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To be published in "Garden of Quanta"
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QED, effective mass, enhanced binding, ground state, dipole approximation, stability of matter
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\title
{Nonrelativistic QED  at  large  momentum of photons} 
\author{Fumio Hiroshima\footnote{Department of Mathematics and Physics, 
Setsunan University,  Osaka 572-8508, Japan. \hspace{5cm} 
 email: hiroshima@mpg.setsunan.ac.jp}   \\
{\small
Dedicated to  Professor  Hiroshi  Ezawa  
on the occasion of his seventieth
birthday}}
\date {}

\begin{document}

\maketitle

\begin{abstract}
An interacting  system 
of one spinless electron 
minimally coupled to a quantized radiation field is  considered. 
On the Hamiltonian, $\hl$, of this system,  
an ultraviolet cutoff $\Lambda>0$ 
and an infrared cutoff $\lambda>0$ are 
imposed. 
We do {\it not} assume the existence of a ground state 
of the decoupled Hamiltonian. 
It is proven, however,  that 
there exists $\Lambda_\ast$ such that 
for all $\Lambda>\Lambda_\ast$, 
 $\hl$ has a unique ground state. 
Furthermore we investigate the asymptotic behavior 
of $g(\Lambda)=\is (\hl)$ 
without external potentials as $\Lambda\rightarrow \infty$.
\end{abstract}
%\newpage
%\tableofcontents
%\newpage



\setlength{\baselineskip}{18pt}
\section{Introduction}
It is a pleasure to dedicate this article to 
Professor Hiroshi Ezawa on the occasion of his seventieth 
birthday. 
He was a source of encouragement in mathematical physics. 
Especially he suggested physical inspirations  to myself 
in seminars in the Gakushuin University, 
in ICMP '97 held in Australia, 
and in Les Houches '98 ``Ecole de Physique des Houches'' 
in France. 
He also invited me as one of lecturers in the summer school of mathematical physics 2000 organized by H. Araki and himself. 
I wish to express my deep thanks to  him 
for stimulating my works and for innumerable,   
unbounded and 
helpful  conversations. 







The aim of this paper is to consider a system of one spinless 
electron minimally coupled to a quantized 
radiation field. 
It is  the so-called  Pauli-Fierz model \cite{pafi} 
 in nonrelativistic quantum electrodynamics. 
The Pauli-Fierz Hamiltonian is defined as a self-adjoint operator acting in the tensor product of $\LR$ and a boson Fock space $\fff$, 
$\LR\otimes \fff$, 
which is given by 
$$\lk \mpp+V\rk\otimes 1+1\otimes \hf$$ 
with $p\otimes 1=-i\nabla_x\otimes 1$ 
replaced by $p\otimes 1
+\A(x)$, i.e., 
$$\frac{1}{2m}(p\otimes 1+\A(x))^2+V\otimes 1+1\otimes \hf,$$ 
where $V$ denotes an external potential, 
$\hf$  the free Hamiltonian in a Fock space and $\A(x)$ 
a quantized radiation field with a ultraviolet cutoff 
$\Lambda>0$. 
Details will be seen later. 
Under some suitable conditions, 
it is established that a ground state of the Pauli-Fierz Hamiltonian 
exists if $\mpp+V$ 
has a  ground state. 
See  \cite{bfs3, fr1, fr2, gll,  hi8, hi18}.  

In this  paper 
we take 
the dipole approximation, i.e., 
we replace as 
$$\A(x)\longrightarrow 1\otimes \A(0),$$
and 
assume that on the  quantized radiation field   
an infrared cutoff is  imposed. 
Then the Hamiltonian under consideration is 
$$\hl=\frac{1}{2m}(p\otimes 1+1\otimes \A(0))^2+V\otimes 1+1\otimes \hf.$$
It is  proven in \cite{hi3} that 
$\hl$ also has a ground state even when no infrared cutoff is imposed, 
if $\mpp+V$ has a ground state. 

A more difficult, but physically reasonable, question is about enhanced binding. 
One may expect that a coupling with a quantized radiation field enhances binding. 
In this paper, 
we  suppose  that the electron is governed by a sufficiently 
shallow nonpositive external potential $V$, 
 and then $\mpp+V$ 
has {\it no ground states}. 
We prove, however,  that $\hl$  has a ground state $\gr$ for 
sufficiently large~$\Lambda$. 
The physical reasoning behind such a result is simple. 
On a  heuristic level one argues that 
a coupling with  a  quantized radiation field 
amounts to renormalizing   bare mass $m$ to  effective  mass: 
$$ 
m\rightarrow \mass=m+\delta m(\Lambda)
$$ 


Since effective mass $m(\Lambda)$ is increasing in $\Lambda$, 
a bound state could 
be produced for a sufficiently large 
$\Lambda$. 
Roughly speaking, 
 $\hl$ may be replaced by 
\eq{sim}
\hl\sim \eff=\lk \frac{\p^2}{2\mass}+V\rk\otimes 1+1\otimes \hf, 
\en 
which binds for sufficiently large 
$\Lambda$.
Needless to say \kak{sim} has 
no  mathematical meaning, we show, 
however, the associated phenomena in this paper.�@
\begin{figure}[ht]

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\begin{picture}(10,9)(-1,5)

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\put(5.7, 8.8){{\footnotesize $m$}}
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\put(6, 10){\vector(-1,0){3.5}}



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\put(18.3, 8){{\footnotesize  $m+\delta m(\Lambda)$}}

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%\linethickness{1mm}
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\caption{Effective mass}
\end{figure}




This kind of investigation has been done in Hiroshima and Spohn \cite{hisp}, 
and 
developed by 
e.g., \cite{ak,  ch, cvv, hvv}. 
Arai and Kawano  \cite{ak} proved the similar result as ours in a general framework. 
Chen, Vougalter and Vugalter \cite{cvv}, Catto and Hainzl \cite{ch}, 
Hainzl, Vougalter and Vugalter \cite{hvv} study the case of 
no dipole approximation under some conditions. 

 
This  paper is organized as follows. 
In Section 2 we define the Pauli-Fierz Hamiltonian and its 
self-adjointness 
is stated. 
In Section 3 we diagonalize the Pauli-Fierz Hamiltonian. 
In Section 4 we prove the main theorem.
Section 5 is devoted to showing 
the multiplicity and localization of a ground state. 
In Section 6 we investigate 
the asymptotic behavior  of the ground state energy of $\hl$ 
with no external potential. 



