication by $x$. Both are
self-adjoint operators on ${L^{2}(\mathbb{R})}$. It satisfies that
$[N,T]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$. The commutator $[A,B]$ of linear
operators $A$ and $B$ is a linear operator defined by

$[A,B]=AB-BA$

on $\mathrm{\mathrm{D}}(AB)\cap\mathrm{\mathrm{D}}(BA)$. Here $\mathrm{D}(A)$
denotes the domain of $A$. In physics, the relationship between position and
momentum and that between energy and time are believed to be formally
complementary. Since the quantized position and momentum, $q$ and $p$, satisfy
the canonical commutation relation $[q,p]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$,
if $N$ is regarded as the energy of a quantum system, $T$ seems to be a
quantization of time. Then $T$ is often times called the time operator
associated with $N$. This paper does not however give any philosophical
discussions on time. It should be emphasized that time operator is just a name
of convenience. We give mathematical definitions of time operators and
conjugate operators in Definition 1.1.

In the previous paper [10] we consider two special conjugate operators, angle
operator $T_{A}$ and Galapon operator $T_{G}$, of $N$ from a mathematical
point of view. $T_{A}$ and $T_{G}$ have been mainly investigated so far from a
physical point of view, e.g., [13, 1, 18, 7, 6, 11, 12, 17] and as far as we
know, there is no firm and robust results on relationships between $T_{A}$ and
$T_{G}$ in a purely mathematical setting. In this paper we comprehensively
investigate conjugate operators including $T_{A}$ and $T_{G}$, and give a
relationship between $T_{A}$ and $T_{G}$. Furthermore we classify conjugate
operators by the pair of parameters $(\omega,m)\in\mathbb{C}\times\mathbb{N}$
and define disjoint three classes of conjugate operators.

###  1.1 Angle operators and Galapon operators

Define the annihilation operator $a$ and the creation operator $a^{\ast}$ in
${L^{2}(\mathbb{R})}$ by

$\displaystyle
a=\frac{1}{\sqrt{2}}(q+ip),\quad{a^{\ast}}=\frac{1}{\sqrt{2}}(q-ip),$

respectively. The canonical commutation relation (CCR)

$\displaystyle[a,{a^{\ast}}]={\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}},\quad[a,a]=0=[{a^{\ast}},{a^{\ast}}]$ (1.1)

holds true on a dense subspace of ${L^{2}(\mathbb{R})}$. The number operator
is defined by ${a^{\ast}}a$ which is actually the harmonic oscillator

${a^{\ast}}a=N,$

and $N$ is self-adjoint on
$\mathrm{D}(N)=\mathrm{D}(p^{2})\cap\mathrm{D}(q^{2})$. The normalized ground
state of $N$ is given by

$\displaystyle v(x)=\pi^{-1/4}e^{-x^{2}/2}.$

Note that $av=0$ and $Nv=0$. The normalized eigenvectors $v_{n}$ are given by

$\displaystyle v_{n}=\frac{1}{\sqrt{n!}}{{a^{\ast}}}^{n}v\quad n\geq 1.$ (1.2)

Here we write $a^{\ast}{}^{n}$ for $(a^{\ast})^{n}$. It satisfies that
$Nv_{n}=nv_{n}$ and ${\rm Spec}(N)=\mathbb{N}$. Here ${\rm Spec}(N)$ denotes
the spectrum of $N$. Note that notation $\mathbb{N}$ describes
$\\{0,1,2,\ldots,\\}$ including zero in this paper.

We recall the definitions of conjugate operators and time operators.

######  Definition 1.1 (Conjugate operators and time operators)

If a self-adjoint operator $A$ in Hilbert space $\mathcal{H}$ admits an
operator $B$ satisfying the canonical commutation relation:

$\displaystyle[A,B]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ (1.3)

on $D_{A,B}\subset\mathrm{D}(AB)\cap\mathrm{D}(BA)$ but $D_{A,B}\neq\\{0\\}$,
then $B$ is called a conjugate operator of $A$ and $D_{A,B}$ a CCR-domain of
(1.3). Furthermore if $B$ is symmetric, then $B$ is called a time operator of
$A$.

###### Remark 1.2

By the definitions, conjugate operators are not necessarily densely defined,
but the time operators are densely defined.

We shall construct a conjugate operator $T$ of $N$, i.e.,

$\displaystyle[N,T]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ (1.4)

on some domain. In the previous paper [10] we investigate the angle operator
e.g., [15] given by

$\displaystyle
T_{A}=-\frac{1}{2}\left(\arctan\left(q^{-1}p\right)\oplus\arctan\left(pq^{-1}\right)\right)$

and the so-called Galapon operator [5, 3, 9, 8] defined by

$\displaystyle T_{G}=i\sum_{n=0}^{\infty}\left(\sum_{m\neq
n}\frac{(v_{m},\cdot)}{n-m}v_{n}\right).$

It is known that $T_{G}$ is a bounded self-adjoint operator, and hence $T_{G}$
is a time operator of $N$. However $T_{A}$ is unbounded and not densely
defined. In particular it is not symmetric. Moreover
$[N,T_{G}]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ holds on a dense domain, but
$[N,T_{A}]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ on a non-dense domain.

Lots of conjugate operators of $N$ appeared in this paper are not densely
defined and hence not symmetric. So, we extend relation (1.4) to the so-called
ultra-weak canonical commutation relation. See Definition 4.6. For the sesqui-
linear form $\mathcal{T}$, the ultra-weak canonical commutation relation is
defined by

$\displaystyle\mathcal{T}[N\psi,\varphi]-\overline{\mathcal{T}[N\varphi,\psi]}=-i(\psi,\varphi).$
(1.5)

Here $\mathcal{T}[\psi,\varphi]$ is linear in $\varphi$ and anti-linear in
$\psi$, and $\mathcal{T}$ is not necessarily densely defined on
${L^{2}(\mathbb{R})}\oplus{L^{2}(\mathbb{R})}$. We can define an ultra-weak
time operator $\mathcal{T}_{A}$ associated with $T_{A}$, and it is shown in
[10] that $\mathcal{T}_{A}$ satisfies (1.5) on a dense domain. See (4.6).

###  1.2 Outline of the paper

In this paper, for notational simplicity, we use symbol $\ell^{2}$ to denote
the space $\ell^{2}(\mathbb{N})$ of square-summable complex-valued sequences
on $\mathbb{N}$. We also identify ${L^{2}(\mathbb{R})}$ with $\ell^{2}$ and
consider conjugate operators of $N$ not on ${L^{2}(\mathbb{R})}$ but on
$\ell^{2}$. The key ingredients are shift operators $L$ and $L^{\ast}$ on
$\ell^{2}$. Here $L$ denotes the left shift, and $L^{\ast}$ is the adjoint of
$L$ and hence the right shift on $\ell^{2}$. The main part of this paper is
Sections 4-7, where we investigate (1)-(5) below.
(1) We show that $T_{A}$ and $T_{G}$ can be represented in terms of $L$,
$L^{\ast}$ and $N$ as

$\displaystyle T_{G}=i\left\\{\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)+\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast})\right\\},$
$\displaystyle
T_{A}=\frac{i}{2}\left\\{\log\left(\sqrt{\frac{N+2{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}}L^{2}\right)\oplus\log\left(\sqrt{\frac{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{{N+2{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}}}L^{2}\right)\right\\}$

in Theorems 4.9 and 5.4.
(2) We generalize $T_{G}$ and $T_{A}$ as

$\displaystyle i\left\\{\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-g_{N}L)+\log\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}g_{N}^{-1}\right)\right\\},$
$\displaystyle\frac{i}{2}\left\\{\log\left(g_{N+2}L^{2}\right)\oplus\log\left(\tilde{g}_{N+2}L^{2}\right)\right\\},$

respectively, in Sections 4 and 5. Here $g_{.}$ is a map from $\mathbb{N}$ to
$\mathbb{C}$.
(3) We construct a general class of conjugate operators in Section 6. Let

$\displaystyle L_{\omega,m}=\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-{L}^{m},\quad(\omega,m)\in\mathbb{C}\times\mathbb{N}.$

We define $T_{\omega,m}$ by

$\displaystyle T_{\omega,m}=\frac{i}{m}\log L_{\omega,m}.$ (1.6)

We can see that $T_{\omega,m}$ is conjugate operators of $N$. Let $\mathbb{D}$
be the open unit disc on the complex plan $\mathbb{C}$. We can see that
$T_{\omega,m}$ with $\omega\in\partial\mathbb{D}$ is bounded and there exists
a dense CCR-domain, and we say that

$\displaystyle T_{\omega,m},\quad m\geq 1,\ \omega\in\partial\mathbb{D}$

are general Galapon operators. It is actually seen that
$T_{G}=T_{1,1}+T_{1,1}^{\ast}$. On the other hand general angle operators are
introduced by

$\displaystyle T_{\omega,m},\quad m\geq 1,\ \omega=0.$ (1.7)

Let

$\displaystyle\mathscr{T}=\\{T_{\omega,m}\mid\omega\in\overline{\mathbb{D}},m\geq
1\\}.$

We divide $\mathscr{T}$ into three disjoint families by
$\omega\in\overline{\mathbb{D}}$. We have

$\displaystyle\mathscr{T}=\mathscr{T}_{\\{0\\}}\cup\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}\cup\mathscr{T}_{\partial\mathbb{D}}.$

Here

$\displaystyle\mathscr{T}_{\\{0\\}}=\\{T_{\omega,m}\mid\omega=0,m\geq 1\\},$
$\displaystyle\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}=\\{T_{\omega,m}\mid
0<|\omega|<1,m\geq 1\\},$
$\displaystyle\mathscr{T}_{\partial\mathbb{D}}=\\{T_{\omega,m}\mid|\omega|=1,m\geq
1\\}.$

We study CCR-domains for each conjugate operators in $\mathscr{T}$.
(5) Let $H$ be a self-adjoint operator. Then the strong time operator $T_{H}$
is defined by the weak Weyl relation

$T_{H}e^{-itH}\supset e^{-itH}(T_{H}+t),\quad t\in\mathbb{R}.$

It is established that if $T_{H}$ is a strong time operator for $H$, then the
spectrum of $H$ is purely continuous spectrum. We can show that conjugate
operators in $\mathscr{T}$ satisfies a discrete version of weak Weyl relations
in Section 7. Furthermore we show that
$T_{\omega,m}(t)=e^{itN}T_{\omega,m}e^{-itN}$ is periodic with period
$2\pi/m$.

##  2 Technical tools

###  2.1 Super coherent vectors

We define the exponent operator $e^{A}$ for a non self-adjoint operator $A$ in
a Hilbert space $\mathcal{H}$ by the geometric series:

$\displaystyle\mathrm{D}\left(e^{A}\right)=\left\\{f\in\bigcap_{k=0}^{\infty}\mathrm{D}(A^{k})\
\middle|\ \lim_{K\to\infty}\sum_{k=0}^{K}\frac{1}{k!}A^{k}f\
\text{exists}\right\\},$ $\displaystyle
e^{A}f=\sum_{k=0}^{\infty}\frac{1}{k!}A^{k}f,\quad
f\in\mathrm{D}\left(e^{A}\right).$

It should be written as $\sum_{k\in\mathbb{N}}\frac{1}{k!}A^{k}$ for $e^{A}$,
but we write $e^{A}$ for the notational simplicity. We define the set
$\mathcal{C}$ consisting of coherent vectors by

$\displaystyle\mathcal{C}=\mathrm{LH}\left\\{e^{\beta{a^{\ast}}}v\ \Big{|}\
\beta\in\mathbb{C}\right\\}.$

Note that $e^{\beta{a^{\ast}}}$ is an unbounded operator, but
$v\in\mathrm{D}(e^{\beta{a^{\ast}}})$ for any $\beta\in\mathbb{C}$, and it is
given by

$\displaystyle
e^{\beta{a^{\ast}}}v(x)=\pi^{-1/4}e^{\beta^{2}/2}\exp\left(-\frac{(x-\sqrt{2}\beta)^{2}}{2}\right).$

It is well known that $\mathcal{C}$ is dense in ${L^{2}(\mathbb{R})}$. We set
$w_{\beta}=e^{-|\beta|^{2}/2}e^{\beta{a^{\ast}}}v$, where $e^{-|\beta|^{2}/2}$
is the normalize constant such that $\|w_{\beta}\|=1$. It is shown that
$w_{\beta}\in\mathrm{D}({a^{\ast}}^{n})$ and
${{a^{\ast}}}^{n}w_{\beta}=e^{-|\beta|^{2}/2}e^{\beta{a^{\ast}}}{{a^{\ast}}}^{n}v$.
Moreover one obtains that

$\displaystyle aw_{\beta}=\beta w_{\beta}.$

Thus $w_{\beta}$ is an eigenvector of $a$ with eigenvalue
$\beta\in\mathbb{C}$. We also see that $e^{\gamma
a}w_{\beta}=e^{\gamma\beta}w_{\beta}$ for any $\beta,\gamma\in\mathbb{C}$.

Next let us consider vectors of the form $e^{\beta{{a^{\ast}}}^{2}/2}v$.
Exponent of ${{a^{\ast}}}^{2}$ is also unbounded. Although
$e^{\beta{a^{\ast}}}v\in{L^{2}(\mathbb{R})}$ for any $\beta\in\mathbb{C}$, it
is not clear that $e^{\beta{{a^{\ast}}}^{2}/2}v\in{L^{2}(\mathbb{R})}$.

###### Lemma 2.1

Let $\beta\in\mathbb{C}$. Then
$v\in\mathrm{D}\left(e^{\beta{a^{\ast}}^{2}/2}\right)$ if and only if
$|\beta|<1$. Moreover,
$e^{\beta{a^{\ast}}^{2}/2}v\in\mathrm{D}({a^{\ast}}^{n})$ for any $|\beta|<1$
and $n\in\mathbb{N}$.

Proof: Let $|\beta|<1$. By the Rodrigues formula and orthogonality of the
Hermite polynomials $H_{k}$,

$\displaystyle\left\|\sum_{k=0}^{K}\frac{1}{k!}\left(\frac{\beta}{2}\right)^{k}{{a^{\ast}}}^{2k}v\right\|^{2}=\left\|\sum_{k=0}^{K}\frac{\beta^{k}}{2^{2k}k!}H_{2k}v\right\|^{2}=\sum_{k=0}^{K}\frac{|\beta|^{2k}}{(2^{2k}k!)^{2}}2^{2k}(2k)!=\sum_{k=0}^{K}\frac{(2k-1)!!}{(2k)!!}|\beta|^{2k}.$

Let $K\to\infty$. Then
$\|e^{\beta{a^{\ast}}^{2}/2}v\|^{2}=(1-|\beta|^{2})^{-1/2}$ holds. This shows
that $v\in\mathrm{D}\left(e^{\beta{a^{\ast}}^{2}/2}\right)$ if and only if
$|\beta|<1$. The statement that
$e^{\beta{a^{\ast}}^{2}/2}v\in\mathrm{D}\left({a^{\ast}}^{n}\right)$ is
similarly proven. $\Box$

We call $e^{\beta{{a^{\ast}}}^{2}/2}v$ a super coherent vector and denote the
linear hull of super coherent vectors by

$\displaystyle\mathcal{S}=\mathrm{LH}\left\\{e^{\beta{{a^{\ast}}}^{2}/2}v\
\Big{|}\ |\beta|<1\right\\}.$

The vectors contained in $\mathcal{S}$ play an important role for studying the
angle operator.

