text:
\section{Introduction}

In this paper, we consider a three-body quantum system
with time-periodic pair interactions.
Since we would like to introduce some notation in many body scattering
theory, we denote the number of particles in the system by
$N$ for a while. Of course, we mainly consider the case where $N=3$.
The system under consideration
is governed by the following Schr\"odinger operator with
time-periodic potentials
\begin{equation}
\tilde{H}(t)=\sum_{j=1}^N\left(-\frac{1}{2m_j}\Delta_j\right)+V(t),\quad
V(t)=\sum_{1\le j<k\le N}V_{jk}(t,r_j-r_k)
\label{1.1}
\end{equation}
acting on $L^2(\boldsymbol{R}^{d\times N})$,
where $m_j$ and $r_j\in\boldsymbol{R}^d$ are
the mass and position vector of the $j$-th particle, respectively,
$$\Delta_j=\sum_{l=1}^d\partial_{r_{j,l}}^2$$
is the Laplacian with
respect to $r_j$, and $V_{jk}(t,r_j-r_k)$'s are pair potentials.
We suppose that $V_{jk}(t,y)$'s are real-valued functions
on $\boldsymbol{R}\times\boldsymbol{R}^d$
which are periodic in $t$ with a period $T>0$:
\begin{equation}
V_{jk}(t+T,y)=V_{jk}(t,y),\quad(t,y)\in\boldsymbol{R}\times
\boldsymbol{R}^d.
\label{1.2}
\end{equation}
We would like to watch the motion of the system in the
center-of-mass frame. To this end, we will introduce the
following configuration spaces:
We equip $\boldsymbol{R}^{d\times N}$ with the metric
$$r\cdot\tilde{r}=\sum_{j=1}^Nm_j\langle r_j,\tilde{r}_j
\rangle,\quad r=(r_1,\ldots,r_N),\,
\tilde{r}=(\tilde{r}_1,\ldots,\tilde{r}_N)
\in\boldsymbol{R}^{d\times N},$$
where $\langle\cdot,\cdot\rangle$ is the standard inner product
on $\boldsymbol{R}^d$.
We usually write $r\cdot r$
as $r^2$. We put $|r|=\sqrt{r^2}$.
We define two subspaces $X$ and $X_\mathrm{cm}$ of $\boldsymbol{R}^{d\times N}$
as
\begin{align*}
&X=\left\{r\in\boldsymbol{R}^{d\times N}\,\middle|\,\sum_{j=1}^Nm_jr_j=0\right\},\\
&X_\mathrm{cm}=\left\{r\in\boldsymbol{R}^{d\times N}\,\middle|\,r_1=\cdots=r_N=0\right\}.
\end{align*}
Then $X$ and $X_\mathrm{cm}$ are perpendicular to each other,
and satisfy $\boldsymbol{R}^{d\times N}=X\oplus X_\mathrm{cm}$.
$\pi:\boldsymbol{R}^{d\times N}\to X$ and
$\pi_\mathrm{cm}:\boldsymbol{R}^{d\times N}\to X_\mathrm{cm}$
denote the orthogonal projections
onto $X$ and $X_\mathrm{cm}$, respectively. We put
$x=\pi r$ and $x_\mathrm{cm}=\pi_\mathrm{cm}r$ for
$r\in\boldsymbol{R}^{d\times N}$.
Now we introduce the time-dependent Hamiltonian
\begin{equation}
H(t)=-\frac{1}{2}\Delta+V(t)
\label{1.3}
\end{equation}
acting on $\mathscr{H}=L^2(X)$. Then $\tilde{H}(t)$ is represented
as
$$\tilde{H}(t)=H(t)\otimes\mathrm{Id}+\mathrm{Id}\otimes\left(-\frac{1}{2}\Delta_\mathrm{cm}\right)$$
on $L^2(\boldsymbol{R}^{d\times N})=\mathscr{H}\otimes L^2(X_\mathrm{cm})$.
Here $\Delta$ and $\Delta_\mathrm{cm}$ are
the Laplace-Beltrami operators on $X$ and $X_\mathrm{cm}$, respectively.
By introducing the velocity operators $p$ and $p_\mathrm{cm}$ on $X$
and $X_\mathrm{cm}$, respectively, $-\Delta$ and $-\Delta_\mathrm{cm}$
can be represented as
$$-\Delta=p^2,\quad -\Delta_\mathrm{cm}=(p_\mathrm{cm})^2.$$
We would like to study some scattering problems for this Hamiltonian $H(t)$
with $N=3$.

A non-empty subset of the set $\{1,\ldots,N\}$ is called
a cluster. Let $C_j$, $1\le j\le m$, be clusters.
If $\cup_{1\le j\le m}C_j=\{1,\ldots,N\}$ and
$C_j\cap C_k=\emptyset$ for $1\le j<k\le m$,
$a=\{C_1,\ldots,C_m\}$ is called a cluster decomposition.
$\#(a)$ denotes the number of clusters in $a$.
Let $\mathscr{A}$ be the set of all cluster decompositions. Suppose $a$,
$b\in\mathscr{A}$. If $b$ is obtained as a refinement of $a$,
that is, if each cluster in $b$ is a subset of a cluster in $a$,
we say $b\subset a$, and
its negation is denoted by $b\not\subset a$. Any $a$ is
regarded as a refinement of itself.
The one and $N$-cluster decompositions are denoted
by $a_\mathrm{max}$ and $a_\mathrm{min}$, respectively.
The pair $(j,k)$ is identified with the $(N-1)$-cluster
decomposition $\{(j,k),(1),\ldots,(\widehat{j}),\ldots,(\widehat{k}),\ldots,(N)\}$. If $N=3$, then $\{(1,2),\,(1,3),\,(2,3)\}$ is the set of all $2$-cluster
decompositions.

Let $a\in\mathscr{A}$. We introduce two subspaces $X^a$ and $X_a$ of
$X$:
\begin{align*}
&X^a=\left\{r\in X\,\middle|\,\sum_{j\in C}m_jr_j=0\ \text{ for each cluster }
\ C\ \text{ in }\ a\right\},\\
&X_a=\left\{r\in X\,\middle|\,r_j=r_k\ \text{ for each pair }\ (j,k)\subset a\right\}.
\end{align*}
$\pi^a:X\to X^a$ and
$\pi_a:X\to X_a$ denote the orthogonal projections onto $X^a$ and $X_a$,
respectively. We put $x^a=\pi^a\,x$ and $x_a=\pi_a\,x$ for $x\in X$.
Since $X^{(j,k)}$ is identified with the
configuration space for the relative position of $j$-th and $k$-th
particles, one can put
$$V_{(j,k)}(t,x^{(j,k)})=V_{jk}(t,r_j-r_k).$$
We now define the cluster Hamiltonian
$$H_a(t)=-\frac{1}{2}\Delta+V^a(t),\quad
V^a(t)=\sum_{(j,k)\subset a}V_{(j,k)}(t,x^{(j,k)}),$$
which governs the motion of the system broken into non-interacting
clusters of particles.
Then $H_a(t)$ is represented as
$$H_a(t)=H^a(t)\otimes \mathrm{Id}+\mathrm{Id}\otimes\left(-\dfrac{1}{2}\Delta_a\right);\quad
H^a(t)=-\frac{1}{2}\Delta^a+V^a(t)$$
on $\mathscr{H}=\mathscr{H}^a\otimes \mathscr{H}_a=L^2(X^a)\otimes
L^2(X_a)$, where
$\Delta^a$ and $\Delta_a$ are the Laplace-Beltrami operators on $X^a$ and
$X_a$, respectively.
By introducing the velocity operators $p^a$ and $p_a$ on $X^a$
and $X_a$, respectively, $-\Delta^a$ and $-\Delta_a$
can be represented as
$$-\Delta^a=(p^a)^2,\quad -\Delta_a=(p_a)^2.$$
The intercluster potential $I_a(t)$ is given by
$$I_a(t,x)=V(t,x)-V^a(t,x)=\sum_{(j,k)\not\subset a}V_{(j,k)}(t,x^{(j,k)}).$$
\medskip

Under some suitable conditions
on $V_{jk}(t)$, the existence and uniqueness of
the unitary propagator $U(t,s)$ generated by $H(t)$
can be guaranteed, even if $N\ge3$ (see e.g. Yajima~\cite{Ya2,Ya3}).
In the study of the asymptotic behavior of
$U(t,s)\phi$, $\phi\in\mathscr{H}$,
as $t\to\pm\infty$, we will frequently
utilize the so-called Floquet Hamiltonian
$K$ associated with $H(t)$: Let
$\boldsymbol{T}=\boldsymbol{R}/(T\boldsymbol{Z})$ be the torus.
Set $\mathscr{K}=L^2(\boldsymbol{T};\mathscr{H})\cong
L^2(\boldsymbol{T})\otimes \mathscr{H}$,
and introduce a strongly continuous one-parameter unitary group
$\{\hat{U}(\sigma)\}_{{\sigma\in\boldsymbol{R}}}$
on $\mathscr{K}$ given by
\begin{equation}
(\hat{U}(\sigma)\Phi)(t)=U(t,t-\sigma)\Phi(t-\sigma)
\label{1.4}
\end{equation}
for $\Phi\in\mathscr{K}$. By virtue of Stone's theorem,
$\hat{U}(\sigma)$ is written as
\begin{equation}
\hat{U}(\sigma)=e^{-i\sigma K}
\label{1.5}
\end{equation}
with a unique self-adjoint operator $K$ on $\mathscr{K}$.
$K$ is called the Floquet Hamiltonian associated with $H(t)$,
and is equal to the natural self-adjoint realization of
$-i\partial_t+H(t)$.
Here we denote by $D_t$ the operator $-i\partial_t$
with domain $AC(\boldsymbol{T})$, which is the
space of absolutely continuous functions on $\boldsymbol{T}$
with their derivatives being square integrable (following
the notation in Reed-Simon~\cite{RS}).
As is well-known, $D_t$ is self-adjoint on $L^2(\boldsymbol{T})$,
and its spectrum $\sigma(D_t)$ is equal to $\omega\boldsymbol{Z}$
with $\omega=2\pi/T$.

In this paper, we would like to propose the definition of
a conjugate operator for $K$ with $N=3$. First
we recall known results in the case where $N=2$
for reference.
Yokoyama~\cite{Yo} introduced the self-adjoint operator
\begin{equation}
\tilde{A}_1=\frac{1}{2}\{x\cdot p(1+p^2/2)^{-1}+(1+p^2/2)^{-1}p\cdot x\}
\label{1.6}
\end{equation}
on $\mathscr{K}$ as a conjugate operator for $K$.
For the sake of brevity, we will use the notation
$\mathrm{Re}\,T$ for an operator on $\mathscr{K}$ in this paper,
which is defined by
$$\mathrm{Re}\,T=\frac{1}{2}(T+T^*).$$
Then $\tilde{A}_1$ can be written as
$\mathrm{Re}\,((1+p^2/2)^{-1}p\cdot x)$.
Roughly speaking, $\tilde{A}_1$ is defined
by multiplying the generator of dilations
\begin{equation}
\hat{A}_0=\frac{1}{2}(x\cdot p+p\cdot x)=\mathrm{Re}\,(p\cdot x)
\label{1.7}
\end{equation}
and the resolvent $(1+p^2/2)^{-1}$ of $p^2/2=-\Delta/2$.
He established the following Mourre estimate
under some suitable conditions
on $V$: Put
$$d_0(\lambda)=\mathrm{dist}(\lambda,\omega\boldsymbol{Z}),\quad
d_1(\lambda)=\mathrm{dist}(\lambda,\omega\boldsymbol{Z}\cap(-\infty,\lambda])$$
for $\lambda\in\boldsymbol{R}$.
Suppose $\lambda_0\in\boldsymbol{R}\setminus\omega\boldsymbol{Z}$ and $0<\delta<\mathrm{dist}(\lambda_0,\omega\boldsymbol{Z})=d_0(\lambda_0)$.
Then, for any $f_\delta\in C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
supported in $[-\delta,\delta]$, the Mourre estimate
\begin{equation}
f_\delta(K-\lambda_0)i[K,\tilde{A}_1]f_\delta(K-\lambda_0)
\ge\frac{2(d_1(\lambda_0)-\delta)}{1+(d_1(\lambda_0)-\delta)}f_\delta(K-\lambda_0)^2+C_{1,\lambda_0,f_\delta}
\label{1.8}
\end{equation}
holds with some compact operator $C_{1,\lambda_0,f_\delta}$ on $\mathscr{K}$.
This estimate \eqref{1.8} is slightly better than
the one obtained in \cite{Yo}
$$
f_\delta(K-\lambda_0)i[K,\tilde{A}_1]f_\delta(K-\lambda_0)
\ge\frac{2(d_0(\lambda_0)-\delta)}{1+(d_0(\lambda_0)-\delta)}f_\delta(K-\lambda_0)^2+C_{1,\lambda_0,f_\delta}'
$$
with some compact operator $C_{1,\lambda_0,f_\delta}'$ on $\mathscr{K}$,
since $d_0(\lambda_0)\le d_1(\lambda_0)$.
Here we note that the positive constant of the Mourre estimate \eqref{1.8}
depends on $\lambda_0$ strictly but the conjugate
operator $\tilde{A}_1$ is independent of $\lambda_0$.
However, its extension to the case where $N\ge3$ has not
been obtained yet, as far as we know (see also M{\o}ller-Skibsted~\cite{MS}).
Recently, Adachi-Kiyose~\cite{AK} proposed
an alternative conjugate operator for $K$ with $N=2$
at a non-threshold energy $\lambda_0$:
Let $\lambda_0\in\boldsymbol{R}\setminus\omega\boldsymbol{Z}$.
Then there exists a unique $n_{\lambda_0}\in\boldsymbol{Z}$
such that $\lambda_0\in I_{n_{\lambda_0}}$.
Take $\delta$ as $0<\delta<\mathrm{dist}(\lambda_0,\omega\boldsymbol{Z})$.
Since $\lambda_0-\delta\in I_{n_{\lambda_0}}$,
it is obvious that $\lambda_0-\delta
\in\boldsymbol{R}\setminus\omega\boldsymbol{Z}\subset\rho(D_t)$. Then
we introduce the self-adjoint operator
\begin{equation}
A_{\lambda_0,\delta}=(\lambda_0-\delta-D_t)^{-1}\otimes \hat{A}_0
\label{1.9}
\end{equation}
on $\mathscr{K}\cong L^2(\boldsymbol{T})\otimes \mathscr{H}$,
by multiplying $\hat{A}_0$ and the resolvent $(\lambda_0-\delta-D_t)^{-1}$ of $D_t$. Here we note that $(\lambda_0-\delta-D_t)^{-1}$
is bounded and self-adjoint.
Then the Mourre estimate
\begin{equation}
f_\delta(K-\lambda_0)i[K,A_{\lambda_0,\delta}]f_\delta(K-\lambda_0)
\ge2f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta}
\label{1.10}
\end{equation}
holds with some compact operator $C_{\lambda_0,f_\delta}$ on $\mathscr{K}$.
Here we note that the positive constant of the Mourre estimate \eqref{1.9}
is independent of $\lambda_0$ but the conjugate
operator $A_{\lambda_0,\delta}$ depends on $\lambda_0$ strictly.
Its extension to the case where $N\ge3$ has not
been obtained generally yet, except in the case where all the pair potentials
are independent of $t$.

