Content-Type: multipart/mixed; boundary="-------------0111050943986" This is a multi-part message in MIME format. ---------------0111050943986 Content-Type: text/plain; name="01-408.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-408.keywords" scaling limits, Pauli-Fierz Hamiltonian, dipole approximation, weak coupling limit, effective potential, Symplectic structure, Bogoliubov transformation ---------------0111050943986 Content-Type: application/x-tex; name="1104.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="1104.tex" \documentstyle[12pt,two side]{article} \topmargin=0truein \oddsidemargin=0truein \evensidemargin=0truein \textheight=21cm \textwidth=15cm \catcode`\@=11 \def\newsymbol#1#2#3#4#5{\let\next@\relax% \ifnum#2=\@ne\else% \ifnum#2=\tw@\let\next@\msyfam@\fi\fi% \mathchardef#1="#3\next@#4#5} \def\mathhexbox@#1#2#3{\relax% \ifmmode\mathpalette{}{\m@th\mathchar"#1#2#3} \else\leavevmode\hbox{$\m@th\mathchar"#1#2#3$}\fi} \def\hexnumber@#1{\ifcase#1 0\or 1\or 2\or 3\or 4\or 5\or 6\or 7\or 8% \or 9\or A\or B\or C\or D\or E\or F\fi} \font\tenmsy=msbm10 \font\sevenmsy=msbm7 \font\fivemsy=msbm5 \newfam\msyfam \textfont\msyfam=\tenmsy \scriptfont\msyfam=\sevenmsy \scriptscriptfont\msyfam=\fivemsy \edef\msyfam@{\hexnumber@\msyfam} \def\Bbb#1{\fam\msyfam\relax#1} \catcode`\@=\active \load{\footnotesize}{\sf} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{remark}[theorem]{Remark} \newtheorem{assumption}[theorem]{Assumption} \newcommand{\cmp}[5]{{#1}, {#2}, {\it Commun. 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\newcommand{\ijk}{\hat{\lambda}_i(k')\hat{\lambda}_j(k')} \newcommand{\ij}{\la_i\la_j} \newcommand{\bea}[1] {\begin{array}{#1}} \newcommand{\ena}{\end{array}} \newcommand{\eq}[1]{\begin{equation} \label{#1}} \newcommand{\en}{\end{equation}} \newcommand{\kak}[1]{(\ref{#1})} \newcommand{\bt}[1]{\begin{theorem} \label{#1}} \newcommand{\et}{\end{theorem}} \newcommand{\bc}[1]{\begin{corollary} \label{#1}} \newcommand{\ec}{\end{corollary}} \newcommand{\bl}[1]{\begin{lemma} \label{#1}} \newcommand{\el}{\end{lemma}} \newcommand{\bp}[1]{\begin{proposition} \label{#1}} \newcommand{\ep}{\end{proposition}} \newcommand{\bi}{\begin{description}} \newcommand{\ei}{\end{description}} \newcommand{\bog}{\Sigma_{\rm B}} \newcommand{\bogg}{\Sigma_{\rm B real}} \makeatletter \@addtoreset{equation}{section} \makeatother \def\theequation{\arabic{section}.\arabic{equation}} \begin{document} \setlength{\baselineskip}{18pt} \pagestyle{myheadings} \markboth{Observable effect}{Observable effect} \title {Observable effects and parametrized scaling limits of a model in nonrelativistic quantum electrodynamics} \author{ Fumio Hiroshima \thanks{ Department of Mathematics and Physics, Setsunan University, Ikeda-naka-machi 17-8, 572-8508, Osaka, Japan.}} \maketitle \begin{abstract} Scaling limits of the Hamiltonian $H$ of a system of $N$ charged particles coupled to a quantized radiation field are considered. Ultraviolet cutoffs, $\la_1,....,\la_N$, are imposed on the radiation field and the Coulomb gauge is taken. It is so called the Pauli-Fierz model in nonrelativistic quantum electrodynamics. We mainly consider two cases: (i) all the ultraviolet cutoffs are identical, $\la_1=\cdots=\la_N$, (ii) supports of ultraviolet cutoffs have no intersection, ${\rm supp}\la_i\cap{\rm supp}\laj=\emptyset$, $i\not= j$. The Hamiltonian acts on $\LR\otimes\fff$, where $\fff$ is a symmetric Fock space and has the form $H=\ele\otimes\I+B+\I\otimes\fq$. Here $\ele$ denotes a particle Hamiltonian, $\fq$ a quadratic field operator, and $B$ an interaction term. The scaling is introduced as $H(\k)=\ele\otimes\I+\k^l B+\k^2\I\otimes\fq$, where $\k$ is a scaling parameter and $l\leq 2$ a parameter of the scaling. Performing a mass renormalization we consider the scaling limit of $H(\k)$ as $\k\rightarrow \infty$ in the strong resolvent sense. Then effective Hamiltonians $\eff$ in $\LR$ infected with reaction of effect of the radiation field is derived. In particular (1) effective Hamiltonians with an effective potential for $l=2$, and (2) effective Hamiltonians with an observed mass for $l=1$, are obtained. \end{abstract} \newpage \tableofcontents \newpage \section{Preliminaries} \subsection{Observable effects and scaling limits} In this paper we consider the Hamiltonian $H$ of a system of $N$ nonrelativistic charged particles interacting with a quantized radiation field. Ultraviolet cutoffs are imposed on the radiation field and recoil of the particles and the radiation field (photons) are excluded, i.e., the dipole approximation is made. Taking scaling limits of the Hamiltonian, we derive effective Hamiltonians. The main idea is to diagonalize $H$ by a canonical transform derived from a symplectic structure. The \sc Hamiltonian of one charged particle with external potential $V$ is of the form \eq{sc} \frac{1}{2m}p^2+V, \en where $p$ denotes the momentum operator and $m$ the bare mass of the particle. To interpret the Lamb shift \cite{bet} intuitively, T. A. Welton \cite{we} formally derived an effective \sc Hamiltonian taking into account of effects of a radiation field. He expected that the position fluctuation of the particle by the radiation field will effectively modify the external potential $V$. The fluctuation was thought of as a Gaussian random variable $\Delta x$, then the effective potential is formally given by the mean value of $V(x+\Delta x)$; $$\veff(x):=\langle V(x+\Delta x)\rangle_{\rm AVE} =(2\pi)^{-3/2}\int_{\RR^3}\hat{V}(k)e^{ikx}\langle e^{ik\Delta x} \rangle_{\rm AVE}$$ $$=(2\pi C)^{-3/2} \int_{\RR^3} V(y)e^{-|x-y|^2/2C}dy,$$ with a certain positive constant $C$. Then \kak{sc} effectively turns out to be the effective Hamiltonian, Welton's Hamiltonian, given by \eq{v} \frac{1}{2m}p^2+\veff. \en He gave an interpretation of the Lamb shift as the difference of the spectrum between the original Hamiltonian and the effective one. On the other hand it was found in \cite{pafi} that photons enhances the bare mass $m$. Then $m$ amounts to the observed mass $m+\delta m$, $\delta m>0$, through the coupling to a radiation filed. Then \kak{sc} formally turns out to be \eq{m} \frac{1}{2(m+\delta m)}p^2+V. \en In this paper performing a mass renormalization to eliminate a divergence contribution, we present a rigorous derivation of \kak{v} and \kak{m} by a parametrized scaling limits of Hamiltonians describing an interaction between particles and radiation fields. Thus we interpolate \kak{v} and \kak{m}. We suppose that $N$ particles move in the $d$-dimensional space and a radiation field has the helicity $d-1$. Throughout this paper we assume $$d\geq 3.$$ $\fff$ denotes the Boson Fock space over $W:=\oplus^{d-1}\LL$, which is defined by $$\fff:=\oplus_{n=0}^\infty [\otimes_s^nW],$$ where $\otimes_s^nW$ denotes the $n$-fold symmetric tensor product of $W$, and $\otimes_s^0W:=\BC$. The bare vacuum $\Omega$ is defined by $$\Omega:=\{1,0,0,\cdots\}.