###  2.2 Wiener-Itô decomposition

As is mentioned in Section 1 we shall investigate conjugate operators on
$\ell^{2}$ for ${L^{2}(\mathbb{R})}$. The first task is to make the
identification between $\ell^{2}$ and ${L^{2}(\mathbb{R})}$ clear, and we
transform $a$ and $a^{\ast}$ on ${L^{2}(\mathbb{R})}$ to operators on
$\ell^{2}$. By the Wiener-Itô decomposition we have

$\displaystyle{L^{2}(\mathbb{R})}=\bigoplus_{n=0}^{\infty}L_{n},$

where $L_{n}=\mathrm{LH}\\{v_{n}\\}$ is the one-dimensional linear subspace
spanned by eigenvectors $v_{n}$ of $N$ with eigenvalue $n\in\mathbb{N}$. For
each $f\in{L^{2}(\mathbb{R})}$, there exists
$(c_{n})_{n\in\mathbb{N}}\in\ell^{2}$ so that
$f=\sum_{n=0}^{\infty}c_{n}v_{n}$. Actually it is given by $c_{n}=(v_{n},f)$.
We often use $(c_{0},c_{1},c_{2},\ldots)$ for $(c_{n})_{n\in\mathbb{N}}$. By
the unitary map

$Uf=(c_{n})_{n\in\mathbb{N}},$ (2.1)

we can identify ${L^{2}(\mathbb{R})}$ with $\ell^{2}$. We set
$\xi_{n}=Uv_{n}\in\ell^{2}$. It is seen that

$\displaystyle UaU^{\ast}\colon\xi_{n}\mapsto\sqrt{n}\xi_{n-1},\quad n\geq 1,$
$\displaystyle
U{a^{\ast}}U^{\ast}\colon\xi_{n}\mapsto\sqrt{n+1}\xi_{n+1},\quad n\geq 0.$

For the notational convenience, we write $a$ for $UaU^{\ast}$. Furthermore
$Uv$ is denoted by $\Omega$. Then $\Omega=(1,0,0,\ldots)$ and
$\xi_{n}=(0,\ldots,0,\stackrel{{\scriptstyle n+1_{th}}}{{1}},0,\ldots)$ for
$n\geq 0$. Note that $\xi_{0}=\Omega$. In the following, our investigation is
transformed on $\ell^{2}$ instead of on ${L^{2}(\mathbb{R})}$ and we fix
$\\{\xi_{n}\\}_{n\in\mathbb{N}}$ as a complete orthonormal system of
$\ell^{2}$. Let $\ell_{\mathrm{fin}}^{2}$ be the finite particle subspace
defined by

$\displaystyle\ell_{\mathrm{fin}}^{2}=\left\\{\sum_{n=0}^{m}c_{n}\xi_{n}\
\middle|\ m\in\mathbb{N},\ c_{n}\in\mathbb{C}\right\\}.$

It is dense, and commutator $[a,{a^{\ast}}]$ and polynomials in $a$ and
$a^{\ast}$ are well defined on this space. Thus algebraic computations on $a$
and $a^{\ast}$ can be done on $\ell_{\mathrm{fin}}^{2}$.

###  2.3 Shift operators $\bm{L}$ and $\bm{L^{\ast}}$ on $\bm{\ell^{2}}$

Let $L$ be the left-shift and the adjoint $L^{\ast}$ the right-shift on
$\ell^{2}$, which are defined by

$\displaystyle L\xi_{n}=\begin{cases}\xi_{n-1}&n\geq 1,\\\ 0&n=0,\end{cases}$
$\displaystyle L^{\ast}\xi_{n}=\xi_{n+1}.$

We can see that

$\displaystyle LL^{\ast}={\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}},$ $\displaystyle
L^{\ast}L={\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-P_{\\{\Omega\\}},$

where $P_{\\{\Omega\\}}$ is the projection to the one-dimensional subspace
spanned by $\Omega$. In general an arbitrary closed operator $A$ can be
decomposed as $A=V|A|$, where $|A|=(A^{\ast}A)^{1/2}$ is a positive self-
adjoint operator and $V$ a partial isometry such that the initial space is
$(\mathop{\mathrm{Ker}}A)^{\perp}$ and the final space $\overline{{\rm
Ran}A}$. $A=V|A|$ is called the polar decomposition of $A$. $V$ is uniquely
determined by these properties together with the addition condition
$\mathop{\mathrm{Ker}}|A|=\mathop{\mathrm{Ker}}V$. In terms of $L$, $L^{\ast}$
and $N$, both of $a$ and $a^{\ast}$ can be represented as

$\displaystyle a=L\sqrt{N}=\sqrt{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}L,$
$\displaystyle{a^{\ast}}=L^{\ast}\sqrt{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}=\sqrt{N}L^{\ast}.$

Note that $N={a^{\ast}}a$ and $N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}=a{a^{\ast}}$. These
are just the polar decompositions of closed operators $a$ and ${a^{\ast}}$.
Note that

$\displaystyle\mathop{\mathrm{Ker}}a=\\{c\,\Omega\mid
c\in\mathbb{C}\\}=\mathop{\mathrm{Ker}}\sqrt{N},$
$\displaystyle\mathop{\mathrm{Ker}}{a^{\ast}}=\\{0\\}=\mathop{\mathrm{Ker}}\sqrt{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}.$

$L$ is a partial isometry such that the initial space is
$(\mathop{\mathrm{Ker}}a)^{\perp}=\\{c\,\Omega\mid
c\in\mathbb{C}\\}^{\perp}\cong\bigoplus_{n=1}^{\infty}L_{n}$ and the final
space $\ell^{2}$. We also see that

$\displaystyle[N,L^{\ast}]\subset L^{\ast},\quad[N,L]\subset-L.$ (2.2)

Then $[N,{L^{\ast}}^{k}]\subset k{L^{\ast}}^{k}$ and $[N,L^{k}]\subset-kL^{k}$
hold true. Let $\mathbb{D}=\\{z\in\mathbb{C}\mid|z|<1\\}$ be the open unit
disc in $\mathbb{C}$. Notice that

$e^{\alpha
NL^{\ast}}\Omega=(1,\alpha,\alpha^{2},\alpha^{3},\ldots),\quad\alpha\in\mathbb{D}.$

It can be seen that

$\displaystyle Le^{\alpha NL^{\ast}}\Omega=\alpha e^{\alpha
NL^{\ast}}\Omega,\quad\alpha\in\mathbb{D}.$ (2.3)

Thus $e^{\alpha NL^{\ast}}\Omega$ is an eigenvector of $L$ with the eigenvalue
$\alpha$.

###### Lemma 2.2

We have ${\rm Spec}(L)={\rm Spec}(L^{\ast})=\overline{\mathbb{D}}$, ${\rm
Spec}_{\rm p}(L)=\mathbb{D}$ and ${\rm Spec}_{\rm p}(L^{\ast})=\emptyset$.

Proof: Since $\|L\|\leq 1$, ${\rm Spec}(L)\cup{\rm
Spec}(L^{\ast})\subset\overline{\mathbb{D}}$. By (2.3), we see that
$\mathbb{D}\subset{\rm Spec}_{\rm p}(L)$ and ${\rm Spec}(L)={\rm
Spec}(L^{\ast})=\overline{\mathbb{D}}$. Let $|\alpha|=1$. The relation
$L\varphi=\alpha\varphi$ implies $\varphi=(\alpha^{n})_{n\in\mathbb{N}}$ but
$\varphi\not\in\ell^{2}$. Hence ${\rm Spec}_{\rm p}(L)=\mathbb{D}$. Let
$|\alpha|\leq 1$ and $L^{\ast}\varphi=\alpha\varphi$. Then
$(L^{\ast}\varphi)_{0}=0=\alpha\varphi_{0}$ and hence
$(L^{\ast}\varphi)_{1}=0$. Repeating this procedure, we see that $\varphi=0$.
Thus we have ${\rm Spec}_{\rm p}(L^{\ast})=\emptyset$. $\Box$

Let $k\in\mathbb{N}$. Similarly to the proof of Lemma 2.2 we can see that, for
any $k\alpha\in\mathbb{D}$, $e^{\alpha N{L^{\ast}}^{k}}\Omega$ is an
eigenvector of $L^{k}$. Since

$\displaystyle e^{\alpha
N{L^{\ast}}^{k}}\Omega=(1,\underbrace{0,\ldots,0}_{k-1},k\alpha,\underbrace{0,\ldots,0}_{k-1},(k\alpha)^{2},\underbrace{0,\ldots,0}_{k-1},(k\alpha)^{3},\ldots),$

it is actually given by

$\displaystyle L^{k}e^{\alpha N{L^{\ast}}^{k}}\Omega=k\alpha e^{\alpha
N{L^{\ast}}^{k}}\Omega.$ (2.4)

We can also extend this to a general version. Let $f$ be a complex-valued
function on $\mathbb{N}$. Set $f(N)=f_{N}$. Then

$\displaystyle e^{\alpha
f_{N}{L^{\ast}}^{k}}\Omega=\biggl{(}1,\underbrace{0,\ldots,0}_{k-1},f(k)\alpha,\underbrace{0,\ldots,0}_{k-1},\frac{f(2k)f(k)}{2!}\alpha^{2},\underbrace{0,\ldots,0}_{k-1},\frac{f(3k)f(2k)f(k)}{3!}\alpha^{3},\ldots\biggr{)}.$
(2.5)

For a given function $f$, we shall find $g$ such that $e^{\alpha
f_{N}{L^{\ast}}^{k}}\Omega$ is an eigenvector of $g_{N+k}L^{k}$ in Sections 4
and 5.

##  3 Phase operators

We give the definition of $\log A$ for a linear operator $A$. It is emphasised
that $A$ is not necessarily self-adjoint.

######  Definition 3.1 ($\log A$)

Let $A$ be a linear operator on a Hilbert space $\mathcal{H}$. We define $\log
A$ by

$\displaystyle\mathrm{D}(\log
A)=\left\\{f\in\bigcap_{k=0}^{\infty}\mathrm{D}(A^{k})\ \middle|\
\lim_{K\to\infty}\sum_{k=1}^{K}\frac{1}{k}({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A)^{k}f\text{ exists }\right\\},$ $\displaystyle\log
Af=-\sum_{k=1}^{\infty}\frac{1}{k}({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-A)^{k}f,\quad
f\in\mathrm{D}(\log A).$

Note that in general

$\displaystyle\log(AB)\neq\log A+\log B,$ (3.1) $\displaystyle\log\alpha
A\neq\log\alpha+\log A,\quad\alpha\in\mathbb{C}.$ (3.2)

We define the so-called phase operator $\rho$ on $\ell^{2}$ by

$\displaystyle\rho=-\frac{i}{2}(\log a-\log{a^{\ast}}).$ (3.3)

The phase operator $\rho$ has been studied in the large number of literatures
e.g., [18, 11, 16] as the conjugate operator of $N$, but it does not appear to
be obvious in the definition of $\log a$ and $\log a^{\ast}$.

###### Lemma 3.2

We have $\mathrm{D}(\log a)\cap\ell_{\mathrm{fin}}^{2}=\\{0\\}$ and
$\mathrm{D}(\log{a^{\ast}})=\\{0\\}$. In particular,
$\mathrm{D}(\rho)=\\{0\\}$.

Proof: Let $\psi=\sum_{l=0}^{m}c_{l}\xi_{l}\in\ell_{\mathrm{fin}}^{2}$ with
$c_{m}\neq 0$. Since

$\displaystyle\lim_{K\to\infty}\left|\left(\xi_{m},\sum_{k=1}^{K}\frac{1}{k}({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-a)^{k}\psi\right)\right|=\lim_{K\to\infty}\left|\sum_{k=1}^{K}\frac{1}{k}\left(({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-a^{\ast})^{k}\xi_{m},\psi\right)\right|=\lim_{K\to\infty}\left|(\xi_{m},\psi)\right|\sum_{k=1}^{K}\frac{1}{k}=\infty,$

we see that $\mathrm{D}(\log a)\cap\ell_{\mathrm{fin}}^{2}=\\{0\\}$. Let
$\psi^{\prime}=\sum_{l=m^{\prime}}^{\infty}c_{l}\xi_{l}\in\ell^{2}$ with
$c_{m^{\prime}}\neq 0$. Similarly, it follows that

$\displaystyle\lim_{K\to\infty}\left|\left(\xi_{m^{\prime}},\sum_{k=1}^{K}\frac{1}{k}({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-a^{\ast})^{k}\psi^{\prime}\right)\right|=\lim_{K\to\infty}\left|(\xi_{m^{\prime}},\psi^{\prime})\right|\sum_{k=1}^{K}\frac{1}{k}=\infty.$

Hence $\mathrm{D}(\log{a^{\ast}})=\\{0\\}$. $\Box$

From Lemma 3.2 we can see that $\rho$ is not well defined on
$\ell_{\mathrm{fin}}^{2}$. This fact is fatal to consider the phase operator
$\rho$ on $\ell^{2}$. Therefore, when we investigate $\rho$, we have to
consider the domain of $\rho$ carefully.

##  4 Angle operators

###  4.1 Angle operators by $\bm{a}$ and $\bm{a^{\ast}}$

It is said that a relationship between the angle operator $T_{A}$ and the
phase operator $\rho$ is given as $T_{A}=(\pi/2){\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-\rho+G(N)$ in e.g., [11, (32)]. However it seems to be formal
and as was seen in Lemma 3.2 the phase operator $\rho$ is not well defined on
$\ell_{\mathrm{fin}}^{2}$.

In order to see a rigorous relationship between phase operator $\rho$ and the
angle operator $T_{A}$, we express $T_{A}$ in terms of the creation operators
and the annihilation operators. Define

$\displaystyle\phi=\frac{1}{\sqrt{2}}(a+{a^{\ast}}).$

We also define two disjoint subspaces of ${L^{2}(\mathbb{R})}$ by

$\displaystyle\mathcal{C}_{0}=\mathrm{LH}\left\\{e^{-\alpha x^{2}/2}\ \Big{|}\
\alpha\in(0,1)\right\\},$
$\displaystyle\mathcal{C}_{1}=\mathrm{LH}\left\\{xe^{-\alpha x^{2}/2}\
\Big{|}\ \alpha\in(0,1)\right\\}.$

Note that $\mathcal{C}_{0}\perp\mathcal{C}_{1}$ and
$\mathcal{C}_{0}+\mathcal{C}_{1}$ is dense. In this paper, $\\#$ denotes
$0\mbox{ or }1$. We transform $\mathcal{C}_{\\#}$ to the subspaces of
$\ell^{2}$ by $U$ defined by (2.1).