The aim of this paper is that we will introduce
a conjugate operator for $K$ with $N=3$ by utilizing the above
conjugate operators for $K$ with $N=2$ due to both \cite{Yo} and \cite{AK}.
As is pointed out by M{\o}ller-Skibsted~\cite{MS}, it is important in obtaining the
Mourre estimates for time-independent many body Schr\"odinger operators
that the generator of dilations
$\hat{A}_0$ in \eqref{1.7} can be decomposed into the sum
$$(\hat{A}_0)^a\otimes\mathrm{Id}+\mathrm{Id}\otimes(\hat{A}_0)_a$$
acting on $\mathscr{H}\cong\mathscr{H}^a\otimes\mathscr{H}_a$, for $a\in\mathscr{A}$, where
\begin{equation}
\begin{split}
&(\hat{A}_0)^a=\frac{1}{2}(x^a\cdot p^a+p^a\cdot x^a)=\mathrm{Re}\,(p^a\cdot x^a),\\
&(\hat{A}_0)_a=\frac{1}{2}(x_a\cdot p_a+p_a\cdot x_a)=\mathrm{Re}\,(p_a\cdot x_a).
\end{split}\label{1.11}
\end{equation}
Unfortunately, the conjugate operator
$\tilde{A}_1$ in \eqref{1.6} does not have such a property.
This seems one of the reasons why its extension to the
case where $N\ge3$ has not been given yet.
On the other hand, the conjugate operator $A_{\lambda_0,\delta}$ in
\eqref{1.10} can be decomposed into the sum
$$(\lambda_0-\delta-D_t)^{-1}\otimes\{(\hat{A}_0)^a\otimes\mathrm{Id}+\mathrm{Id}\otimes(\hat{A}_0)_a\}$$
acting on $\mathscr{K}\cong L^2(\boldsymbol{T})\otimes\mathscr{H}^a\otimes\mathscr{H}_a$, for $a\in\mathscr{A}$. If $N=3$ and $a\in\mathscr{A}$ is a pair,
then one can recognize the operator
$(\lambda_0-\delta-D_t)^{-1}\otimes(\hat{A}_0)^a$ as a
conjugate operator for $K^a=D_t+H^a(t)$
acting on
$\mathscr{K}^a=L^2(\boldsymbol{T};\mathscr{H}^a)\cong L^2(\boldsymbol{T})\otimes\mathscr{H}^a$,
by virtue of a result of Adachi-Kiyose~\cite{AK}. $K^a$ is the Floquet Hamiltonian
associated with the subsystem Hamiltonian $H^a(t)$.
However, we cannot interpret the operator $(\lambda_0-\delta-D_t)^{-1}
\otimes(\hat{A}_0)_a$ as a conjugate operator for the
intercluster Hamiltonian $-\Delta_a/2$ acting on $\mathscr{H}_a$,
unfortunately. We think that this is
one of the reasons why any extension of $A_{\lambda_0,\delta}$ to the
case where $N\ge3$ has not been given yet.
In order to overcome the difficulty mentioned above, we will
recognize the operator
$$\tilde{A}_{1,a}=\mathrm{Re}\,((1+(p_a)^2/2)^{-1}p_a\cdot x_a)$$
acting on $\mathscr{H}_a$ as a conjugate operator for $-\Delta_a/2$, and the
sum
$$(\lambda_0-\delta-D_t)^{-1}\otimes(\hat{A}_0)^a\otimes\mathrm{Id}+
\mathrm{Id}\otimes\mathrm{Id}\otimes\tilde{A}_{1,a}$$
as a conjugate operator $A_a$ for $K_a=D_t+H_a(t)$ acting on $\mathscr{K}$.
$K_a$ is the Floquet
Hamiltonian associated with the cluster Hamiltonian $H_a(t)$.
We call $K_a$ a cluster Floquet Hamiltonian.
After introducing $A_a$'s, we will glue these together by using a
partition of unity of $X$.
This is our strategy of introducing a conjugate operator $A$ for $K$ with $N=3$.

Now we will give the precise definition of $A$. We first note that
without loss of generality,
we may assume that a non-threshold energy $\lambda_0$ belongs to
the interval $[0,\omega)$, because the spectrum $\sigma(K)$
of $K$ is $\omega$-periodic, as is well-known.
Let $\delta\in(0,\omega/4)$ and $a\in\mathscr{A}$.
We define a conjugate operator $A^a$
for $K^a=D_t+H^a(t)$ by
\begin{equation}
A^a=(3\omega/2-D_t)^{-1}\otimes(\hat{A}_0)^a
\label{1.12}
\end{equation}
acting on $L^2(\boldsymbol{T})\otimes\mathscr{H}^a$
(see \eqref{1.7} and \eqref{1.11} as for $\hat{A}_0$ and $(\hat{A}_0)^a$).
Here we note that $A^a$ is independent of $\lambda_0\in[0,\omega)$,
unlike $A_{\lambda_0,\delta}$ in \eqref{1.9},
and that $K^{a_\mathrm{min}}=D_t$ and $A^{a_\mathrm{min}}=0$.
We also define a conjugate operator
$\tilde{A}_{\omega/4,a}$ for $-\Delta_a/2$ by
\begin{equation}
\tilde{A}_{\omega/4,a}=\mathrm{Re}\,((\omega/4+(p_a)^2/2)^{-1}p_a\cdot x_a)
\label{1.13}
\end{equation}
acting on $\mathscr{H}_a$. Here we note that $\tilde{A}_{\omega/4,a_\mathrm{max}}=0$.
Finally we put
\begin{equation}
A_a=A^a\otimes\mathrm{Id}+\mathrm{Id}\otimes\tilde{A}_{\omega/4,a}
\label{1.14}
\end{equation}
acting on $\mathscr{K}\cong\mathscr{K}^a\otimes\mathscr{H}_a$.
$A_a$'s are self-adjoint.
In order to glue $A_a$'s together, we will introduce a Graf partition
of unity of $X$ (see e.g. Graf~\cite{Gr}, Skibsted~\cite{Sk}
and Derezi\'nski-G\'erard~\cite{DG}): Given $\kappa_0>1$. Then
there exist $r_0,\,r_1>0$ and $\{j_a\}_{a\in\mathscr{A}}\subset
C^\infty(X;\boldsymbol{R})$ such that the following is satisfied;
$j_a$'s are all bounded smooth functions on $X$ with bounded derivatives
satisfying $0\le j_a(x)\le1$ and
$$\sum_{a\in\mathscr{A}}j_a(x)^2\equiv1.$$
On $\mathrm{supp}\,j_a$, $|x^a|\le r_0$ holds, and
$|x^{(j,k)}|\ge r_1$ holds if $(j,k)\not\subset a$.
If $\kappa\ge \kappa_0$, then $j_a(x)j_b(\kappa x)=0$ for $b\not\subset a$,
and
$$j_a(x)=j_a(x)\sum_{b\subset a}j_b(\kappa x)^2.$$
For the sake of brevity, we put
$j_{a,R}(x)=j_a(x/R)$
for a parameter $R\ge1$.
By using $\{j_{a,R}\}_{a\in\mathscr{A}}$, we define
\begin{equation}
A(R)=\sum_{a\in\mathscr{A}}\bar{A}_a(R),\qquad
\bar{A}_a(R)=j_{a,R}A_aj_{a,R},\quad a\in\mathscr{A},
\label{1.15}
\end{equation}
with $j_{a,R}=j_{a,R}(x)$.
The self-adjointness of $A(R)$ can be guaranteed
by Nelson's commutator theorem (see Theorem \ref{thm2.1} in \S2).
We will see later that $A(R)$ with sufficiently large $R$ is a conjugate
operator for $K$.

Now we impose the following
condition $(V)_3$ on $V$ under consideration:
\medskip

\noindent
$(V)_3$ $V_{jk}(t,y)$, $(j,k)\in\mathscr{A}$, is a real-valued
function on $\boldsymbol{R}\times\boldsymbol{R}^d$,
is $T$-periodic in $t$, belongs to $C^2(\boldsymbol{R}\times
\boldsymbol{R}^d)$,
and satisfies the decaying conditions
\begin{equation}
\begin{split}
&\sup_{t\in\boldsymbol{R}}|(\partial_y^\alpha
V_{jk})(t,y)|\le C\langle y\rangle^{-\rho-|\alpha|},\quad
|\alpha|\le2,\\
&\sup_{t\in\boldsymbol{R}}|(\partial_t\partial_y^\alpha
V_{jk})(t,y)|\le C\langle y\rangle^{-1-\rho-|\alpha|},\quad
|\alpha|\le1,\\
&\sup_{t\in\boldsymbol{R}}|(\partial_t^2V_{jk})(t,y)|\le C\langle y\rangle^{-1-\rho}
\end{split}\label{1.16}
\end{equation}
with some $\rho>0$.
\medskip

\noindent
Here we give some remarks on the condition $(V)_3$.
By a certain technical reason, which will be stated in \S3,
we do not allow $V_{jk}$'s
to have any local singularity, although when $N=2$,
some local singularity of $V_{12}$ can be allowed in \cite{AK}
(see $(V)_2$ stated in \S2).
In keeping
the application to some scattering problems under the
AC Stark effect in mind,
we mainly suppose that
$V_{jk}$'s are given as
$$V_{jk}(t,y)=\bar{V}_{jk}(y+c_{jk}(t)),$$
where $\bar{V}_{jk}\in C^2(\boldsymbol{R}^d;\boldsymbol{R})$
satisfying the decaying conditions
$$|(\partial_y^\alpha\bar{V}_{jk})(y)|\le C\langle y\rangle^{-\rho-|\alpha|},\quad|\alpha|\le2,$$
and $c_{jk}\in C^2(\boldsymbol{T};\boldsymbol{R}^d)$
(see \S4 for details). By simple calculation, we have
\begin{align*}
&(\partial_tV_{jk})(t,y)=(\langle \dot{c}_{jk}(t),\nabla_y\rangle
\bar{V}_{jk})(y+c_{jk}(t)),\\
&(\partial_t^2V_{jk})(t,y)=(\langle \ddot{c}_{jk}(t),\nabla_y\rangle
\bar{V}_{jk})(y+c_{jk}(t))+(\langle \dot{c}_{jk}(t),\nabla_y\rangle^2
\bar{V}_{jk})(y+c_{jk}(t)).
\end{align*}
Hence it is obvious that $V_{jk}(t,y)$'s satisfy
\eqref{1.16}. In \cite{AK}, the third condition in \eqref{1.16} with
$(j,k)=(1,2)$ is replaced by
\begin{equation}
\sup_{t\in\boldsymbol{R}}|(\partial_t^2V_{jk})(t,y)|\le C\langle y\rangle^{-2-\rho}.
\label{1.17}
\end{equation}
This condition
is stronger that the third one in \eqref{1.16}. This causes that when
we apply the results of \cite{AK} without modification
to two-body scattering problems
under the AC Stark effect, the short-range condition $\rho>1$ has to
be assumed.
\medskip

Now we state the main results of this paper.

\begin{thm}\label{thm1.1}
Suppose $N=3$. Assume $V$ satisfies $(V)_3$.
Put
$$\varTheta=\bigcup_{a\in\mathscr{A}\setminus\{a_\mathrm{max}\}}\sigma_{\mathrm{pp}}(K^a),\quad
\widehat{\varTheta}=\bigcup_{a\in\mathscr{A}}\sigma_{\mathrm{pp}}(K^a)=
\varTheta\cup\sigma_\mathrm{pp}(K),
$$
and
\begin{align*}
&d_0(\lambda)=\mathrm{dist}(\lambda,\varTheta),\quad
d_1(\lambda)=\mathrm{dist}(\lambda,\varTheta\cap(-\infty,\lambda]),\\
&\widehat{d}_0(\lambda)=\mathrm{dist}(\lambda,\widehat{\varTheta})
\end{align*}
for $\lambda\in\boldsymbol{R}$. Define $A(R)$ by \eqref{1.15}.
Then the following hold:

\noindent
$(1)$ Let $\lambda_0\in[0,\omega)$, $\epsilon>0$ and $0<\delta_0<\omega/4$.
Then there exists $R_\epsilon\ge1$ and $0<\delta_{0,\epsilon}<\delta_0$
such that the following holds: Take $\delta$ such that
$0<\delta<\delta_{0,\epsilon}$. If $\delta_0\le\lambda_0\le\omega-\delta_0$,
then for any
$f_\delta\in C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
supported in $[-\delta,\delta]$,
\begin{equation}
\begin{split}
&f_\delta(K-\lambda_0)i[K,A]f_\delta(K-\lambda_0)\\
\ge{}&\frac{2(d_1(\lambda_0)-\delta_0)-\epsilon}{3\omega/2}f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta,\epsilon}
\end{split}\label{1.18}
\end{equation}
holds with $A=A(R)$ for $R\ge R_\epsilon$ and
some compact operator $C_{\lambda_0,f_\delta,\epsilon}$ on $\mathscr{K}$.
On the other hand, if $\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$,
then for any
$f_\delta\in C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
supported in $[-\delta,\delta]$,
\begin{equation}
f_\delta(K-\lambda_0)i[K,A]f_\delta(K-\lambda_0)\\
\ge\frac{-\epsilon}{3\omega/2}f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta,\epsilon}
\label{1.19}
\end{equation}
holds. In particular, when $\lambda_0\not\in\varTheta$,
by taking $\epsilon$ such that $\epsilon<d_1(\lambda_0)$,
and $\delta_0$ such that $2\delta_0\le\epsilon$,
the Mourre estimate
\begin{equation}
f_\delta(K-\lambda_0)i[K,A]f_\delta(K-\lambda_0)
\ge\frac{2(d_1(\lambda_0)-\epsilon)}{3\omega/2}
f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta,\epsilon}
\label{1.20}
\end{equation}
can be obtained. Hence, for any $\hat{\delta}$ such that
$0<\hat{\delta}<\delta$,
$\sigma_\mathrm{pp}(K)\cap I_{\lambda_0,\hat{\delta}}$
is finite, and the eigenvalues of $K$
in $I_{\lambda_0,\hat{\delta}}$ are of finite multiplicity.
Here we denote by $I_{\lambda,\delta'}$ the open interval
$(\lambda-\delta',\lambda+\delta')\subset\boldsymbol{R}$
centered around $\lambda\in\boldsymbol{R}$ with the radius $\delta'>0$.

\noindent
$(2)$ In addition, assume $\lambda_0\not\in\sigma_\mathrm{pp}(K)$.
Take $\epsilon$ such that $2\epsilon<d_1(\lambda_0)$,
and $\delta_0$ such that $2\delta_0\le\epsilon$ and
$\delta_0\le\widehat{d}_0(\lambda_0)$, which implies
$\delta_0\le\lambda_0\le\omega-\delta_0$.
Then there exists a small $\delta_{1,\epsilon}>0$ such that
$\delta_{1,\epsilon}<\delta_{0,\epsilon}$ and
\begin{equation}
f_{\delta_{1,\epsilon}}(K-\lambda_0)i[K,A]f_{\delta_{1,\epsilon}}(K-\lambda_0)\ge \frac{2(d_1(\lambda_0)-2\epsilon)}{3\omega/2}f_{\delta_{1,\epsilon}}(K-\lambda_0)^2
\label{1.21}
\end{equation}
holds. Suppose $s>1/2$ and $0<\hat{\delta}<\delta_{1,\epsilon}$. Then
\begin{equation}
\sup_{\substack{\mathrm{Re}\,z\in\overline{I_{\lambda_0,\hat{\delta}}}\\
\mathrm{Im}\,z\not=0}}\|
\langle A\rangle^{-s}(K-z)^{-1}\langle A\rangle^{-s}
\|_{\boldsymbol{B}(\mathscr{K})}<\infty
\label{1.22}
\end{equation}
holds, where $\overline{I_{\lambda,\delta'}}=[\lambda-\delta',\lambda+\delta']$.
Moreover, $\langle A\rangle^{-s}(K-z)^{-1}\langle A\rangle^{-s}$
is a $\boldsymbol{B}(\mathscr{K})$-valued $\theta(s)$-H\"older continuous
function on $z\in S_{\lambda_0,\hat{\delta},\pm}$ with some $0<\theta(s)<1$, where
$$
S_{\lambda_0,\hat{\delta},\pm}=\bigl\{\zeta\in\boldsymbol{C}\bigm|
\mathrm{Re}\,\zeta\in\overline{I_{\lambda_0,\hat{\delta}}},\
0<\pm\mathrm{Im}\,\zeta\le1\bigr\}.
$$
And, there exist the norm limits
$$\langle A\rangle^{-s}(K-(\lambda\pm i0))^{-1}\langle A\rangle^{-s}=
\lim_{\varepsilon\to+0}\langle A\rangle^{-s}(K-(\lambda\pm i\varepsilon))^{-1}\langle A\rangle^{-s}$$
in $\boldsymbol{B}(\mathscr{K})$ for any
$\lambda\in\overline{I_{\lambda_0,\hat{\delta}}}$.
$\langle A\rangle^{-s}(K-(\lambda\pm i0))^{-1}\langle A\rangle^{-s}$
are also $\theta(s)$-H\"older continuous in $\lambda$.
\end{thm}

\begin{cor}\label{cor1.2}
Assume $V$ satisfies $(V)_3$. Then the following hold:

\noindent
$(1)$ The eigenvalues of $K$ in $\boldsymbol{R}\setminus\varTheta$
can accumulate only at $\varTheta$.
Moreover, $\widehat{\varTheta}$ is a countable
closed set.