$$ Hamiltonians under consideration are of the form $$H:=\ele\otimes \I +B+\I \otimes \fq$$ acting on the Hilbert space $$\hhh:=\LR\otimes \fff. $$ Here $\ele$ describes the particle Hamiltonian of the form $$\ele:=\jjj \mhalf \vp^2+V.$$ Here $\vp:=-i\hbar \vec{\nabla}_j$ denotes the momentum operator of the $j$th particle canonically conjugate to the position operator $x_j$, $V:\BRN\rightarrow \RR$ an external potential, $\hbar$ the Planck constant divided by $2\pi$, $\mj$ the mass of the $j$th particle. $\fq$ a quadratic field operator in $\fff$, and $B$ an interaction term. We scale $H$ as follows: $$H(\k):=\ele\otimes \I +\k^l B+\kt \I \otimes \fq.$$ Davies \cite{da1,da2} considered the scaling limit $\k\rightarrow \infty$ of the case of $$l=1$$ for certain simple models to derive $N$-body \sc Hamiltonians. He called this scaling limit {\it the weak coupling limit}. See also \cite{du,pa}. On the other hand Arai \cite{ar1} considered the case of $$l=2$$ for the one particle Pauli-Fierz model in the dipole approximation without the self-interacting term $A^2$. Then he derived effective Hamiltonians with an effective potential. In this paper we consider scaling limits of $$l\leq 2$$ for the $N$-particle Pauli-Fierz model in the dipole approximation. In particular, performing a mass renormalization, subtracting from $H(\k)$ some energy reset, and taking $\k\rightarrow \infty$ in the resolvent sense, we derive, for $l=2$, Welton's Hamiltonians, and for $l=1$, Hamiltonians with an observed mass. \subsection{Nonrelativistic models} Let $\add {}(f)$ and $\aaa {} (f)$, $f\in W$, denote the smeared creation and annihilation operators acting on $\fff$, respectively. $\ass {}$ stands for $a$ or $\add{}$. We set $$\ass r(f):=\add {}( \underbrace{0\oplus0\cdots \oplus \stackrel{r_{\rm th}}{f}\oplus \cdots 0\oplus 0}_{d-1\ \mbox{{\footnotesize times}}}).$$ They satisfy the canonical commutation relations, $$[\aaa s(g), \add r(f)]=\delta_{rs}\int_\BR f(k) g(k) dk,$$ $$[\add r(f),\add s(g)]=[\aaa r(f),\aaa s(g)]=0$$ on the finite particle subspace of $\fff$. We take the Coulomb gauge. The radiation field with the ultraviolet cutoff $\laj \in\LL$ is defined by $$A_{j\mu}(x_j):={A}_\mu (x_j,\laj ):=\sh \rrr \lkk \add r(\emr e^{-ikx_j}\widetilde{\la}_j )+\aaa r(\emr e^{ikx_j}\laj ) \rkk,$$ where $e^r(k)=(e^r_1(k),...,e^r_d(k))$ denotes polarization vectors satisying $e^r(k)\cdot e^s(k)=\delta_{rs}$ and $e^r(k)\cdot k=0$, $$\widetilde{\la}_j (k):=\laj (-k),$$ and $\hat{f}$ denotes the Fourier transform of $f$. We assume $$\laj(k)=\laj(-k) $$ to ensure that $A_{j\mu}(x_j)$ is symmetric for each $x\in\BR$. The physically reasonable choice of the ultraviolet cutoff is $$\laj (k)=\frac{\sqrt{c}\hbar \hat{\rho}_j(k)}{\sqrt{(2\pi)^d \omega(k)}}, $$ where $\hat{\rho}_j$ denotes the Fourier transform of the charge distribution $\rho_j$ of the $j$th particle and $c$ denotes the speed of the light. The dispersion relation $\omega(k)$ is given by $$\omega(k):=|k|.$$ The free Hamiltonian $\hf$ in $\fff$ is defined by $$\hf:=\hbar c \rrr \int\omega(k)\add r (k)\aaa r(k) dk,$$ where $\add r(k)$ and $\aaa r(k)$ denote the kernels of $\add r(f)$ and $\aaa r(f)$, respectively. The total Hamiltonian is given by $$H_{\rm total}:=\AAAx+\I\otimes \hf+V\otimes \I,$$ where $\al$'s are non zero coupling constants. We take the dipole approximation, i.e., we replace $A_j(x_j)$ with $$A_j:=A_j(0).$$ Then the Hamiltonian in the dipole approximation is given by $$H:=\AAA+\I\otimes \hf+V\otimes \I.