###### Lemma 4.1

Suppose that $0<\alpha<1$. Then

$\displaystyle Ue^{-\alpha
x^{2}/2}=\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega,$
$\displaystyle Uxe^{-\alpha
x^{2}/2}=\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}{\frac{\sqrt{2}}{1+\alpha}}{a^{\ast}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega.$

Proof: We set $\beta=(1-\alpha)/2$. Since $UxU^{\ast}=\phi$ and
$Ue^{-x^{2}/2}=\pi^{1/4}\Omega$, we have

$\displaystyle Ue^{-\alpha x^{2}/2}$
$\displaystyle=Ue^{-(\alpha-1)x^{2}/2}U^{\ast}Ue^{-x^{2}/2}=\pi^{1/4}\sum_{n=0}^{\infty}\frac{1}{n!}\beta^{n}\phi^{2n}\Omega$
$\displaystyle=\pi^{1/4}\sum_{k=0}^{\infty}\left(\sum_{n\geq
k}\frac{1}{(n-k)!}\frac{(2n-1)!!}{2^{n-k}(2k-1)!!}\beta^{n}\right)\frac{1}{k!}\frac{{a^{\ast}}^{2k}}{2^{k}}\Omega$
$\displaystyle=\pi^{1/4}\sum_{k=0}^{\infty}\frac{1}{\sqrt{1-\beta}}\left(\frac{\beta}{1-\beta}\right)^{k}\frac{1}{k!}\frac{{a^{\ast}}^{2k}}{2^{k}}\Omega$
$\displaystyle=\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega.$

Hence we have

$\displaystyle Uxe^{-\alpha
x^{2}/2}=\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\phi\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega=\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\frac{\sqrt{2}}{1+\alpha}{a^{\ast}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega.$

Then the lemma is proven. $\Box$

Disjoint sets of super coherent vectors is defined by

$\displaystyle\mathcal{S}_{0}=\mathrm{LH}\left\\{e^{\beta{a^{\ast}}^{2}/2}\Omega\
\Big{|}\ \beta\in(0,1)\right\\},$
$\displaystyle\mathcal{S}_{1}=\mathrm{LH}\left\\{{a^{\ast}}e^{\beta{a^{\ast}}^{2}/2}\Omega\
\Big{|}\ \beta\in(0,1)\right\\}.$

Note that $\mathcal{S}_{0}\perp\mathcal{S}_{1}$ and
$\mathcal{S}_{0}\cup\mathcal{S}_{1}$ is dense. Lemma 4.1 also shows that
$U\mathcal{C}_{\\#}=\mathcal{S}_{\\#}$. In order to express $T_{A}$ in terms
of $a$ and $a^{\ast}$, we need the inverse of ${a^{\ast}}$. The operator
${a^{\ast}}^{-1}$ is defined by

$\displaystyle\mathrm{D}\left({a^{\ast}}^{-1}\right)=\left\\{\sum_{n=0}^{\infty}c_{n}\xi_{n}\in\ell^{2}\
\bigg{|}\ c_{0}=0\right\\},$
$\displaystyle{a^{\ast}}^{-1}\sum_{n=1}^{\infty}c_{n}\xi_{n}=\sum_{n=1}^{\infty}\frac{c_{n}}{\sqrt{n}}\xi_{n-1}.$

We consider $\log({a^{\ast}}^{-1}a)$ and $\log(a{a^{\ast}}^{-1})$. Note that
${a^{\ast}}^{-1}$ is well defined on ${\rm Ran}L^{\ast}$.

###### Lemma 4.2

We have (1) and (2).

  * (1)

$\mathcal{S}_{0}\subset\mathrm{D}\left(\log({a^{\ast}}^{-1}a)\right)$ and

$\displaystyle\log\left({a^{\ast}}^{-1}a\right)e^{\beta{a^{\ast}}^{2}/2}\Omega=(\log\beta)e^{\beta{a^{\ast}}^{2}/2}\Omega,\quad
0<\beta<1.$

  * (2)

$\mathcal{S}_{1}\subset\mathrm{D}\left(\log(a{a^{\ast}}^{-1})\right)$ and

$\displaystyle\log\left(a{a^{\ast}}^{-1}\right){a^{\ast}}e^{\beta{a^{\ast}}^{2}/2}\Omega=(\log\beta){a^{\ast}}e^{\beta{a^{\ast}}^{2}/2}\Omega,\quad
0<\beta<1$

Proof: Since $e^{\beta{a^{\ast}}^{2}/2}\Omega$ is an eigenvector of
${a^{\ast}}{}^{-1}a$ with the eigenvalue $\beta$, for all $n\in\mathbb{N}$, we
see that $({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-{{a^{\ast}}}^{-1}a)^{n}e^{\beta{a^{\ast}}^{2}/2}\Omega=(1-\beta)^{n}e^{\beta{a^{\ast}}^{2}/2}\Omega$.
Hence by the definition of $\log\left({a^{\ast}}^{-1}a\right)$ we obtain that
$\log\left({a^{\ast}}^{-1}a\right)e^{\beta{a^{\ast}}^{2}/2}\Omega=(\log\beta)e^{\beta{a^{\ast}}^{2}/2}\Omega$.
The proof of (2) is the same as that of (1). $\Box$

###### Lemma 4.3

Both $\log({a^{\ast}}^{-1}a)$ and $\log(a{a^{\ast}}^{-1})$ are unbounded.

Proof: By Lemma 4.2, we can see that $e^{\beta{a^{\ast}}^{2}/2}\Omega$ is an
eigenvector of $\log\left({a^{\ast}}^{-1}a\right)$ with eigenvalues
$\log\beta$. Similarly ${a^{\ast}}e^{\beta{a^{\ast}}^{2}/2}\Omega$ is an
eigenvector of $\log\left(a{a^{\ast}}^{-1}\right)$ with eigenvalues
$\log\beta$. Since $0<\beta<1$, we have

$\displaystyle{\rm Spec}\left(\log\left({a^{\ast}}^{-1}a\right)\right)\cap{\rm
Spec}\left(\log\left(a{a^{\ast}}^{-1}\right)\right)\supset(-\infty,0).$

Hence the lemma follows. $\Box$

We recall a definition of $\arctan(A)$ for a linear operator $A$.

###### Definition 4.4

Let $A$ be a linear operator on a Hilbert space $\mathcal{H}$. We define a
linear operator $\arctan(A)$ as follows:

$\displaystyle\mathrm{D}(\arctan(A))=\left\\{\varphi\in\bigcap_{k=0}^{\infty}\mathrm{D}\left(A^{2k+1}\right)\
\middle|\
\lim_{K\to\infty}\sum_{k=0}^{K}\frac{(-1)^{k}}{2k+1}A^{2k+1}\varphi\mbox{
exists}\right\\},$
$\displaystyle\arctan(A)\varphi=-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}A^{2k+1}\varphi,\quad\varphi\in\mathrm{D}(\arctan(A)).$

Now we transform $T_{A}$ to an operator on $\ell^{2}$. Let

$\displaystyle\mathcal{M}_{0}$
$\displaystyle=\mathrm{LH}\left\\{x^{2n}e^{-\alpha
x^{2}/2}\in{L^{2}(\mathbb{R})}\ \Big{|}\ n\in\mathbb{N},\
\alpha\in(0,1)\right\\},$ $\displaystyle\mathcal{M}_{1}$
$\displaystyle=\mathrm{LH}\left\\{x^{2n+1}e^{-\alpha
x^{2}/2}\in{L^{2}(\mathbb{R})}\ \Big{|}\ n\in\mathbb{N},\
\alpha\in(0,1)\right\\}.$

Subspaces $\mathcal{M}_{\\#}$ are useful subspaces to consider the commutation
relation between the harmonic oscillator and the angle operator:

$\displaystyle\left[N,-\arctan\left(q^{-1}p\right)\right]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}\quad\textrm{ on }\mathcal{M}_{0},$
$\displaystyle\left[N,-\arctan\left(pq^{-1}\right)\right]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}\quad\textrm{ on }\mathcal{M}_{1}.$

We define $\mathcal{N}_{\\#}\subset\ell^{2}$ by

$\displaystyle\mathcal{N}_{0}$
$\displaystyle=\mathrm{LH}\left\\{a^{\ast}{}^{2n}e^{-\beta
a^{\ast}{}^{2}/2}\Omega\in\ell^{2}\ \Big{|}\ n\in\mathbb{N},\
\beta\in(0,1)\right\\},$ $\displaystyle\mathcal{N}_{1}$
$\displaystyle=\mathrm{LH}\left\\{a^{\ast}{}^{2n+1}e^{-\beta
a^{\ast}{}^{2}/2}\Omega\in\ell^{2}\ \Big{|}\ n\in\mathbb{N},\
\beta\in(0,1)\right\\}.$

It is immediate to see that $U\mathcal{M}_{\\#}=\mathcal{N}_{\\#}$. We show
algebraic relations we used often in this paper. Let $A$ and $B$ be linear
operators. Then the algebraic relation

$\displaystyle ABf=BAf-\operatorname{ad}_{B}(A)f=(B-\operatorname{ad}_{B})Af$

holds for $f\in\mathrm{D}(AB)\cap\mathrm{D}(BA)$. Hence
$AB^{2}=(B-\operatorname{ad}_{B})^{2}A$ on
$\mathrm{D}(AB^{2})\cap\mathrm{D}(BAB)\cap\mathrm{D}(B^{2}A)$ and then for any
$n\in\mathbb{N}$ and any $f\in\bigcap_{k=0}^{n}\mathrm{D}(B^{k}AB^{n-k})$,

$\displaystyle AB^{n}f=(B-\operatorname{ad}_{B})^{n}Af.$ (4.1)

###### Lemma 4.5

The following relations hold:

$\displaystyle U\arctan\left(q^{-1}p\right)U^{\ast}$
$\displaystyle=-\frac{i}{2}\log\left({a^{\ast}}^{-1}a\right)\hskip 5.0pt\mbox{
on }\mathcal{N}_{0},$ $\displaystyle U\arctan\left(pq^{-1}\right)U^{\ast}$
$\displaystyle=-\frac{i}{2}\log\left(a{a^{\ast}}^{-1}\right)\hskip 5.0pt\mbox{
on }\mathcal{N}_{1}.$

Proof: We see that by [10, Lemma 3.6]

$\displaystyle\arctan\left(q^{-1}p\right)x^{2n}e^{-\alpha
x^{2}/2}=\frac{i}{2}\left\\{\left(x^{2}-2\frac{d}{d\alpha}\right)^{n}\log\left(\frac{1+\alpha}{1-\alpha}\right)\right\\}e^{-\alpha
x^{2}/2}.$

Then, by Lemma 4.1,

$\displaystyle U\arctan\left(q^{-1}p\right)x^{2n}e^{-\alpha x^{2}/2}$
$\displaystyle=-\frac{i}{2}\left\\{\left(\phi^{2}-2\frac{d}{d\alpha}\right)^{n}\log\left(\frac{1-\alpha}{1+\alpha}\right)\right\\}\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega.$
(4.2)

Let $Z=a^{\ast}{}^{-1}a$. Since $\operatorname{ad}_{\phi^{2}}(Z)=-(1+Z)^{2}$
on $\mathcal{N}_{0}$, we have for any $m\in\mathbb{N}$ and any analytic
function $F$,

$\displaystyle\operatorname{ad}_{\phi^{2}}^{m}\left(F(Z)\right)=\left\\{\left(-(1+x)^{2}\frac{d}{dx}\right)^{m}F(x)\right\\}\bigg{|}_{x=Z}$

on $\mathcal{N}_{0}$. Since
$Z\exp\left(\beta{a^{\ast}}^{2}/2\right)\Omega=\beta\exp\left(\beta{a^{\ast}}^{2}/2\right)\Omega$,
we see that $\exp\left(\beta{a^{\ast}}^{2}/2\right)\Omega$ is also an
eigenvector of $\operatorname{ad}_{\phi^{2}}^{m}\left(F(Z)\right)$ and

$\displaystyle\left(\operatorname{ad}_{\phi^{2}}^{m}F(Z)\right)\exp\left(\frac{\beta}{2}{a^{\ast}}^{2}\right)\Omega=\left\\{\left(-(1+\beta)^{2}\frac{d}{d\beta}\right)^{m}F(\beta)\right\\}\exp\left(\frac{\beta}{2}{a^{\ast}}^{2}\right)\Omega.$

In particular

$\left(\operatorname{ad}_{\phi^{2}}^{m}F(Z)\right)\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega=\left\\{\left(2\frac{d}{d\alpha}\right)^{m}F\left(\frac{1-\alpha}{1+\alpha}\right)\right\\}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega.$

Moreover replacing $F(Z)$ with $\log Z$, it is obtained that

$\displaystyle\left(\operatorname{ad}_{\phi^{2}}^{m}\log
Z\right)\exp\left(\frac{1-\alpha}{1+\alpha}\frac{\
{a^{\ast}}^{2}}{2}\right)\Omega=\left\\{\left(2\frac{d}{d\alpha}\right)^{m}\log\left(\frac{1-\alpha}{1+\alpha}\right)\right\\}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{\
{a^{\ast}}^{2}}{2}\right)\Omega.$ (4.3)

By (4.2) and (4.3), we have

$\displaystyle U\arctan\left(q^{-1}p\right)x^{2n}e^{-\alpha
x^{2}/2}=-\frac{i}{2}\left\\{\left(\phi^{2}-\operatorname{ad}_{\phi^{2}}\right)^{n}\log
Z\right\\}\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega.$

From (4.1), we can furthermore see that

$\displaystyle U\arctan\left(q^{-1}p\right)x^{2n}e^{-\alpha x^{2}/2}=$
$\displaystyle-\frac{i}{2}(\log
Z)\phi^{2n}\pi^{1/4}\sqrt{\frac{2}{1+\alpha}}\exp\left(\frac{1-\alpha}{1+\alpha}\frac{{a^{\ast}}^{2}}{2}\right)\Omega$
$\displaystyle=$ $\displaystyle-\frac{i}{2}(\log Z)U\left(x^{2n}e^{-\alpha
x^{2}/2}\right).$

Then the first equality is proven. The second equality is similarly proven.
$\Box$

Let

$\displaystyle S_{0}=\frac{i}{2}\log\left({a^{\ast}}^{-1}a\right),$
$\displaystyle S_{1}=\frac{i}{2}\log\left(a{a^{\ast}}^{-1}\right).$

It is clear that $\mathrm{D}(S_{0})\cap\mathrm{D}(S_{1})=\\{0\\}$, and hence
$\mathrm{D}(S_{0}+S_{1})=\\{0\\}$. Then we define the ultra-weak time operator
of $S_{0}+S_{1}$. For the self-consistency we show the definition of ultra-
weak time operators below.

######  Definition 4.6 (Ultra-weak time operator [4])

Let $H$ be a self-adjoint operator on $\mathcal{H}$ and $D_{1}$ and $D_{2}$ be
non-zero subspaces of $\mathcal{H}$. A sesqui-linear form

$\displaystyle\mathcal{T}\colon D_{1}\times D_{2}\to\mathbb{C},\quad
D_{1}\times
D_{2}\ni(\varphi,\psi)\mapsto\mathcal{T}[\varphi,\psi]\in\mathbb{C}$

with domain $\mathrm{D}(\mathcal{T})=D_{1}\times D_{2}$
($\mathcal{T}[\varphi,\psi]$ is antilinear in $\varphi$ and linear in $\psi$)
is called an ultra-weak time operator of $H$ if there exist non-zero subspaces
$D$ and $E$ of $D_{1}\cap D_{2}$ such that (1)–(3) below hold:

  * (1)

$E\subset\mathrm{D}(H)\cap D$.

  * (2)

$\overline{\mathcal{T}[\varphi,\psi]}=\mathcal{T}[\psi,\varphi]$ for all
$\varphi,\psi\in D$.

  * (3)

$HE\subset D_{1}$ and, for all $\psi,\varphi\in E$,

$\displaystyle\mathcal{T}[H\varphi,\psi]-\overline{\mathcal{T}[H\psi,\varphi]}=-i(\varphi,\psi).$
(4.4)

We call $E$ an ultra-weak CCR-domain and $D$ a symmetric domain of
$\mathcal{T}$.