\noindent
$(2)$ Let $I$ be a compact interval in
$\boldsymbol{R}\setminus\widehat{\varTheta}$.
Suppose $1/2<s\le1$. Then
\begin{equation}
\sup_{\substack{\mathrm{Re}\,z\in I\\
\mathrm{Im}\,z\not=0}}\|
\langle x\rangle^{-s}(K-z)^{-1}\langle x\rangle^{-s}
\|_{\boldsymbol{B}(\mathscr{K})}<\infty
\label{1.23}
\end{equation}
holds. Moreover, $\langle x\rangle^{-s}(K-z)^{-1}\langle x\rangle^{-s}$
is a $\boldsymbol{B}(\mathscr{K})$-valued $\theta(s)$-H\"older continuous
function on $z\in S_{I,\pm}$, where
$$S_{I,\pm}=\bigl\{\zeta\in\boldsymbol{C}\bigm|
\mathrm{Re}\,\zeta\in I,\
0<\pm\mathrm{Im}\,\zeta\le1\bigr\}.$$
And, there exist the norm limits
$$\langle x\rangle^{-s}(K-(\lambda\pm i0))^{-1}\langle x\rangle^{-s}=
\lim_{\varepsilon\to+0}\langle x\rangle^{-s}(K-(\lambda\pm i\varepsilon))^{-1}
\langle x\rangle^{-s}$$
in $\boldsymbol{B}(\mathscr{K})$ for $\lambda\in I$.
$\langle x\rangle^{-s}(K-(\lambda\pm i0))^{-1}
\langle x\rangle^{-s}$ are also $\theta(s)$-H\"older continuous in $\lambda$.
\end{cor}

In order to obtain Corollary \ref{cor1.2}, we use
the argument of Perry-Sigal-Simon~\cite{PSS},
and the boundedness of
$$A(R)(K-\lambda_0-i)^{-1}\langle x\rangle^{-1},$$
which can be given by that
$\langle D_t\rangle^{-1}(K-\lambda_0-i)^{-1}\langle p\rangle^2$ is bounded.
By virtue of this, one can show that
$$A(R)(K-\lambda_0-i)^{-1}\langle p\rangle\langle x\rangle^{-1},\quad
A(R)(K-\lambda_0-i)^{-1}\langle D_t\rangle^{1/2}
\langle x\rangle^{-1}
$$
are also bounded. Then the limiting absorption
principle
$$\sup_{\substack{\mathrm{Re}\,z\in I\\
\mathrm{Im}\,z\not=0}}\|
\langle x\rangle^{-s}\mathcal{D}^s(K-z)^{-1}\mathcal{D}^s
\langle x\rangle^{-s}
\|_{\boldsymbol{B}(\mathscr{K})}<\infty
$$
may be expected as mentioned in \cite{AK},
where $\mathcal{D}=\langle p\rangle+\langle D_t\rangle^{1/2}$ is equivalent to
$\mathscr{D}^{1/2}=(\langle p\rangle^4+\langle D_t\rangle^2)^{1/4}$
as weights,
which was introduced in Kuwabara-Yajima~\cite{KuY}
for the sake of obtaining a refined limiting absorption principle for
$K$. But this has not been given by our analysis yet.
It is caused by the unboundedness of
$$(K-\lambda_0-i)^{-1}\langle p\rangle\langle x\rangle^{-1},\quad
(K-\lambda_0-i)^{-1}\langle D_t\rangle^{1/2}\langle x\rangle^{-1}.$$
Instead of the above limiting absorption principle,
one can obtain
\begin{equation}
\sup_{\substack{\mathrm{Re}\,z\in I\\
\mathrm{Im}\,z\not=0}}\|
\langle D_t\rangle^{-s/2}
\langle x\rangle^{-s}\langle p\rangle^s(K-z)^{-1}\langle p\rangle^s
\langle x\rangle^{-s}\langle D_t\rangle^{-s/2}
\|_{\boldsymbol{B}(\mathscr{K})}<\infty
\label{1.24}
\end{equation}
from \eqref{1.22}, as in \cite{AK}.
As for general $N$-body Floquet Hamiltonians,
a refined limiting absorption principle for $K$
$$\sup_{\substack{\mathrm{Re}\,z\in I\\
\mathrm{Im}\,z\not=0}}\|
\langle x\rangle^{-s}\langle p\rangle^r(K-z)^{-1}\langle p\rangle^r
\langle x\rangle^{-s}\|_{\boldsymbol{B}(\mathscr{K})}<\infty$$
with $0\le r<1/2<s\le1$ was obtained by M\o ller-Skibsted~\cite{MS}.
They used an extended Mourre theory due to
Skibsted~\cite{Sk}, and took a conjugate operator for $K$ in
the extended Mourre theory as $\hat{A}_0$.
However, we would like to stick to find
a candidate of a conjugate operator for $K$ not in an extended
but in the standard Mourre theory, because it seems much easier
to obtain some useful propagation estimates for $K$ as will be
seen in \S4.

The plan of this paper is as follows: In \S2, we will revisit
the case where $N=2$. The construction of $A(R)$ in \eqref{1.15}
is based on the arguments and results in \S2.
In \S3, we will give
the proof of Theorem \ref{thm1.1}, in particular, \eqref{1.18}
and \eqref{1.19}.
In \S4, we will make some
remarks on our results.
\medskip

\leftline{\textbf{Acknowledgement}}
The first author
is partially supported by the Grant-in-Aid for Scientific Research
(C) \#17K05319 from JSPS.

\section{The two-body case revisited}

In this section, we revisit the proof of the Mourre estimate
for $K$ with $N=2$. So we suppose $N=2$ throughout this section.
We impose the following
condition $(V)_2$ on $V$ under consideration:
\medskip

\noindent
$(V)_2$ $V_{12}(t,y)$ is a real-valued
function on $\boldsymbol{R}\times\boldsymbol{R}^d$,
is $T$-periodic in $t$,
and is decomposed into the sum of $V_{12}^\mathrm{sing}(t,y)$
and $V_{12}^\mathrm{reg}(t,y)$, which are also
$T$-periodic in $t$. If $d<3$, then $V_{12}^\mathrm{sing}=0$.
If $d\ge3$, then $V_{12}^\mathrm{sing}(t,\cdot)$ belongs to
$C(\boldsymbol{R},L^{q_0}(\boldsymbol{R}^d))$ with
some $q_0>d$, and $\mathrm{supp}\,V_{12}^\mathrm{sing}(t,\cdot)$'s
are included in a common compact subset of $\boldsymbol{R}^d$.
$(\partial_tV_{12}^\mathrm{sing})(t,\cdot)$ and
$|(\nabla V_{12}^\mathrm{sing})(t,\cdot)|$ belong to
$C(\boldsymbol{R},L^{q_1}(\boldsymbol{R}^d))$
with some $q_1>d/2$, where if $d=3$, then we define $q_1$ by
$1/q_1=1/(2q_0)+1/2$. On the other hand,
$V_{12}^\mathrm{reg}(t,y)$ belongs to $C^2(\boldsymbol{R}\times
\boldsymbol{R}^d)$,
and satisfies the decaying conditions
\begin{equation}
\begin{split}
&\sup_{t\in\boldsymbol{R}}|(\partial_y^\alpha
V_{12}^\mathrm{reg})(t,y)|\le C\langle y\rangle^{-\rho-|\alpha|},\quad
|\alpha|\le2,\\
&\sup_{t\in\boldsymbol{R}}|(\partial_t\partial_y^\alpha
V_{12}^\mathrm{reg})(t,y)|\le C\langle y\rangle^{-1-\rho-|\alpha|},\quad
|\alpha|\le1,\\
&\sup_{t\in\boldsymbol{R}}|(\partial_t^2V_{12}^\mathrm{reg})(t,y)|\le C\langle y\rangle^{-1-\rho}
\end{split}\label{2.1}
\end{equation}
with some $\rho>0$.
\medskip

\noindent
As for $V_{12}^\mathrm{sing}(t,y)$, we mainly suppose
that it has a local singularity like $|y|^{-\gamma}$
with $\gamma>0$,
as in \cite{AK} (see also e.g. Adachi-Kimura-Shimizu~\cite{AKS}).
If $d\ge3$, then the local singularity like $|y|^{-1+\epsilon}$
with $0<\epsilon<1$ can be permitted by $(V)_2$.

First we state some properties of $A^{a_\mathrm{max}}=(3\omega/2-D_t)^{-1}
\hat{A}_0$ for
reference, although those of $A_{\lambda_0,\delta}$ were given
in \cite{AK}.
One of the basic properties of $A^{a_\mathrm{max}}$ is that
\begin{equation}
\begin{split}
i[K_0,A^{a_\mathrm{max}}]={}&(3\omega/2-D_t)^{-1}p^2\\
={}&2(3\omega/2-D_t)^{-1}(K_0-D_t),\\
i[i[K_0,A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]={}&4(3\omega/2-D_t)^{-2}(K_0-D_t)
\end{split}\label{2.2}
\end{equation}
hold, where $K_0=K_{a_\mathrm{min}}=D_t+p^2/2$ is the free Floquet Hamiltonian.
This yields the fact that
$$i[K_0,A^{a_\mathrm{max}}]\langle K_0\rangle^{-1},
\quad i[i[K_0,A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]
\langle K_0\rangle^{-1}$$
are bounded. Under the condition $(V)_2$,
$\langle K_0\rangle^{-1/2}i[V,A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$
with $V=V_{(1,2)}$
is bounded. In fact, as for the regular part $V_{(1,2)}^\mathrm{reg}$ of
$V_{(1,2)}$, it follows from
\begin{align*}
i[V_{(1,2)}^\mathrm{reg},A^{a_\mathrm{max}}]
={}&
-(3\omega/2-D_t)^{-1}((x\cdot\nabla)V_{(1,2)}^\mathrm{reg})\\
&\quad-(3\omega/2-D_t)^{-1}(\partial_t V_{(1,2)}^\mathrm{reg})(3\omega/2-D_t)^{-1}\hat{A}_0
\end{align*}
that
$i[V_{(1,2)}^\mathrm{reg},A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$
is bounded, where $\nabla=ip$.
Here we used the fact that
$\langle D_t\rangle^{-1/2}
\langle p\rangle\langle K_0\rangle^{-1}$
is bounded, which can be shown in the same way as in the case of Stark
Hamiltonians (see e.g. Simon~\cite{S}).
Moreover, we see that $\langle K_0\rangle^{-1}i[V_{(1,2)}^\mathrm{reg},A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$ is compact, by virtue of the local compactness
property of $K_0$.
On the other hand, as for the singular part $V_{(1,2)}^\mathrm{sing}$ of $V_{(1,2)}$,
by using the fact that for each $t\in\boldsymbol{R}$
$$\langle p\rangle^{-1}((x\cdot\nabla)V_{(1,2)}^\mathrm{sing}(t))\langle p\rangle^{-1},\quad\langle p\rangle^{-1}
(\partial_tV_{(1,2)}^\mathrm{sing}(t))\langle p\rangle^{-1}
$$
are bounded on $\mathscr{H}$, one can show firstly that
$\langle K_0\rangle^{-1/2}i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$ is bounded. Moreover, we see that $\langle K_0\rangle^{-1}i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$ is compact.
Next, by identifying
$i[i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]$
with
$$i(i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}]A^{a_\mathrm{max}}-
A^{a_\mathrm{max}}i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}]),$$
one can show that
$\langle K_0\rangle^{-1}i[i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$ is also bounded.
However, one cannot show generally that $i[i[V_{(1,2)}^\mathrm{reg},A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]\langle K_0\rangle^{-1}$
is bounded, except in the case where $\rho\ge1$.
In fact, a simple calculation yields
\begin{align*}
&i[i[V_{(1,2)}^\mathrm{reg},A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]\\
={}&
(3\omega/2-D_t)^{-2}((x\cdot\nabla)^2V_{(1,2)}^\mathrm{reg})\\
&\quad+2(3\omega/2-D_t)^{-2}(\partial_t(x\cdot\nabla)V_{(1,2)}^\mathrm{reg})(3\omega/2-D_t)^{-1}\hat{A}_0\\
&\qquad+(3\omega/2-D_t)^{-2}(\partial_t^2V_{(1,2)}^\mathrm{reg})(3\omega/2-D_t)^{-2}\hat{A}_0^2.
\end{align*}
The first two terms of the right-hand side of this equality are
$K_0$-bounded. But, in showing the $K_0$-boundedness of the
third term of the right-hand side of this equality generally,
the condition $\rho\ge1$ is needed. In \cite{AK}, a stronger condition
\eqref{1.17} is assumed for the sake of avoiding this difficulty.
In this paper, in order to avoid this difficulty, we will replace
$A^{a_\mathrm{max}}$ by
$$\bar{A}_{a_\mathrm{max}}(R)=j_{a_\mathrm{max},R}A_{a_\mathrm{max}}j_{a_\mathrm{max},R}$$
with $A_{a_\mathrm{max}}=A^{a_\mathrm{max}}+\tilde{A}_{\omega/4,a_\mathrm{max}}=A^{a_\mathrm{max}}$.
Here we note that
on $\mathrm{supp}\,j_{a_\mathrm{max},R}$, $|x|\le r_0R$ holds.
By virtue of this, one can show that
\begin{align*}
&i[K_0,\bar{A}_{a_\mathrm{max}}(R)]\langle K_0\rangle^{-1},
\quad i[i[K_0,\bar{A}_{a_\mathrm{max}}(R)],\bar{A}_{a_\mathrm{max}}(R)]
\langle K_0\rangle^{-1},\\
&\langle K_0\rangle^{-1/2}i[V,\bar{A}_{a_\mathrm{max}}(R)]\langle K_0\rangle^{-1},
\quad
\langle K_0\rangle^{-1}i[i[V,\bar{A}_{a_\mathrm{max}}(R)],\bar{A}_{a_\mathrm{max}}(R)]\langle K_0\rangle^{-1}
\end{align*}
are all bounded, and that
$\langle K_0\rangle^{-1}i[V,\bar{A}_{a_\mathrm{max}}(R)]\langle K_0\rangle^{-1}$ is compact. Here we used
$i[V,\bar{A}_{a_\mathrm{max}}(R)]=j_{a_\mathrm{max},R}i[V,A^{a_\mathrm{max}}]j_{a_\mathrm{max},R}$,
\begin{equation}
i[j_{a,R},p^2/2]=-\mathrm{Re}\,\{(\nabla j_{a,R})\cdot p\},\quad a\in\mathscr{A},
\label{2.3}
\end{equation}
and that $\langle D_t\rangle^{-1/2}\langle p\rangle\langle K_0\rangle^{-1}$ is bounded.