\section{Nonrelativistic QED}
\subsection{The Pauli-Fierz Hamiltonian}
The boson Fock space $\fff$ over $L^2(\BR\times\{1,2\})$ 
is defined by 
$$\fff=\fff(L^2(\BR\times\{1,2\}))=
\bigoplus_{n=0}^\infty \fff^{(n)}(L^2(\BR\times\{1,2\})),$$ 
where 
$$\fff^{(n)}(L^2(\BR\times\{1,2\}))=
\otimes_s^n L^2(\BR\times\{1,2\})$$ 
denotes the $n$-fold symmetric tensor product of 
$L^2(\BR\times\{1,2\})$ with 
$$\otimes_s^0 L^2(\BR\times\{1,2\})
=\CC. $$ 
The Fock vacuum is defined by 
$$\Omega=\{1,0,0,...\}\in\fff.$$
$\{a^\ast(k,j), a(k,j)\}_{j=1,2}$ 
are bose fields with commutation relations
$$
[\add(k,j),a(k',j')]=\delta_{jj'}\delta(k-k'),
$$ 
$$
[\add(k,j),\add (k',j')]=0,
$$ 
$$
[a(k,j),a(k',j')]=0.
$$ 
The quantized radiation field,  
$\A(x), x\in\BR$,  is defined by 
$$\A(x)=\sum_{j=1,2}\int   \frac{\phk }{\sqrt{2\omega(k)}}\ejk
\lkk \add(k,j)e^{-ik\cdot x}+ a(k,j)e^{ik\cdot x}\rkk d^3k, $$
where $\ph$ denotes the momentum cutoff given by 
$$\phk=\lkk\begin{array}{ll}0,&|k|< \lambda,\\
1,&\lambda\leq |k|\leq \Lambda,\\
0,&|k|> \Lambda.
\end{array}
\right.$$
We fix an infrared cutoff, $\lambda>0$,  and assume that 
$$\lambda\leq \Lambda.$$
$e_j(k)=(e_j^1(k), e_j^2(k), e_j^3(k)), j=1,2,$ are polarization vectors, i.e., 
 $$\{e_1(k),  e_2(k),  k/|k|\}$$
 forms a right-handed system, i.e.,
$$e_j\cdot k=0, \ \  e_i\cdot e_j=\delta_{ij},\ \ e_1\times e_2=k/|k|.$$
The dispersion relation is given by 
$$ 
\omega(k)=|k| 
$$ 
and therefore the free Hamiltonian of $\fff$ is defined by 
$$\hf=\sij \int \omega(k)\add(k,j)a(k,j) d^3k.$$
Let us denote the spectrum (resp.  discrete spectrum,  essential spectrum) of 
$T$ by 
$\sigma(T)$ (resp. $\sigma_{\rm disc}(T)$, $\sigma_{\rm ess}(T)$).
It is well known that 
$$\sigma(\hf)=[0,\infty),\ \ \ \sigma_{\rm disc}(\hf)=\{0\},$$
and 
$$\hf \Omega=0.$$
The  Hilbert space of our system is 
$$\hhh=\LR\otimes \fff.$$
The decoupled Hamiltonian is given by 
$$H_0=\lk\mpp+V\rk\otimes 1+1\otimes \hf.$$
Here $m$ denotes the bare mass of an electron, 
$\p=-i\nabla_x$ the momentum operator canonically conjugate to the position 
operator $\x$ in $\LR$ and 
$V$  an external potential. 
Thus 
$$
\sigma(H_0)=\{\lambda_1+\lambda_2|\lambda_1\in \sigma(\mpp+V),\ 
\lambda_2\in\sigma(\hf)\}.$$
We take the dipole approximation, i.e., 
we replace as 
$$\A(x)\longrightarrow 1\otimes \A(0).$$ 
Set $$\A=\A(0).$$
The total Hamiltonian, $\hl$, 
 is defined by $H_0$ with the replacement 
$$p\otimes 1\longrightarrow p\otimes 1+1\otimes \A,$$
i.e., 
$$ 
\hl =\frac{1}{2m}\left(\p\otimes 1+  1 \otimes \A\right)^2+V\otimes 1+1\otimes \hf.
$$ 
For notational convenience we mostly omit 
the tensor notation $\otimes$ unless confusions may arise, e.g., 
we simply write as $p$, $V$, $\A$, $\hf$, 
 instead of $p\otimes 1$, $V\otimes 1$, $1\otimes \A$
 and $1\otimes \hf$, respectively. 
Then  $\hl$ is simply written as 
$$\hl=\frac{1}{2m}(p+\A)^2+V+\hf.$$
\bp{111}
Suppose that 
$D(p^2)\subset D(V)$ and 
$$\|Vf\|\leq a\|\mpp  f\|+b\|f\|$$
with some $0\leq a<1$ and  $b\geq 0$. 
Then 
$\hl$ is self-adjoint on $D(p^2) \cap D(\hf)$ and 
bounded from below. Moreover it is essentially self-adjoint on any core of
$\mpp +\hf$.
\ep
\proof See \cite{hi11,hi17}.
\qed
\begin{remark}
The Hamiltonian without the dipole approximation, i.e.,
$$\frac{1}{2m}(p+\A(x))^2+V+\hf,$$
is also self-adjoint on $D(p^2)\cap D(\hf)$.
See \cite{hi11,hi17}.
\end{remark}

\subsection{Enhanced binding and the main theorem}
In this paper we consider the case where the external potential $V$ 
is sufficiently shallow. 
By the Lieb-Thirring inequality \cite{lith},  
$$\#\{\mbox{bound states of } \mpp+V\}\leq 
L_{0,3} \int  |mV_-(x)|^{3/2} d^3x  $$
$$(V_- :\ \mbox{the negative part of } V)$$
with some constant $L_{0,3}$ independent of $V$, 
we see that for a sufficiently shallow nonpositive potential $V$, 
$\mpp+V$ has no bound states. In particular it has no ground state. 
Thus the decoupled Hamiltonian $\mpp+V+\hf$ also has no ground state. 
Our contribution  is to show, however, 
 that 
$\hl=\frac{1}{2m}(p+\A)^2+V+\hf$ has 
a ground state for sufficiently large $\Lambda$. 

Throughout this paper we assume Hypothesis~(V). 

\noindent
{\bf Hypothesis (V)} \vspace{-1cm}\\
{\it \bi
\item[(a)] $ V\in\ccc.$ 
\item[(b)] $V\leq 0.$
\item[(c)] There exists $\mu_0>1$ and $r>0$ such that 
for $\mu>\mu_0$,  
$$ \is (\mpp+\mu V)<-r.$$
\ei
}
Since $V$ is relatively compact with respect to $\mpp$, 
it holds that  
$$\sigma_{\rm ess}(\mpp+\mu V)=[0,\infty).$$
Hence $\mpp+\mu V$, $\mu>\mu_0$,  has a ground state. 
\begin{remark}
\label{re}
We do not assume the existence of ground states of $\mpp+V$. 
\end{remark}
The main result of this paper is as follows. 
\bt{main0}
There exists $\Lambda_\ast$ such that 
for $\Lambda>\Lambda_\ast$, $\hl$ has a 
ground state. 
\et


\section{Canonical transformations}
The following proposition is  a key in this paper. 
\begin{proposition}
There  exists a unitary operator $U$ such that 
$U$ maps $D(\p^2)\cap D(\hf)$ onto itself and 
$$U\f \hl U=\frac{1}{2\mass} p^2+ V(\cdot-K/\mass ) +\hf + 
\frac{1}{\pi} \int_{-\infty}^\infty 
\frac{t^2 \|\la/(t^2+\omega^2)\|^2}
{m+  \mmm\|\la/\sqrt{t^2+\omega^2}\|^2}dt,$$ 
where 
$$\mass=m+\frac{8\pi}{3}(\Lambda-\lambda),$$ 
\eq{KK}
K=\jjj   \int \frac{\ph(k)e_j(k)}{\sqrt{2\omega(k)}} 
\lkk \frac{\add(k,j) }{\omega(k) D_-(\omega(k)^2)}
 + \frac{a(k,j)}{\omega(k) D_+(\omega(k)^2)} \rkk \ddk, 
\en 
and 
$$D_{\pm}(s)=m-\frac{4\pi}{3}\lk \lim_{\epsilon\downarrow 0}\int_{|s-x|<\epsilon}
\frac{\sqrt x \ph(\sqrt x)}{s-x} dx\mp 2\pi i \ph(\sqrt s)\sqrt s\rk\not=0.$$
\ep
\proof 
This is proven by a slight  modification of \cite[Appendix]{hisp}. 
\qed
\begin{remark}
In \cite{hisp} it is obtained that 
$$\mass=m+\mmm \|\la/{\omega}\|^2.$$
\end{remark}
\begin{remark}
We see that 
$D_\pm(s)=m$ for $0\leq s <\lambda^2 $ or $\Lambda^2<s$, 
and $\Im D_\pm(s)=2\pi \sqrt s$ for $\lambda^2\leq s\leq \Lambda^2$. 
In particular $D_\pm(s)\not=0$ for $s\geq 0$. 
\end{remark}
We set 
$$\Kp(\cdot,j) =(\Kp_1(\cdot,j),  \Kp_2(\cdot,j), \Kp_3(\cdot,j)),$$ 
where 
$$\Kp_\mu (k, j) =\frac{e_j ^\mu (k) \ph(k)}{\omega^{3/2}(k) D_+(\omega(k)^2)}, \ \ \ \mu=1,2,3.$$
Then we rewrite \kak{KK}, $K=(K_1,K_2,K_3)$,  as 
$$K_\mu =\jjj\sh\int\lkk \Kp_\mu(k,j) \add(k,j)+ \Kp_\mu(k,j)^\ast(k)a(k,j)\rkk\ddk$$
and 
$\Kp_\mu(\cdot,j)$ satisfies that  
\eq{w}
\|\omega^{n/2}\Kp_\mu(\cdot,j) \|\leq C_1
\|\omega^{{(n-3)}/{2}}\la\|
\en 
with some constant $C_1$. 
Let 
$$\phi(f)=\jjj\int \lkk f(k) \add(k,j)+f^\ast(k) a(k,j)\rkk \ddk.$$
Then the following inequality is well known for $\Psi\in D(\hf^\han)$, 
$$\|\phi(f)\Psi\|\leq (\|f/\omega^\han\|+\|f\|)(2\|\hf^\han\Psi\|+\|\Psi\|).$$
From \kak{w} 
it follows that 
\eq{aef}
\|\aef_\mu \Psi\|\leq
C_2
(\|\la/\omega^2\| +\|\la/\omega^{3/2}\|)
(\|\hf^\han\Psi\|+\|\Psi\|)
\en
with some constant 
$C_2$. 
We set 
$$\dv=V(\cdot-K/\mass )-V,$$
$$\eff=\frac{1}{2\mass}p^2+V,$$
$$g(\Lambda) =
\frac{1}{\pi} \int_{-\infty}^\infty 
\frac{t^2 \|\la/(t^2+\omega^2)\|^2}
{m+  \mmm\|\la/\sqrt{t^2+\omega^2}\|^2}dt,$$ 
and 
\eq{34}
\th=U\f \hl U=\eff+\dv +\hf+g(\Lambda).
\en 
We investigate a spectral property of $\th$ instead of $\hl$ in what follows.
\bl{eff}
Let  $\Lambda$ be such that 
$$\Lambda>\frac{3}{8\pi}(\mu_0-1)m.$$
Then $\eff$ has a ground state, and moreover 
\eq{lo}
\#\{\mbox{bound states of } \eff\}\leq L_{0,3} \lk m+\frac{8}{3}\pi(\Lambda-\lambda)\rk^{3/2}\int 
|V(x)|^{3/2} d^3x.
\en 
\el
\proof
By Hypothesis (V),  
$$\eff = \frac{1}{2\mass}p^2+V=\frac{m}{\mass}\lk\mpp+\frac{\mass}{m} V\rk$$
implies that 
if 
\eq{exp}
\frac{\mass}{m}>\mu_0,
\en 
then $\eff$ has a ground state. 
\kak{exp} is identical with 
$\Lambda>\frac{3}{8\pi}(\mu_0-1)m.$ 
\kak{lo} follows from the Lieb-Thirring inequality. 
Then the lemma follows.
\qed