Let us define an ultra-weak time operator of $N$ through $S_{0}$ and $S_{1}$.
We define

$\displaystyle\mathcal{T}_{0}[\varphi,\psi]=\frac{1}{2}\left\\{(S_{0}\varphi,\psi)+(\varphi,S_{0}\psi)\right\\},\quad\varphi,\psi\in\mathcal{N}_{0},$
$\displaystyle\mathcal{T}_{1}[\varphi,\psi]=\frac{1}{2}\left\\{(S_{1}\varphi,\psi)+(\psi,S_{1}\varphi)\right\\},\quad\varphi,\psi\in\mathcal{N}_{1}$

and

$\displaystyle\mathcal{T}=\mathcal{T}_{0}\oplus\mathcal{T}_{1}$ (4.5)

with symmetric domain
$\left(\mathrm{D}(S_{0})\times\mathrm{D}(S_{0})\right)\oplus\left(\mathrm{D}(S_{1})\times\mathrm{D}(S_{1})\right)$.
On the other hand let us define the ultra-weak time operator $\mathcal{T}_{A}$
associated with angle operator $T_{A}$ below. Let
$h_{0}=\arctan\left(q^{-1}p\right)$ and $h_{1}=\arctan\left(pq^{-1}\right)$.
We define

$\displaystyle\mathcal{T}_{A0}[\varphi,\psi]=\frac{1}{2}\left\\{(h_{0}\varphi,\psi)+(\varphi,h_{0}\psi)\right\\},\quad\varphi,\psi\in\mathcal{N}_{0},$
$\displaystyle\mathcal{T}_{A1}[\varphi,\psi]=\frac{1}{2}\left\\{(h_{1}\varphi,\psi)+(\psi,h_{1}\varphi)\right\\},\quad\varphi,\psi\in\mathcal{N}_{1}$

and

$\displaystyle\mathcal{T}_{A}=\mathcal{T}_{A0}\oplus\mathcal{T}_{A1}$ (4.6)

with symmetric domain
$\left(\mathrm{D}(h_{0})\times\mathrm{D}(h_{0})\right)\oplus\left(\mathrm{D}(h_{1})\times\mathrm{D}(h_{1})\right)$.

###### Theorem 4.7

The ultra-weak time operator $\mathcal{T}_{A}$ of the harmonic oscillator on
$L^{2}(\mathbb{R})$ is unitary equivalent to $\mathcal{T}$ on $\ell^{2}$:

$\displaystyle\mathcal{T}_{A}[\varphi,\psi]=\mathcal{T}[U\varphi,U\psi].$
(4.7)

Proof: This follows from Lemma 4.5. $\Box$

###### Remark 4.8

(4.7) in Theorem 4.9 can be interpreted as a rigorous justification of

$\frac{1}{2}\left(\arctan\left(q^{-1}p\right)+\arctan\left(pq^{-1}\right)\right)\cong\frac{i}{2}\left\\{\log\left({a^{\ast}}^{-1}a\right)+\log\left(a{a^{\ast}}^{-1}\right)\right\\}.$

In some literature however the following description is stated:

$\frac{1}{2}\left(\arctan\left(q^{-1}p\right)+\arctan\left(pq^{-1}\right)\right)\cong\frac{i}{2}\left\\{\log
a+\log{a^{\ast}}\right\\}.$

The right-hand side can not be however defined as an operator by Lemma 3.2.
Even if it is a formal argument, it is dangerous to base the argument on this
equivalent relation.

###  4.2 Angle operator by shift operators

Let $L$ be the left shift operator defined in Section 2.3. The angle operator
$T_{A}$ can be represented by $L$ and $N$. Let us define

$\displaystyle L_{A0}=\frac{i}{2}\log\left(\sqrt{\frac{N+2{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}}L^{2}\right),$ $\displaystyle
L_{A1}=\frac{i}{2}\log\left(\sqrt{\frac{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}{{N+2{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}}}L^{2}\right).$

###### Theorem 4.9

We have

$\displaystyle S_{0}=L_{A0}\quad\mbox{on }\mathcal{N}_{0},$ (4.8)
$\displaystyle S_{1}=L_{A1}\quad\mbox{on }\mathcal{N}_{1}.$ (4.9)

In particular

$\displaystyle U\arctan\left(q^{-1}p\right)U^{\ast}$
$\displaystyle=L_{A0}\hskip 5.0pt\mbox{ on }\mathcal{N}_{0},$ (4.10)
$\displaystyle U\arctan\left(pq^{-1}\right)U^{\ast}$
$\displaystyle=L_{A1}\hskip 5.0pt\mbox{ on }\mathcal{N}_{1}.$ (4.11)

Proof: We see that

${a^{\ast}}^{-1}a=\left(\sqrt{N}L^{\ast}\right)^{-1}\sqrt{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}L=L\sqrt{\frac{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{N}}L=\sqrt{\frac{N+2{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}}L^{2}$

on ${\rm Ran}({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-P_{\\{{a^{\ast}}\Omega\\}})$. Similarly

$a{a^{\ast}}^{-1}=L\sqrt{N}\left(L^{\ast}\sqrt{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}\right)^{-1}=L\sqrt{\frac{N}{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}}L=\sqrt{\frac{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}{{N+2{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}}}L^{2}$

on ${\rm Ran}({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-P_{\\{\Omega\\}})$. From this we have
(4.8) and (4.9). Relations (4.10) and (4.11) follow from Theorem 4.5. $\Box$

###  4.3 Generalization of angle operator $\bm{T_{A}}$

We generalize $L_{A0}$ and $L_{A1}$ to the form of
$i\log\left(g(N)L^{2}\right)$. A fundamental idea is to find eigenvectors
$\varphi$ of linear operator
$\operatorname{ad}_{N}\bigl{(}\log\left(g(N)L^{2}\right)\bigr{)}$ so that
$\operatorname{ad}_{N}\bigl{(}\log\left(g(N)L^{2}\right)\bigr{)}\varphi=a\varphi$
with $a\not=0$. Then it follows that

$\displaystyle\left[N,-\frac{i}{a}\log\left(g(N)L^{2}\right)\right]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$

on CCR-domain $\mathrm{LH}\\{\mbox{eigenvectors }\varphi\\}$. Hereafter we
write $g_{N}$ for $g(N)$. Let $P_{\geq 2}$ be the projection onto
$\overline{\mathrm{LH}\\{\xi_{n}\mid n\geq 2\\}}$, and $P_{0}$ the projection
onto $\overline{\mathrm{LH}\\{\xi_{2n}\mid n\in\mathbb{N}\\}}$.

###### Lemma 4.10

Let $f$ and $g$ be complex-valued functions on $\mathbb{N}$, and
$\alpha\in\mathbb{C}$. Suppose that there exists a constant
$\beta\in\mathbb{C}$ such that

$\displaystyle g_{N+2}f_{N+2}-g_{N}f_{N}P_{\geq 2}=\beta{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}\quad\text{on }P_{0}\ell_{\mathrm{fin}}^{2}.$ (4.12)

Then, for all $n\in\mathbb{N}$, we see that
$\operatorname{ad}_{g_{N+2}L^{2}}\left(f_{N}{L^{\ast}}^{2}\right)^{n}=n\beta\left(f_{N}{L^{\ast}}^{2}\right)^{n-1}$
on $P_{0}\ell_{\mathrm{fin}}^{2}$.

Proof: The relation (4.12) is equivalent to
$\bigl{[}g_{N+2}L^{2},f_{N}{L^{\ast}}^{2}\bigr{]}=\beta{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$
on $P_{0}\ell_{\mathrm{fin}}^{2}$. This implies that
$\operatorname{ad}_{g_{N+2}L^{2}}\left(f_{N}{L^{\ast}}^{2}\right)^{n}=n\beta\left(f_{N}{L^{\ast}}^{2}\right)^{n-1}$
on $P_{0}\ell_{\mathrm{fin}}^{2}$. $\Box$

###### Lemma 4.11

Let $f$ be a complex-valued function on $\mathbb{N}$. Suppose that $|f(2n)|>0$
for all natural number $n\geq 1$ and there exists a limit (including infinity)

$\displaystyle M_{f}=\lim_{n\to\infty}\frac{n}{|f(2n)|}\leq\infty.$

Then, for all $l,m\in\mathbb{N}$ and $\alpha\in\mathbb{C}$ such that
$|\alpha|<M_{f}$,

$\displaystyle\Omega\in\mathrm{D}\left(N^{l}\left(f_{N}L^{\ast}{}^{2}\right)^{m}e^{\alpha
f_{N}L^{\ast}{}^{2}}\right).$

Proof: Note that

$\displaystyle\left\|N^{l}\left(f_{N}L^{\ast}{}^{2}\right)^{m}e^{\alpha
f_{N}L^{\ast}{}^{2}}\Omega\right\|^{2}$
$\displaystyle=\sum_{n=0}^{\infty}\frac{|\alpha|^{2n}}{(n!)^{2}}\left\|N^{l}\left(f_{N}L^{\ast}{}^{2}\right)^{(n+m)}\Omega\right\|^{2}.$

Since

$N^{l}\left(f_{N}L^{\ast}{}^{2}\right)^{(n+m)}\Omega=(2(n+m))^{l}\prod_{j=1}^{n+m}f(2j)L^{\ast}{}^{2(n+m)}\Omega,$

the radius of the convergence of above infinite series is given by
$\lim_{n\to\infty}n/|f(2n+2m))|=M_{f}$. Then the lemma follows. $\Box$

###### Lemma 4.12

Let $f$ and $g$ be complex-valued functions on $\mathbb{N}$ such that (4.12)
is satisfied. Then, for all $l,m,n\in\mathbb{N}$ and $\alpha\in\mathbb{C}$
such that $|\alpha|<M_{f}$,

$\displaystyle\Omega\in\mathrm{D}\left(\left(g_{N+2}L^{2}\right)^{l}N^{m}\left(f_{N}L^{\ast}{}^{2}\right)^{n}e^{\alpha
f_{N}L^{\ast}{}^{2}}\right).$

Moreover, $e^{\alpha f_{N}{L^{\ast}}^{2}}\Omega$ is an eigenvector of
$g_{N+2}L^{2}$ such that

$\displaystyle g_{N+2}L^{2}e^{\alpha f_{N}{L^{\ast}}^{2}}\Omega=\alpha\beta
e^{\alpha f_{N}{L^{\ast}}^{2}}\Omega.$ (4.13)

Proof: We have

$\displaystyle\left\|\left(g_{N+2}L^{2}\right)^{l}N^{m}\left(f_{N}L^{\ast}{}^{2}\right)^{m}e^{\alpha
f_{N}L^{\ast}{}^{2}}\Omega\right\|^{2}=\sum_{n=0}^{\infty}\frac{|\alpha|^{2n}}{(n!)^{2}}\left\|\left(g_{N+2}L^{2}\right)^{l}N^{m}\left(f_{N}L^{\ast}{}^{2}\right)^{(n+m)}\Omega\right\|^{2}.$

Since

$\displaystyle\left(g_{N+2}L^{2}\right)^{l}N^{m}\left(f_{N}L^{\ast}{}^{2}\right)^{(n+m)}\Omega$
$\displaystyle=\begin{cases}\left(\prod_{j=n+m-l+1}^{n+m}g_{2j}\right)(2(n+m))^{m}\left(\prod_{j=1}^{n+m}f_{2j}\right)L^{\ast}{}^{2(n+m-l)}\Omega,&n+m\geq
l,\\\ 0,&n+m<l,\end{cases}$

the right-hand side above converges for $\alpha\in\mathbb{C}$ such that
$|\alpha|<M_{f}$. From Lemma 4.10, we can also see that $e^{\alpha
f_{N}{L^{\ast}}^{2}}\Omega$ is an eigenvector of $g_{N+2}L^{2}$ and (4.13)
follows. $\Box$

We define the super coherent vector $\xi_{\alpha,f}$ by

$\xi_{\alpha,f}=e^{\alpha f_{N}{L^{\ast}}^{2}}\Omega.$

We set

$\displaystyle\mathbb{D}_{f,\beta}=\left\\{\alpha\in\mathbb{C}\ \Big{|}\
|1-\alpha\beta|<1,\ |\alpha|<M_{f}\right\\}.$ (4.14)

###### Lemma 4.13

Let $f$ and $g$ be complex-valued functions on $\mathbb{N}$ such that (4.12)
is satisfied. Then for all $n\in\mathbb{N}$ and
$\alpha\in\mathbb{D}_{f,\beta}$, we have
$\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}\in\mathrm{D}\left(\log\big{(}g_{N+2}L^{2}\big{)}\right)$
and

$\displaystyle\log\left(g_{N+2}L^{2}\right)\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}=\left\\{\left({f_{N}L^{\ast}}^{2}+\frac{d}{d\alpha}\right)^{n}(\log\alpha\beta)\right\\}\xi_{\alpha,f}.$

Proof: By Lemma 4.12 we have
$\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}\in\bigcap_{m=0}^{\infty}\mathrm{D}\left(\big{(}1-g_{N+2}L^{2}\big{)}^{m}\right)$.
From (4.1) and (4.3) it follows that

$\displaystyle\log\left(g_{N+2}L^{2}\right)\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}=-\sum_{k=1}^{K}\frac{1}{k}\left(1-g_{N+2}L^{2}\right)^{k}\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}$
$\displaystyle\quad=-\sum_{k=1}^{K}\frac{1}{k}\left\\{\left(f_{N}{L^{\ast}}^{2}-\operatorname{ad}_{f_{N}L^{\ast}{}^{2}}\right)^{n}\left(1-g_{N+2}L^{2}\right)^{k}\right\\}\xi_{\alpha,f}$
$\displaystyle\quad=-\sum_{k=1}^{K}\frac{1}{k}\left\\{\left(f_{N}{L^{\ast}}^{2}+\frac{d}{d\alpha}\right)^{n}\left(1-\alpha\beta\right)^{k}\right\\}\xi_{\alpha,f}\to\left\\{\left(f_{N}{L^{\ast}}^{2}+\frac{d}{d\alpha}\right)^{n}(\log\alpha\beta)\right\\}\xi_{\alpha,f}$

as $K\to\infty$. Then the proof is complete. $\Box$

The next theorem is a generalization of (1) of Lemma 4.2.

###### Theorem 4.14

Let $f$ and $g$ be complex-valued functions on $\mathbb{N}$ such that (4.12)
is satisfied. Then,

$\displaystyle\left[N,\frac{i}{2}\log\left(g_{N+2}L^{2}\right)\right]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$
(4.15)

holds on CCR-domain
$\mathrm{LH}\left\\{\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}\
\big{|}\ n\in\mathbb{N},\ \alpha\in\mathbb{D}_{f,\beta}\right\\}$.