Next we state some properties of
$\tilde{A}_{\omega/4,a_\mathrm{min}}=\mathrm{Re}\,
\{(\omega/4+p^2/2)^{-1}p\cdot x\}$
for reference (see also \cite{Yo}).
One of the basic properties of $\tilde{A}_{\omega/4,a_\mathrm{min}}$ is that
\begin{equation}
\begin{split}
&i[K_0,\tilde{A}_{\omega/4,a_\mathrm{min}}]\\
={}&(\omega/4+p^2/2)^{-1}p^2=2\{1-(\omega/4)(\omega/4+p^2/2)^{-1}\},\\
&i[i[K_0,\tilde{A}_{\omega/4,a_\mathrm{min}}],\tilde{A}_{\omega/4,a_\mathrm{min}}]\\
={}&\omega(\omega/4+p^2/2)^{-3}p^2=2\omega(\omega/4+p^2/2)^{-2}\{1-(\omega/4)(\omega/4+p^2/2)^{-1}\}
\end{split}\label{2.4}
\end{equation}
hold. Obviously these are bounded. Under the condition $(V)_2$,
$i[V,\tilde{A}_{\omega/4,a_\mathrm{min}}]$
with $V=V_{(1,2)}$
is bounded. In fact, as for the regular part $V_{(1,2)}^\mathrm{reg}$ of
$V_{(1,2)}$, it follows from
$$i[V_{(1,2)}^\mathrm{reg},\tilde{A}_{\omega/4,a_\mathrm{min}}]
=\mathrm{Re}\,(V_{(1,2)}^\mathrm{reg})_{a_\mathrm{min},\mathrm{Y}}'$$
with
\begin{align*}
(V_{(1,2)}^\mathrm{reg})_{a_\mathrm{min},\mathrm{Y}}'={}&
(\omega/4+p^2/2)^{-1}(\mathrm{Re}\,(\nabla V_{(1,2)}^\mathrm{reg}\cdot p))
(\omega/4+p^2/2)^{-1}p\cdot x\\
&\quad-
(\omega/4+p^2/2)^{-1}((x\cdot\nabla)V_{(1,2)}^\mathrm{reg})
\end{align*}
that
$i[V_{(1,2)}^\mathrm{reg},\tilde{A}_{\omega/4,a_\mathrm{min}}]$
is bounded. By virtue of the local compactness
property of $K_0$, we also see that
$\langle K_0\rangle^{-1}i[V_{(1,2)}^\mathrm{reg},\tilde{A}_{\omega/4,a_\mathrm{min}}]\langle K_0\rangle^{-1}$ is compact.
Similarly one can show that $i[i[V_{(1,2)}^\mathrm{reg},\tilde{A}_{\omega/4,a_\mathrm{min}}],\tilde{A}_{\omega/4,a_\mathrm{min}}]$ is bounded.
On the other hand, as for the singular part $V_{(1,2)}^\mathrm{sing}$ of $V_{(1,2)}$,
by using the fact that
$$|\nabla V_{(1,2)}^\mathrm{sing}(t)|\langle p\rangle^{-2},\quad\langle p\rangle^{-1}|\nabla V_{(1,2)}^\mathrm{sing}(t)|\langle p\rangle^{-1},\quad((x\cdot\nabla)V_{(1,2)}^\mathrm{sing}(t))\langle p\rangle^{-2},
$$
are bounded on $\mathscr{H}$, one can show firstly that
$i[V_{(1,2)}^\mathrm{sing},\tilde{A}_{\omega/4,a_\mathrm{min}}]$ is bounded. We also
see that $\langle K_0\rangle^{-1}i[V_{(1,2)}^\mathrm{sing},\tilde{A}_{\omega/4,a_\mathrm{min}}]\langle K_0\rangle^{-1}$ is compact.
One can also show that
$i[i[V_{(1,2)}^\mathrm{sing},\tilde{A}_{\omega/4,a_\mathrm{min}}],\tilde{A}_{\omega/4,a_\mathrm{min}}]$ is also bounded,
in the same way as in the case of
$i[i[V_{(1,2)}^\mathrm{sing},A^{a_\mathrm{max}}],A^{a_\mathrm{max}}]$.
However, in this paper, we will replace
$\tilde{A}_{\omega/4,a_\mathrm{min}}$ by
$$\bar{A}_{a_\mathrm{min}}(R)=j_{a_\mathrm{min},R}A_{a_\mathrm{min}}j_{a_\mathrm{min},R}$$
with $A_{a_\mathrm{min}}=A^{a_\mathrm{min}}+\tilde{A}_{\omega/4,a_\mathrm{min}}=\tilde{A}_{\omega/4,a_\mathrm{min}}$.
Here we note that we do not have to deal with the local singularity
of $V_{(1,2)}$ in the calculation of $i[K,\bar{A}_{a_\mathrm{min}}(R)]$ and
$i[i[K,\bar{A}_{a_\mathrm{min}}(R)],\bar{A}_{a_\mathrm{min}}(R)]$ for
large $R$, since on
$\mathrm{supp}\,j_{a_\mathrm{min},R}$, $|x|\ge r_1R$ holds.
By virtue of this, one can show that
\begin{align*}
&i[K_0,\bar{A}_{a_\mathrm{min}}(R)]\langle K_0\rangle^{-1},
\quad i[i[K_0,\bar{A}_{a_\mathrm{min}}(R)],\bar{A}_{a_\mathrm{min}}(R)]
\langle K_0\rangle^{-1},\\
&\langle K_0\rangle^{-1/2}i[V,\bar{A}_{a_\mathrm{min}}(R)]\langle K_0\rangle^{-1},
\quad
\langle K_0\rangle^{-1}i[i[V,\bar{A}_{a_\mathrm{min}}(R)],\bar{A}_{a_\mathrm{min}}(R)]\langle K_0\rangle^{-1}
\end{align*}
are all bounded, and
$$\langle K_0\rangle^{-1}i[V,\bar{A}_{a_\mathrm{min}}(R)]\langle K_0\rangle^{-1}=O(R^{-\min\{\rho,1\}})$$
as $R\to\infty$. Here we used
$i[V,\bar{A}_{a_\mathrm{min}}(R)]=j_{a_\mathrm{min},R}i[V,\tilde{A}_{\omega/4,a_\mathrm{max}}]j_{a_\mathrm{min},R}$ and
\eqref{2.3}.

Now we will introduce
\begin{equation}
A(R)=\sum_{a\in\mathscr{A}}\bar{A}_a(R)=\bar{A}_{a_\mathrm{max}}(R)+\bar{A}_{a_\mathrm{min}}(R)
\label{2.5}
\end{equation}
as in \eqref{1.15}. The following Nelson's commutator theorem
guarantees the self-adjointness of $A(R)$
(as for the proof, see e.g. Reed-Simon~\cite{RS} and G{\'e}rard-{\L}aba~\cite{GL}).

\begin{thm}\label{thm2.1} Let $\mathcal{K}$ be a Hilbert space.
Suppose that $N_0\ge c>0$ is a self-adjoint operator
on $\mathcal{K}$ and $A$ is a symmetric operator on $\mathcal{K}$
such that $D(N_0)\subset D(A)$ and
there exists a constant $C>0$ such that
\begin{align*}
&\|Au\|\le C\|N_0u\|\quad\textrm{for}\,\,u\in D(N_0),\\
&|(Au,N_0u)-(N_0u,Au)|\le C\|{N_0}^{1/2}u\|^2\quad\textrm{for}\,\,
u\in D(N_0)
\end{align*}
hold. Then $A$ is essentially self-adjoint on $D(N_0)$.
Denoting by $\bar{A}$ the unique self-adjoint extension
of $A$, if $u\in D(\bar{A})$, then $(1+i\epsilon N_0)^{-1}u$
converges to $u$ in the graph topology of $D(\bar{A})$
as $\epsilon\to0$.
\end{thm}

Applying Theorem \ref{thm2.1} with $\mathcal{K}=\mathscr{K}$,
$N_0=\langle D_t\rangle+p^2/2+x^2/2$
and $A=A(R)$, we see that $A(R)$ has its unique
self-adjoint extension, which is also denoted by $A(R)$.
Here we used \eqref{2.3} and
$$i[x^2/2,\hat{A}_0]=-x^2,\quad
i[x^2/2,\tilde{A}_{\omega/4,a_\mathrm{min}}]
=\mathrm{Re}\,(x^2/2)_{a_\mathrm{min},\mathrm{Y}}'
$$
with
\begin{align*}
(x^2/2)_{a_\mathrm{min},\mathrm{Y}}'={}&
(\omega/4+p^2/2)^{-1}(\mathrm{Re}\,(p\cdot x))(\omega/4+p^2/2)^{-1}p\cdot x\\
&\quad-
(\omega/4+p^2/2)^{-1}x^2.
\end{align*}
By virtue of the properties of $\{\bar{A}_a\}_{a\in\mathscr{A}}$, we see that
\begin{align*}
&i[K_0,A(R)]\langle K_0\rangle^{-1},
\quad i[i[K_0,A(R)],A(R)]
\langle K_0\rangle^{-1},\\
&\langle K_0\rangle^{-1/2}i[V,A(R)]\langle K_0\rangle^{-1},
\quad
\langle K_0\rangle^{-1}i[i[V,A(R)],A(R)]\langle K_0\rangle^{-1}
\end{align*}
are all bounded, and
$$\langle K_0\rangle^{-1}i[V,A(R)]\langle K_0\rangle^{-1}=O(R^{-\min\{\rho,1\}})+C_R$$
with some compact operator $C_R$ on $\mathscr{K}$.

In the usual proof of the Mourre estimate for $K$, one of the points
to be checked is that the condition
\begin{equation}
\sup_{|\kappa|\le1}\|Ke^{i\kappa A(R)}(K+i)^{-1}\|_{\mathscr{B}(\mathscr{K})}<\infty
\label{2.6}
\end{equation}
is satisfied by a conjugate operator $A(R)$ (see e.g. Mourre~\cite{M}).
However, it seems not easy to verify directly that $A(R)$ defined by \eqref{2.5}
satisfies \eqref{2.6}.
In order to overcome this difficulty,
we need the following proposition
(see e.g. Lemma 3.2.2 and Proposition 3.2.3 of \cite{GL};
see also Amrein-Boutet de Monvel-Georgescu~\cite{ABG}):

\begin{prop}\label{prop2.2} Let $\mathcal{K}$ be a Hilbert space.
Suppose that $K$, $K_0$ and $N_0$ are self-adjoint operators
on $\mathcal{K}$ such that $N_0\ge c>0$, $D(K)=D(K_0)$ as Banach spaces,
and for $z\in\boldsymbol{C}\setminus\sigma(K)$,
$(K-z)^{-1}$ preserves $D(N_0)$.
Let $A$ be a symmetric operator on $\mathcal{K}$. Suppose
that $K_0$ and $A$ satisfy
$D(N_0)\subset D(K_0)$, $D(N_0)\subset D(A)$,
\begin{align*}
&\|K_0u\|\le C\|N_0u\|\quad\textrm{for}\,\,u\in D(N_0),\\
&|(K_0u,N_0u)-(N_0u,K_0u)|\le C\|{N_0}^{1/2}u\|^2\quad\textrm{for}\,\,
u\in D(N_0),\\
&\|Au\|\le C\|N_0u\|\quad\textrm{for}\,\,u\in D(N_0),\\
&|(Au,N_0u)-(N_0u,Au)|\le C\|{N_0}^{1/2}u\|^2\quad\textrm{for}\,\,
u\in D(N_0).
\end{align*}
Denote the unique self-adjoint extension of $A$ also by $A$.
Assume moreover that
$$|(Au,Ku)-(Ku,Au)|\le C(\|Ku\|^2+\|u\|^2)\quad\textrm{for}\,\,u\in D(N_0)
$$
holds. Then the following hold:

\noindent
$(1)$ $D(N_0)$ is dense in $D(K)\cap D(A)$
with the norm $\|Ku\|+\|Au\|+\|u\|$.

\noindent
$(2)$ The commutator $i[K,A]$, defined as a quadratic form
on $D(K)\cap D(A)$, is the unique extension of the quadratic form
$i[K,A]$ on $D(N_0)$.

\noindent
$(3)$ $K\in C^1(A)$, that is, for some $z\in\boldsymbol{C}
\setminus\sigma(K)$, the map
$$\boldsymbol{R}\ni \kappa\mapsto e^{i\kappa A}
(K-z)^{-1}e^{-i\kappa A}\in \mathscr{B}(\mathcal{K})$$
is $C^1$ in the strong topology of $\mathscr{B}(\mathcal{K})$,
which is the algebra of bounded linear operators in $\mathcal{K}$.

\noindent
$(4)$ $D(K)\cap D(A)$ is a core for $K$, and the quadratic
form $i[K,A]$ on $D(K)\cap D(A)$ extends uniquely to a
bounded operator from $D(K)$ to its dual space $D(K)^*$,
which is denoted also by $i[K,A]$.

\noindent
$(5)$ The virial relation holds:\ \ For any $\lambda\in\boldsymbol{R}$,
$$E_K(\{\lambda\})i[K,A]E_K(\{\lambda\})=0$$
holds. Here $E_K(S)$ stands for the
spectral projection for $K$ onto $S\subset\boldsymbol{R}$.

\noindent
$(6)$ For $z\in\boldsymbol{C}\setminus\sigma(K)$,
$i[(K-z)^{-1},A]=-(K-z)^{-1}i[K,A](K-z)^{-1}$ holds.

\noindent
$(7)$ For $z\in\boldsymbol{C}\setminus\sigma(K)$,
$(K-z)^{-1}$ preserves $D(A)$.
\end{prop}

By virtue of Proposition \ref{prop2.2} with $\mathcal{K}=\mathscr{K}$,
$N_0=\langle D_t\rangle+p^2/2+x^2/2$ and $A=A(R)$, one can
show the following theorem and corollary without using
\eqref{2.6}:

\begin{thm}\label{thm2.3}
Suppose $N=2$. Assume $V$ satisfies $(V)_2$.
Put
\begin{align*}
&\varTheta=\bigcup_{a\in\mathscr{A}\setminus\{a_\mathrm{max}\}}\sigma_{\mathrm{pp}}(K^a)=\sigma_{\mathrm{pp}}(D_t)=\omega\boldsymbol{Z},\\
&\widehat{\varTheta}=\bigcup_{a\in\mathscr{A}}\sigma_{\mathrm{pp}}(K^a)=
\varTheta\cup\sigma_\mathrm{pp}(K),
\end{align*}
and
\begin{align*}
&d_0(\lambda)=\mathrm{dist}(\lambda,\varTheta),\quad
d_1(\lambda)=\mathrm{dist}(\lambda,\varTheta\cap(-\infty,\lambda]),\\
&\widehat{d}_0(\lambda)=\mathrm{dist}(\lambda,\widehat{\varTheta})
\end{align*}
for $\lambda\in\boldsymbol{R}$. Define $A(R)$ by \eqref{2.5}.
Then the following hold:

\noindent
$(1)$ Let $\lambda_0\in[0,\omega)$, $\epsilon>0$ and
$0<\delta_0<\omega/4$. Then there exists $R_\epsilon\ge1$ such that
the following holds: Take $\delta$ such that
$0<\delta<\delta_0$.
If $\delta_0\le\lambda_0\le\omega-\delta_0$, then
for any $f_\delta\in C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
supported in $[-\delta,\delta]$,
\begin{equation}
\begin{split}
&f_\delta(K-\lambda_0)i[K,A]f_\delta(K-\lambda_0)\\
\ge{}&\frac{2(d_1(\lambda_0)-\delta_0)-\epsilon}{3\omega/2}
f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta,\epsilon}
\end{split}\label{2.7}
\end{equation}
holds with $A=A(R)$ for $R\ge R_\epsilon$ and
some compact operator $C_{\lambda_0,f_\delta,\epsilon}$ on $\mathscr{K}$.
On the other hand, if $\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$,
then for any $f_\delta\in C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
supported in $[-\delta,\delta]$,
\begin{equation}
f_\delta(K-\lambda_0)i[K,A]f_\delta(K-\lambda_0)
\ge\frac{-\epsilon}{3\omega/2}
f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta,\epsilon}
\label{2.8}
\end{equation}
holds with $A=A(R)$ for $R\ge R_\epsilon$
and some compact operator $C_{\lambda_0,f_\delta,\epsilon}$ on $\mathscr{K}$.
In particular, when $\lambda_0\not\in\varTheta$,
by taking $\epsilon$ such that $\epsilon<d_1(\lambda_0)$,
and $\delta_0$ such that $2\delta_0\le\epsilon$,
the Mourre estimate
\begin{equation}
f_\delta(K-\lambda_0)i[K,A]f_\delta(K-\lambda_0)
\ge\frac{2(d_1(\lambda_0)-\epsilon)}{3\omega/2}
f_\delta(K-\lambda_0)^2+C_{\lambda_0,f_\delta,\epsilon}
\label{2.9}
\end{equation}
can be obtained. Hence, for any $\hat{\delta}$ such that
$0<\hat{\delta}<\delta_0$,
$\sigma_\mathrm{pp}(K)\cap I_{\lambda_0,\hat{\delta}}$
is finite, and that the eigenvalues of $K$
in $I_{\lambda_0,\hat{\delta}}$ are of finite multiplicity.