\section{Existence of ground states} 
\subsection{Lattice approximation} 
We introduce an artificial mass of photon, $\nu>0$, 
 and 
define 
$$\thn=\eff+\dvn+\hfn+g(\Lambda),$$ 
where 
$\dvn$ and  $\hfn$ are defined by 
$\dv$ and  $\hf$ with $\omega$ replaced by 
$\omega+\nu$, respectively. 
Using 
a  momentum lattice approximation we will prove that $\thn$ 
has 
a ground state. 
Let $\Gamma(l,a)$, $l=(l_1,l_2,l_3)\in{\Bbb Z}^3$, $a>0$,  
be the momentum  lattice with spacing $1/a$, i.e.,   
$$\Gamma(l,a)=
[\frac{l_1}{a}, \frac{(l_1+1)}{a})
\times 
[\frac{l_2}{a}, \frac{(l_2+1)}{a}) 
\times 
[\frac{l_3}{a}, \frac{(l_3+1)}{a}),$$ and 
$$\PH(k) =\lkk\begin{array}{ll}0,&k\not\in \Gamma(l,a),\\
a^{3/2} ,&k\in\Gamma(l,a).
\end{array}\right.$$
For $L>0$ we define the momentum-lattice-approximated Hamiltonian
by 
$$\thn\K=\eff+\hfn\K+\dvn\K+g(\Lambda),$$
where 
$$\hfn\K=\jjj\int  \LLL\PH(k) (\omega(l)+\nu)
 \add(k,j)a(k,j)\ddk,$$
$$\dvn\K=V(\cdot- K^\nu\K/\mass)$$
and $K^\nu\K=(K_1^\nu\K, K_2^\nu\K, K_3^\nu\K)$ with 
$$K_\mu^\nu\K=\jjj\frac{1}{\sqrt 2} 
 \int \LLL\PH(k)(\Kp_\mu^\nu(l,j)\add(k,j)+
\Kp_\mu^\nu(l,j)^\ast a(k,j)) \ddk.$$
In the similar way as  \kak{aef} we can show that 
$$\|K_\mu^\nu\K \Psi\|\leq C_2'
\lk  
\left\|\frac{\LLL\PH \ph(l)}{\sqrt{\omega}(\LLL\PH\omega(l))^{3/2}}\right\|
+\left\| \frac{\LLL\PH \ph(l)}
{(\LLL\PH\omega(l))^{3/2}}\right\|\rk$$
\eq{aef2}
\times 
(\|\hfn\K^\han\Psi\|+\|\Psi\|)
\en 
with some constant $C_2'$. 
Here we used 
$$c_1\|\hfn\K^\han\Psi\|\leq \|{\hfn}^\han  \Psi\|\leq c_2\|\hfn\K^\han\Psi\|$$
with some constants $c_1$ and $c_2$. 
\bl{l7}
We have 
$$\lim_{L\rightarrow \infty}\lim_{a\rightarrow \infty} \thn\K=\thn$$
in the operator norm in the sense of resolvent.
\el
\proof
It is well known that 
there exists $\epsilon(a,L)$ such that 
$$\|(\hfn\K-\hfn)\Psi\|\leq \epsilon(a,L)\|\hfn\Psi\|$$
and 
$$\limla \epsilon(a,L)=0.$$
Moreover
$$\|(\dvn\K-\dvn)\Psi\|=\|(V(\cdot-K^\nu\K/\mass)-V(\cdot-K^\nu/\massn))\Psi\|$$
$$\leq \sum_{\mu=1}^3 \frac{\|\nabla_\mu V\|_\infty}{\mass}
\|(K_\mu^\nu-K_\mu^\nu\K)\Psi\|.$$ 
Since 
 $$\|(K_\mu^\nu-K_\mu^\nu\K)\Psi\|
\leq 
\jjj C 
\lk \left\|\frac{1}{\sqrt\omega} 
\lkk 
\frac{
\Kp_\mu^\nu(\cdot,j)}{\omega^{3/2}}-\frac{\LLL \PH \Kp_\mu^\nu(l,j)}{
(\LLL\PH\omega(l))^{3/2}}
\rkk \right\|\right.$$
$$\left.+
\left\|\frac{\Kp_\mu^\nu(\cdot,j)}{\omega^{3/2}}-\frac{
\LLL\PH \Kp_\mu^\nu(l,j)}{
(\LLL\PH\omega(l))^{3/2}}  \right \|\rk
(\|\hfn^\han\Psi\|+\|\Psi\|)
$$
$$=\epsilon(a,L)' (\|\hfn^\han\Psi\|+\|\Psi\|)$$
with some constant $C$,  and 
$\epsilon(a,L)'$ satisfies that 
$$\limla \epsilon(a,L)'=0,$$
we have 
$$\|(\thn-z)\f\Psi-(\thn\K-z)\f\Psi\|$$
$$\leq \|(\thn\K-z)\f\| 
\|(\thn\K- \thn)(\thn-z)\f  \Psi\|$$
$$\leq \frac{\mv  \epsilon(a,L)'}{|\Im z|\mass } 
\lk 
 \|\hfn^\han (\thn-z)\f \Psi\|+\| (\thn-z)\f \Psi\|\rk$$
$$+\frac{\epsilon(a,L)}{|\Im z|} 
 \|\hfn^\han(\thn-z)\f \Psi\|.$$
Since 
$$\|\hfn (\thn-z)\f \Psi\|\leq C'\|\Psi\|$$
with some constant $C'$, we have 
$$\|(\thn-z)\f\Psi-(\thn\K-z)\f\Psi\|\leq \epsilon''(a,L)\|\Psi\|$$
with $\epsilon''(a,L)$ such that 
$$\lim_{L\rightarrow \infty}
\lim_{a\rightarrow \infty} \epsilon''(a,L)=0.$$
Hence the lemma follows. 
\qed