Proof: In this proof we set $X=f_{N}L^{\ast}{}^{2}$ and $Y=g_{N+2}L^{2}$. From
Lemmas 4.11, 4.12 and 4.13, we see that, for all $n\in\mathbb{N}$,

$\displaystyle X^{n}\xi_{\alpha,f}\in\mathrm{D}\left(N(\log
Y)\right)\cap\mathrm{D}\left((\log Y)N\right).$

Since $\beta N=2\overline{XY}$ on $P_{0}\mathrm{D}(N)$, we have

$\displaystyle(\log Y)NX^{n}\xi_{\alpha,f}$
$\displaystyle=\frac{2}{\beta}(\log Y)\left(YX-\beta{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}\right)X^{n}\xi_{\alpha,f}$
$\displaystyle=\frac{2}{\beta}\left(Y(\log Y)X^{n+1}-\beta(\log
Y)X^{n}\right)\xi_{\alpha,f}.$

By Lemma 4.13 we have

$\displaystyle(\log Y)NX^{n}\xi_{\alpha,f}$
$\displaystyle=\frac{2}{\beta}\left\\{(Y(X-\operatorname{ad}_{X})-\beta)(\log
Y)X^{n}\right\\}\xi_{\alpha,f}$
$\displaystyle=\left\\{\left(N-\frac{2}{\beta}Y\operatorname{ad}_{X}\right)(\log
Y)X^{n}\right\\}\xi_{\alpha,f}$ $\displaystyle=\left\\{N(\log
Y)X^{n}-\frac{2}{\beta}Y(X-\operatorname{ad}_{X})^{n}\operatorname{ad}_{X}(\log
Y)\right\\}\xi_{\alpha,f}$ $\displaystyle=\left\\{N(\log
Y)X^{n}-\frac{2}{\beta}(X-\operatorname{ad}_{X})^{n}Y\operatorname{ad}_{X}(\log
Y)\right\\}\xi_{\alpha,f}.$

Here we used
$Y(X-\operatorname{ad}_{X})Z\varphi=(X-\operatorname{ad}_{X})YZ\varphi$ for
any linear operator $Z$ and any
$\varphi\in\mathrm{D}(XYZ)\cap\mathrm{D}(YXZ)\cap\mathrm{D}(YZX)$. In a
similar way to (4.3), we can obtain that

$\displaystyle(\log Y)NX^{n}\xi_{\alpha,f}$ $\displaystyle=\left\\{N(\log
Y)X^{n}+\frac{2}{\beta}\left(X+\frac{d}{d\alpha}\right)^{n}\alpha\beta\frac{d}{d\alpha}\log\left(\alpha\beta\right)\right\\}\xi_{\alpha,f}$
$\displaystyle=\left(N(\log Y)+2{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}\right)X^{n}\xi_{\alpha,f}.$

Hence $[N,\frac{i}{2}\log Y]X^{n}\xi_{\alpha,f}=-iX^{n}\xi_{\alpha,f}$ follows
and (4.15) is proven. $\Box$

We can also see a generalization of (2) of Lemma 4.2.

###### Theorem 4.15

Let $\tilde{h}$ be a complex-valued function on $\mathbb{N}$ such that
$\tilde{h}_{N+1}^{-1}$ is bounded and $\tilde{f}$ and $\tilde{g}$ be complex-
valued functions on $\mathbb{N}$ such that

$\displaystyle\tilde{h}_{N+1}^{-1}\tilde{g}_{N+3}\tilde{h}_{N+3}\tilde{f}_{N+2}-\tilde{h}_{N-1}^{-1}\tilde{g}_{N+1}\tilde{h}_{N+1}\tilde{f}_{N}P_{\geq
2}=\tilde{\beta}{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}\quad$

on $P_{0}\ell_{\mathrm{fin}}^{2}$ with some $\tilde{\beta}\in\mathbb{C}$. Let

$\displaystyle\mathbb{D}_{\tilde{f},\tilde{h},\tilde{\beta}}=\left\\{\alpha\in\mathbb{C}\
\Bigg{|}\ \alpha\in\mathbb{D}_{\tilde{f},\tilde{\beta}},\
\xi_{\alpha,\tilde{f}}\in\bigcap_{n=0}^{\infty}\mathrm{D}\left(\tilde{g}_{N+3}\tilde{h}_{N+3}\left(\tilde{f}_{N}{L^{\ast}}^{2}\right)^{n}\right)\right\\}.$
(4.16)

Then $\tilde{h}_{N}L^{\ast}\xi_{\alpha,\tilde{f}}$ is an eigenvector of
$\tilde{g}_{N+2}L^{2}$ such that

$\displaystyle\tilde{g}_{N+2}L^{2}\tilde{h}_{N}L^{\ast}\xi_{\alpha,\tilde{f}}$
$\displaystyle=\alpha\tilde{\beta}\tilde{h}_{N}L^{\ast}\xi_{\alpha,\tilde{f}}.$
(4.17)

Furthermore it follows that

$\displaystyle\left[N,\frac{i}{2}\log\left(\tilde{g}_{N+2}L^{2}\right)\right]$
$\displaystyle=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ (4.18)

on CCR-domain
$\mathrm{LH}\left\\{\tilde{h}_{N}L^{\ast}\left(\tilde{f}_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,\tilde{f}}\
\middle|\ n\in\mathbb{N},\
\alpha\in\mathbb{D}_{\tilde{f},\tilde{h},\tilde{\beta}}\right\\}$.

Proof: By the following equalities

$\displaystyle\tilde{g}_{N+2}L^{2}\tilde{h}_{N}L^{\ast}$
$\displaystyle=\tilde{g}_{N+2}\tilde{h}_{N+2}L=\tilde{h}_{N}\tilde{h}_{N}^{-1}\tilde{g}_{N+2}\tilde{h}_{N+2}L^{\ast}L^{2}=\tilde{h}_{N}L^{\ast}\left(\tilde{h}_{N+1}^{-1}\tilde{g}_{N+3}\tilde{h}_{N+3}L^{2}\right)$

on $P_{0}\ell^{2}\cap\mathrm{D}(\tilde{g}_{N+1}\tilde{h}_{N+1})$, one can show
(4.17) in a similar manner to the proof of (4.12) by replacing $g_{N+2}$ of
(4.12) with $\tilde{h}_{N+1}^{-1}\tilde{g}_{N+3}\tilde{h}_{N+3}$. The proof of
(4.18) is also similar to those of Lemma 4.12 and Theorem 4.14. $\Box$

By Theorems 4.14 and 4.15 we can also construct an ultra-weak time operator of
$N$. This is a generarization of $\mathcal{T}$ in (4.5). Let $f$, $\tilde{f}$,
$g$, $\tilde{g}$ and $h$ be functions on $\mathbb{N}$ given in Theorems 4.14
and 4.15. Let

$\displaystyle\mathcal{K}_{0}=\mathrm{LH}\left\\{\left(f_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,f}\
\Big{|}\ n\in\mathbb{N},\ \alpha\in\mathbb{D}_{f,\beta}\right\\},$
$\displaystyle\mathcal{K}_{1}=\mathrm{LH}\left\\{\tilde{h}_{N}L^{\ast}\left(\tilde{f}_{N}{L^{\ast}}^{2}\right)^{n}\xi_{\alpha,\tilde{f}}\
\Big{|}\ n\in\mathbb{N},\
\alpha\in\mathbb{D}_{\tilde{f},\tilde{h},\tilde{\beta}}\right\\}.$

Note that $\mathcal{K}_{0}\oplus\mathcal{K}_{1}$ is dense if
$\mathbb{D}_{f,\beta}\neq\emptyset$ and
$\mathbb{D}_{\tilde{f},\tilde{h},\tilde{\beta}}\neq\emptyset$. Let

$\displaystyle S_{0g}=\frac{i}{2}\log\left(g_{N+2}L^{2}\right),$
$\displaystyle
S_{1\tilde{g}}=\frac{i}{2}\log\left(\tilde{g}_{N+2}L^{2}\right).$

These are generalization of $S_{0}$ and $S_{1}$. Define

$\displaystyle\mathcal{T}_{0g}[\varphi,\psi]=\frac{1}{2}\\{(S_{0g}\varphi,\psi)+(\varphi,S_{0g}\psi)\\},\quad\varphi,\psi\in\mathcal{K}_{0},$
$\displaystyle\mathcal{T}_{1\tilde{g}}[\varphi,\psi]=\frac{1}{2}\\{(S_{1\tilde{g}}\varphi,\psi)+(\varphi,S_{1\tilde{g}}\psi)\\},\quad\varphi,\psi\in\mathcal{K}_{1}$

and set

$\displaystyle\mathcal{T}_{g,\tilde{g}}=\mathcal{T}_{0g}\oplus\mathcal{T}_{1\tilde{g}}.$
(4.19)

###### Theorem 4.16

Suppose that $f,\tilde{f},g,\tilde{g}$ and $\tilde{h}$ satisfy the assumptions
given in Theorem 4.14 and Lemma 4.15. We also assume that
$\mathbb{D}_{f,\beta}\neq\emptyset$ and
$\mathbb{D}_{\tilde{f},\tilde{h},\tilde{\beta}}\neq\emptyset$. Then
$\mathcal{T}_{g,\tilde{g}}$ is an ultra-weak time operator of $N$ with a dense
CCR-domain $\mathcal{K}_{0}\oplus\mathcal{K}_{1}$.

Proof: This follows from (4.15) and (4.18). $\Box$

###  4.4 Conjugate operators of the form $\bm{\log p(L)$}$

In the previous section we consider the conjugate operator of the form
$\log(g_{N}L^{2})$. Another direction of generalizations is to consider
conjugate operators of the form $\log(g_{N}L^{n})$. In special cases it can be
reduced to consider conjugate operators of $\log(g_{N}L^{n})$ to those of
$\log(L^{n})$. See Lemma 4.17 below. Let $f$ be a function on $\mathbb{N}$ and

$\displaystyle
f!_{k}(N)\xi_{n}=\left(\prod_{m=0}^{[n/k]}f_{n-km}\right)\xi_{n},\quad
n\in\mathbb{N}$

We extend $f!_{k}(N)$ to $\ell_{\rm fin}^{2}$ by the linearity and denote the
closure of $f!_{k}(N)\lceil_{\ell_{\rm fin}^{2}}$ by the same symbol
$f!_{k}(N)$.

###### Lemma 4.17

Let $f$ and $g$ be complex-valued functions on $\mathbb{N}$ such that
$f_{n}g_{n}=n$ for all $n\in\mathbb{N}$. Suppose that $\alpha\in(0,1)$ and
$\Omega\in\mathrm{D}(e^{\alpha f_{N}L^{\ast}{}^{k}})$. Then

$\displaystyle\log\left(g_{N+k}L^{k}\right)e^{\alpha
f_{N}{L^{\ast}}^{k}}\Omega=f!_{k}(N)\log\left((N+k{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})L^{k}\right)e^{\alpha{L^{\ast}}^{k}}\Omega.$ (4.20)

In particular, setting $h_{n}=n$ for all $n\in\mathbb{N}$, we have

$\displaystyle\left(\log L^{k}\right)e^{\alpha
N{L^{\ast}}^{k}}\Omega=h!_{k}(N)\log\left((N+k{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})L^{k}\right)e^{\alpha{L^{\ast}}^{k}}\Omega,$ (4.21)
$\displaystyle\log\left(g_{N+k}L^{k}\right)e^{\alpha
f_{N}{L^{\ast}}^{k}}\Omega=h!_{k}(N)^{-1}f!_{k}(N)\left(\log
L^{k}\right)e^{\alpha N{L^{\ast}}^{k}}\Omega.$ (4.22)

Proof: It is straightforward to see that

$\displaystyle\log\left(g_{N+k}L^{k}\right)e^{\alpha
f_{N}{L^{\ast}}^{k}}\Omega$
$\displaystyle=\log\left(g_{N+k}L^{k}\right)f!_{k}(N)e^{\alpha{L^{\ast}}^{k}}\Omega$
$\displaystyle=f!_{k}(N)\log\left(g_{N+k}f_{N+k}L^{k}\right)e^{\alpha{L^{\ast}}^{k}}\Omega$
$\displaystyle=f!_{k}(N)\log\left((N+k{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})L^{k}\right)e^{\alpha{L^{\ast}}^{k}}\Omega.$

Putting $g_{n}=1$ for all $n\in\mathbb{N}$ in (4.20), we can derive (4.21).
Combining (4.20) and (4.21) we can also see (4.22). Then the lemma follows.
$\Box$

By (4.22), $\log\big{(}g_{N+k}L^{k}\big{)}$ can be represented as $C(N)\log
L^{k}$ on $\mathrm{LH}\big{\\{}e^{\alpha N{L^{\ast}}^{k}}\Omega\big{\\}}$ with

$C(N)=h!_{k}(N)^{-1}f!_{k}(N).$

Since the operator $C(N)$ commutes with $N$, the investigation of conjugate
operators of the form $\log\big{(}g_{N+k}L^{k}\big{)}$ can be reduced to that
of $\log L^{k}$ on $\mathrm{LH}\big{\\{}L^{\ast}{}^{nk}e^{\alpha
N{L^{\ast}}^{k}}\Omega\ \big{|}\ n\in\mathbb{N},\ |\alpha|<1\big{\\}}$.

###### Example 4.18

We see that $T_{A}\cong L_{A0}\oplus L_{A1}$ with
$L_{A0}=\frac{i}{2}\log\left(\sqrt{\frac{N+2{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}}L^{2}\right)$ and
$L_{A1}=\frac{i}{2}\log\left(\sqrt{\frac{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{{N+2{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}}}L^{2}\right)$. By Lemma 4.17 to
investigate the CCR-domain of $L_{A0}$ and $L_{A1}$ can be reduced to
investigate the CCR-domain of $\frac{i}{2}\log L^{2}$.

Thus from now on we study the conjugate operator of the form $\log p(L)$. The
idea to find a CCR-domain of $\log p(L)$ is similar to that of the previous
subsection. Formally

$[N,\log p(L)]\subset-p(L)^{-1}p^{\prime}(L)L,$

and $e^{\alpha NL^{\ast}}\Omega$ is an eigenvector of $L$ and that of $p(L)$.
Thus $[N,\log p(L)]=-p(\alpha)^{-1}p^{\prime}(\alpha)\alpha$ on
$\mathrm{LH}\\{e^{\alpha NL^{\ast}}\Omega\\}$. To solve $[N,\log p(L)]=c$ we
study the algebraic equation of the form

$\displaystyle\alpha p^{\prime}(\alpha)+cp(\alpha)=0$

with some $c\in\mathbb{C}$. Then we set

$\displaystyle\mathrm{D}(p,c)$
$\displaystyle=\mathrm{LH}\\{\varphi\in\mathrm{D}(N)\mid\exists\alpha\in\mathbb{C}\
s.t.\ |p(\alpha)-1|<1,\ \alpha p^{\prime}(\alpha)+cp(\alpha)=0,\
\varphi\in\mathop{\mathrm{Ker}}(L-\alpha{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})\\}.$

It may be possibly $\mathrm{D}(p,c)=\\{0\\}$.

###### Theorem 4.19

Let $p$ be a polynomial and $c\in\mathbb{C}\setminus\\{0\\}$. If there exists
$\alpha\in\mathbb{C}$ such that $|\alpha|<1$, $\alpha
p^{\prime}(\alpha)+cp(\alpha)=0$ and $|p(\alpha)-1|<1$, then
$\mathrm{D}(p,c)\not=\\{0\\}$ and

$\displaystyle\left[N,-\frac{i}{c}\log p(L)\right]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$

with CCR-domain $\mathrm{D}(p,c)$. Moreover, in $\mathrm{D}\big{(}\sum
k^{-1}N({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-p(L))^{k}\big{)}$, there is no infinite
dimensional CCR-domain for $N$ and $c\log p(L)$ unless $p(x)=cx^{m}$ for some
$c\in\mathbb{C}$ and $m\in\mathbb{N}$.