$$ We work with the unit $c=\hbar =1$ in what follows. Since $$[\vp_\mu\otimes 1, 1\otimes \vA_\nu]=0,$$ we have $$H=\ele\otimes \I -\jjj \frac{\al}{\mj}\vp\otimes \vA+\I\otimes\lk \jjj \frac{\at}{2\mj}\vA^2+\hf\rk.$$ It is of mathematical interest to see $A^2_j$-corrections in scaling limits. Then we define, for $\e\geq 0$, $$\he:=\ele\otimes\I -\jjj\frac{\al}{\mj}\vp\otimes \vA+ \I\otimes\lk \jjj \frac{\e \at }{2\mj}A^2_j+\hf\rk$$ with the domain $$D(\he)=D(\ele\otimes\I)\cap D(\I\otimes\hf)\cap D(\I\otimes \jjj \frac{\epsilon\alpha_j^2}{2m_j} A_j^2).$$ Actually, for $\e<1$, $\he$ turns out to be an unphysical Hamiltonian for sufficiently large coupling constants, since in this case $\he$ is unbounded from below. See Remark \ref{999} for details. Let $$\k>1$$ be a scaling parameter. We introduce the scaled Hamiltonian $\hk$ as follows: $$\hk:=\ele\otimes\I -\k^l\jjj \frac{\al }{\mj}\vp\otimes \vA+\k^2 \I\otimes \lk \jjj \frac{\e \at }{2\mj} A^2_j+\hf\rk,$$ where $$l\leq 2.$$ In particular we set $$H(\kappa):=H_1(\k).$$ Unless confusions arise we omit $\otimes \I$ and $\I\otimes$ for simplicity in what follows. Throughout this paper we suppose that $V$ satisfies Hypothesis \vvvv: \bi \item[Hypothesis \vvvv:] $V$ is infinitesimally small with respect to $\jjj\frac{p_j^2}{2m_j}$. \ei First of all we have to establish self-adjointness of $\hk$. Note that if we assume that $\laj/\so,\laj,\omega\laj\in\LL$, then $$ \|\lk -\jjj\frac{\al}{\mj}\vp \vA+ \jjj \frac{\e \at }{2\mj}A^2_j\rk\Psi\|\leq a' \|\lk\jjjj+\hf\rk \Psi\|+b'\|\Psi\|$$ for $\Psi\in D(\hf)$ with some $a'$ and $b'$. Let $$\ddd:=\domain.$$ \bp{self6} Let $\e=1$. We suppose that $\laj/\so,\laj,\omega\laj\in\LL$. (1) $H_{\rm total}$ and $H$ are self-adjoint on $\ddd$ and bounded from below. (2) Let $l=1$. Then $H(\k)$ is self-adjoint on $\ddd$ and bounded from below for all $\k$. \ep \proof See \cite{hi3,hi4} for (1). In the case of $l=1$, $H(\k)$ is derived from the scaling $\al\rightarrow \k\al$ and $\omega\rightarrow \k\omega$ in $H$. Thus the self-adjointness follows from (1) \qed \bp{eself6} Let $\e\not=1$ and $V=0$. Suppose $\laj/\so,\laj,\omega\laj\in\LL$. Then $\hk$ is essentially self-adjoint on any core of $\xxx+\hf$. \ep \proof The proof is due to Arai \cite[Theorem 2.2]{ar2}. Let $L:=\xxx+\hf+\I$. Then we see that $$\|\hk\Psi\|\leq c\| L\Psi\|,$$ and $$|(\hk\Psi, L\Psi)-(L\Psi,\hk\Psi)|\leq c'\|L^\han \Psi\|\|L^\han \Psi\|$$ with some $c$ and $c'$. Thus, by the Nelson commutator theorem \cite[Theorem X.36]{rs2}, $\hk$ is essentially self-adjoint on any core of $L$. Thus the proposition follows. \qed Self-adjointness of $\hk$ for $\e<1$ or $l\not=1$ is discussed later (Theorem \ref{self4}). We denote the self-adjoint extensions of $H_\epsilon$ and $H_\epsilon(\k)$ by the same symbols. \subsection{Problems} For the case of $N\geq 2$ we have to take care of the choice of ultraviolet cutoffs. For simplicity we mainly consider two cases: identical case, $\la_1=\cdots=\la_N:=\la$, and independent case, $\supp \laj\cap \supp \la_i=\emptyset $ for $i\not=j$. Arai \cite{ar1} considered the scaling limit of the case of $l=2$, $\epsilon=0$ and $N=1$. Moreover in \cite{hi1,hi2} distinct scaling limits were studied, but unfortunately $A^2$-corrections did not appear in scaling limits. In this paper we want to improve \cite{hi1,hi2}, and furthemore to see the relationship between Davies' scaling limit and Arai's one. As was seen in subsection 1.1, Davies' scaling limits correspond to $l=1$. We set $$\hz:=\he\lceil_{V=0}.