\subsection{Artificial  mass} 
Let $f\in\LR$. We identify 
$$\{f(l)\}_{l\in{\Bbb Z}^3}\in\ell^2({\Bbb Z}^3)$$ 
with 
$$ \sum_{l\in{\Bbb Z}^3}f(l) \PH(\cdot)\in \LR. $$ 
By this identification we regard $\ell^2({\Bbb Z}^3)$ as 
$$\ell^2=\ell^2({\Bbb Z}^3)\subset \LR.$$
Let 
$$\hhh_a=\LR\otimes \fff(\ell^2\otimes\CC^2).$$
By the following fundamental identification
$$
\fff(L^2(\BR\times\{1,2\}))
\cong 
\fff(\LR\otimes\CC^2)
=
\fff([\ell^2\oplus \ell^{2\perp}]\otimes\CC^2)$$
$$
\cong \fff(\ell^2\otimes\CC^2)\otimes \fff( \ell^{2\perp}\otimes\CC^2)
\cong 
\fff( \ell^2 \otimes\CC^2)
\otimes 
\left[
\bigoplus_{n=0}^\infty \fff^{(n)} 
(\ell^{2\perp}\otimes\CC^2) 
\right]
$$
$$=
\fff( \ell^2 \otimes\CC^2)
\otimes 
\left[
\bigoplus_{n=1}^\infty \fff^{(n)} 
(\ell^{2\perp}\otimes\CC^2) \oplus \CC
\right]
$$
$$
\cong 
\lk 
\fff(\ell^2\otimes\CC^2) \otimes 
\left[ \bigoplus_{n=1}^\infty \fff^{(n)} 
(\ell^{2\perp}\otimes\CC^2) 
\right]
\rk
\oplus
\fff( \ell^2 \otimes\CC^2) 
, 
$$
 we have 
$$\hhh\cong \lk \hhh_a\otimes \left[
\bigoplus_{n=1}^\infty \fff^{(n)} 
(\ell^{2\perp}\otimes\CC^2) 
\right] \rk \oplus \hhh_a.$$
In particular 
we see that 
\eq{l3}
\hhh_a^\perp\cong \hhh_a\otimes 
\left[
\bigoplus_{n=1}^\infty \fff^{(n)} 
(\ell^{2\perp}\otimes\CC^2) 
\right].
\en 
The following  lemma is easily proven. 
\bl{l1}
$\thn\K$ is reduced by $\hhh_a$ and,  
under identification \kak{l3},  we have 
$$\left.\thn\K\right\lceil_{\hhh_a^\perp}
\cong \lk \left.\thn\K\right\lceil_{\hhh_a}\rk\otimes 1+1\otimes
\lc{\hfn\K}{\left[
\oplus_{n=1}^\infty \fff^{(n)} 
(\ell^{2\perp}\otimes\CC^2) 
\right]
}.$$
In particular 
$$\left.\is (\thn\K\right\lceil_{\hhh_a^\perp})\geq 
\is(\hfn\K)+\nu.$$
\el
Note that $\lc{\hfn\K}{\hhh_a}$ 
has a purely discrete spectrum, i.e., 
$$\sigma(\lc{\hfn\K}{\hhh_a})=\sigma_{\rm disc}(\lc{\hfn\K}{\hhh_a}).$$





\begin{lemma}
\label{42}
Let $\Psi\in D(\p^2)\cap D(\hf^{\han})$. 
Then 
\bi
\item[(1)] $\Psi\in D(\dvn)$ and 
\eq{bb1}
\|\dvn\Psi\|\leq \C
(\|\hfn^{1/2}\Psi\|+\|\Psi\|),
\en 
where 
$$\C=\frac{C\|\nabla V\|_\infty}
{\mass}(\|\ph/\omega^2\|+\|\ph/\omega^{3/2}\|)$$
with some constant $C$.
\item[(2)]
 $\Psi\in D(\dvn\K)$ and 
\begin{equation}
\label{b1}
\|\dvn\K \Psi\|\leq \CA 
(\|\hfn\K^{1/2}\Psi\|+\|\Psi\|), 
\end{equation}
where 
$$\hspace{-5cm}
\CA= \frac{C' \|\nabla V\|_\infty}{\massn} 
$$
$$\times 
\lk 
\left \|\frac{1}{\sqrt\omega}\lkk 
\frac{\LLL\PH \la(l)}{(\LLL\PH\omega(l))^{3/2}}
\rkk
\right\| 
+
\left \|
\frac{\LLL\PH \la(l)}{
(\LLL\PH\omega(l))^{3/2}}
\right\|\rk 
$$
with some constant $C'$. 
\ei
\end{lemma}
\proof
We have 
$$\|\dvn\Psi\|\leq \sum_{\mu=1}^3\frac{\|\nabla_\mu V\|_\infty}{\mass} 
 \|K_\mu \Psi\|.$$
\kak{bb1} follows from \kak{aef}. 
\kak{b1} is similarly proven. 
\qed

\begin{remark}
$\C$ and $\CA$ are independent of $\nu$. 
\end{remark}
We write 
$A\leq B$, if $D(B)\subset D(A)$ and 
$(\psi, A\psi)\leq(\psi, B\psi)$ for 
$\psi\in D(B)$. 

We have 
$$|(\Psi, \dvn \Psi)|\leq 
\C
\lkk \|\Psi\| ( \|\hfn^\han\Psi\|+\|\Psi\| )\rkk $$
$$\leq 
\C\lk \|\Psi\|^2+\half \|\Psi\|^2+\half \|\hfn^\han \Psi\|^2\rk 
=(\Psi,\C(\mmmm+\half \hfn)\Psi).$$
Thus 
we  conclude 
\eq{23}
-\C \left(\frac{1}{2}\hfn+\frac{3}{2}\right)
\leq \dvn
\leq 
\C \left(
\frac{1 }{2}\hfn+\frac{3}{2}\right).
\en
Hence for $f\in\ccc$, 
$$\is (\thn)\leq (f\otimes \Omega, \thn f\otimes \Omega)\leq 
(f,(\eff+g(\Lambda)+\mmmm\C) f).$$ 
In particular, since $\ccc$ is a core of $\eff$, 
 we have
$$ 
\is(\thn)\leq \is(\eff) + \frac{3\C}{2}+ g(\Lambda).
$$ 
Similarly we have 
\eq{25}
-\CA \left(\frac{1}{2}\hfn\K+\frac{3}{2}\right)
\leq \dvn\K
\leq 
\CA \left(
\frac{1 }{2}\hfn\K +\frac{3}{2}\right)
\en
and hence 
$$ 
\is(\thn\K)\leq \is(\eff) + \frac{3\CA}{2}+ \ggg. 
$$ 
We summarize above statements in the following lemma.
\bl{ground}
We have 
$$ 
\is(\thn)\leq \is(\eff) + \frac{3\C}{2}+ g(\Lambda)
$$ 
and 
$$
\is(\thn\K)\leq \is(\eff) + \frac{3\CA}{2}+ g(\Lambda).$$
\el
We  see that as $\Lambda\rightarrow \infty$ 
$$
\|\ph/\omega^l\|\sim \Lambda^{-l+\frac{3}{2}},\ \ \ 
 l\not=\frac{3}{2} , $$
$$
  \|\ph/\omega^l\|\sim \sqrt{\log \Lambda},\ \ \ 
 l=\frac{3}{2}.
$$ 
From this it follows that as $\Lambda\rightarrow\infty$ 
\eq{r1}
\mass=m+\frac{8\pi}{3}(\Lambda-\lambda)  \sim \Lambda,
\en 
\eq{r2}
\|\ph/\omega^2\|\sim \Lambda^{-\han},
\en 
\eq{r3}
\|\ph/\omega^{3/2}\|\sim \sqrt{\log\Lambda}.
\en 
\bl{LLL}
It follows that 
$$ 
\lim_{\Lambda\rightarrow\infty}\C=0
$$ 
and 
$$
\lim_{\Lambda\rightarrow\infty}\CA=0.
$$ 
\el
\proof
These are follows from \kak{r1}, \kak{r2}, \kak{r3} 
and the definitions of $\C$ and $\CA$.
\qed