Proof: We can see that

$\displaystyle[N,\log
p(L)]=-\sum_{k=1}^{\infty}\frac{1}{k}\left[N,\left({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L)\right)^{k}\right]=-\sum_{k=1}^{\infty}\left({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L)\right)^{k-1}Lp^{\prime}(L)$ (4.23)

on $\mathrm{D}(\sum k^{-1}N({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L))^{k})\cap\mathrm{D}(N\log p(L))\cap\mathrm{D}(\log
p(L)N)$. Let $\varphi\in\mathop{\mathrm{Ker}}(L-\alpha{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})$. Then

$\displaystyle[N,\log p(L)]\varphi=-\alpha
p^{\prime}(\alpha)p(\alpha)^{-1}\varphi.$

By $\alpha p^{\prime}(\alpha)+cp(\alpha)=0$ we see that

$\displaystyle[N,\log p(L)]\varphi=c\varphi.$

Next we shall show that the dimension of the CCR-domain in ${\rm Ran}(p(L))$
is finite. For any

$\varphi\in\mathrm{D}\left(\sum_{k\geq 1}\frac{1}{k}N({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L))^{k}\right)$

in the CCR-domain, there exists $c\in\mathbb{C}$ such that $[N,\log
p(L)]\varphi=c\varphi$. From (4.23), we see that
$\lim_{k\to\infty}({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-p(L))^{k}Lp^{\prime}(L)\varphi=0$.
Therefore,

$\displaystyle cp(L)\varphi=p(L)[N,\log
p(L)]\varphi=-\sum_{k=1}^{\infty}p(L)\left({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L)\right)^{k-1}Lp^{\prime}(L)\varphi=-Lp^{\prime}(L)\varphi.$

Thus $\varphi\in\mathop{\mathrm{Ker}}(Lp^{\prime}(L)+cp(L))$. This implies
that the CCR-domain is at most finite dimensional space, since
$\dim\mathop{\mathrm{Ker}}(L-\lambda{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})^{k}\leq k$ for all
$\lambda\in\mathbb{C}$ and $k\in\mathbb{N}$, and $p$ is a polynomial with
$Lp^{\prime}(L)+cp(L)\neq 0$. This fact can be derived from a fundamental
lemma below. Then the theorem follows. $\Box$

###### Lemma 4.20

Let $A$ be a bounded operator on a Banach space $\mathcal{K}$. Let
$\\{n_{1},\ldots,n_{k}\\}\subset\mathbb{N}$,
$\\{\lambda_{1},\ldots,\lambda_{k}\\}\subset\mathbb{C}$ such that
$\lambda_{i}\neq\lambda_{j}$ for $i\neq j$. Then it follows that

$\displaystyle\mathop{\mathrm{Ker}}\left(\prod_{j=1}^{k}(\lambda_{j}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A)^{n_{j}}\right)=\bigoplus_{j=1}^{k}\mathop{\mathrm{Ker}}(\lambda_{j}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A)^{n_{j}}.$

Proof: We refer to [2, Lemma 1.76]. $\Box$

##  5 Galapon operators

###  5.1 Galapon operators by shift operators

In this section we investigate Galapon operator which is a bounded self-
adjoint time operator. Recall that ${\rm Spec}(N)=\\{n\\}_{n\in\mathbb{N}}$
and $\\{v_{n}\\}_{n\in\mathbb{N}}$ the set of normalized eigenvectors of $N$
in $L^{2}(\mathbb{R})$. We define $T_{G}$ by

$\displaystyle\mathrm{D}(T_{G})=\mathrm{LH}\\{v_{n}\mid n\in\mathbb{N}\\},$
$\displaystyle T_{G}\varphi=i\sum_{n=0}^{\infty}\left(\sum_{m\neq
n}\frac{(v_{m},\varphi)}{n-m}v_{n}\right),\quad\varphi\in\mathrm{D}(T_{G}).$
(5.1)

It is known that $T_{G}$ is bounded and

$\displaystyle[N,T_{G}]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ (5.2)

on a dense CCR-domain $\mathrm{LH}\\{v_{n}-v_{m}\mid n,m\in\mathbb{N}\\}$ (see
[5]). In particular, $T_{G}$ is not equal to the angle operator $T_{A}$. The
angle operator $T_{A}$ is expressed on $\ell^{2}$ by Theorem 4.9. On the other
hand the Galapon operator is given by (5.1). As it is, no comparison of
$T_{A}$ and $T_{G}$ can be made, so we shall transform $T_{G}$ to an operator
on $\ell^{2}$. Let us consider $\log\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L\right)$ and
$\log\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast}\right)$.

###### Lemma 5.1

We have $\ell_{\mathrm{fin}}^{2}\subset\mathrm{D}\left(\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)\right)\cap\mathrm{D}\left(\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\right)$.

Proof: By the definition of $\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L)$, we see that

$\displaystyle\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)\xi_{n}=-\sum_{k=1}^{\infty}\frac{1}{k}L^{k}\xi_{n}=-\sum_{k=1}^{n}\frac{1}{k}\xi_{n-k}.$

Thus $\xi_{n}\in\mathrm{D}\left(\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L)\right)$ and hence
$\ell_{\mathrm{fin}}^{2}\subset\mathrm{D}(\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L))$. We also see that

$\displaystyle\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\xi_{n}=-\sum_{k=1}^{\infty}\frac{1}{k}{L^{\ast}}^{k}\xi_{n}=-\sum_{k=1}^{\infty}\frac{1}{k}\xi_{n+k}$

and

$\displaystyle\left\|\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\xi_{n}\right\|^{2}=\sum_{k=1}^{\infty}\frac{1}{k^{2}}<\infty.$

Thus $\xi_{n}\in\mathrm{D}\left(\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast})\right)$ and
$\ell_{\mathrm{fin}}^{2}\subset\mathrm{D}\left(\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\right)$. $\Box$

###### Remark 5.2

By the proof of Lemma 5.1 we see that

  * (1)

$\ell_{\mathrm{fin}}^{2}\subset\mathrm{D}(\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}))$, but $\mathrm{D}(\log
L^{\ast})\cap\ell_{\mathrm{fin}}^{2}=\\{0\\}$,

  * (2)

$\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)\ell_{\mathrm{fin}}^{2}\subset\ell_{\mathrm{fin}}^{2}$.

We consider the following operator $L_{G}$.

###### Definition 5.3

We define a linear operator $L_{G}$ on $\ell^{2}$ by

$\displaystyle\mathrm{D}(L_{G})=\left\\{\varphi\in\ell^{2}\ \middle|\
\lim_{K\to\infty}\sum_{k=1}^{K}\frac{1}{k}\left(L^{\ast}{}^{k}-L^{k}\right)\varphi\text{
exists}\right\\},$ $\displaystyle
L_{G}\varphi=i\sum_{k=1}^{\infty}\frac{1}{k}\left(L^{\ast}{}^{k}-L^{k}\right)\varphi,\quad\varphi\in\mathrm{D}(L_{G}).$

Note that $\ell^{2}_{\mathrm{fin}}\subset\mathrm{D}(L_{G})$ and
$i\\{\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L)-\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\\}\subset L_{G}$.

###### Theorem 5.4

  1. (1)

It holds that $L_{G}\cong T_{G}$ on $\ell^{2}_{\mathrm{fin}}$.

  2. (2)

$L_{G}$ is a bounded operator.

Proof: (1) Let $U$ be the unitary operaotr defined by $Uv_{n}=\xi_{n}$ for
each $n\in\mathbb{N}$. For arbitrary $\varphi\in\mathrm{D}(T_{G})$, we see
that

$\displaystyle UT_{G}\varphi$
$\displaystyle=i\sum_{n=0}^{\infty}\left(\sum_{m<n}\frac{\left(\xi_{m},U\varphi\right)}{n-m}+\sum_{m>n}\frac{\left(\xi_{m},U\varphi\right)}{n-m}\right)\xi_{n}$
$\displaystyle=i\sum_{n=0}^{\infty}\left(\sum_{m<n}\frac{\left(L^{n-m}\xi_{n},\varphi\right)}{n-m}-\sum_{m>n}\frac{\left(L^{\ast}{}^{m-n}\xi_{n},U\varphi\right)}{m-n}\right)\xi_{n}$
$\displaystyle=i\sum_{n=0}^{\infty}\left(\sum_{k=1}^{\infty}\frac{\left(L^{k}\xi_{n},U\varphi\right)}{k}-\sum_{k=1}^{\infty}\frac{\left(L^{\ast}{}^{k}\xi_{n},U\varphi\right)}{k}\right)\xi_{n}.$

From Lemma 5.1, we have
$U\mathrm{D}(T_{G})=\ell^{2}_{\mathrm{fin}}\subset\mathrm{D}(L_{G})$. Then

$\displaystyle UT_{G}\varphi$
$\displaystyle=i\sum_{n=0}^{\infty}\left(\sum_{k=1}^{\infty}\frac{1}{k}\left(L^{k}-L^{\ast}{}^{k}\right)\xi_{n},U\varphi\right)\xi_{n}=\sum_{n=0}^{\infty}\left(\xi_{n},i\sum_{k=1}^{\infty}\frac{1}{k}\left(L^{\ast}{}^{k}-L^{k}\right)U\varphi\right)\xi_{n}.$

This implies that $UT_{G}\varphi=L_{G}U\varphi$ for
$\varphi\in\mathrm{D}(T_{G})$.

(2) From the Hilbert inequality, we see that, for any
$\varphi=\sum_{n}c_{n}\xi_{n}\in\mathrm{D}(L_{G})$,

$|(\varphi,L_{G}\varphi)|=\left|\sum_{n=0}^{\infty}\sum_{m\neq
n}\frac{c_{n}c_{m}}{n-m}\right|\leq\pi\|\varphi\|^{2}.$

Thus we have $\|L_{G}\varphi\|\leq\pi\|\varphi\|$. Then the theorem is proven.
$\Box$

###### Theorem 5.5

We have $[N,L_{G}]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ on
$\mathrm{D}(NL_{G})\cap\mathrm{D}(L_{G}N)$.

Proof: We obtain that, for any
$\varphi\in\mathrm{D}(NL_{G})\cap\mathrm{D}(L_{G}N)$,

$\displaystyle[N,L_{G}]\varphi$
$\displaystyle=\sum_{n=0}^{\infty}\left(\xi_{n},(NL_{G}-L_{G}N)\varphi\right)\xi_{n}$
$\displaystyle=i\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k}\left(\xi_{n},\left(\left[N,{L^{\ast}}^{k}\right]-\left[N,L^{k}\right]\right)\varphi\right)\xi_{n}$
$\displaystyle=i\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},\left({L^{\ast}}^{k}+L^{k}\right)\varphi\right)\xi_{n}.$
(5.3)

Let $\varphi=\sum_{n=0}^{\infty}c_{n}\xi_{n}$. Since
$\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},\left({L^{\ast}}^{k}+L^{k}\right)\varphi\right)\xi_{n}=\sum_{n=0}^{\infty}(\sum_{m\neq
n}c_{m})\xi_{n}$, we have

$\infty>\|[N,L_{G}]\varphi\|^{2}=\sum_{n=0}^{\infty}|(\sum_{m\neq
n}c_{m})|^{2}.$

In particular

$\left\|\sum_{n=0}^{\infty}\left(\sum_{m\neq
n}c_{m}\right)\xi_{n}+\varphi\right\|^{2}=\sum_{n=0}^{\infty}|\sum_{m=0}^{\infty}c_{m}|^{2}<\infty,$

which implies that
$\varphi\in\big{\\{}\sum_{n=0}^{\infty}c_{n}\xi_{n}\in\ell^{2}\ \big{|}\
\sum_{n=0}^{\infty}c_{n}=0\big{\\}}$, and

$\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},\left({L^{\ast}}^{k}+L^{k}\right)\varphi\right)\xi_{n}=-\varphi.$

Hence $[N,L_{G}]\varphi=-i\varphi$ and the theorem is proven. $\Box$

It is immediate to see that $[N,T_{G}]\phi=-i\phi$ does not hold true for
$\phi=v_{n}$, $n\in\mathbb{N}$. It is however shown in [9] that a CCR-domain
of $T_{G}$ is $\mathrm{LH}\\{v_{n}-v_{m}\mid n,m\in\mathbb{N}\\}$. Note that

$U\left(\mathrm{LH}\\{v_{n}-v_{m}\mid
n,m\in\mathbb{N}\\}\right)=({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\ell^{2}_{\mathrm{fin}}.$

Hence $[N,L_{G}]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ holds on $({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\ell^{2}_{\mathrm{fin}}$ and $({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\ell^{2}_{\mathrm{fin}}\subset\mathrm{D}(NL_{G})\cap\mathrm{D}(L_{G}N)$.

###### Remark 5.6

To our knowledge, it was previously unknown that $[N,L_{G}]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$
holds on a strictly larger subspace than $({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\ell^{2}_{\mathrm{fin}}$. It can be shown however
that the CCR-domain $\mathrm{D}(NL_{G})\cap\mathrm{D}(L_{G}N)$ of $L_{G}$ is
strictly larger than $({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\ell^{2}_{\mathrm{fin}}$. It can be actually seen
that $({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast})e^{\alpha
L^{\ast}}\Omega\in(\mathrm{D}(L_{G}N)\cap\mathrm{D}(NL_{G}))\setminus({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\ell^{2}_{\mathrm{fin}}$ for $\alpha\neq 0$.

###### Remark 5.7

In (5.3), if $\varphi\in{\rm Ran}({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast})$, then we
have

$\displaystyle[N,L_{G}]\varphi=i\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},\left({L^{\ast}}^{k}+L^{k}\right)\varphi\right)\xi_{n}=i\left(({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})^{-1}+({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)^{-1}-2{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}\right)\varphi.$ (5.4)

(5.4) can be extended in (5.8).

###  5.2 Generalization of Galapon operator $\bm{T_{G}}$

In Theorem 5.4 we show that $T_{G}$ can be expressed as
$i\\{\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L)-\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})\\}$ on $\ell^{2}_{\mathrm{fin}}$. We shall
generalize this. We set

$\displaystyle L_{g}=i\left\\{\log\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-g_{N}L\right)-\log\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}g_{N}^{-1}\right)\right\\}.$

###### Lemma 5.8

Let $g$ be a complex-valued function on $\mathbb{N}$ such that $|g_{n}|=1$ for
all $n\in\mathbb{N}$. Then $L_{g}$ is unitary equivalent to $L_{G}$ on
$\ell^{2}_{\mathrm{fin}}$.

Proof: We can construct the unitary operator $V$ on $\ell^{2}$ such that
$g_{N}L=V^{\ast}LV$. It is actually given by

$\displaystyle
V(c_{0},c_{1},c_{2},c_{3},\ldots,)=(c_{0},g_{0}c_{1},g_{0}g_{1}c_{2},g_{0}g_{1}g_{2}c_{3},\ldots,).$

Since $|g_{n}|=1$, $V$ is unitary. Thus $L_{g}$ is unitary equivalent to
$L_{G}$. $\Box$

Let $\ell_{1}^{2}(g)=\big{\\{}\sum_{n=0}^{\infty}c_{n}\xi_{n}\in\ell^{2}\
\big{|}\ \sum_{n=0}^{\infty}g!(n)c_{n}=0\big{\\}}$, where $g!(n)=\prod_{k\leq
n}g_{k}$.

###### Theorem 5.9

Let $g$ be a complex-valued function on $\mathbb{N}$.

  1. (1)

Suppose that $|g_{n}|>0$ for all $n\in\mathbb{N}$. Then
$\left[N,L_{g}\right]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ on
$\mathrm{D}(NL_{g})\cap\mathrm{D}(L_{g}N)$.

  2. (2)

Suppose that $|g_{n}|=1$ for all $n\in\mathbb{N}$. Then $L_{g}$ is a bounded
time operator with CCR-domain $\mathrm{D}(NL_{g})\cap\mathrm{D}(L_{g}N)$.