\noindent
$(2)$ In addition, assume $\lambda_0\not\in\sigma_\mathrm{pp}(K)$.
Take $\epsilon$ such that $2\epsilon<d_1(\lambda_0)$,
and $\delta_0$ such that $2\delta_0\le\epsilon$ and
$\delta_0\le\widehat{d}_0(\lambda_0)$, which implies
$\delta_0\le\lambda_0\le\omega-\delta_0$.
Then there exists a small $\delta_{1,\epsilon}>0$ such that
$\delta_{1,\epsilon}<\delta_0$ and
\begin{equation}
f_{\delta_{1,\epsilon}}(K-\lambda_0)i[K,A]f_{\delta_{1,\epsilon}}(K-\lambda_0)
\ge\frac{2(d_1(\lambda_0)-2\epsilon)}{3\omega/2}f_{\delta_{1,\epsilon}}(K-\lambda_0)^2
\label{2.10}
\end{equation}
holds. Suppose $s>1/2$ and $0<\hat{\delta}<\delta_{1,\epsilon}$. Then
\begin{equation}
\sup_{\substack{\mathrm{Re}\,z\in\overline{I_{\lambda_0,\hat{\delta}}}\\
\mathrm{Im}\,z\not=0}}\|
\langle A\rangle^{-s}(K-z)^{-1}\langle A\rangle^{-s}
\|_{\boldsymbol{B}(\mathscr{K})}<\infty
\label{2.11}
\end{equation}
holds. Moreover, $\langle A\rangle^{-s}(K-z)^{-1}\langle A\rangle^{-s}$
is a $\boldsymbol{B}(\mathscr{K})$-valued $\theta(s)$-H\"older continuous
function on $z\in S_{\lambda_0,\hat{\delta},\pm}$ with some $0<\theta(s)<1$.
And, there exist the norm limits
$$\langle A\rangle^{-s}(K-(\lambda\pm i0))^{-1}\langle A\rangle^{-s}=
\lim_{\varepsilon\to+0}\langle A\rangle^{-s}(K-(\lambda\pm i\varepsilon))^{-1}\langle A\rangle^{-s}$$
in $\boldsymbol{B}(\mathscr{K})$ for any
$\lambda\in\overline{I_{\lambda_0,\hat{\delta}}}$.
$\langle A\rangle^{-s}(K-(\lambda\pm i0))^{-1}\langle A\rangle^{-s}$
are also $\theta(s)$-H\"older continuous in $\lambda$.
\end{thm}

\begin{cor}\label{cor2.4}
Assume $V$ satisfies $(V)_2$. Then the following hold:

\noindent
$(1)$ The eigenvalues of $K$ in $\boldsymbol{R}\setminus\varTheta$
can accumulate only at $\varTheta$.
Moreover, $\widehat{\varTheta}$ is a countable
closed set.

\noindent
$(2)$ Let $I$ be a compact interval in
$\boldsymbol{R}\setminus\widehat{\varTheta}$.
Suppose $1/2<s\le1$. Then
\begin{equation}
\sup_{\substack{\mathrm{Re}\,z\in I\\
\mathrm{Im}\,z\not=0}}\|
\langle x\rangle^{-s}(K-z)^{-1}\langle x\rangle^{-s}
\|_{\boldsymbol{B}(\mathscr{K})}<\infty
\label{2.12}
\end{equation}
holds. Moreover, $\langle x\rangle^{-s}(K-z)^{-1}\langle x\rangle^{-s}$
is a $\boldsymbol{B}(\mathscr{K})$-valued $\theta(s)$-H\"older continuous
function on $z\in S_{I,\pm}$.
And, there exist the norm limits
$$\langle x\rangle^{-s}(K-(\lambda\pm i0))^{-1}\langle x\rangle^{-s}=
\lim_{\varepsilon\to+0}\langle x\rangle^{-s}(K-(\lambda\pm i\varepsilon))^{-1}
\langle x\rangle^{-s}$$
in $\boldsymbol{B}(\mathscr{K})$ for $\lambda\in I$.
$\langle x\rangle^{-s}(K-(\lambda\pm i0))^{-1}
\langle x\rangle^{-s}$ are also $\theta(s)$-H\"older continuous in $\lambda$.
\end{cor}

We will sketch the proof of the estimates \eqref{2.7} and \eqref{2.8}
only. \eqref{2.7} yields the Mourre estimate \eqref{2.9}. Thus
Theorem \ref{thm2.3} and Corollary \ref{cor2.4} can be shown
by the standard argument in the Mourre theory.
In particular, for the proof of Corollary \ref{cor2.4}, we use
the argument due to Perry-Sigal-Simon~\cite{PSS},
and the boundedness of
$$A(R)(K-\lambda_0-i)^{-1}\langle x\rangle^{-1},$$
which follows from that
$\langle D_t\rangle^{-1}(K-\lambda_0-i)^{-1}\langle p\rangle^2$ is bounded.

\begin{proof}[Proof of \eqref{2.7} and \eqref{2.8}]

Let $\lambda_0\in[0,\omega)$,
$\epsilon>0$ and $0<\delta_0<\omega/4$.
Denote by $f_{\delta_0}$
any function in $C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
such that $\mathrm{supp}\,f_{\delta_0}\subset[-\delta_0,\delta_0]$.
For the sake of simplicity, we write $f_{\delta_0}(K-\lambda_0)$
as $f_{\delta_0,\lambda_0}(K)$.
By the assumption $(V)_2$, we see that
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)i[K,A(R)]f_{\delta_0,\lambda_0}(K)\\
={}&f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{max},R}i[K_0,A_{a_\mathrm{max}}]j_{a_\mathrm{max},R}f_{\delta_0,\lambda_0}(K)\\
&\quad+f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}i[K_0,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\qquad+O(R^{-\min\{\rho,1\}})+C_{R,1}
\end{split}\label{2.13}
\end{equation}
holds with some compact operator $C_{R,1}$ on $\mathscr{K}$.
By \eqref{2.2} and the IMS localization formula
$$p^2=\sum_{a\in\mathscr{A}}j_{a,R}p^2j_{a,R}-\sum_{a\in\mathscr{A}}|\nabla j_{a,R}|^2,$$
we have
$$i[K_0,A_{a_\mathrm{max}}]=\sum_{a\in\mathscr{A}}j_{a,R}i[K_0,A_{a_\mathrm{max}}]j_{a,R}+O(R^{-2}).$$
Here we note that $(3\omega/2-D_t)^{-1}$ does commute
with $j_{a,R}$'s.
Since $f_{\delta_0,\lambda_0}(K)-f_{\delta_0,\lambda_0}(K_0)$ is compact by the assumption $(V)_2$, we obtain
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)i[K,A(R)]f_{\delta_0,\lambda_0}(K)\\
={}&f_{\delta_0,\lambda_0}(K)i[K_0,A_{a_\mathrm{max}}]f_{\delta_0,\lambda_0}(K)\\
&\quad+f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\qquad+O(R^{-\min\{\rho,1\}})+C_{R,1}\\
={}&f_{\delta_0,\lambda_0}(K_0)i[K_0,A_{a_\mathrm{max}}]f_{\delta_0,\lambda_0}(K_0)\\
&\quad+f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)\\
&\qquad+O(R^{-\min\{\rho,1\}})+C_{R,2}
\end{split}\label{2.14}
\end{equation}
with some compact operator $C_{R,2}$ on $\mathscr{K}$.
Here we note that $\langle D_t\rangle^{-1/2}\langle p\rangle\langle K_0\rangle^{-1}$
and $\langle D_t\rangle^{-1}\langle p\rangle^2\langle K_0\rangle^{-1}$
are bounded as mentioned above, and
\begin{equation}
\begin{split}
&i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}]\\
={}&2\{(\omega/4+H_0)^{-1}-(3\omega/2-D_t)^{-1}\}H_0\\
={}&2(3\omega/2-D_t)^{-1}(5\omega/4-K_0)(\omega/4+H_0)^{-1}H_0
\end{split}\label{2.15}
\end{equation}
with $H_0=p^2/2$. $f_{\delta_0,\lambda_0}(K_0)i[K_0,A_{a_\mathrm{max}}]f_{\delta_0,\lambda_0}(K_0)$ can be decomposed into the
direct integral
$$\bigoplus_{n\in\boldsymbol{Z}}
\frac{2H_0}{3\omega/2-n\omega}f_{\delta_0,\lambda_0-n\omega}(H_0)^2=
\bigoplus_{n\le1}
\frac{2H_0}{3\omega/2-n\omega}f_{\delta_0,\lambda_0-n\omega}(H_0)^2
$$
with $f_{\delta_0,\lambda_0-n\omega}(H_0)=f_{\delta_0}(n\omega+H_0-\lambda_0)$.
Here we note that when $n\ge 2$,
$f_{\delta_0,\lambda_0-n\omega}(H_0)=0$ holds since $\lambda_0-n\omega+\delta_0\le\lambda_0-2\omega+\delta_0<0$.
When $n\le-1$,
\begin{align*}
\frac{H_0}{3\omega/2-n\omega}f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\ge{}&\frac{\lambda_0-n\omega-\delta_0}{3\omega/2-n\omega}f_{\delta_0,\lambda_0-n\omega}(H_0)^2\\
\ge{}&\frac{\lambda_0+\omega-\delta_0}{5\omega/2}f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\end{align*}
holds. We will consider the case where $n=0$. If $\lambda_0-\delta_0<0$,
that is, $\lambda_0<\delta_0$,
then
$$\frac{H_0}{3\omega/2}f_{\delta_0,\lambda_0}(H_0)^2
\ge0;$$
while, if $\lambda_0\ge\delta_0$, then
$$\frac{H_0}{3\omega/2}f_{\delta_0,\lambda_0}(H_0)^2
\ge\frac{\lambda_0-\delta_0}{3\omega/2}f_{\delta_0,\lambda_0}(H_0)^2.$$
We will consider the case where $n=1$.
If $\lambda_0-\omega+\delta_0>0$, that is, $\lambda_0>\omega-\delta_0$,
then
$$\frac{H_0}{\omega/2}f_{\delta_0,\lambda_0-\omega}(H_0)^2
\ge0.$$
On the other hand, if $\lambda_0-\omega+\delta_0\le0$, that
is, $\lambda_0\le\omega-\delta_0$, then $f_{\delta_0,\lambda_0-\omega}(H_0)=0$. By combining these, we see that if $\delta_0\le\lambda_0\le\omega-\delta_0$,
then
$$f_{\delta_0,\lambda_0}(K_0)i[K_0,A_{a_\mathrm{max}}]f_{\delta_0,\lambda_0}(K_0)
\ge\frac{2(\lambda_0-\delta_0)}{3\omega/2}f_{\delta_0,\lambda_0}(K_0)^2;$$
while, if  $0\le\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$, then
$$f_{\delta_0,\lambda_0}(K_0)i[K_0,A_{a_\mathrm{max}}]f_{\delta_0,\lambda_0}(K_0)
\ge0.$$
We next consider
$$f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0).$$
Noting $[f_{\delta_0,\lambda_0}(K_0),j_{a_\mathrm{min},R}]\langle D_t\rangle^{-1/2}=O(R^{-1})$,
\begin{align*}
&f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)\\
={}&j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}\\
&\quad+O(R^{-1})
\end{align*}
can be shown easily by \eqref{2.15}.
Then we would like to use the estimate
\begin{align*}
&f_{\delta_0,\lambda_0}(K_0)(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])f_{\delta_0,\lambda_0}(K_0)\\
={}&\bigoplus_{n\le1}\{(\omega/4+H_0)^{-1}
-(3\omega/2-n\omega)^{-1}\}(2H_0)f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\\
={}&\bigoplus_{n\le1}\frac{5\omega/4-n\omega-H_0}{3\omega/2-n\omega}(\omega/4+H_0)^{-1}(2H_0)f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\\
\ge{}&\bigoplus_{n\le1}\frac{5\omega/4-n\omega-(\lambda_0-n\omega+\delta_0)}{3\omega/2-n\omega}(\omega/4+H_0)^{-1}(2H_0)f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\\
={}&\bigoplus_{n\le1}\frac{5\omega/4-(\lambda_0+\delta_0)}{3\omega/2-n\omega}(\omega/4+H_0)^{-1}(2H_0)f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\\
\ge{}&0
\end{align*}
since $\lambda_0+\delta_0<5\omega/4$. This estimate yields
$$f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}(i[K_0,A_{a_\mathrm{min}}]-i[K_0,A_{a_\mathrm{max}}])j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)\ge
O(R^{-1}).$$
It follows from these and \eqref{2.14}
that if $\delta_0\le\lambda_0\le\omega-\delta_0$, then
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)i[K,A(R)]f_{\delta_0,\lambda_0}(K)\\
\ge{}&\frac{2(\lambda_0-\delta_0)}{3\omega/2}f_{\delta_0,\lambda_0}(K_0)^2+O(R^{-\min\{\rho,1\}})+C_{R,2}\\
={}&\frac{2(\lambda_0-\delta_0)}{3\omega/2}f_{\delta_0,\lambda_0}(K)^2+O(R^{-\min\{\rho,1\}})+C_{R,3}
\end{split}\label{2.16}
\end{equation}
with some compact operator $C_{R,3}$ on $\mathscr{K}$.
Now we will take $\delta$ such that
$0<\delta<\delta_0$.
By sandwiching \eqref{2.16} in two $f_{\delta,\lambda_0}(K)$'s,
one can obtain
\begin{equation}
\begin{split}
&f_{\delta,\lambda_0}(K)i[K,A(R)]f_{\delta,\lambda_0}(K)\\
\ge{}&\frac{2(\lambda_0-\delta_0)}{3\omega/2}f_{\delta,\lambda_0}(K)^2
+f_{\delta,\lambda_0}(K)O(R^{-\min\{\rho,1\}})f_{\delta,\lambda_0}(K)+C_{R,4}
\end{split}\label{2.17}
\end{equation}
with some compact operator $C_{R,4}$ on $\mathscr{K}$,
because of the arbitrariness of $f_{\delta_0,\lambda_0}$.
\eqref{2.17} yields \eqref{2.7}, by taking $R\ge1$ sufficiently large.
Here we note $d_1(\lambda_0)=\lambda_0$.
Similarly, if $0\le\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$,
\begin{align*}
&f_{\delta,\lambda_0}(K)i[K,A(R)]f_{\delta,\lambda_0}(K)\\
\ge{}&f_{\delta,\lambda_0}(K)O(R^{-\min\{\rho,1\}})f_{\delta,\lambda_0}(K)+C_{R,4},
\end{align*}
which yields \eqref{2.8} immediately.
\end{proof}

\section{Proof of Theorem \ref{thm1.1}}

In this section, we prove Theorem \ref{thm1.1}. As in \S2,
we will show the estimates \eqref{1.18} and \eqref{1.19} only.
\eqref{1.18} yields the Mourre estimate \eqref{1.20}. Thus
Theorem \ref{thm1.1} and Corollary \ref{cor1.2} can be shown
by the standard argument in the Mourre theory.

\begin{proof}[Proof of \eqref{1.18} and \eqref{1.19}]