$$ Since $\hz$ commutes with $p_{j\mu}$, $\he$ is decomposable in the spectrum of the particle momentum operators; $$\hz=\int_{\BRN}^\oplus \heq dq.$$ $\heq$ acts on $\fff$. Its ground state energy is denoted by $$E(q):=\inf\sigma(\heq).$$ Then we call $$m^\ast(\mu,\nu,j):=\lk\left.\frac{\partial^2}{\partial q_{j\mu}\partial q_{j\nu}} E(q)\right\lceil_{p=0}\rk\f$$ an {\it observed mass}. In this paper we explicitly give ground state energy of $\hz$ and its observed mass. Thus the list of problems considered in this paper is epitomized as follows. \begin{itemize} \item[(1)] Explicit ground state energy, $\inf\sigma(\hz)$. \item[(2)] Observed mass, $m^\ast(\mu,\nu,j)$. \item[(3)] Deriving effective Hamiltonians $\eff$ through scaling limits of $\hk$ subtracting a divergent contribution $\er(\k)$, $$s-\limk(\hk-\er(\k)-z)\f=(\eff-z)\f\otimes P_\e$$ for $\zcr$, where $P_\e$ denotes a projection to the subspace spanned by ground states of $\xx+\hf$. \item[(4)] What is a reasonable choice of $\er(\k)$? \item[(5)] $l$-dependence of $\eff$. \item[(6)] $A^2_j$-dependence (i.e., $\e$-dependence) of $\eff$. \item[(7)] Ultraviolet cutoff dependence of $\eff$. \end{itemize} We give answers for (1) in Lemma \ref{form}, (2) in Corollary \ref{bbc}, (3) in Theorems \ref{m21}-\ref{m24}, (4) in \kak{we}, (5), (6) and (7) in \kak{star}. This paper is organized as follows. In Section 2 we discuss symplectic group structures of $\hz$. In Section 3 we construct the canonical transformation and diagonalize $\hz$. In Section 4 we derive effective Hamiltonians by scaling limits. In Section 5 we give a remark on $g$-factor shift. Sections 6 and 7 are appendixes. \section{Symplectic group structures} \subsection{Unitary representations} Let $B(W)$ denote the set of bounded operators on $W$. We define for $K\in B(W)$ $$\ov{K}f:=\ov{K\ov{f}}.$$ We introduce a notation. For $A,B,C,D\in B(W)$ we define $$ \mat A B C D \v{a(f)}{\add {}(g)} := \v{a(Af)+\add{}(Bg)}{a(Cf)+\add{}(Dg)}.$$ Set $$ \Tmat A B C D:=\mat A C B D.$$ \begin{definition} Let $S,T\in B(W)$ and set $$\xi:=\mat{S}{\ov T}{ T}{\ov S}:W\oplus W\rightarrow W\oplus W.$$ \begin{itemize} \item [(1)] We say that $\xi$ is of symplectic group $\Sigma$ if \eq{xi} \xi^\ast J \xi=\xi J\xi^\ast =J, \en where $$\xi^\ast=\mat{S^\ast}{T^\ast}{\ov{T}^\ast}{\ov{S}^\ast},\ \ \ J:=\mat{I}{0}{0}{-I}.$$ \item[(2)] We say that $\xi$ is of $\bog$ if (i) $\xi\in\Sigma$ and (ii) $T$ is a Hilbert-Schmidt operator. \end{itemize} \end{definition} $\bog$ is a subgroup of $\Sigma$. We define $$\hm:=L^2(\RR^m)\otimes\fff\cong \int_\rrm^\oplus \fff dq.$$ Suppose that $$\xi=\mat {S} {\ov{T}}{ T} {\ov{S}}\in\bog,$$ and $$L=(L_1,...,L_m)\in\oplus^m W.$$ For $f\in W$ we let $$\v{b(f)}{\bdd(f)}:= \Tmat {S} {\ov{T}}{ T} {\ov{S}} \v{a(f)}{\add{}(f)}$$ and $$B(f):=\I\otimes b(f)+ \nnn (\L, f) p_l\otimes\I,$$ $$\Bdd(f):=\I\otimes \bdd(f)+\nnn (\ov{\L},f) p_l\otimes\I,$$ where $p_l$, $l=1,...,m$, denote the momentum operators in $L^2(\rrm)$. It follows that $$[b(f),\bdd(g)]=(\ov f, g),$$ $$ [b(f),b(g)]=[\bdd(f),\bdd(g)]=0.$$ Moreover since $$\xi\f=J\xi^\ast J=\mat {S^\ast} {-T^\ast} {-\ov{T}^\ast} {\ov{S}^\ast}, $$ we have \eq{aa} \v{a(f)}{\add {}(f)} = \Tmat {S^\ast} {-T^\ast} {-\ov{T}^\ast} {\ov{S}^\ast} \v{b(f)}{\bdd(f)} \en and \eq{aaa} \v{\I\otimes a(f)}{\I\otimes \add {}(f)} = \Tmat {S^\ast} {-T^\ast} {-\ov{T}^\ast} {\ov{S}^\ast} \v{B(f)}{\Bdd(f)} -\nnn \v{ (S \L-\ov T \ov \L, f) p_l\otimes\I} {(\ov S \ov \L- T \L, f) p_l\otimes \I}. \en Define $$\R:=T\L-\ov{S\L},$$ and $$\theta(q,\xi,L):=\exp\lkk\nnn q_l\lkk \add {}(R_l)-a(\ov R_l)\rkk \rkk,\ \ \ q=(q_1,\cdots,q_m)\in\rrm.