Proof: As in the proof of Theorem 5.5, we see that
$\mathrm{D}(NL_{g})\cap\mathrm{D}(L_{g}N)\subset\ell_{1}^{2}(g)$. For any
$\varphi\in\mathrm{D}(NL_{g})\cap\mathrm{D}(L_{g}N)$, we have

$\displaystyle[N,L_{g}]\varphi=i\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},\left(\left(L^{\ast}g_{N}^{-1}\right)^{k}+\left(g_{N}L\right)^{k}\right)\varphi\right)\xi_{n}=-i\varphi.$

Then (1) is proven. Since $L_{G}$ is self-adjoint and bounded,
$\overline{L_{g}}$ is also self-adjoint and bounded from Lemma 5.8. By (1)
$L_{g}$ satisfies the canonical commutation relation. $\Box$

$L_{g}$ is no longer a symmetric operator when $|g|\neq 1$. Therefore, in
order to construct a time operator from $L_{g}$ with $|g|\neq 1$, we need to
symmetrize it.

###### Corollary 5.10

Let $g$ be a complex-valued function on $\mathbb{N}$. Suppose that there exist
$n_{0}\in\mathbb{N}$, $c_{1},\ c_{2}\in(0,\infty)$ and $d_{1},\
d_{2}\in(0,1/2)$ such that, for all $n>n_{0}$,

$\displaystyle c_{1}n^{-d_{1}}\leq\prod_{k=0}^{n}|g_{k}|\leq c_{2}n^{d_{2}}.$
(5.5)

Then $(L_{g}+L_{\bar{g}^{-1}})/2$ is a time operator of $N$.

Proof: The assumption $c_{1}n^{-d_{1}}\leq\prod_{k\leq n}|g_{k}|$ implies that
for any $m\in\mathbb{N}$

$\displaystyle\left\|\sum_{k\geq
1}\frac{1}{k}\left(L^{\ast}g_{N}^{-1}\right)^{k}\xi_{m}\right\|$
$\displaystyle=\left\|\sum_{k\geq
1}\frac{1}{k}\left(\prod_{l=0}^{k-1}g_{m+l}\right)^{-1}\\!\\!\xi_{m+k}\right\|$
$\displaystyle\leq\left(\sum_{1\leq k\leq
n_{0}}\\!\\!\\!\\!\frac{1}{k^{2}}\left|\prod_{l=0}^{k-1}g_{m+l}\right|^{-2}\\!\\!+\left|c_{1}^{-1}g!(m-1)\right|^{2}\sum_{k>n_{0}}k^{-2(1-d_{1})}\right)^{1/2}<\infty.$

Thus $\xi_{m}\in\mathrm{D}(\log\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}g_{N}^{-1}\right))$ and
$\xi_{m}\in\mathrm{D}(L_{g})$. In the same way, $\prod_{k=0}^{n}|g_{k}|\leq
c_{2}n^{d_{2}}$ implies that $\xi_{n}\in\mathrm{D}(\log\left({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}\bar{g}_{N}\right))$ and then
$\xi_{n}\in\mathrm{D}(L_{\bar{g}^{-1}})$. Thus (5.5) implies that
$\xi_{n}\in\mathrm{D}(L_{g})\cap\mathrm{D}(L_{\bar{g}^{-1}})$. Since
$L_{g}^{\ast}\supset-i\left\\{\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}\bar{g}_{N})-\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-\bar{g}_{N}^{-1}L)\right\\}=L_{\bar{g}^{-1}}$, the operator
$L_{g}+L_{\bar{g}^{-1}}$ is symmetric. The canonical commutation relation
$\big{[}N,(L_{g}+L_{\bar{g}^{-1}})/2\big{]}=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$
is proven in the same way as Theorem 5.9. $\Box$

###  5.3 Conjugate operators of the form $\bm{\log({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L))-\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-p(L^{\ast}))}$

In Section 4.4 conjugate operators of the form $\log p(L)$ is considered. Let
$Y_{p}={\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-p(L)$. In this section we consider
conjugate operators of the form

$\displaystyle X_{p}=\log(Y_{p})-\log(Y_{p}^{\ast}),$

under some conditions on polynomials $p$. $X_{p}$ is a generalization of
$L_{G}$. Note that $\mathop{\mathrm{Ker}}(Y_{p}^{\ast})=\\{0\\}$ if $p\neq 1$.
Hence the inverse of $Y_{p}^{\ast}$ exists.

###### Lemma 5.11

Let $p$ be a polynomial. Suppose that $Y_{p}$ is injective and
$\lim_{k\to\infty}p(L)^{k}=0$. Then

$\displaystyle\left[N,\log Y_{p}\right]=Lp^{\prime}(L)Y_{p}^{-1}$ (5.6)

on ${\rm Ran}(Y_{p})\cap\mathrm{D}\left(N\log
Y_{p}\right)\cap\mathrm{D}\left(\log(Y_{p})N\right)$ and

$\displaystyle\left[N,\log(Y_{p}^{\ast})\right]=-L^{\ast}p^{\prime}(L^{\ast})(Y_{p}^{\ast})^{-1}$
(5.7)

on ${\rm
Ran}(Y_{p}^{\ast})\cap\mathrm{D}\left(N\log(Y_{p}^{\ast})\right)\cap\mathrm{D}\left(\log(Y_{p}^{\ast})N\right)$.

Proof: Let $\varphi\in\mathrm{D}\left(N\log
Y_{p}\right)\cap\mathrm{D}\left(\log(Y_{p})N\right)$ and $\varphi=Y_{p}\psi$.
Inserting the expansion $\varphi=\sum_{n=0}^{\infty}(\xi_{n},\varphi)\xi_{n}$
to $[N,\log Y_{p}]\varphi$, we see that

$\displaystyle\left[N,\log
Y_{p}\right]\varphi=-\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k}\left(\xi_{n},\left[N,p(L)^{k}\right]\varphi\right)\xi_{n}.$

Since $[N,p(L)^{k}]\subset-kLp^{\prime}(L)p(L)^{k-1}$, we have

$\displaystyle\left[N,\log Y_{p}\right]\varphi$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},Lp^{\prime}(L)p(L)^{k-1}\varphi\right)\xi_{n}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},Lp^{\prime}(L)p(L)^{k-1}({\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-p(L))\psi\right)\xi_{n}.$

By assumption $\lim_{k\to\infty}p(L)^{k}=0$, we have

$\displaystyle\left[N,\log
Y_{p}\right]\varphi=-\sum_{n=0}^{\infty}\lim_{k\to\infty}\left(\xi_{n},Lp^{\prime}(L)p(L)^{k}\psi\right)\xi_{n}+\sum_{n=0}^{\infty}\left(\xi_{n},Lp^{\prime}(L)\psi\right)\xi_{n}=Lp^{\prime}(L)Y_{p}^{-1}\varphi.$

Then (5.6) follows. (5.7) can be shown in a similar way to (5.6). $\Box$

From Lemma 5.11, if
$Lp^{\prime}(L)Y_{p}^{-1}+L^{\ast}p^{\prime}(L^{\ast})(Y_{p}^{\ast})^{-1}$ has
a non-zero eigenvalue, then $N$ has a conjugate operator on the eigenvector
space. Let $\SS^{1}=\\{\alpha\in\mathbb{C}\mid|\alpha|=1\\}$. Thus, we have
the following theorem.

###### Theorem 5.12

Let $p$ be a polynomial with degree $m$ such that

$\displaystyle(1)\ {\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-p(L)\mbox{ is injective},$
$\displaystyle(2)\ \lim_{k\to\infty}p(L)^{k}=0,$ $\displaystyle(3)\
\\{z\in\mathbb{C}\mid 1-p(z)=0\\}\subset\SS^{1}.$

Let $\alpha_{1},\ldots,\alpha_{m}\in\mathbb{C}$ be all roots of
$1-p(\alpha)=0$ within the multiplicity. Then

$\displaystyle\left[N,\frac{i}{m}X_{p}\right]=-i{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$

with CCR-domain ${\rm
Ran}\big{(}\prod_{i=1}^{m}(\alpha_{i}L^{\ast}-{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})\big{)}\cap\mathrm{D}(X_{p}N)\cap\mathrm{D}(NX_{p})$.

Proof: By Lemma 5.11 we have

$\displaystyle[N,X_{p}]$
$\displaystyle=Lp^{\prime}(L)Y_{p}^{-1}+L^{\ast}p^{\prime}(L^{\ast})(Y_{p}^{\ast})^{-1}$
$\displaystyle=\sum_{i=1}^{m}\left(\alpha_{i}(\alpha_{i}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)^{-1}+\alpha_{i}(\alpha_{i}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})^{-1}-2{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}\right).$ (5.8)

Note that (5.8) corresponds to (5.4). It suffices to show that the right-hand
side above has an eigenvector in
$\mathrm{D}\left(NX_{p}\right)\cap\mathrm{D}\left(X_{p}N\right)$. Since
$\alpha_{i}\in\SS^{1}$,we see that

$\displaystyle\sum_{i=1}^{m}\left(\alpha_{i}(\alpha_{i}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)^{-1}+\alpha_{i}(\alpha_{i}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})^{-1}\right)=\sum_{i=1}^{m}\left(\alpha_{i}(\alpha_{i}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)^{-1}+\alpha_{i}^{-1}(\alpha_{i}^{-1}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})^{-1}\right).$

Let $\psi\in\ell^{2}\setminus\\{0\\}$ and $\varphi=(\alpha
L^{\ast}-{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})\psi$. Since

$\displaystyle\left(\alpha(\alpha{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)^{-1}+\alpha^{-1}(\alpha^{-1}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})^{-1}\right)\varphi$ $\displaystyle=\alpha
L^{\ast}\psi-\psi=\varphi,$

$\varphi$ is an eigenvector of $\alpha(\alpha{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)^{-1}+\alpha^{-1}(\alpha^{-1}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast})^{-1}$ with an eigenvalue $1$. Hence
$\prod_{i=1}^{m}(\alpha_{i}L^{\ast}-{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})\psi$ is an
eigenvector of
$Lp^{\prime}(L)Y_{p}^{-1}+L^{\ast}p^{\prime}(L^{\ast})(Y_{p}^{\ast})^{-1}$
with the eigenvalue $-1$ and

$[N,X_{p}]\prod_{i=1}^{m}(\alpha_{i}L^{\ast}-{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})\psi=-m\prod_{i=1}^{m}(\alpha_{i}L^{\ast}-{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}})\psi.$

Then a CCR-domain is given by ${\rm
Ran}(\prod_{i=1}^{m}(\alpha_{i}L^{\ast}-{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}))\cap\mathrm{D}(X_{p}N)\cap\mathrm{D}(NX_{p})$, and the
theorem is proven. $\Box$

###### Corollary 5.13

Let $|\omega|=1$. Then $im^{-1}\\{\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-\omega
L^{m})-\log({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-\omega^{\ast}L^{\ast}{}^{m})\\}$ is a
time operator of $N$.

Proof: Let $p(z)=\omega z^{m}$. Then the corollary follows from Theorem 5.12.
$\Box$

##  6 Classification of conjugate operators

In this section we focus on investigating conjugate operators of the form

$T_{\omega,m}=\frac{i}{m}\log(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})\quad(\omega,m)\in\overline{\mathbb{D}}\times(\mathbb{N}\setminus\\{0\\}).$

This type of operators have appeared in Lemma 5.1 and Corollary 5.13.
Furthermore note that $T_{A}\cong L_{A0}\oplus L_{A1}$. In Example 4.18 we
mentioned that to investigate the CCR-domain of $L_{A0}$ and $L_{A1}$ can be
reduced to investigate the CCR-domain of

$\frac{i}{2}\log L^{2}=T_{0,2}.$

Let us define the family of operators with parameters $\omega$ and $m$ by

$\mathscr{T}=\\{T_{\omega,m}\mid\omega\in\overline{\mathbb{D}},m\geq 1\\}.$

We divide $\mathscr{T}$ into three disjoint families:

$\displaystyle\mathscr{T}=\mathscr{T}_{\\{0\\}}\cup\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}\cup\mathscr{T}_{\partial\mathbb{D}}.$

Here

$\displaystyle\mathscr{T}_{\\{0\\}}=\\{T_{\omega,m}\mid\omega=0,m\geq 1\\},$
$\displaystyle\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}=\\{T_{\omega,m}\mid
0<|\omega|<1,m\geq 1\\},$
$\displaystyle\mathscr{T}_{\partial\mathbb{D}}=\\{T_{\omega,m}\mid|\omega|=1,m\geq
1\\}.$

In what follows we discuss CCR-domains of $T_{\omega,m}$.

We define

$\displaystyle\ell_{m}^{2}=\left\\{\varphi\in\ell^{2}\ \middle|\
\lim_{k\to\infty}({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{m})^{k}\varphi=0\right\\},$

######  Theorem 6.1 ($\mathscr{T}_{\\{0\\}}$)

We have

$\left[N,\frac{i}{m}\log L^{m}\right]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$

on $\ell_{m}^{2}\cap\mathrm{\mathrm{D}}(N\log
L^{m})\cap\mathrm{\mathrm{D}}(\log(L^{m})N)$, and
$\dim\left(\ell_{m}^{2}\cap\mathrm{\mathrm{D}}(N\log
L^{m})\cap\mathrm{\mathrm{D}}(\log(L^{m})N)\right)=\infty$.

Proof: This is a special case of Theorem 4.19. Let $p(z)=z^{m}$. Then
$zp^{\prime}(z)-mp(z)=0$ for all $z\in\mathbb{C}$. In a similar way of the
proof of Theorem 4.19, for any
$\varphi\in\ell_{m}^{2}\cap\mathrm{\mathrm{D}}(N\log
L^{m})\cap\mathrm{\mathrm{D}}(\log(L^{m})N)$, we see that $[N,\log
L^{m}]\varphi=-m\sum_{k=1}^{\infty}\left({\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m}\right)^{k-1}L^{m}\varphi=-m\varphi$. Then the theorem
follows. $\Box$

Next we consider $\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}$. Let
$c\in\mathbb{C}$ and we recall that

$\displaystyle\mathrm{D}(\omega-z^{m},c)$
$\displaystyle=\mathrm{LH}\\{\varphi\mid\exists\alpha\in\mathbb{C}\ s.t.\
|1-\omega+\alpha^{m}|<1,\ \alpha^{m}=c(m+c)^{-1}\omega,\
\varphi\in\mathop{\mathrm{Ker}}(L-\alpha{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})\\}.$

######  Theorem 6.2 ($\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}$)

Let $\omega\in\mathbb{D}\setminus\\{0\\}$. Then for
$c\in\mathbb{C}\setminus\\{0\\}$ such that $|c|$ is sufficiently small, we see
that $\mathrm{D}(\omega-z^{m},c)\neq\\{0\\}$,
$\dim\mathrm{D}(\omega-z^{m},c)<\infty$ and

$\displaystyle\left[N,-\frac{i}{c}\log(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})\right]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$

on $\mathrm{D}(\omega-z^{m},c)$. Moreover, for any
$c\in\mathbb{C}\setminus\\{0\\}$, there is no infinite dimensional CCR-domain
for $N$ and $c\log(\omega-L^{m})$

Proof: Let $p(z)=\omega-z^{m}$ with $|\omega|<1$ and $|\omega-1|<1$. Suppose
that $c$ is a sufficiently small positive number. Then
$zp^{\prime}(z)+cp(z)=-(c+m)z^{m}+c\omega=0$ has the roots

$\alpha_{k}=\left|\frac{c}{c+m}\right|\omega^{1/m}e^{2\pi ik/m},\quad
k=0,\ldots,m-1.$

By Theorem 4.19, since $|p(\alpha_{k})-1|=|\omega-1-\alpha_{k}^{m}|<1$ and
$|\alpha_{k}|<1$, $k=0,\ldots,m-1$, we have
$[N,\frac{i}{c}\log(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})]=i{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ on
$\mathop{\mathrm{Ker}}(L-\alpha_{k}{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}})$ for
$k=0,\ldots,m-1$. Moreover the dimension of the CCR-domain
$\mathrm{LH}\\{\mathop{\mathrm{Ker}}(L-\alpha_{k}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}),k=0,\ldots,m-1\\}$ is finite. $\Box$

We consider conjugate operators of the form $i\\{\log(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)-\log(\bar{\omega}{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast})\\}$ with
$\omega\in\partial\mathbb{D}$. We consider
$\mathop{\mathrm{Log}}(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-A)$ for
$\log(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-A)$ to avoid a singularity.