We set 
$$\Sigma=\Sigma(\Lambda)=\is(\eff).$$
It is easily seen that 
\eq{E}
\lim_{\Lambda\rightarrow\infty} \Sigma =\inf_x V(x)<0.
\en 
For a self-adjoint operator $M$, 
the spectral projection of $M$ 
on a Borel set ${\cal B}\subset \RR$ is denoted by 
$$E^M_{\cal B}.$$ 
\begin{lemma}
\label{gr}
Let $a$,  $L$ and $\Lambda$  be sufficiently large such that 
\eq{I}
\JJ . 
\en
Then for $\nu$ such that $|\Sigma|>3\CA+\nu$, 
$$\thn\K\lceil_{\hhh_a} -\is( \thn\K\lceil_{\hhh_a} )-\nu
$$
$$\geq \X\otimes\lk(1-\frac{\CA}{2})\hfn\K-3\CA-\nu\rk.$$
\end{lemma}
\proof 
In this proof we set 
$$H=\thn\K\lceil_{\hhh_a}$$
as an operator  in $\hhh_a$
and 
$$\T=\CA.$$
We directly see that 
$$H-\is(H)-\nu$$
$$= \eff+\dvn\K+\hfn\K+\ggg -\is(H)-\nu$$
$$\geq \eff+\dvn\K+\hfn\K+\ggg -\ggg -\mmmm\T -\Sigma -\nu$$
$$\geq \eff+(1-\frac{\T}{2}) \hfn\K -\mmmm\T-\mmmm\T-\Sigma -\nu$$
$$= (\eff-\Sigma) +(1-\frac{\T}{2}) \hfn\K -3\T -\nu$$
$$\geq |\Sigma|\Y\otimes 1 
 -\T'(\X+\Y)\otimes 1$$
$$  + (1-\frac{\T}{2})(\X+\Y)\otimes  \hfn\K,$$
where $\T'=3\T +\nu$. 
Then 
$$H-\is(H)-\nu 
\geq (|\Sigma|-\T') \Y \otimes 1
+ (1-\frac{\T}{2})\Y \otimes  \hfn\K 
$$
$$+\X\otimes \lk (1-\frac{\T}{2}) \hfn\K-\T'\rk.$$
Since by \kak{I}, 
$$|\Sigma|-\T'=|\Sigma|-3\theta-\nu>0$$
and 
$$1-\frac{\T}{2}>0,$$
we have 
$$
H-\is(H)-\nu \geq 
\X\otimes \lk (1-\frac{\T}{2}) \hfn\K-\T'\rk.$$
Thus the lemma follows.
\qed

Set 
$$T={\thn\K} -\is({\thn\K})-\nu.$$
Define 
$$\hhh_a(+)=E^T_{[0,\infty)}\hhh_a,$$
$$\hhh_a(-)=E^T_{[-\nu,0)}\hhh_a.$$
\bl{l5}
Suppose that 
$ \JJ$. 
Then 
for $\nu$ such that 
$|\Sigma| > 3 \CA + \nu$, 
$\lc{T}{\hhh_a(-)} $ has a purely discrete spectrum, i.e., 
$$
\sigma(\lc{\thn\K}{\hhh_a})\cap [\inf\sigma(\lc{\thn\K}{\hhh_a}),
\inf\sigma(\lc{\thn\K}{\hhh_a})+\nu)$$
$$\hspace{5cm}\subset \sigma_{\rm disc}(\lc{\thn\K}{\hhh_a}).
$$
\el
\proof
Let $\{\phi_n\}_n$ be a complete orthonormal system  of $\hhh_a(-)$ 
and $\{\psi_m\}_m$ a complete orthonormal system  of $\hhh_a(+)$. 
We see that by Lemma \ref{gr},  
$$0\geq \tr \lc{T}{\hhh_a(-)}=  \sum_n(\phi_n, T\phi_n)
\geq\sum_n(\phi_n, T'\phi_n),$$
where 
$$T'=\X\otimes\lk(1-\frac{\CA}{2})\hfn\K-3\CA-\nu\rk.$$
Set 
$$T'_-=T'E^{T'}_{(-\infty, 0)}.$$
Then 
$$0\geq \tr \lc{T}{\hhh_a(-)}\geq \sum_n(\phi_n, T'_-\phi_n)$$
$$\geq \sum_n(\phi_n, T'_-\phi_n)+
\sum_m(\psi_m, T'_-\psi_m)
=
\tr T_-'.$$
Hence we obtain that 
$$\left| 
\tr \lc{T}{\hhh_a(-)}\right|
\leq \left |\tr T_-'\right|$$
$$=\tr \X\times 
\left| 
\tr \lc{\lk (1-\frac{\CA}{2})\hfn\K-3\CA-\nu\rk_-}{\hhh_a}
\right|,$$
where 
$\lk \cdots \rk_-$ denotes the negative  part of 
the spectrum of $\lk \cdots \rk$. 
Since 
$$\sigma(\lc{\hfn\K}{\hhh_a})=
\sigma_{\rm disc}(\lc{\hfn\K}{\hhh_a})$$ and 
$$\left|\tr\X\right|<\infty, $$ 
it follows that 
$$\left| 
\tr \lc{T}{\hhh_a(-)}\right|<\infty
$$
Thus the lemma follows.
\qed
\bl{l6}
Suppose that 
$\JJ$. 
Then 
for $\nu$ such that 
$|\Sigma| > 3 \CA + \nu$, 
$$\H{\thn\K}.$$
\el
\proof
We have by Lemmas \ref{l1} and \ref{l5},  
$$\sigma(\thn\K)=\sigma(\lc{\thn\K}{\hhh_a^\perp})\cup 
\sigma(\lc{\thn\K}{\hhh_a}),$$
$$\sigma(\lc{\thn\K }{\hhh_a^\perp})
\subset[\is(\th\K)+\nu,\infty),$$
and 
$$
\sigma(\lc{\thn\K}{\hhh_a})\cap [\inf\sigma(\lc{\thn\K}{\hhh_a}),
\inf\sigma(\lc{\thn\K}{\hhh_a})+\nu)$$
$$\hspace{5cm}\subset \sigma_{\rm disc}(\lc{\thn\K}{\hhh_a}).
$$
Hence the lemma follows. 
\qed
\bl{l8}
Suppose that 
$\JJJ$. Then 
$$\H{\thn}.$$
In particular $\thn$ has a ground state.
\el
\proof 
Note that 
$$\lim_{L\rightarrow \infty}\lim_{a\rightarrow \infty}\CA=\C$$
Then by Lemmas \ref{l7} and \ref{l6}, the lemma follows from \cite{b}. 
\qed



\subsection{Binding} 
The number operator $N$ in $\fff$ is defined by 
$$N=\jjj\int \add(k,j)a(k,j)\ddk.$$
A ground state of  $\thn$ is denoted by $\grn$. 
\begin{lemma}
\label{N}
Suppose that $\JJJ$.
Then 
for $\nu$ such that 
$|\Sigma| > 3 \C + \nu$, 
$$\frac{\|\nf^\han\n\|}{\|\n\|}
\leq 
\frac{C \|\ph/\omega^{5/2}\| \mv }{\massn}$$
with some constant $C$. 
\end{lemma}
\proof 
We set 
$$E=\is(\thn).$$
Since 
$$[\thn, a(k,j)]=-\omega(k)-\nu+[\dvn,a(k,j)],$$
we have 
$$\hspace{-2cm}
\thn  a(k,j)\grn=-(\omega(k)+\nu) a(k,j)\grn$$
$$\hspace{2cm}
+Ea(k,j)\grn+[\dvn, a(k,j)]\grn.$$
Hence 
$$(\thn-E+\omega(k)+\nu)a(k,j) \grn=[\dvn, a(k,j)]\grn.$$
Here we see that 
$$
[\dvn, a(k,j)]=[V(\cdot-K^\nu/\massn),a(k,j)]=[\apm V\ap,a(k,j)]$$
$$\hspace{2cm}
=\apm[V,\ap a(k,j)\apm]\ap.$$
Since 
$$\ap a(k,j)\apm=a(k,j)-\frac{i}{\sqrt 2\massn} p  \rrr,$$
it follows that 
$$[\dvn,a(k,j)]=\apm[V,
-\frac{i}{\sqrt 2\massn} p  \rrr]\ap$$
$$=\apm\lk   \frac{1}{\sqrt 2\massn} (\nabla V) \rrr  \rk\ap.$$
Thus we obtain a pull-through formula:  
$$a(k,j)\grn=
(\thn-E+\omega(k)+\nu)\f \times $$
\eq{id}
\times 
\apm\lk \frac{1}{\sqrt 2\massn} (\nabla V) \rrr  \rk\ap
\grn.
\en 
Using identity \kak{id} we see that 
$$(N^\han \grn, N^\han \grn)=\jjj\int\|a(k,j)\grn\|^2 \ddk$$
$$\hspace{-2cm}=
\left. \jjj\int\right \| 
(\thn-E+\omega(k)+\nu)\f \times  $$ 
$$ \left. \hspace{3cm}
\times \apm\lk   \frac{1}{\sqrt 2\massn} (\nabla V)\rrr  \rk\ap
\grn 
\right\|^2\ddk $$
$$\leq 
3\sum_{\mu=1}^3 
\jjj\int \lk\frac{1}{\omega(k)}\|\nabla_\mu  V\|_\infty\rk^2 
\left| 
  \frac{1}{\sqrt 2\massn}  \Kp_\mu^\nu(k,j) 
  \right |^2 
\ddk
\|\grn \|^2 $$
$$\leq 
\frac{3}{2} 
\sum_{\mu=1}^3
\lk \frac{\|\nabla_\mu V\|_\infty}{\massn}\rk^2 
 \jjj\|\Kp^\nu(\cdot,j) /\omega\|^2\|\grn\|^2$$
$$\leq 
C \lk \frac{\mv }{\massn} \rk^2
 \|\ph/\omega^{5/2}\|^2 \|\n\|^2.$$
Hence the lemma follows. 
\qed