$$ Furthermore set $$\th:=\int_\rrm^\oplus \theta(q,\xi,L)dq= \exp\lkk\nnn p_l\otimes\lkk \add {}(R_l)-a(\ov R_l)\rkk \rkk.$$ \bp{sym} Suppose $\xi=\mat{S}{\ov{T}}{T}{\ov{S}}\in\bog$. Then there exists a unitary operator $U(\xi)$ of $\fff$ such that for all $f\in W$ \begin{enumerate} \item[(1)] $ U(\xi)\f\bss (f)U(\xi)=\ass{}(f)$, \item[(2)] $U(\xi)\f \th \f \Bss(f)\th U(\xi)=\ass{}(f).$ \end{enumerate} \ep \proof See \cite{be,kmp, ru1} for details \footnote{We easily see that for $\xi_1:=\mat {S} {\ov{T}} {T} {\ov{S}} \in\bog$ and $\xi_2:=\mat U {\ov{V}} V {\ov{U}} \in\bog$, $$U(\xi_2)U(\xi_1) a(f) U(\xi_1)\f U(\xi_2)\f= a\lk(U S+\ov{V}T)f\rk+\add{}\lk(VS+\ov{U}T)f\rk,$$ $$U(\xi_2)U(\xi_1) \add{}(f) U(\xi_1)\f U(\xi_2)\f= a\lk(U\ov{T}+\ov{V}\ov{S})f\rk+\add{}\lk(V\ov{T}+\ov{U}\ov{S})f\rk,$$ and $$\xi_2\cdot \xi_1=\mat {US+\ov V T}{U\ov T +\ov V \ov S}{VS+\ov U T}{V\ov T+\ov U \ov S}. $$ Then $U(\xi_2)U(\xi_1)=U(\xi_2\xi_1).$ Hence the map $\bog\ni\xi\mapsto U(\xi)$ defines a unitary representation of $\bog$.}. \qed We set $$\cu(q,\xi,L):=\theta(q,\xi,L) U(\xi):\fff\rightarrow\fff,$$ and $$\cu(\xi,L):=\th U(\xi):\hhh_m\rightarrow\hhh_m.$$ We call $\cu(\xi,L)$ the canonical transformation associated with $\{\xi, L\}\in\bog\times \oplus^m W$. \subsection{Average of fluctuations} Define $$\pi(f):=i\lkk \add{ }(\bar{f})-a(f)\rkk,\ \ \ f\in W.$$ We want to obtain the average of fluctuation $F_{\xi L}$ of a function $F:\RR^m\rightarrow \BC$; $$F_{\xi L}:=\cu(\xi,L)\f F \cu(\xi,L)$$ $$= U(\xi)\f F(x_1+\pi(R_1),\cdots,x_m+\pi(R_m)) U(\xi),$$ with respect to the bare vacuum. Let $\Gamma$ be the $m\times m$ symmetric matrix defined by $$\Gamma:=\lk ({L}_\mu,{L}_\nu)_W \rk_{1\leq \mu,\nu\leq m},$$ and $\ab{a,b}$ denote the scalar product on $\BC^m$. Suppose $${\rm Rank}\ \Gamma=n\leq m$$ and $T$ is a unitary matrix such that $$T\Gamma T\f={\rm diag} \{\mu_1,\cdots,\mu_n,0,\cdots,0\},\ \ \ \mu:=\mu_1\cdots\mu_n\not=0.$$ Let $$\ft(x):=F(T\f x),\ \ \ x\in \rrm.$$ We define $$\efff(x):=(2\pi \mu)^{-m/2} \int_{\RR^n}\ft(y_1,\cdots,y_n,(Tx)_{n+1},\cdots,(Tx)_m)\times$$ $$\times e^{-\sum_{j=1}^n|(Tx)_j-y_j|^2/(2\mu)}dy.$$ In particular if ${\rm Rank}\ \Gamma=n$, then $$\efff(x)= (2\pi{\rm det}\Gamma )^{-m/2} \int_\rrm F(y)e^{-|x-y|^2/(2{\rm det}\Gamma)} dy.$$ \bl{potential} Let $\xi= \mat {S} {\ov T} {{T}} {\ov{S}} \in\bog$ and $(L_1,\cdots,L_m)\in\oplus^m W$. Suppose that $F$ is such that $$ \int_{\RR^n}|\ft| (y_1,\cdots,y_n,(Tx)_{n+1},\cdots,(Tx)_m)e^{-\sum_{j=1}^n|(Tx)_j-y_j|^2/(2\mu)}dy \in L_{\rm loc}^1(\rrm,dx).$$ Then for $f\in L^2(\RR^m)$, and $g\in L^2(\rrm)$ such that $\cu(\xi,L)g\otimes\Omega\in D(F)$, we see that $g\in D(\efff)$ and \eq{in} (f\otimes \Omega, F_{\xi L} g\otimes \Omega)_\hm =(f,\efff g)_{L^2(\rrm)}. \en \el \proof We see that by \kak{aa} $$a(\ov \R )=b(S^\ast \ov \R )-\bdd (\ov{T}^\ast \ov \R ),$$ $$\add{}(\R)=-b(T^\ast \R )+\bdd(\ov{S}^\ast \R ).$$ Thus we have $$\add {}(\R)-a(\ov \R)=\bdd(\ov{S}^\ast \R +\ov{T}^\ast \ov \R )- b(T^\ast \R +S^\ast \ov \R )$$ $$:=\bdd (\R')-b(\ov{\R'}).$$ It follows from $\xi^\ast J\xi=\xi J\xi^\ast$ that $$\R'=\ov{S}^\ast(T\L-\ov{S\L})+\ov{T}^\ast(\ov{T\L}-S\L)$$ $$=(\ov{S}^\ast T-\ov{T}^\ast S)\L+(\ov{T}^\ast \ov T-\ov{S}^\ast\ov S)\ov \L =-\ov \L.$$ Hence $$\pi(\R)=i\lkk \add {}(\R)-a(\ov\R)\rkk =i\lkk b(\L)-\bdd(\ov \L)\rkk.