Let $\lambda_0\in[0,\omega)$, $\epsilon>0$ and $0<\delta_0<\omega/4$.
Denote by
$f_{\delta_0}$ any function in $C_0^\infty(\boldsymbol{R};\boldsymbol{R})$
such that $\mathrm{supp}\,f_{\delta_0}\subset[-\delta_0,\delta_0]$.
As in \S2, we write $f_{\delta_0}(K-\lambda_0)$
as $f_{\delta_0,\lambda_0}(K)$.
First of all, we note that the estimate
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)i[K,A(R)]f_{\delta_0,\lambda_0}(K)\\
={}&\sum_{a\in\mathscr{A}}f_{\delta_0,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta_0,\lambda_0}(K)+O(R^{-1})\\
={}&\sum_{a\in\mathscr{A}\setminus\{a_\mathrm{max}\}}f_{\delta_0,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta_0,\lambda_0}(K)+O(R^{-1})+C_{R,1}.
\end{split}\label{3.1}
\end{equation}
holds with some compact operator $C_{R,1}$ on $\mathscr{K}$.
Here we used the compactness of
$f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{max},R}i[K,A_{a_\mathrm{max}}]j_{a_\mathrm{max},R}f_{\delta_0,\lambda_0}(K)$. Put
\begin{align*}
&\widehat{\varTheta}_a=\bigcup_{b\subset a}\sigma_\mathrm{pp}(K^b),\\
&\widehat{d}_{0,a}(\lambda)=\mathrm{dist}(\lambda,\widehat{\varTheta}_a),\quad
\widehat{d}_{1,a}(\lambda)=\mathrm{dist}(\lambda,\widehat{\varTheta}_a\cap(-\infty,\lambda])
\end{align*}
for $a\in\mathscr{A}\setminus\{a_\mathrm{max}\}$.
We first estimate
$$f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}i[K,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K).$$
Here we note
\begin{align*}
j_{a_\mathrm{min},R}i[K,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}={}&j_{a_\mathrm{min},R}i[K_0,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}+O(R^{-\rho}),\\
i[K_0,A_{a_\mathrm{min}}]={}&2(\omega/4+H_0)^{-1}H_0,
\end{align*}
where $K_0=K_{a_\mathrm{min}}=D_t+p^2/2$ and $H_0=T_{a_\mathrm{min}}=p^2/2$.
In fact, it is easy to see
$$j_{a_\mathrm{min},R}i[I_{a_\mathrm{min}},A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}=j_{a_\mathrm{min},R}i[V,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}=O(R^{-\rho}).$$
Now we will treat $i[K_0,A_{a_\mathrm{min}}]$ with a partition
of unity with respect to $H_0$.
Let $\eta$ be a function in $C^\infty(\boldsymbol{R};\boldsymbol{R})$ such that
$\mathrm{supp}\,\eta\subset[1,\infty)$, $0\le\eta(\tau)\le1$, $\eta(\tau)=1$ on $[2,\infty)$, and
$$\eta(\tau)^2+\bar{\eta}(\tau)^2=1,$$
where $\bar{\eta}(\tau)=1-\eta(\tau)$. For the sake of brevity,
we put $\eta_s(\tau)=\eta(\tau/s)$ and $\bar{\eta}_s(\tau)=\bar{\eta}(\tau/s)$ for $s>0$. Thus $\eta_s(\tau)$ and $\bar{\eta}_s(\tau)$ satisfy
$0\le\eta_s(\tau)\le1$, $0\le\bar{\eta}_s(\tau)\le1$,
$$\eta_s(\tau)^2+\bar{\eta}_s(\tau)^2=1,\quad
\eta_s(\tau)=\begin{cases} 1 & (\tau\ge 2s)\\
0 &(\tau\le s)\end{cases},\quad
\bar{\eta}_s(\tau)=\begin{cases} 0 & (\tau\ge 2s)\\
1 &(\tau\le s)\end{cases}.$$
By the partition of unity
$\{\eta_\omega(H_0),\,\bar{\eta}_\omega(H_0)\}$,
$(\omega/4+H_0)^{-1}H_0$
can be decomposed into the sum
$$\eta_\omega(H_0)(\omega/4+H_0)^{-1}H_0\eta_\omega(H_0)+
\bar{\eta}_\omega(H_0)(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0).$$
Using the estimate
$$\eta_\omega(H_0)(\omega/4+H_0)^{-1}H_0\eta_\omega(H_0)\ge
\frac{\omega}{\omega/4+\omega}\eta_\omega(H_0)^2,$$
we have
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}i[K,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
={}&f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}i[K_0,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)
+O(R^{-\rho})\\
\ge{}&\frac{2\omega}{\omega/4+\omega}f_{\delta_0,\lambda_0}(K)
j_{a_\mathrm{min},R}\eta_\omega(H_0)^2j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\quad+f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}\bar{\eta}_\omega(H_0)i[K_0,A_{a_\mathrm{min}}]\bar{\eta}_\omega(H_0)j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\qquad+O(R^{-\rho}).
\end{split}\label{3.2}
\end{equation}
Noting
\begin{equation}
\{f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}-
j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)\}\bar{\eta}_\omega(H_0)=O(R^{-\min\{\rho,1\}})
\label{3.3}
\end{equation}
because of the boundedness of $\langle p\rangle\bar{\eta}_\omega(H_0)$,
we see that
\begin{align*}
&f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}\bar{\eta}_\omega(H_0)i[K_0,A_{a_\mathrm{min}}]\bar{\eta}_\omega(H_0)j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
={}&j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)\bar{\eta}_\omega(H_0)i[K_0,A_{a_\mathrm{min}}]\bar{\eta}_\omega(H_0)f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}\\
&\quad+O(R^{-\min\{\rho,1\}})
\end{align*}
holds. We will use the direct integral
\begin{align*}
&f_{\delta_0,\lambda_0}(K_0)\bar{\eta}_\omega(H_0)(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0)f_{\delta_0,\lambda_0}(K_0)\\
={}&\bigoplus_{n\le1}
(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\end{align*}
as in \S2. When $n\le-1$,
\begin{align*}
&(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0-n\omega}(H_0)^2\\
\ge{}&\frac{\lambda_0-n\omega-\delta_0}{\omega/4+(\lambda_0-n\omega-\delta_0)}\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0-n\omega}(H_0)^2\\
\ge{}&\frac{\lambda_0+\omega-\delta_0}{\omega/4+(\lambda_0+\omega-\delta_0)}\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0-n\omega}(H_0)^2
\end{align*}
holds. Here we note that when $n\le-3$, $\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0-n\omega}(H_0)^2=0$ holds. We will consider the case where
$n=0$. In the same way as in \S2, we see that if $\lambda_0<\delta_0$, then
$$(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0}(H_0)^2\ge0;$$
while, if $\lambda_0\ge\delta_0$, then
\begin{align*}
&(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0}(H_0)^2\\
\ge{}&\frac{\lambda_0-\delta_0}{\omega/4+(\lambda_0-\delta_0)}\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0}(H_0)^2.
\end{align*}
We will also consider the case where $n=1$. If $\lambda_0\le\omega-\delta_0$, then $f_{\delta_0,\lambda_0-\omega}(H_0)=0$. On the other hand,
if $\lambda_0>\omega-\delta_0$, then
$$
(\omega/4+H_0)^{-1}H_0\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0-\omega}(H_0)^2\ge0.
$$
By combining these and using \eqref{3.3},
we see that if $\delta_0\le\lambda_0\le\omega-\delta_0$, then
\begin{align*}
&f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}\bar{\eta}_\omega(H_0)i[K_0,A_{a_\mathrm{min}}]\bar{\eta}_\omega(H_0)j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
\ge{}&\frac{2(\lambda_0-\delta_0)}{\omega/4+(\lambda_0-\delta_0)}j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K_0)\bar{\eta}_\omega(H_0)^2f_{\delta_0,\lambda_0}(K_0)j_{a_\mathrm{min},R}\\
&\quad+O(R^{-\min\{\rho,1\}})\\
={}&\frac{2(\lambda_0-\delta_0)}{\omega/4+(\lambda_0-\delta_0)}f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}\bar{\eta}_\omega(H_0)^2j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\quad+O(R^{-\min\{\rho,1\}}).
\end{align*}
This and \eqref{3.2} yield
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}i[K,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
\ge{}&\frac{2\omega}{\omega/4+\omega}f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}\eta_\omega(H_0)^2j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\quad+\frac{2(\lambda_0-\delta_0)}{\omega/4+(\lambda_0-\delta_0)}f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}\bar{\eta}_\omega(H_0)^2j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\\
&\qquad+O(R^{-\min\{\rho,1\}})\\
\ge{}&\frac{2(\widehat{d}_{1,a_\mathrm{min}}(\lambda_0)-\delta_0)}{3\omega/2}f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}^2f_{\delta_0,\lambda_0}(K)\\
&\quad+O(R^{-\min\{\rho,1\}}).
\end{split}\label{3.4}
\end{equation}
Here we used $5\omega/4>\omega>\lambda_0-\delta_0$
and $\widehat{d}_{1,a_\mathrm{min}}(\lambda_0)=\lambda_0$.
Similarly, we see that if $0\le\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$, then
\begin{equation}
f_{\delta_0,\lambda_0}(K)j_{a_\mathrm{min},R}i[K,A_{a_\mathrm{min}}]j_{a_\mathrm{min},R}f_{\delta_0,\lambda_0}(K)\ge
O(R^{-\min\{\rho,1\}}).
\label{3.5}
\end{equation}