\begin{lemma}
\label{Q}
Let $P_\Omega$ be the projection onto 
$\{\alpha\Omega\ |\ \alpha\in\CC\}$ 
and 
$$Q=E_{[\Sigma+\delta, \infty)}^{\eff}\otimes P_\Omega$$ 
with some $\delta>0$ such that 
$$\delta>\mmmm\C.$$ 
Suppose that 
$\JJJ$.
Then 
for $\nu$ such that 
$|\Sigma| > 3 \C + \nu$, 
$$\frac{\|Q\n\|}{\|\n\|}
\leq 
\sqrt{\frac{\C}{\delta -\mmmm\C}}.
$$
\end{lemma}
\proof 
Since $(\n, Q(\thn-\is(\thn))\n)=0$, we have 
$$(\n, Q(\eff-\is(\thn)+ g(\Lambda))\n)=-(\n, Q\dvn  \n).$$
The left-hand side above is estimated as 
$$
(\n, Q(\eff-\is(\thn)+ g(\Lambda))\n) $$
$$\geq 
 (\Sigma+\delta -\is(\thn)+ g(\Lambda)) (\n, Q\n).$$
Note that 
$$ \Sigma+\delta +g(\Lambda)-\is(\thn)\geq 
\Sigma+\delta+ g(\Lambda)-\Sigma- \mmmm\C-g(\Lambda)$$
$$=\delta -\mmmm\C>0.$$
Then 
$$ (-\is(\thn)+ g(\Lambda)) (\n, Q\n)\geq 
(\delta -\mmmm\C)\|Q\n\|^2>0.$$ 
Moreover 
$$|(\n, Q\dvn\n)|=|(\dvn Q\n, \n)|\leq 
\|\dvn Q\n\|\|\n\|$$
$$\leq \C\lk \|\hf^\han Q\n\|+\|Q\n\|\rk\|\n\|$$
$$=\C\|Q\n\|\|\n\|\leq \C\|\n\|^2.$$
Hence we have 
$$0< (  \delta -\mmmm\C)\|Q\n\|^2\leq \C\|\n\|^2.$$
The lemma follows.
\qed

\subsection{Proof of the main theorem}
We normalize $\grn$, i.e., $\|\grn\|=1$. 
 Take a subsequence $\nu'$ such that  
$\nn$ weakly converges to a vector $\gr$ as $\nu'\rightarrow \infty$. 
It is established in \cite{arhi} 
that if $\gr\not=0$, then $\gr$ is a ground state of $\hl$. 

{\it Proof of Theorem \ref{main0}}


\noindent
It is enough to prove $\gr\not=0.$
Note that on $\fff$ 
$$N+P_\Omega\geq 1.$$
Then on $\hhh$ 
$$1\otimes N+1 \otimes  P_\Omega\geq 1.$$
Hence
 $$1\otimes N+(\XX+\YY)\otimes P_\Omega$$
$$=1\otimes N+\XX\otimes P_\Omega+Q\geq 1,$$
and 
\eq{st}
\XX\otimes P_\Omega\geq 1-1\otimes N-Q.
\en 
Suppose that 
$\JJJ$ and  $\delta>\mmmm\C$. 
Then 
for $\nu'$ such that 
$|\Sigma| > 3 \C + \nu'$, 
we have by \kak{st},  
$$(\nn, \XX\otimes P_\Omega\nn)$$
$$\geq \|\nn\|^2-(\nn, N\nn)-(\nn,Q\nn)$$
$$\geq 
\|\nn\|^2-\frac{C\|\ph/\omega^{5/2}\|}{\massn}
\mv \|\nn\|^2-
\frac{\C}{\delta -\mmmm\C}\|\nn\|^2$$
$$=
1-\frac{C\|\ph/\omega^{5/2}\|}{\massn} \mv - 
\frac{\C}{\delta -\mmmm\C}.$$
Note that 
$$ 
\lim_{\Lambda\rightarrow\infty} 
\frac{\|\ph/\omega^{5/2}\|}{\mass}=0,
$$ 
$$\lim_{\Lambda\rightarrow\infty} \frac{\C}{\delta-\mmmm\C} =0.$$
Hence for sufficiently large $\Lambda$, 
$$(\nn, \XX\otimes P_\Omega\nn)>\epsilon$$
uniformly in $\nu'$ with some $\epsilon>0$.  
Take $\nu'\rightarrow \infty$ on the both sides above. 
Since $\XX\otimes P_\Omega$ is a finite rank operator, 
we have 
$$(\gr, \XX\otimes P_\Omega \gr)>\epsilon.$$
In particular $\gr\not=0$. Then $\gr$ is a ground state of $\th$. 
\qed
\begin{remark}
$\gr$ satisfies that $\gr\in D(N^\han)$. 
Actually 
in the similar manner  as \kak{id} 
we have 
$$a(k,j)\gr=
(\hl-\is(\hl) +\omega(k))\f $$
$$ 
\times 
\appm\lk \frac{1}{\sqrt 2\massn} (\nabla V) \rrrr  \rk\app
\gr.
$$ 
Using this we can obtain that  
$$(N^\han\gr,N^\han\gr)=\jjj\int\|a(k,j)\gr\|^2\ddk<\infty.$$
\end{remark}
\section{Remarks} 
\subsection{Multiplicity and localization}
In Section 3 we established that 
$\hl$ has a ground state for $\Lambda>\Lambda_\ast$ with some $\Lambda_\ast$ 
even in the case where 
the decoupled Hamiltonian $\mpp+V+\hf$ has no ground state. 
The uniqueness of  ground states is established 
by means  of a functional integral representation of $\eht$. 
There exists a unitary operator  such that  
$$\U: \hhh=\LR\otimes\fff 
\longrightarrow L^2(\BR\times {\bf S},dx\otimes d\mu),$$
where $d\mu$ is a Gaussian probability measure on 
the following direct sum of the real-valued Schwartz distribution, 
$$
{\bf S}=
{\cal S}'_{\rm real}(\BR)\times 
{\cal S}'_{\rm real}(\BR)\times {\cal S}'_{\rm real}(\BR).$$
Using  the identification implemented by $\U$,  
$$\hhh\cong  L^2(\BR\times {\bf S},dx\otimes d\mu),$$
we established in \cite{hi9} that 
$$\Theta(t)=e^{i\frac{\pi}{2}N}
\eht e^{-i\frac{\pi}{2}N}$$ 
is positivity improving, i.e., 
$$(\Psi,  \U \Theta(t) \U\f \Phi)>0$$
for $\Psi,\Phi\in L^2(\BR\times {\bf S},dx\otimes d\mu)$ 
with 
$\Psi,\Phi\geq 0$ but $\Psi\not\equiv0,\Phi\not\equiv0$.
This implies that a ground state (if it exists) of $\hl$ is unique 
up to a phase factor. 


In Section 3 we also see that 
$\gr\in D(N^\han).$
Modifying a method in \cite{hi20} slightly, 
we can also establish that 
\eq{hk}
\gr\in \bigcap_{k=1}^\infty D(N^{k/2}).
\en 
\bp{ezawa}
Let $\Lambda>\Lambda_\ast$. 
Then the ground state, $\gr$,  of $\hl$ is unique and 
\kak{hk} is satisfied. 
\ep

\subsection{$\A^2$ term}
$\A^2$ term in $\hl$ is a subjective  part for large $\Lambda$. 
It is established in \cite[Remark 3.10]{hi21} that 
the Hamiltonian without $\A^2$, 
$$\hl'=\mpp +V+\frac{1}{m} p\cdot \A+\hf,$$  
is unbounded from below for $\Lambda$ 
such that 
$$\|\ph/\omega\|^2>m.$$
In particular $\hl'$ has no ground state for such $\Lambda$. 