######  Definition 6.3 ($\mathop{\mathrm{Log}}(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A)$)

Let $A$ be a linear operator on a Hilbert space $\mathcal{H}$. We define

$\displaystyle\mathrm{D}(\mathop{\mathrm{Log}}(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A))=\left\\{\varphi\in\bigcap_{k=0}^{\infty}\mathrm{D}(A^{k})\
\middle|\
\lim_{K\to\infty}\sum_{k=1}^{K}\frac{1}{k}\left(\frac{1}{\omega}A\right)^{k}\\!\\!\\!\varphi\text{
exists}\right\\},$ $\displaystyle\mathop{\mathrm{Log}}(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A)\varphi=\log(\omega)\varphi-\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{\omega}A\right)^{k}\\!\\!\\!\varphi,\quad\varphi\in\mathrm{D}(\mathop{\mathrm{Log}}(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-A)).$ (6.1)

######  Theorem 6.4 ($\mathscr{T}_{\partial\mathbb{D}}$)

Let $\omega\in\partial\mathbb{D}$. Then
$L_{m,\omega}=im^{-1}(\mathop{\mathrm{Log}}(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})-\mathop{\mathrm{Log}}(\bar{\omega}{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{\ast}{}^{m}))$ is a time operator with the dense CCR-
domain $\mathrm{D}(NL_{m,\omega})\cap\mathrm{D}(L_{m,\omega}N)$.

Proof: As in the proof of Theorem 5.5, we see that

$\mathrm{D}(NL_{m,\omega})\cap\mathrm{D}(L_{m,\omega}N)\subset\left.\left\\{\sum_{n=0}^{\infty}c_{n}\xi_{n}\right|\sum_{n=0}^{\infty}\bar{\omega}^{n}c_{l+mn}=0\text{
for all }l<m\right\\}$

and for any
$\varphi\in\mathrm{D}(NL_{m,\omega})\cap\mathrm{D}(L_{m,\omega}N)$,

$\displaystyle[N,L_{m,\omega}]\varphi$
$\displaystyle=i\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\left(\xi_{n},\left((\bar{\omega}L^{m})^{k}+(\omega
L^{\ast}{}^{m})^{k}\right)\varphi\right)\xi_{n}=-i\varphi.$

Then the theorem follows. $\Box$

| $\mathscr{T}_{\\{0\\}}$ | $\mathscr{T}_{\mathbb{D}\setminus\\{0\\}}$ | $\mathscr{T}_{\partial\mathbb{D}}$
---|---|---|---
$T_{\omega,m}$ | $\omega=0$ | $\omega\in\mathbb{D}\setminus\\{0\\}$ | $\omega\in\partial\mathbb{D}$
boundedness | unbounded | unbounded | bounded
CCR-domain | infinite dim. | finite dim. | dense
example | $T_{A}$ |  | $T_{G}=T_{1,1}+T_{1,1}^{\ast}$
$\log$ | Def 3.1 | Def 3.1 | Def 6.3
Table 1: Classification of $T_{\omega,m}$

We can also introduce operator $T_{\omega,m}$ with $|\omega|>1$ by a Dunford
integral. See (A.1) in Appendix.

##  7 Weak Wyle relations for $\bm{N}$

In this section we consider the time evolution of conjugate operators. The
Weyl relation [19]

$e^{-isp}e^{-itq}=e^{ist}e^{-itq}e^{-isp},\quad s,t\in\mathbb{R}.$

holds true. From this we can derive the so-called weak Weyl relation [14]:

$\displaystyle qe^{-itp}=e^{-itp}(q+t),\quad t\in\mathbb{R},$ (7.1)

on $\mathrm{D}(q)$. The strong time operator $T$ is defined as an operator
satisfying (7.1) with $q$ and $p$ replaced by $T$ and Hamiltonian $H$,
respectively. More precisely, we explain the weak Weyl relation (7.1) as
follows.

###### Definition 7.1

We say that the pair $(H,A)$ consisting of a self-adjoint operator $H$ and a
symmetric operator $A$ on a Hilbert space $\mathcal{H}$ obeys the weak Weyl
relation if for all $t\in\mathbb{R}$,

  * (1)

$e^{-itH}\mathrm{D}(A)\subset\mathrm{D}(A)$;

  * (2)

$Ae^{-itH}\varphi=e^{-itH}(A+t)\varphi$ for all $\varphi\in\mathrm{D}(A)$.

Here $A$ is referred to as a strong time operator associated with $H$. Note
that a strong time operator is not unique. If strong time operator $A$ is
self-adjoint, then it is known that

$e^{-isA}e^{-itH}=e^{-ist}e^{-itH}e^{-isA},\quad s,t\in\mathbb{R}$

holds. In particular when Hilbert space is separable, by the von Neumann
uniqueness theorem the Weyl relation above implies that $H$ and $A$ are
unitarily equivalent to $\oplus^{n}p$ and $\oplus^{n}q$ with some
$n\leq\infty$, respectively. As was mentioned above, although a strong time
operator is automatically a time operator, the converse is not true. It is
remarkable that when the pair $(H,A)$ obeys the weak Weyl relation, $H$ has
purely absolutely continuous spectrum. For example since the spectrum of $N$
is purely discrete, there is no strong time operator associated with $N$. Weak
Weyl relation can be understood as the time evolution of $A$:

$\displaystyle e^{itH}Ae^{-itH}\supset A+t$ (7.2)

We can see the time evolution of the Galapon operator directly. It can be seen
that

$T_{G}(t)\varphi=i\sum_{n=0}^{\infty}\sum_{m\neq
n}\frac{e^{+it(n-m)}(v_{m},\varphi)}{n-m}v_{n}.$

Hence $T_{G}(t)$ is periodic so that

$T_{G}(t+2\pi n)=T_{G}(t),\quad n\in\mathbb{N},t\in\mathbb{R}.$

Let us consider the time evolutions of general conjugate operators. We define

$T_{\omega,m}(t)=e^{itN}T_{\omega,m}e^{-itN}.$

###### Theorem 7.2

Let $T_{\omega,m}\in\mathscr{T}$. Then

$T_{\omega,m}(t)=\frac{i}{m}\log(\omega-e^{-itm}L^{m})$

and $T_{\omega,m}(t)$ is periodic with period $2\pi/m$, i.e.,

$T_{\omega,m}(t)=T_{\omega,m}(t+(2\pi/m)n),\quad
n\in\mathbb{N},t\in\mathbb{R}.$

Proof: Since $e^{itN}Le^{-itN}=e^{-it}L$, we see that
$e^{itN}\log(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})e^{-itN}=\log(\omega-e^{-itm}L^{m})$. $\Box$

###### Example 7.3

It is shown that the angle operator $T_{A}=\frac{1}{2}(\arctan
q^{-1}p\oplus\arctan pq^{-1})$ can be represented as

$\displaystyle\frac{1}{2}\arctan\left(q^{-1}p\right)\cong\frac{i}{2}\log\left(\sqrt{\frac{N+2{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{N+{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}}L^{2}\right),$
$\displaystyle\frac{1}{2}\arctan\left(pq^{-1}\right)\cong\frac{i}{2}\log\left(\sqrt{\frac{N+{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}}{{N+2{\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}}}}L^{2}\right)$

in Theorem 4.9. Thus the period of the time evolutions of both of
$\arctan\left(q^{-1}p\right)$ and $\arctan\left(pq^{-1}\right)$ are $\pi$.

Finally we show an example of (3.2). Let $\omega=0$ and consider the time
evolution of $T_{0,m}(t)$.

###### Corollary 7.4

Let $m\in\mathbb{N}$ and $\arg e^{-itm}\neq 0$. Then

$\displaystyle\log(-e^{-itm}L^{m})\neq\log\left(e^{-itm}\right)+\log(-L^{m}).$

Proof: Suppose that
$\log(-e^{-itm}L^{m})=\log\left(e^{-itm}\right)+\log(-L^{m})$. Then we see
that

$\displaystyle e^{itN}T_{0,m}e^{-itN}$
$\displaystyle=\frac{i}{m}\log(-e^{-itm}L^{m})=-\frac{1}{m}\arg
e^{-itm}+\frac{i}{m}\log(-L^{m})=-\frac{1}{m}\arg e^{-itm}+T_{0,m}.$

Hence we see that for any eigenvecotr $v_{n}$ of $N$

$\displaystyle(v_{n},T_{0,m}v_{n})=(v_{n},e^{itN}T_{0,m}e^{-itN}v_{n})=-\frac{1}{m}(v_{n},\arg
e^{-itm}v_{n})+(v_{n},T_{0,m}v_{n}).$ (7.3)

Since $(v_{n},\arg e^{-itm}v_{n})\neq 0$, (7.3) is contradiction. Then the
corollary follows. $\Box$

##  Appendix A CCR domains for $\mathscr{T}_{\overline{\mathbb{D}}^{c}}$

We consider conjugate operators of the form $i\log(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})$ with $|\omega|>1$. To avoid a singularity, we need to
modify the definition of $i\log(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{m})$. Let
$r\mathbb{D}=\\{rz\mid z\in\mathbb{D}\\}$ for $r>0$ and $f$ be an analytic
function such that $f(\overline{r\mathbb{D}})\not\ni 0$ for some $r>1$. Then
$\log f$ is analytic on $r\mathbb{D}$. Let $\|A\|<1$. Since $(z-A)^{-1}$ is
bounded for any $z\not\in\overline{\mathbb{D}}$, it can be seen that the
Dunford integral $\oint_{r\partial\mathbb{D}}\log f(z)(z-A)dz$ defines the
sesquilinear form

$\displaystyle(\varphi,\psi)\mapsto
Q(\varphi,\psi)=\oint_{r\partial\mathbb{D}}\log
f(z)(\varphi,(z-A)^{-1}\psi)dz.$ (A.1)

(A.1) also defines the bounded operator such that
$Q(\varphi,\psi)=(\varphi,B\psi)$. $B$ is also denoted by the same symbol
$\log f(A)$ as that of Definition 3.1. Set $f(z)=\omega-z^{m}$ and $A=L$. Then
we can define $\log(\omega{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{m})$ by (A.1).

###### Lemma A.1

There is no polynomial $p$ satisfying the following conditions:

  1. (1)

$0\not\in{\rm Spec}\left(p(L)\right)$,

  2. (2)

$[N,i\\{\log p(L)-\log p(L^{\ast})\\}]=-i{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ on some infinite
dimenstional subspace $\mathcal{D}$.

Proof: Let $p(x)=\sum_{k=0}^{n}a_{k}x^{k}$ be a real-valued polynomial with
$a_{n}\neq 0$. We see that

$p(L)p(L^{\ast}){L^{\ast}}^{n}=\left(\sum_{k=0}^{n}a_{k}L^{\ast}{}^{n-k}\right)\left(\sum_{k=0}^{n}a_{k}{L^{\ast}}^{k}\right)=\sum_{k,l\geq
0}^{n}a_{k}a_{l}L^{\ast}{}^{n-l+k}.$ (A.2)

It is easy to see that

$[N,\log p(L)-\log p(L^{\ast})]\subset-
Lp^{\prime}(L)p(L)^{-1}-L^{\ast}p^{\prime}(L^{\ast})p(L^{\ast})^{-1}.$ (A.3)

Note that ${\rm
Ran}({L^{\ast}}^{n})=\\{(c_{k})_{k\in\mathbb{N}}\in\ell^{2}\mid
c_{0}=\cdots=c_{n-1}=0\\}$. Hence $\mathcal{D}\cap{\rm
Ran}({L^{\ast}}^{n})=\\{0\\}$ implies that $\dim\mathcal{D}<\infty$. Then
$\mathcal{D}\cap{\rm Ran}({L^{\ast}}^{n})\neq\\{0\\}$, since
$\dim\mathcal{D}=\infty$. Notice that, for any $k\leq n$,
$L^{k}{L^{\ast}}^{k}={L^{\ast}}^{k}L^{k}={\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}$ on ${\rm
Ran}({L^{\ast}}^{n})$. Thus we also have the following relation on the non-
trivial domain by (A.3):

$\displaystyle p(L)p(L^{\ast}){L^{\ast}}^{n}$
$\displaystyle=-p(L)p(L^{\ast})[N,\log p(L)-\log p(L^{\ast})]{L^{\ast}}^{n}$
$\displaystyle=Lp^{\prime}(L)p(L^{\ast}){L^{\ast}}^{n}+L^{\ast}p^{\prime}(L^{\ast})p(L){L^{\ast}}^{n}$
$\displaystyle=\left(\sum_{k=1}^{n}ka_{k}L^{\ast}{}^{n-k}\right)\left(\sum_{k=0}^{n}a_{k}{L^{\ast}}^{k}\right)+\left(\sum_{k=1}^{n}ka_{k}{L^{\ast}}^{k}\right)\left(\sum_{k=0}^{n}a_{k}L^{\ast}{}^{n-k}\right)$
$\displaystyle=\sum_{k,l\geq
1}^{n}(ka_{k}a_{l}+la_{l}a_{k})L^{\ast}{}^{n-l+k}+\sum_{k=1}^{n}ka_{k}a_{0}L^{\ast}{}^{n-k}+\sum_{k=1}^{n}ka_{k}a_{0}L^{\ast}{}^{n+k}.$
(A.4)

Let $L^{\ast}{}^{n}\varphi\in\mathcal{D}\setminus\\{0\\}$ and
$n_{0}=\inf\mathop{\mathrm{supp}}(\varphi)$. Comparing the coefficients of
$L^{\ast}{}^{0}$ in $\eqref{p1}$ and $\eqref{p2}$, we have
$na_{0}a_{n}=a_{0}a_{n}$. From (1) we have $a_{0}\neq 0$. Since
$a_{0}a_{n}\neq 0$, we see that $n=1$. Hence $p(x)=ax+b$. By (A.4) we also
have $a=\pm b$, and then $p(x)=a(1\pm x)$. We conclude that
$p(L)=a({\mathchoice{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}\pm L)$ and ${\rm Spec}(p(L))=p({\rm
Spec}(L))\ni 0$. This contradicts $0\not\in{\rm Spec}\left(p(L)\right)$. Then
the lemma follows. $\Box$

###### Theorem A.2

Let $|\omega|>1$. The operator $i\\{\log(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L^{m})-\log(\bar{\omega}{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast}{}^{m})\\}$
has no infinite dimensional CCR-domain for $N$.

Proof: Let $p(z)=\omega-z^{m}$ with $|\omega|>1$. Then ${\rm
Spec}(p(L))\not\ni 0$. Thus the theorem follows from Lemma A.1. $\Box$

###### Remark A.3

The existence of a non zero CCR domain of $i\\{\log(\omega{\mathchoice{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm
1\mskip-5.0mul}}-L)-\log(\bar{\omega}{\mathchoice{\rm 1\mskip-4.0mul}{\rm
1\mskip-4.0mul}{\rm 1\mskip-4.5mul}{\rm 1\mskip-5.0mul}}-L^{\ast})\\}$ with
$|\omega|>1$ is unknown.

Acknowledgements: FH is financially supported by JSPS KAKENHI 20K20886 and
JSPS KAKENHI 20H01808.

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