We next estimate
$f_{\delta_0,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta_0,\lambda_0}(K)$
with $a\in\mathscr{A}\setminus\{a_\mathrm{max},a_\mathrm{min}\}$.
We first note
\begin{align*}
j_{a,R}i[K,A_a]j_{a,R}={}&j_{a,R}i[K_a,A_a]j_{a,R}+O(R^{-\rho}),\\
i[K_a,A_a]={}&i[K^a,A^a]+2(\omega/4+T_a)^{-1}T_a,\\
i[K_0^a,A^a]={}&(3\omega/2-D_t)^{-1}(p^a)^2=2(3\omega/2-D_t)^{-1}(K_0^a-D_t),\\
i[V^a,A^a]={}&-(3\omega/2-D_t)^{-1}((x^a\cdot\nabla^a)V^a)\\
&\quad-(3\omega/2-D_t)^{-1}(\partial_t V^a)(3\omega/2-D_t)^{-1}(\hat{A}_0)^a
\end{align*}
and $|x^a|\le r_0R$ holds on $\mathrm{supp}\,j_{a,R}$.
Here $T_a=(p_a)^2/2$ and $K_0^a=D_t+(p^a)^2/2$. In fact, it is easy to see
$$j_{a,R}i[I_a,A_a]j_{a,R}=O(R^{-\rho}).$$
Now we will treat $i[K_a,A_a]$ with a partition of unity with respect to
$-D_t$. For each $L\in\boldsymbol{N}$,
we introduce a partition of unity
$\{\eta_{L\omega}(-D_t),\,\bar{\eta}_{L\omega}(-D_t)\}$.
Then $i[K_0,A_a]$
can be decomposed into the sum
$$\eta_{L\omega}(-D_t)i[K_0,A_a]\eta_{L\omega}(-D_t)+
\bar{\eta}_{L\omega}(-D_t)i[K_0,A_a]\bar{\eta}_{L\omega}(-D_t).$$
On the other hand, as for $i[V^a,A_a]=i[V^a,A^a]$, the relation
\begin{equation}
\begin{split}
i[V^a,A_a]={}&\eta_{L\omega}(-D_t)i[V^a,A_a]\eta_{L\omega}(-D_t)\\
&\quad+\bar{\eta}_{L\omega}(-D_t)i[V^a,A_a]\bar{\eta}_{L\omega}(-D_t)
+O(L^{-1})
\end{split}\label{3.6}
\end{equation}
as $L\to\infty$ can be obtained generally, in virtue of that
$i[i[V^a,A^a],D_t]$ is bounded by the assumption $(V)_3$.
Except in the case where $V^a$ is time-independent,
we have to deal with the above error term $O(L^{-1})$.
This is one of the technical reasons why we also need some
regularity of second derivatives of $V^a$, as mentioned in \S1.
Since
$$i[K_0^a,A^a]=2+2(3\omega/2-D_t)^{-1}(K_0^a-3\omega/2)$$
and $(\omega/4+T_a)^{-1}T_a\ge0$,
there exist a large $L_0\in\boldsymbol{N}$ such that
if $L\ge L_0$, then
\begin{align*}
&f_{\delta_0,\lambda_0}(K)j_{a,R}\eta_{L\omega}(-D_t)i[K_a,A_a]\eta_{L\omega}(-D_t)j_{a,R}f_{\delta_0,\lambda_0}(K)\\
&{}\ge\frac{2\omega}{\omega/4+\omega}
f_{\delta_0,\lambda_0}(K)j_{a,R}\eta_{L\omega}(-D_t)^2
j_{a,R}f_{\delta_0,\lambda_0}(K)
\end{align*}
holds. Here we used that $K_0^a$ and $(3\omega/2-D_t)i[V^a,A^a]j_{a,R}$ are $K$-bounded, and
$$\frac{2\omega}{\omega/4+\omega}<2.$$
Since on $\mathrm{supp}\,j_{a,R}$, $|x^a|\le r_0R$ holds,
$f_{\delta_0,\lambda_0}(K)j_{a,R}i[K^a,A^a]j_{a,R}f_{\delta_0,\lambda_0}(K)$
can be recognized as
$$f_{\delta_0,\lambda_0}(K)j_{a,R}i[K^a,A_R^a]j_{a,R}f_{\delta_0,\lambda_0}(K)+
O(R^{-1}),$$
where $A_R^a$ is the conjugate operator for $K^a$ defined as in \S2.
Hence we have
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta_0,\lambda_0}(K)\\
\ge{}&\frac{2\omega}{\omega/4+\omega}
f_{\delta_0,\lambda_0}(K)j_{a,R}\eta_{L\omega}(-D_t)^2
j_{a,R}f_{\delta_0,\lambda_0}(K)\\
&{}\quad+f_{\delta_0,\lambda_0}(K)j_{a,R}\bar{\eta}_{L\omega}(-D_t)B_{a,R}
\bar{\eta}_{L\omega}(-D_t)j_{a,R}f_{\delta_0,\lambda_0}(K)\\
&\qquad+f_{\delta_0,\lambda_0}(K)j_{a,R}O(L^{-1})j_{a,R}f_{\delta_0,\lambda_0}(K)+O(R^{-\min\{\rho,1\}})
\end{split}\label{3.7}
\end{equation}
as $L\to\infty$, where
$$B_{a,R}=i[K^a,A_R^a]+2(\omega/4+T_a)^{-1}T_a.$$
Now we will treat
$$f_{\delta_0,\lambda_0}(K)j_{a,R}\bar{\eta}_{L\omega}(-D_t)B_{a,R}\bar{\eta}_{L\omega}(-D_t)j_{a,R}f_{\delta_0,\lambda_0}(K).
$$
Since $\langle p\rangle\bar{\eta}_{L\omega}(-D_t)(K_0+i)^{-1}=O(L^{1/2})$,
we see that
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)(j_{a,R}\bar{\eta}_{L\omega}(-D_t))-
(j_{a,R}\bar{\eta}_{L\omega}(-D_t))f_{\delta_0,\lambda_0}(K_a)\\
={}&O(L^{-1})+O(L^{1/2}R^{-1})+O(R^{-\rho})
\end{split}\label{3.8}
\end{equation}
holds by the assumption $(V)_3$. By using \eqref{3.8}, we have
\begin{equation}
\begin{split}
&f_{\delta_0,\lambda_0}(K)j_{a,R}\bar{\eta}_{L\omega}(-D_t)B_{a,R}
\bar{\eta}_{L\omega}(-D_t)j_{a,R}f_{\delta_0,\lambda_0}(K)\\
={}&j_{a,R}\bar{\eta}_{L\omega}(-D_t)f_{\delta_0,\lambda_0}(K_a)B_{a,R}
f_{\delta_0,\lambda_0}(K_a)\bar{\eta}_{L\omega}(-D_t)j_{a,R}\\
&\quad+O(L^{-1})+O(L^{1/2}R^{-1})+O(R^{-\rho}).
\end{split}\label{3.9}
\end{equation}
Now, by following the argument of Froese-Herbst~\cite{FH},
we will show that there exists a small $\delta_{0,\epsilon}^a$
such that $0<\delta_{0,\epsilon}^a<\delta_0$, and
\begin{equation}
\begin{split}
&\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)B_{a,R}f_{\delta,\lambda_0}(K_a)\bar{\eta}_{L\omega}(-D_t)\\
\ge{}&\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}
\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)^2\bar{\eta}_{L\omega}(-D_t)
\end{split}\label{3.10}
\end{equation}
for any $0<\delta\le\delta_{1,\epsilon}^a$.
The left-hand side of \eqref{3.10}
can be decomposed into the direct integral
\begin{align*}
&\int_{[0,\infty)}^\oplus
\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K^a+\lambda_a)\left(i[K^a,A_R^a]+\frac{2\lambda_a}{\omega/4+\lambda_a}\right)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times
f_{\delta,\lambda_0}(K^a+\lambda_a)\bar{\eta}_{L\omega}(-D_t)\,d\lambda_a.
\end{align*}
By using the decomposition
$$[0,\infty)=\bigsqcup_{n=0}^\infty J_n;\quad
J_0=[0,\lambda_0],\quad
J_n=(\lambda_0+(n-1)\omega,\lambda_0+n\omega],\quad n\in\boldsymbol{N},
$$
we have
\begin{align*}
&\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)B_{a,R}
f_{\delta,\lambda_0}(K_a)\bar{\eta}_{L\omega}(-D_t)\\
={}&\sum_{n=0}^\infty\int_{J_n}^\oplus \bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0-\lambda_a}(K^a)
\left(i[K^a,A_R^a]+\frac{2\lambda_a}{\omega/4+\lambda_a}\right)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times
f_{\delta,\lambda_0-\lambda_a}(K^a)\bar{\eta}_{L\omega}(-D_t)\,d\lambda_a.
\end{align*}
For a while we will treat
$$f_{\delta,\lambda_0-\lambda_a}(K^a)
i[K^a,A_R^a]f_{\delta,\lambda_0-\lambda_a}(K^a)$$
for $\lambda_a\in J_n$ with some $n\in\boldsymbol{N}_0=\{0\}\cup\boldsymbol{N}$.
It follows from the results in \S2 that
there exists a large $R_\epsilon^a\ge1$ and a small $\delta_{1,\epsilon}^a$
such that for any $R\ge R_\epsilon^a$ and $0<\delta\le\delta_{1,\epsilon}^a$,
the following holds: When $n\ge1$, $J_n$ is decomposed as
$J_n=J_{n,1}\sqcup J_{n,2}$ with
$$J_{n,1}=[\lambda_0+(n-1)\omega+\delta_0,\lambda_0+n\omega-\delta_0],\quad
J_{n,2}=J_n\setminus J_{n,1}.$$
On the other hand, $J_0$ is decomposed as
$J_0=J_{0,1}\sqcup J_{0,2}$ with
$$J_{0,1}=[\delta_0,\omega-\delta_0]\cap J_0,\quad
J_{0,2}=J_0\setminus J_{0,1}.$$
Here we note that if $\lambda_0<\delta_0$,
then $J_{0,1}=\emptyset$. If $\lambda_a\in J_{n,1}$, then
\begin{equation}
\begin{split}
&f_{\delta,\lambda_0-\lambda_a}(K^a)
i[K^a,A_R^a]f_{\delta,\lambda_0-\lambda_a}(K^a)\\
\ge{}&\frac{2(\widehat{d}_{1,a}(\lambda_0-\lambda_a)-\delta_0)-\epsilon/4}{3\omega/2}
f_{\delta,\lambda_0-\lambda_a}(K^a)^2
\end{split}\label{3.11}
\end{equation}
holds  because of $\lambda_0-\lambda_a\in[-n\omega,-(n-1)\omega-\delta_0]$;
while, if $\lambda_a\in J_{n,2}$, then
\begin{equation}
f_{\delta,\lambda_0-\lambda_a}(K^a)
i[K^a,A_R^a]f_{\delta,\lambda_0-\lambda_a}(K^a)
\ge\frac{-\epsilon/4}{3\omega/2}f_{\delta,\lambda_0-\lambda_a}(K^a)^2
\label{3.12}
\end{equation}
holds. Here we emphasize that $R_\epsilon^a$ and $\delta_{1,\epsilon}^a$
can be taken uniformly in $\lambda_a\in[0,\infty)$,
by using the $\omega$-periodicity of $\sigma(K^a)$
and following the argument of \cite{FH}.
If $\lambda_a\in J_{n,1}$ with $n\ge1$, then
$$\frac{2\lambda_a}{\omega/4+\lambda_a}\ge\frac{2\{\lambda_0+(n-1)\omega+\delta_0\}}{
\omega/4+\{\lambda_0+(n-1)\omega+\delta_0\}}\ge\frac{2(\lambda_0+\delta_0)}{\omega/4+\lambda_0+\delta_0}
\ge\frac{2(\lambda_0+\delta_0)}{3\omega/2},$$
which yields
$$\frac{2(\widehat{d}_{1,a}(\lambda_0-\lambda_a)-\delta_0)-\epsilon/4}{3\omega/2}+\frac{2\lambda_a}{\omega/4+\lambda_a}\ge
\frac{2\lambda_0-\epsilon/4}{3\omega/2}$$
because $\widehat{d}_{1,a}(\lambda_0-\lambda_a)\ge0$; while, if $\lambda_0\ge\delta_0$ and
$\lambda_a\in J_{0,1}$, then
$$\frac{2(\widehat{d}_{1,a}(\lambda_0-\lambda_a)-\delta_0)-\epsilon/4}{3\omega/2}+\frac{2\lambda_a}{\omega/4+\lambda_a}\ge
\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2},$$
because $\widehat{d}_{1,a}(\lambda_0-\lambda_a)\ge\widehat{d}_{1,a}(\lambda_0)-\lambda_a$ and $\omega/4+\lambda_a<3\omega/2$.
On the other hand, if $\lambda_a\in J_{n,2}$ with $n\ge1$, then
$$\frac{2\lambda_a}{\omega/4+\lambda_a}\ge\frac{2\{\lambda_0+(n-1)\omega\}}{
\omega/4+\{\lambda_0+(n-1)\omega\}}\ge\frac{2\lambda_0}{\omega/4+\lambda_0}
\ge\frac{2\lambda_0}{3\omega/2},$$
which yields
$$\frac{-\epsilon/4}{3\omega/2}+\frac{2\lambda_a}{\omega/4+\lambda_a}\ge
\frac{2\lambda_0-\epsilon/4}{3\omega/2};$$
while, if $\lambda_a\in J_{0,2}$, then
$$\frac{-\epsilon/4}{3\omega/2}+\frac{2\lambda_a}{\omega/4+\lambda_a}\ge
\frac{-\epsilon/4}{3\omega/2}.$$
Finally we see that if $\delta_0\le\lambda_0\le\omega-\delta_0$, then
\begin{equation}
\begin{split}
&\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)B_{a,R}
f_{\delta,\lambda_0}(K_a)\bar{\eta}_{L\omega}(-D_t)\\
\ge{}&\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}\sum_{n=0}^\infty\int_{J_n}^\oplus \bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0-\lambda_a}(K^a)^2\bar{\eta}_{L\omega}(-D_t)\,d\lambda_a\\
={}&\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)^2\bar{\eta}_{L\omega}(-D_t),
\end{split}\label{3.13}
\end{equation}
because of $\lambda_0\ge\widehat{d}_{1,a}(\lambda_0)$; while, if
$\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$, then
\begin{equation}
\begin{split}
&\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)B_{a,R}
f_{\delta,\lambda_0}(K_a)\bar{\eta}_{L\omega}(-D_t)\\
\ge{}&\frac{-\epsilon/4}{3\omega/2}\bar{\eta}_{L\omega}(-D_t)f_{\delta,\lambda_0}(K_a)^2\bar{\eta}_{L\omega}(-D_t).
\end{split}\label{3.14}
\end{equation}
Now we will consider the case where $\delta_0\le\lambda_0\le\omega-\delta_0$
for a while.
By \eqref{3.7}, \eqref{3.8}, \eqref{3.9} and \eqref{3.13},
the estimate
\begin{equation}
\begin{split}
&f_{\delta_{1,\epsilon}^a,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta_{1,\epsilon}^a,\lambda_0}(K)\\
\ge{}&\frac{2\omega}{\omega/4+\omega}
f_{\delta_{1,\epsilon}^a,\lambda_0}(K)j_{a,R}\eta_{L\omega}(-D_t)^2
j_{a,R}f_{\delta_{1,\epsilon}^a,\lambda_0}(K)\\
&{}+\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}
f_{\delta_{1,\epsilon}^a,\lambda_0}(K)j_{a,R}\bar{\eta}_{L\omega}(-D_t)^2j_{a,R}f_{\delta_{1,\epsilon}^a,\lambda_0}(K)\\
&\quad+O(L^{-1})+O(L^{1/2}R^{-1})+O(R^{-\min\{\rho,1\}})\\
\ge{}&\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}f_{\delta_{1,\epsilon}^a,\lambda_0}(K)j_{a,R}^2f_{\delta_{1,\epsilon}^a,\lambda_0}(K)\\
&\quad+O(L^{-1})+O(L^{1/2}R^{-1})+O(R^{-\min\{\rho,1\}}).
\end{split}\label{3.15}
\end{equation}
can be obtained. Here we used
$$\frac{2\omega}{\omega/4+\omega}>
\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}.$$
By sandwiching \eqref{3.15} in two $f_{\delta,\lambda_0}(K)$
with $0<\delta<\delta_{1,\epsilon}^a$, one can obtain
\begin{equation}
\begin{split}
&f_{\delta,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta,\lambda_0}(K)\\
\ge{}&\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/4}{3\omega/2}f_{\delta,\lambda_0}(K)j_{a,R}^2f_{\delta,\lambda_0}(K)\\
&\quad+f_{\delta,\lambda_0}(K)\{O(L^{-1})+O(L^{1/2}R^{-1})+O(R^{-\min\{\rho,1\}})\}f_{\delta,\lambda_0}(K),
\end{split}\label{3.16}
\end{equation}
because of the arbitrariness of $f_{\delta_{1,\epsilon}^a,\lambda_0}$.
Then one can take $L_\epsilon\in\boldsymbol{N}$ so large that
$L_\epsilon\ge L_0$ and
\begin{equation}
\begin{split}
&f_{\delta,\lambda_0}(K)j_{a,R}i[K,A_a]j_{a,R}f_{\delta,\lambda_0}(K)\\
\ge{}&\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/2}{3\omega/2}f_{\delta,\lambda_0}(K)j_{a,R}^2f_{\delta,\lambda_0}(K)\\
&\quad+f_{\delta,\lambda_0}(K)\{O(L_\epsilon^{1/2}R^{-1})+O(R^{-\min\{\rho,1\}})\}f_{\delta,\lambda_0}(K).
\end{split}\label{3.17}
\end{equation}
By \eqref{3.1}, \eqref{3.4}, \eqref{3.17}, we finally obtain the
estimate
\begin{equation}
\begin{split}
&f_{\delta,\lambda_0}(K)i[K,A(R)]f_{\delta,\lambda_0}(K)\\
\ge{}&\frac{2(\widehat{d}_{1,a_\mathrm{min}}(\lambda_0)-\delta_0)}{3\omega/2}f_{\delta,\lambda_0}(K)j_{a_\mathrm{min},R}^2f_{\delta,\lambda_0}(K)\\
&+\sum_{a\in\mathscr{A}\setminus\{a_\mathrm{max}\}}\frac{2(\widehat{d}_{1,a}(\lambda_0)-\delta_0)-\epsilon/2}{3\omega/2}f_{\delta,\lambda_0}(K)j_{a,R}^2f_{\delta,\lambda_0}(K)\\
&\quad+f_{\delta,\lambda_0}(K)O(R^{-\min\{\rho,1\}})f_{\delta,\lambda_0}(K)+
C_R\\
\ge{}&\frac{2(d_1(\lambda_0)-\delta_0)-\epsilon/2}{3\omega/2}f_{\delta,\lambda_0}(K)^2\\
&\quad+f_{\delta,\lambda_0}(K)O(R^{-\min\{\rho,1\}})f_{\delta,\lambda_0}(K)+
C_R
\end{split}\label{3.18}
\end{equation}
with some compact operator $C_R$ on $\mathscr{K}$
for $0<\delta<\delta_{1,\epsilon}$, where
$$\delta_{1,\epsilon}=\min_{a\in\mathscr{A}\setminus\{a_\mathrm{max},a_\mathrm{min}\}}\delta_{1,\epsilon}^a.$$
Here we used
$$\widehat{d}_{1,a}(\lambda_0)\ge
d_1(\lambda_0),\quad
a\in\mathscr{A}\setminus\{a_\mathrm{max}\}.$$
\eqref{3.18} yields \eqref{1.18}. Similarly one can show that if
$\lambda_0<\delta_0$ or $\lambda_0>\omega-\delta_0$, then
\begin{equation}
\begin{split}
&f_{\delta,\lambda_0}(K)i[K,A(R)]f_{\delta,\lambda_0}(K)\\
\ge{}&\frac{-\epsilon/2}{3\omega/2}f_{\delta,\lambda_0}(K)^2
+f_{\delta,\lambda_0}(K)O(R^{-\min\{\rho,1\}})f_{\delta,\lambda_0}(K)+
C_R
\end{split}\label{3.19}
\end{equation}
holds. \eqref{3.19} yields \eqref{1.19}.
\end{proof}

\begin{rem}
In the above proof, we have used \eqref{3.3} and \eqref{3.8}.
It has been well known since the work of
Froese-Herbst~\cite{FH} that
the estimates like
\begin{equation}
f_{\delta_0,\lambda_0}(K)j_{a,R}-
j_{a,R}f_{\delta_0,\lambda_0}(K_a)=O(R^{-\rho}),\quad
a\in\mathscr{A}
\label{3.20}
\end{equation}
are very useful for the inductive argument
in the proof of the Mourre estimates for Hamiltonians
which govern many body quantum systems.
However, in our case,
we do not know whether \eqref{3.20}
holds or not, as mentioned also
in \cite{MS}. In our analysis, we need cut-offs
like $\bar{\eta}_\omega(H_0)$ or $\bar{\eta}_{L\omega}(-D_t)$.
\end{rem}

\section{Concluding remarks}

Let $N\ge2$, and consider a system of $N$ particles moving in a
given $T$-periodic electric field $\mathscr{E}(t)\in C^0(\boldsymbol{R};\boldsymbol{R}^d)$.
The total Hamiltonian $\hat{H}(t)$ in the center-of-mass frame is given as
$$\hat{H}(t)=-\frac{1}{2}\Delta-E(t)\cdot x+\bar{V},
\quad \bar{V}=\sum_{1\le j<k\le N}\bar{V}_{jk}(x_j-x_k)$$
on $L^2(X)$, where
$$E(t)=\pi((q_1/m_1)\mathscr{E}(t),\ldots,(q_N/m_N)\mathscr{E}(t))
\in C^0(\boldsymbol{R};X)$$
is $T$-periodic, $q_j$ is the charge of the $j$-th particle,
and $\bar{V}_{jk}$'s are time-independent pair potentials.
$q_j/m_j$ is called the specific charge of the $j$-th particle.
Suppose that there exists a pair $(j,k)$ such that
$q_j/m_j\not=q_k/m_k$. Under this assumption, if $\mathscr{E}(t)\not=0$,
then $E(t)\not=0$.
Denote by $\hat{U}(t,s)$ the propagator generated by $\hat{H}(t)$,
and put
$$E_\mathrm{m}=\frac{1}{T}\int_0^TE(s)\,ds\in X.$$
As in M\o ller~\cite{Mo} and Adachi~\cite{A1},
define $X$-valued $T$-periodic
functions $b_0(t)$, $b(t)$ and
$c(t)$ on $\boldsymbol{R}$ by
\begin{align*}
&b_0(t)=\int_0^t(E(s)-E_\mathrm{m})\,ds,\quad
b_{0,\mathrm{m}}=\frac{1}{T}\int_0^Tb_0(s)\,ds,\\
&b(t)=b_0(t)-b_{0,\mathrm{m}},\quad
c(t)=\int_0^tb(s)\,ds,
\end{align*}
and introduce the time-dependent Hamiltonian
$$H(t)=H_0+\bar{V}(x+c(t)),\quad H_0=-\frac{1}{2}\Delta-E_\mathrm{m}\cdot x$$
on $L^2(X)$.
By introducing $\boldsymbol{R}^d$-valued $T$-periodic functions
$\bar{\mathscr{E}}_0(t)$,
$\bar{\mathscr{E}}(t)$ and $\bar{\bar{\mathscr{E}}}(t)$ as
\begin{align*}
&\mathscr{E}_\mathrm{m}=\frac{1}{T}\int_0^T\mathscr{E}(s)\,ds,\quad\bar{\mathscr{E}}_0(t)=\int_0^t(\mathscr{E}(s)-\mathscr{E}_\mathrm{m})\,ds,\\
&\bar{\mathscr{E}}_{0,\mathrm{m}}=\frac{1}{T}\int_0^T\bar{\mathscr{E}}_0(s)\,ds,\quad\bar{\mathscr{E}}(t)=\bar{\mathscr{E}}_0(t)-\bar{\mathscr{E}}_{0,\mathrm{m}},\quad
\bar{\bar{\mathscr{E}}}(t)=\int_0^t\bar{\mathscr{E}}(s)\,ds,
\end{align*}
$b(t)$, $c(t)$ and $\bar{V}(x+c(t))$ can be represented as
\begin{align*}
&b(t)=\pi((q_1/m_1)\bar{\mathscr{E}}(t),\ldots,(q_N/m_N)\bar{\mathscr{E}}(t)),\\
&c(t)=\pi((q_1/m_1)\bar{\bar{\mathscr{E}}}(t),\ldots,(q_N/m_N)\bar{\bar{\mathscr{E}}}(t)),\\
&\bar{V}(x+c(t))=\sum_{(j,k)\in\mathscr{A}}\bar{V}_{jk}(r_j-r_k+c_{jk}(t))
\end{align*}
with $c_{jk}(t)=(q_j/m_j-q_k/m_k)\bar{\bar{\mathscr{E}}}(t)$.
Suppose $V_{jk}$'s belong to $C^2(\boldsymbol{R}^d;\boldsymbol{R})$,
and satisfy the decaying conditions
\begin{equation}
|(\partial_y^\alpha V_{jk})(y)|\le C\langle y\rangle^{-\rho-|\alpha|},
\quad|\alpha|\le2
\label{4.1}
\end{equation}
with some $\rho>0$, and put $V_{jk}(t,y)=\bar{V}_{jk}(y+c_{jk}(t))$.
Then $V_{jk}(t,y)$'s satisfy \eqref{1.16}.