\bp{ezawa2}
$\hl'$ has no ground state for $\Lambda>m/(4\pi)+\lambda$. 
\ep
 


\section{Stability of matter} 
Stability of matter investigated in this section are pointed out in e.g., 
Lieb and Loss \cite{lilo} and references therein.
See also 
Fefferman,   Fr\"ohlich and  Graf 
\cite{ffg}. 
\subsection{Ultraviolet cutoffs} 
In this subsection 
we estimate the asymptotics of $g(\Lambda)$ in \kak{34} 
as $\Lambda\rightarrow \infty$. 
\begin{remark}
In the case of $V=0$, from Proposition \ref{hiro} it follows that 
$$g(\Lambda)=\is(\hl).$$ 
It is seen that $g(\Lambda)$ is monotonously increasing in $\Lambda$. 
\end{remark}
Since 
$$\|\ph/\sqrt{t^2+\omega^2}\|^2
=
4\pi\lkk(\Lambda-\lambda)+t
\lk \arctan\frac{\lambda}{t}-\arctan\frac{\Lambda}{t}\rk \rkk,$$
and 
$$\|\ph/(t^2+\omega^2)\|^2=
4\pi\lkk\frac{1}{2t}
\lk 
\arctan\frac{\Lambda}{t}-\arctan\frac{\lambda}{t}
\rk +\half
\lk 
\frac{\lambda}{t^2+\lambda^2}-\frac{\Lambda}{t^2+\Lambda^2}\rk 
\rkk,$$
we have 
$$g(\Lambda)=\frac{1}{\pi}\int_{-\infty}^\infty
\frac{2\pi t\lkk (\arctan\frac{\Lambda}{t}-\arctan\frac{\lambda}{t})+\lk\frac{\lambda/t}{1+(\lambda/t)^2}-\frac{\Lambda/t}{1+(\Lambda/t)^2}\rk\rkk}{m+\frac{8\pi t}{3}
\lkk\lk\frac{\Lambda}{t}-\frac{\lambda}{t}\rk+\lk \arctan\frac{\lambda}{t}-\arctan\frac {\Lambda}{t}\rk\rkk} dt.$$
Change variable $t$ to  $r=\Lambda/t$. 
We have 
$$g(\Lambda)=4\Lambda^2
\int_0 ^\infty 
\frac{ 
(\arctan r-\frac{r}{1+r^2})
-
\lk \arctan r\LM-\frac{r\LM}{1+r^2\LM^2}\rk 
}{mr+\frac{8\pi}{3}\Lambda 
\lkk (r-\arctan r)-(r\LM-\arctan r\LM)\rkk} \frac{dr}{r^2}.$$

\bt{ezawa22}
Assume that 
\eq{mu6}
m>8\pi\lambda /3. 
\en 
Then 
$$
\frac{8}{3}\lk\frac{3}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}
\leq 
\lim_{\Lambda\rightarrow\infty} 
\frac{g(\Lambda)}{\Lambda^{3/2}}
\leq \frac{8}{3}\lk\frac{9}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}.
$$
\et
\proof 
This  proof is  due to Ezawa \cite{e}. 
We decompose 
$$\frac{g(\Lambda)}{4\Lambda}=
\int_0 ^\infty 
\frac{ 
(\arctan r-\frac{r}{1+r^2})
-
\lk \arctan r\LM-\frac{r\LM}{1+r^2\LM^2}\rk 
}{\frac{m}{\Lambda} r+\frac{8\pi}{3}
\lkk (r-\arctan r)-(r\LM-\arctan r\LM)\rkk} \frac{dr}{r^2}$$
such as 
$$ \frac{g(\Lambda)}{4\Lambda} =
\int_0^{1/\sll}+\int_{1/\sll}^\infty=I_1(\Lambda)+I_2(\Lambda).$$
It is enough to show that 
\eq{mu1}
\frac{2}{3}\lk\frac{3}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2} 
\leq 
\lim_{\Lambda\rightarrow\infty}\frac{ I_1(\Lambda)}{\sl}\leq 
\frac{2}{3}\lk\frac{9}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}, 
\en 
and that 
\eq{mu2}
 \lim_{\Lambda\rightarrow\infty}\frac{I_2(\Lambda)}{\sl}=0.
\en 
Note that 
$$\arctan x=\frac{x}{1+x^2}+\frac{2}{3}\frac{x^3}{(1+x^2)^2}+\frac{5}{4}\frac{3}{2}\frac{x^5}{(1+x^2)^3}+\cdots.$$
We define $f$ and $h$ by 
$$\arctan r-\frac{r}{1+r^2}=\frac{2}{3}\frac{r^3}{(1+r^2)^2}+f(r)$$
and 
$$r-\arctan r=\frac{r^3}{1+r^2}- h(r).$$
It is satisfied that $f(r)\geq 0$,  $h(r)\geq 0$,  
$$\lim_{r\rightarrow 0} \frac{f(r)}{r^3}=0,\ \ \ 
\lim_{r\rightarrow 0} \frac{h(r)}{r^3}=\frac{2}{3}
$$
and 
$$h(r)=
\frac{2}{3}\frac{r^3}{(1+r^2)^2}+f(r)=\arctan r-\frac{r}{1+r^2}.$$
Let us set 
$$\fla(r)= \arctan r\LM-\frac{r\LM}{1+r^2\LM^2}>0 $$
and 
$$\gla(r)= r\LM-\arctan r\LM >0.$$
Then $I_1(\Lambda)$ is written as 
$$I_1(\Lambda) 
=\int_0^{1/\sll}\frac{\frac{2}{3}+\frac{(1+r^2)^2}{r^3}(f(r)-\fla(r))}{
\frac{m}{\Lambda}(1+r^2)+\frac{8\pi}{3}r^2-
\frac{8\pi}{3}
\frac{1+r^2}{r}
(h(r)+\gla(r))}\frac{dr}{1+r^2}.$$
It follows that for $0\leq r\leq 1/\sll$, 
$$
\frac{8\pi}{3}
\frac{1+r^2}{r}
(h(r)+\gla(r))=
\frac{8\pi}{3}
r^2 (1+r^2) 
\frac{h(r)+\gla(r)}{r^3}
$$
$$\leq 
\frac{8\pi}{3}  
\lk 1+\frac{1}{\sl}\rk 
r^2 
\theta(\Lambda)  
:=r^2 \delta(\Lambda),$$ 
where 
$$ 
\theta(\Lambda) :=\sup_{0\leq r \leq 1/\sll}  \frac{h(r)+\gla(r)}{r^3}.
$$ 
Note that 
$$\lim_{r\downarrow 0} \frac{h(r)+\gla(r)}{r^3} =\frac{2}{3}+\frac{1}{3}\LM^3$$
and 
\eq{mu3}
\liml \delta(\Lambda)= \frac{8\pi}{3} \frac{2}{3}. 
\en 
Moreover we have 
$$\frac{(1+r^2)^2}{r^3}(f(r)-\fla(r))\leq 
\sup_{0\leq r \leq (1/\sll)} \frac{(1+r^2)^2}{r^3}
f(r)$$ 
and 
$$\frac{(1+r^2)^2}{r^3}(f(r)-\fla(r))\geq 
-\sup_{0\leq r \leq (1/\sll)} \frac{(1+r^2)^2}{r^3}\fla(r).$$
Set 
$$\epsilon(\Lambda):=\max\lkk\sup_{0\leq r \leq (1/\sll)} \frac{(1+r^2)^2}{r^3}
f(r),  
\sup_{0\leq r \leq (1/\sll)} \frac{(1+r^2)^2}{r^3}\fla(r) \rkk.$$ 
It is trivial that 
\eq{mu4}
\liml\epsilon (\Lambda)=0.
\en 
Hence we have 
$$
\frac{(2/3)-\epsilon(\Lambda)}{1+1/\sl}\int_0^{1/\sll}
\frac{1}{\frac{m}{\Lambda}+
\lk \frac{m}{\Lambda}+\frac{8\pi}{3}\rk r^2} dr
\leq I_1(\Lambda)$$
$$
\leq 
\frac{(2/3)+\epsilon(\Lambda)}{1-1/\sl}
\int_0^{1/\sll}\frac{1}{\frac{m}{\Lambda}+ 
\lk \frac{m}{\Lambda}+\frac{8\pi}{3}   -\delta(\Lambda)  \rk r^2} dr.$$
Together with \kak{mu3} a direct calculation yields that 
$$\liml \frac{1}{\sl} 
\int_0^{1/\sll}\frac{1}{\frac{m}{\Lambda}+ 
\lk \frac{m}{\Lambda}+\frac{8\pi}{3}  - \delta(\Lambda) \rk r^2} dr
$$
$$=\liml 
\frac{1}{\sqrt{m (\frac{m}{\Lambda}+\frac{8\pi}{3}- \delta(\Lambda) )}}
\arctan\sqrt{\frac{\frac{m}{\Lambda}+\frac{8\pi}{3}- \delta(\Lambda)}
{m/\sl}}$$ 
$$=
\lk\frac{9}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}.$$
Similarly  we have 
$$
\liml 
\frac{1}{\sl}\int_0^{1/\sll}\frac{1}{\frac{m}{\Lambda}+ 
\lk \frac{m}{\Lambda}+\frac{8\pi}{3}\rk r^2} dr=
\lk\frac{3}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}.$$
Thus by \kak{mu4} we have 
$$
\frac{2}{3}\lk\frac{3}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}\leq 
\liml\frac{1}{\sl}I_1(\Lambda)
\leq 
\frac{2}{3}\lk\frac{9}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}.$$
Hence 
\kak{mu1} follows. 
Next we show \kak{mu2}. 
Since 
$$ 
(\arctan r-\frac{r}{1+r^2})
-
\lk \arctan r\LM-
\frac{r\LM}{1+r^2\LM^2}\rk $$
$$
\leq \frac{2}{3}
\frac{r^3}{(1+r^2)^2}+
\frac{5}{4}
\frac{3}{2}
\frac{r^5}{(1+r^2)^2}$$
and by the assumption 
$ 
m>8\pi \lambda/3,
$
$$
\frac{m}{\Lambda} r+\frac{8\pi}{3}
\lkk (r-\arctan r)-\lk r\LM-\arctan r\LM \rk \rkk$$ 
$$
> 
\frac{m}{\Lambda} r-\frac{\lambda}{\Lambda} \frac{8\pi}{3}r +\frac{8\pi}{3}
\lkk \frac{r^3}{1+r^2}-
\frac{2}{3}\frac{r^3}{1+r^2}+
\arctan (r\LM)\rkk$$ 
$$
>\frac{8\pi}{9}\frac{r^3}{1+r^2}.$$
 Then 
$$\liml
\frac{1}{\sl}I_2(\Lambda) \leq 
\liml \frac{1}{\sl}\int_{1/\sll}^\infty 
\frac{\frac{2}{3}
\frac{r^3}{(1+r^2)^2}+
\frac{5}{4}
\frac{3}{2}
\frac{r^5}{(1+r^2)^2}}{
\frac{8\pi}{9}\frac{r^3}{1+r^2}
}\frac{dr}{r^2}$$
$$= 
\liml \frac{9}{8\pi} 
\frac{1}{\sl}\int_{1/\sll}^\infty 
\frac{1}{1+r^2}\lk \frac{2}{3}+\frac{15}{8}r^2\rk
\frac{dr}{r^2}=0.$$
Thus \kak{mu2} follows, and then the theorem follows.
\qed


\subsection{$N$ particle system} 
We consider an $N$ particle  system. 
We assume simply that each particle has mass $m$ 
and there is no external potential. 
The Hamiltonian, $\hln$,  is defined 
as  a self-adjoint operator acting on 
$$L^2(\RR^{3N})\otimes\fff,$$
and is given by 
$$\hln=\sum_{j=1}^N\frac{1}{2m}(p_j+\Aj(0))^2+\hf,$$
where 
$$\Aj(0)=\jjjj\int\frac{\chi_{j\Lambda}(k)}{\sqrt{2\omega(k)}}e_{j'}(k)\lkk 
a^\ast(k,j')+a(k,j')\rkk d^3k.$$
Let 
$$\is (\hln)=g(\Lambda, N).$$
We consider the two cases such as
\bi
\item[(1)] 
$\chi_{j\Lambda}(k)=\chi_\Lambda(k),\ \ \ j=1,...,N,$
\item[(2)]
$\chi_{j\Lambda}$, $j=1,...,N$,  are 
characteristic functions on closed sets in $\BR$ 
such as 
$${\rm supp}\chi_{j\Lambda}\cap {\rm supp} \chi_{i\Lambda}\cap \{0\}=\emptyset,\ \ \   
(i\not=j).$$
\ei
We will see below that the asymptotic behavior of $g(\Lambda, N)$ 
as $N\rightarrow \infty$ depends on ultraviolet cutoffs. 
In the case of (2) we intuitively expect that 
$$g(\Lambda, N)\approx N,$$
since $N$  particles 
may have no interaction through quantized radiation fields.  
In \cite{hi21} we established the following proposition.
\bp{hiro} 
In the case of (1),  
$$g(\Lambda, N)=
\frac{N}{\pi}\int_{-\infty}^\infty \frac{t^2
\|\chi_\Lambda/(t^2+\omega^2)\|^2}{m+\frac{2}{3}N\|\chi_\Lambda/
\sqrt{t^2+\omega^2}\|^2}dt,$$
in the case of (2),  
$$g(\Lambda, N)= 
\sum_{j=1}^N \frac{1}{\pi}\int_{-\infty}^\infty \frac{t^2
\|\chi_{j\Lambda}/(t^2+\omega^2)\|^2}
{m+\frac{2}{3}\|\chi_{j\Lambda}/\sqrt{t^2+\omega^2}\|^2}dt.
$$
\ep
In the case of (1) the following theorem holds. 
\bt{ezawa23}
We assume case (1) and 
\eq{mu7}
m> 8\pi \lambda/3. 
\en 
Then 
$$
\frac{8}{3}\lk\frac{3}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}
\leq 
\limln \frac{g(\Lambda, N)}{\sqrt N \Lambda^{3/2}}
\leq 
\frac{8}{3}\lk\frac{9}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}.
$$
\et
\proof 
By Proposition \ref{hiro} we have 
$$\frac{g(\Lambda, N)}{4\Lambda}=
\int_0 ^\infty 
\frac{ 
(\arctan r-\frac{r}{1+r^2})
-
\lk \arctan r\LM-\frac{r\LM}{1+r^2\LM^2}\rk 
}{\frac{m}{N \Lambda} r+\frac{8\pi}{3}
\lkk (r-\arctan r)-(r\LM-\arctan r\LM)\rkk} \frac{dr}{r^2}.$$
Then in the similar way as the proof of Theorem \ref{ezawa23} 
we decompose it such as 
$$\frac{g(\Lambda, N)}{4\Lambda}=\int_0^{1/\sln}+\int_{1/\sln}^\infty=I_1(\Lambda,N)+I_2(\Lambda, N),$$
and it can be seen that 
$$ 
\frac{2}{3}\lk\frac{3}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}\leq 
\limln \frac{I_1(\Lambda, N)}{\sqrt{N\Lambda}}\leq 
\frac{2}{3}\lk\frac{9}{8\pi}\frac{1}{m}\rk^\han\frac{\pi}{2}, 
$$
and that 
$$
\limln \frac{I_2(\Lambda, N)}{\sqrt{N\Lambda}}=0.$$
Then the theorem follows. 
\qed





 

\begin{remark}
In the case of (2), if we adjust $\chi_{j\Lambda}$ such as 
$$\frac{1}{\pi}\int_{-\infty}^\infty \frac{t^2
\|\chi_{j\Lambda}/(t^2+\omega^2)\|^2}
{m+\frac{2}{3}\|\chi_{j\Lambda}/\sqrt{t^2+\omega^2}\|^2}dt=g$$
with some constant $g$ independent of $j$. 
Then 
$$g(\Lambda, N)= N g.$$
\end{remark}

 




{\bf Acknowledgment}
{\footnotesize 
I thank  H. Ezawa for sending 
 his proof of  Theorems \ref{ezawa22} and \ref{ezawa23}  
 to myself. 
I thank  Grant-in-Aid 13740106 for Encouragement of Young Scientists from the Ministry 
of Education, Science, Sports and Culture 
for financial support.}

















{\footnotesize
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