If $E_\mathrm{m}=0$, then
$H_0$ is called the free $N$-body Schr\"odinger operator;
while, if $E_\mathrm{m}\not=0$, then
$H_0$ is called the free $N$-body Stark Hamiltonian.
Denote by $U(t,s)$ the unitary propagator
generated by $H(t)$. As is well-known, the
following Avron-Herbst formula holds:
\begin{equation}
\hat{U}_0(t,s)=\mathscr{T}(t)e^{-i(t-s)H_0}\mathscr{T}(s)^*,\quad
\hat{U}(t,s)=\mathscr{T}(t)U(t,s)\mathscr{T}(s)^*
\label{4.2}
\end{equation}
with
$$
\mathscr{T}(t)=e^{-ia(t)}e^{ib(t)\cdot x}e^{-ic(t)\cdot p},
\quad a(t)=\int_0^t\left(\frac{1}{2}|b(s)|^2-E_\mathrm{m}\cdot c(s)\right)\,ds.
$$

When $E_\mathrm{m}\not=0$, in \cite{A1} and \cite{A2},
the author already obtained the result
of the asymptotic completeness for the system under consideration,
both in the short-range
and the long-range cases,
by introducing the Floquet Hamiltonian $K$
associated with $\hat{H}(t)$.
$$A=\frac{E_\mathrm{m}}{|E_\mathrm{m}|}\cdot p$$
can be taken as a conjugate operator for $K$ in the standard Mourre theory.
Here we emphasize that in the case where $N=2$, in \cite{Mo},
M\o ller proposed this
operator as a conjugate operator for $K$ before \cite{A1}.
On the other hand, when $E_\mathrm{m}=0$, any candidates of a conjugate
operator for $K$ in the standard Mourre theory have not been found up until
now, except in the case where $N=2$.
$\hat{H}(t)$ with $E_\mathrm{m}=0$ is called an $N$-body AC Stark
Hamiltonian. As mentioned in \S1, in the case where
$N=2$, Yokoyama~\cite{Yo}, and Adachi-Kiyose~\cite{AK} proposed
conjugate operators for $K$. Unfortunately, these operators seem not
have any natural extension to $N$-body systems.
M\o ller-Skibsted~\cite{MS} used $\hat{A}_0$ as a conjugate operator
for $K$ in an extended Mourre theory, in order to avoid this difficulty.
Our construction of $A(R)$ in \eqref{1.15} seems the first attempt
to give a conjugate operator for $K$ in the standard Mourre theory
when $N\ge3$.

As for the asymptotic completeness for
$\hat{H}(t)$ with $N=2$,
Yajima~\cite{Ya1} proved it in the short-range case
via the Howland-Yajima method,
and Kitada-Yajima~\cite{KY} proved it in the long-range case
via the Enss method.
On the other hand, for
$\hat{H}(t)$ with $N=3$,
Korotyaev~\cite{Ko} and Nakamura~\cite{N} gave some
partial results on it in the short-range case
via the Howland-Yajima and the Faddeev methods.
As is well-known, the limiting absorption principle
\eqref{1.23} yields the local $K$-smoothness of
$\langle x\rangle^{-s}$ with $s>1/2$
\begin{equation}
\begin{split}
&\int_{-\infty}^\infty\|\langle x\rangle^{-s}e^{-i\sigma K}f_{\delta}(K-\lambda_0)\Phi\|_{\mathscr{K}}^2\,d\sigma\le C\|\Phi\|_{\mathscr{K}}^2
\end{split}\label{4.3}
\end{equation}
for $\lambda_0\not\in\widehat{\varTheta}$.
\eqref{4.3} was already obtained by M{\o}ller-Skibsted
even if $N\ge3$. However, \eqref{4.3} is not enough for the
proof of the asymptotic completeness in the case where $N\ge3$,
unlike in the case where $N=2$.
We expect that the Mourre estimate \eqref{1.20} will
be useful for the proof of the asymptotic
completeness in the case where $N=3$. In fact, for
$\lambda_0\not\in\widehat{\varTheta}$,
the so-called minimal velocity estimate
\begin{equation}
\begin{split}
&\int_1^\infty\bigg\|F\left(\frac{|x|}{\sigma}\le\sqrt{c_0(d_1(\lambda_0)-2\epsilon)}\right)\\
&\qquad\qquad\qquad\qquad\qquad\times e^{-i\sigma K}f_{\delta}(K-\lambda_0)\Phi\bigg\|_{\mathscr{K}}^2\,\frac{d\sigma}{\sigma}\le C\|\Phi\|_{\mathscr{K}}^2
\end{split}\label{4.4}
\end{equation}
with some $c_0>0$ may be yielded by
\begin{equation}
\begin{split}
\int_1^\infty&\bigg\|F\left(-c_2\le\frac{A}{\sigma}-
\frac{2(d_1(\lambda_0)-2\epsilon)}{3\omega/2}\le-c_1\right)\\
&\qquad\qquad\qquad\qquad\qquad\times e^{-i\sigma K}f_{\delta}(K-\lambda_0)\Phi\bigg\|_{\mathscr{K}}^2\,\frac{d\sigma}{\sigma}\le C\|\Phi\|_{\mathscr{K}}^2
\end{split}\label{4.5}
\end{equation}
with some $0<c_1<c_2$.
Here $F(x\in\Omega)$ denotes the characteristic function of the set of
$\Omega$. The minimal velocity estimate is one of the
most important propagation estimates for $N$-body Schr\"odinger
operators, as is well-known (see e.g. Graf~\cite{Gr}).
These propagation estimates can be proved in the same way as
in Sigal-Soffer~\cite{SS}, by virtue of the Mourre estimate \eqref{1.20} or
\eqref{1.21}.
The Mourre estimate for a general $N$-body Floquet Hamiltonian $K$
may be also obtained by our construction of a conjugate operator for $K$.
We would like to study the problem of the asymptotic completeness
for $\hat{H}(t)$ with $N\ge2$ by using some useful propagation estimates like
\eqref{4.4} in future research.

redpajama_set_name: RedPajamaArXiv


text:
Related Expertise: Health Care Payers, Providers, Systems & Services, Biopharma, Medical Devices and Technology
Patients Want to Feel Better. Who Should Pay for That?
09 June 2021 By Meghna Eichelberger, Hannah Hagen, Meg Kedrowski, Nicholas Kransz, Suchita Shah, and Natasha Taylor
How do US payers evaluate quality-of-life therapies to make coverage decisions? Through surveys and in-depth interviews, we found that the willingness to cover these therapies varies by payer type and condition, but payers can be convinced with the right real-world evidence.
• Payers that focus on employer-sponsored health insurance tend to weigh quality-of-life outcomes more heavily than other commercial payers, because employers usually take employees' complaints about coverage quite seriously.
• Payers value patient-reported outcomes more highly when drug makers rely heavily on PROs to measure a therapy's efficacy.
• Pharma and medtech companies need to think about how to ensure patient access to the therapies they develop. That means translating PROs into metrics that demonstrate a therapy's value so that payers understand why covering it makes financial sense.
We've developed a five-step approach to improving coverage. Click here to learn more.
New health therapies that enhance the quality of life are tremendously important to patients. But when these therapies cost more than existing ones, payers are reluctant to cover them without evidence that they will also lower the overall cost of care. For patients as well as for the pharmaceutical and medtech companies that are developing these therapies, determining how to convince payers to expand access to these treatments is a pressing concern.
With this in mind, BCG conducted a survey and in-depth interviews to understand how US payers evaluate quality-of-life therapies to make coverage decisions. We found that the willingness to cover these therapies varies by payer type and condition, but payers can be convinced with the right real-world evidence. In this article, we distill our findings and present a five-step approach that pharma and medtech companies can use to demonstrate therapeutic value.
Payers' Views of Quality-of-Life Benefits
Researchers have used patient-reported outcomes (PROs) to assess improvements in patients' quality of life for decades. Over time, PROs increasingly became important for R&D, clinical trials, and US Food and Drug Administration (FDA) approvals. In cases where PROs were clinically validated—sometimes across hundreds of thousands of patients—the data was quite robust. Consequently, references to PRO data in clinical literature exploded from 2005 through 2020.
Five Lessons from the Pandemic for Health Care Systems
How Hospitals Can Work Together to Improve Value in Health Care
So why didn't this attention to PROs translate into payer coverage of quality-of-life therapies on a significant scale? PRO claims carry more weight with payers when they appear on an FDA label, but in 2010, the FDA published guidance for designing clinical trials that led to fewer PRO-based label claims receiving FDA approval, even as the number of references to PRO data in clinical literature increased. (See Exhibit 1.) As a result, few FDA labels currently carry PRO claims unless they treat a condition that is difficult or impossible to evaluate without PRO data. Examples of such drugs and the conditions they treat include Linzess (irritable bowel syndrome), Jakafi (myelofibrosis), Nucala (asthma), and Nucynta (severe acute pain).
A few therapies have benefited from including PRO data even if it pertains to how the drug improves the symptoms or consequences of the condition, rather than the condition. For example, Tremfya, a drug for psoriatic arthritis, uses PRO data to demonstrate that it increases patients' ability to perform daily activities, such as dressing, washing, eating, and walking. Having this PRO data on its FDA label has helped Tremfya stand out from the crowded market for autoimmune disease therapies.
Even so, providing quality-of-life benefits doesn't always translate into coverage and improved patient access because payers usually care most about clinical efficacy and cost. Payers typically evaluate efficacy using what they consider to be objective measures of endpoints specific to the disease, such as living longer and physiological improvements assessed through diagnostic tests. And because payers are focused on costs, to gain coverage, therapies that enhance patients' quality of life need to demonstrate improvements in the total cost of care, such as by reducing the number of hospital readmissions or by increasing patients' ability to perform tasks independently and avoid nursing care. (See Exhibit 2.)
Since PROs are not generally viewed as objective, many payers do not weigh them heavily in coverage decisions. Indeed, patients and pharma and medtech companies face a high level of skepticism about the value of PROs. "Patient-reported outcomes are very subjective, not very scientific, and rely on patients to tell you how they feel," said a former vice president of pharmacy at a payer. "Frankly, how patients feel doesn't translate into money for payers." That might be an unusually blunt statement, but it's not a unique sentiment.
The Differences Among Payers
It's important for pharma and medtech companies to understand that not all payers approach quality-of-life issues in the same way. (See Exhibit 3.) For example, on the basis of our interviews, payers that focus on employer-sponsored health insurance tend to weigh quality-of-life outcomes more heavily than other commercial payers; that's because employers usually take employees' complaints about coverage quite seriously. Other commercial payers will sometimes provide coverage for therapies that enhance patients' quality of life, but they are more likely to enforce utilization controls, such as higher co-pays, on patients.
Meanwhile, payers whose coverage is funded mainly by the government (such as Medicare and Medicaid) are very focused on cost-effectiveness. This is out of necessity. The government holds payers to traditional metrics that can make covering quality-of-life therapies difficult. Therefore, these payers often do not cover a therapy that costs more than the prior standard of care, despite the quality-of-life benefits.
Payers value PROs more highly when drug makers rely heavily on PROs to measure a therapy's efficacy.
How much payers weigh PROs and patients' quality of life also varies somewhat by the condition. For example, payers value PROs more highly when drug makers rely heavily on PROs to measure a therapy's efficacy. Such therapies include treatments for depression and migraines. Commercial payers—some of which did not cover mental health for all patients until the Affordable Care Act went into effect—consider most evidence in the mental health area very subjective, with PROs the least bad data. "Subjective data gets looked at more heavily when objective data is not possible," said a former deputy chief medical officer at a top-five payer. Payers implement controls, however. Step edits, which require that patients try inexpensive generic therapies first, are widely used.
Government payers, by comparison, have been covering mental health for longer. For example, Medicare ensures mental health coverage by identifying antidepressants and antipsychotic drugs as protected classes of drugs. The government recognizes that it will ultimately bear a greater total cost for patients with untreated mental health issues through disability coverage and other social-support benefits.
Besides these variations by payer types and therapeutic areas, there are also variations among individual decision makers. Some thought leaders in payer organizations are already combining their own clinical judgment with digital tools to weigh quality-of-life benefits more heavily in their coverage decisions. This suggests that appealing to a decision maker's clinical experience could help convert some skeptics.
A Five-Step Approach
Given this backdrop, pharma and medtech companies need to think carefully about how to ensure that patients have access to the therapies they develop. That means translating PROs into metrics that demonstrate a therapy's value so payers understand why covering it makes financial sense. Fortuitously, recent advances in remote digital monitoring as well as data and analytics allow pharma and medtech companies to measure and value their therapy's impact on patients' quality of life more precisely than ever. Innovators developing therapies that improve the quality of life should take the following five-step approach to demonstrating value to payers and increasing patient access.
Target specific stakeholders. The first step is to identify the decision makers who are most likely to appreciate quality-of-life benefits—and then make sure the data reaches them. These stakeholders typically include medical directors, physician associations that create treatment guidelines, and payer and hospital advisors who write treatment pathway protocols for their health care provider networks to follow.
Assess quality-of-life improvements. Early in the R&D process, evaluate how new treatments affect patients' quality of life and that of their caregivers. When possible, use digital tracking and analytics or longitudinal trend analysis, or both, to gather detailed evidence and improve the perceived objectivity of the results. Identify and describe the specific steps that patients and caregivers should take when using the new therapy to maximize quality-of-life outcomes.
Gather quality-of-life outcomes. When conducting clinical trials, use well-established survey tools that are designed to collect PROs and measure quality-of-life outcomes. Such tools will increase the probability that payers will recognize the results. Be sure to comply with the FDA's PRO guidance when designing clinical trials so that PRO-based claims have a greater chance of inclusion on the FDA-approved label. Include specific PROs as primary endpoints for clinical trials to increase payer acceptance.
Translate outcomes into metrics to demonstrate value. To build broad support for a new therapy, translate quality-of-life outcomes into metrics that matter to various stakeholders—payers, providers, and patients.
To build credibility with payers, show how improving a patient's quality of life reduces the cost of care.
To build credibility with payers, in particular, show how improving a patient's quality of life reduces the total cost of care, and then validate the evidence with industry players. For example, run pilot programs with payers to demonstrate that quality-of-life improvements are associated with higher employee productivity and lower absenteeism—results that employers care about deeply. "You need to show that [improving] quality of life translates into a dollar savings or better efficacy or safety outcomes," said a pharmacy director at a top-five payer.
Additionally, it is critical to work with clinical organizations to ensure that findings are incorporated into updated care guidelines. Payers cited peer-reviewed clinical data (74%) and treatment guidelines (87%) as far and away the most important sources of information when evaluating the clinical value of new therapies.
Build a quality-of-life plan into the long-term strategy. Quality-of-life improvements will become more important to payers as value-based health care gains traction over the next decade. Value-based health care rewards providers for helping make patients' lives healthier. It is a marked shift from the fee for service (FFS) model, in which payers reimburse providers for each health care service they render, rewarding them for quantity rather than quality. With this trend in mind, innovators will want to use quality-of-life results to inform their own long-term R&D and reimbursement strategies.
Although the shift from FFS to value-based payments is happening more slowly than many people originally expected, momentum is building. The Centers for Medicare and Medicaid Services is easing best-price regulations to incentivize value-based arrangements; many state governments are pushing for reimbursement models with a value-based component; some big commercial players are investing in value-based capabilities at scale, such as buying physician groups; and payers are increasingly distributing bonuses on the basis of quality scores (for which patient satisfaction is a major contributing factor). "Value-based arrangements will become a major way for innovators to leverage these quality-of-life outcomes," said the former deputy chief medical officer at a top-five payer.
Gaining Greater Traction
While there's been progress across the health care value chain in considering quality-of-life issues, quality-of-life outcomes are still valued less than other health care results that are perceived as more objective. Innovators can accelerate progress and expand access to treatments by quantifying improvements—using PROs, digital tracking, and data analysis—and demonstrating value to payers, providers, and patients.
Longer term, pharma and medtech companies that focus on improving patients' quality of life, in addition to measuring outcomes, will benefit disproportionately as the industry continues to shift toward value-based care. Pharma and medtech companies can look forward to delivering not only immense benefits to patients but also very healthy investment returns to shareholders.
Meghna Eichelberger
Partner & Associate Director
Hannah Hagen
Meg Kedrowski
Nicholas Kransz
Knowledge Expert, Team Manager
ACC – Boston
Suchita Shah
Natasha Taylor
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