$$ Then we have \eq{ll} U(\xi)\f \pi(\R) U(\xi)=i\lkk a(L_l)-\add{}(\ov{L}_l)\rkk =-\pi(\L). \en Let $$F_N(x):= \lkk \bea{cc} F(x),&|x|0,$$ where $\rho_\epsilon(x):=\rho(x/\epsilon)/\epsilon^m$, $\rho(x)\geq 0$, ${\rm supp}\rho\subset\{x\in\rrm||x|\leq 1\}$, and $\int_\rrm \rho(x)dx=1$. \kak{ll} leads to $$(f\otimes \Omega, (F_{N\e})_{\xi L} g\otimes \Omega)$$ $$= (2\pi)^{-m/2}\int_\rrm dk e^{ikx}\check{F}_{N\e}(k) (f\otimes \Omega, U(\xi)\f e^{i\ab{k, x+\pi(R)}} U(\xi) g\otimes \Omega)$$ $$=(2\pi)^{-m/2}\int_\BR dk e^{ikx}\check{F}_{N\e}(k) (f\otimes \Omega, e^{i\ab{k, x-\pi(\ov L)}} g\otimes \Omega)$$ $$ =(2\pi)^{-m/2}\int_\rrm dk e^{ikx}\check{F}_{N\e}(k) \ov{f(x)}g(x) e^{-\ab{k,\Gamma k}/2}$$ $$=(f,(F_{N\e})_{\rm eff} g).$$ Taking $\epsilon\rightarrow 0$, we have $(F_{N\e})_{\rm eff}(x)\rightarrow (F_N)_{\rm eff}(x)$, and $(F_{N\e})_{\xi L}\rightarrow (F_N)_{\xi L}$ strongly. Then the lemma follows for $F_N$. Let $f\in C_0^\infty(\rrm)$. Then taking $N\rightarrow\infty$, we also have $(f,(F_N)_{\xi L} g)\rightarrow (f, F_{\xi L} g)$ and $(f, (F_N)_{\rm eff} g)\rightarrow (f,\efff g)$ by the dominated convergence theorem. Hence the lemma follows. \qed \begin{remark} $\efff$ is independent of $\xi$. \end{remark} \subsection{Physical vacuum} Let $\xi=\mat {S}{\ov T}{T}{\ov S}\in\bog$ and $L\in \oplus^{dN} W$. We define $$\Omega(q,\xi,L):=\cu(q,\xi, L)\Omega.$$ Let $K$ be a Hilbert-Schmidt operator. Then there exist complete orthonormal systems (CONS) $\{\psi_i\}$, $\{\phi_i\}$, and a positive sequence $\{\mu_i\}$ such that $$Kf=\sum_{i=1}^\infty \mu_i (\psi_i,f)\phi_i,\ \ \ f\in\hhh_m,$$ with $$\sumi \mu_i^2=\|K\|^2_{\rm HS},$$ where $\|\cdot\|_{\rm HS}$ denotes the Hilbert-Schmidt norm. We define for a finite particle vector $\Psi$ $$\ab{\add{ }|K|\add{ }}\Psi:=s-\limn \sum_{i=1}^n \mu_i\add{}(\ov\psi_i)\add{}(\phi_i)\Psi.$$ The above strong limit actually exists. It is well known that if $$(1)K\ \mbox{is a Hilbert-Schmidt operator}, \ \ \ (2)\ov K^\ast=K,\ \ \ (3)\|K\|<1,$$ then $$U_0(K):=s-\limN \sum_{n=1}^N \frac{1}{n!}\lk-\half \ab{\add{} |K|\add{}}\rk^n\Psi$$ exists for a finite particle vector $\Psi$. See \cite{ar9}. \bl{pv} We have $$\Omega(q,\xi,L)={\rm det}(\I-(TS\f)^\ast TS\f)^{1/4}\theta(q,\xi,L)U_0(TS\f)\Omega.$$ \el \proof See \cite{hi1} for details. \qed \section{Canonical transformations} \subsection{Fixed particle-momentum Hamiltonians} We define the fixed particle-momentum Hamiltonian by $$\heq:=\jjj \mhalf q_j^2-\jjj \frac{\al}{\mj} q_j\cdot A_j+ \jjj \frac{\e \at}{2\mj}A^2_j+\hf.$$ Here $q$ is a fixed particle-momentum: $$q=(q_1,\cdots,q_N)\in\BRN.$$ \bp{self} Suppose that $\laj/\sqrt{\omega}, \laj, \omega\laj\in\LL$ for $j=1,...,N$. Then for all $\aa$ and all $q\in\BRN$, $\heq$ is self-adjoint on $D(\hf)$ and bounded from below. \ep \proof The proof is similar to Proposition \ref{eself6}. Let $L:=\hf+\I$. Then we easily have $$\|\heq\Psi\|\leq c\|L\Psi\|,$$ and $$|(\heq\Psi,L\Phi)-(L\Psi,\heq\Phi)|\leq c'\|L^\han \Psi\|\|L^\han \Phi\|$$ with some constants $c$ and $c'$ for $\Psi,\Phi\in D(\hf)$. Hence $\heq$ is essentially self-adjoint on $D(\hf)$ by the Nelson commutator theorem. Moreover it is seen that $$c''\lk\|\hf \Psi\|^2+\|\jjj\lk \frac{q^2_j}{2\mj}-\frac{\al}{\mj}q_j\cdot A_j+\frac{\e \al}{2\mj}A_j^2\rk\Psi\|^2\rk \leq \|\heq\Psi\|^2+c'''\|\Psi\|^2$$ with some $c''$ and $c'''$. Then $\heq$ is closed on $D(\hf)$. Thus $\heq$ is self-adjoint on $D(\hf)$. \qed By a canonical transformation $\cu(q)$, we will diagonalize $\heq$ as \eq{h1} \cu(q)\f\heq \cu(q)=E(q)+\hf \en with some constant $E(q)$. In \cite{hi1}, \kak{h1} was established for $N=1$, in \cite{hi2} for $N\geq1$ but sufficiently small $|\al|$'s. In this section we improve it as follows: \bi \item[(1)] we extend \kak{h1} to $N\geq1$ and arbitrary $\al$'s, \item[(2)] we derive an explicit form of $E(q)$, \item[(3)] we show analytic properties of $E(q)$ in each $\al$'s. \ei \subsection{Assumptions on ultraviolet cutoffs and coupling constants} In this subsection we introduce assumptions on $\laj$ and $\al$. We suppose the following assumptions on $\laj$'s. \begin{itemize} \item[(A.1)] $\laj$ is rotation invariant, i.e., there exits a function $\varphi_j:[0,\infty)\rightarrow \RR$ such that $\laj (k)=\varphi_j(|k|)$, \item[(A.2)] $\laj/\omega,\omega\laj \in\LL$, \item[(A.3)] $\tvp(s):=\varphi_i(\sqrt{s})\varphi_j(\sqrt{s}) s^{(d-1)/2}\in L^\delta([0,\infty),ds)$ for some $0<\delta<1$, and there exists $0\eta, x>0} \frac{\tvp(x)}{\omega(k)^2-x}dx-2\pi i \tvp(\omega(k)^2)\rk, \en where $S_{d-1}$ denotes the volume of the $(d-1)$-dimensional unit sphere and $$\dd:=\frac{d-1}{d}.$$ We define $$D:=D(k)=(D_{ij}(k))_{1\leq i,j\leq N}:\oplus^N \LL\rightarrow\oplus^N\LL. $$ We introduce two assumptions of ultraviolet cutoffs, \rIl\ and \rIIl, as follows: \bi \item[Hypothesis \rIl:] $\la_1=\cdots=\la_N:=\la$ and $\Im \la(k)\not= 0$ for all $k\in\BR$. \item[Hypothesis \rIIl:] Each $\laj$ has a compact support such that $\supp \laj\cap \supp \la_i=\emptyset $ for $i\not=j$, and $\left|\frac{\laj(k)}{D_{jj}(k)}\right|\rightarrow 0$ as $k$ converges to the boundary of ${\rm supp}\laj$. \ei In what follows we mainly study above two cases. The next assumptions are needed to see the boundedness of the Hamiltonians from below. \bi \item[Hypothesis $\ie$:] Ultraviolet cutoffs satisfies \rIl, and $\aa$ satisfies that $$\lkk \bea{ll} \jjj \atmm j<\frac{1}{\oo}\frac{1}{1-\e},& 0\leq \e<1,\\ \ &\\ \aa\in\RR^N,& \e\geq 1. \ena\right.$$ \item[Hypothesis $\iie$:] Ultraviolet cutoffs satisfies \rIIl, and $\aa$ satisfies that $$\lkk \bea{ll} \atmm j \leq \frac{1}{\ooj}\frac{1}{1-\e},\ \ \ j=1,...,N,& 0\leq \e<1,\\ \ &\\ \aa\in\RR^N,& \e\geq 1.\ena\right. $$ \ei \subsection{Diagonalization of $\heq$} Since $$D_{ij}(0)=m_{ij}+\e\aij\frac{\dd}{2}S_{d-1}\int_0^\infty \frac{\varphi_{ij}(x)}{x}dx,$$ we easily see that $D(0)\f$ exists as an $N\times N$ matrix for \rIl and \rIIl. Set $$ \lk\bea{c} R_1\\ \vdots\\ R_N \ena \rk := \so D(0)\f \lk\bea{c} \la_1\\ \vdots\\ \la_N \ena \rk. $$ We define $$\theta(q):=\exp\lk \jjj\rrr\mmm \frac{\al q_{j\mu}}{\sqrt{2}} \lkk a^r\lk\ov{ \frac{\emr R_j}{\omega^{3/2}}}\rk -\add{r} \lk \frac{\emr R_j}{\omega^{3/2}}\rk\rkk\rk.$$ \bp{dia} \sup. \bi \item[(1)] There exist $\W\in\bog$ and $Q_j$, $j=1,...,N$, such that $$ \theta(q, \W,L)=\theta(q),$$ where $L=(L_1^1,...,L_1^d,...,L_N^1,...,L_N^d)\in\oplus^{dN}W$ with $$L_j^\mu:=\frac{\al}{\sqrt{2}}\oplus_{r=1}^{d-1}\frac{\emr Q_j}{\omega^{3/2}}.$$ \item[(2)] Let $$\cu(q):=\theta(q) U(\W).$$ Then $\cu(q)$ maps $D(\hf)$ onto itself, and $$ \cu(q) \f \heq\cu(q) =E(q)+\hf$$ with some $E(q)$. \item[(3)] \ $E(q)$ can be analytically continued to a neighborhood ${\cal O}$ of $\RR$ in each $\alpha_j$'s. \ei \ep \proof See Appendix. \qed \subsection{Ground state energy} In this subsection we show an explicit form of $E(q)$ for both cases \rIl\ and \rIIl. In the case of $N=1$ it is shown in \cite{hisp}. For the case of $N\geq 2$ one needs a slight modification of \cite{hisp}. We do not, however, need to go into the minor details for the moment. So a proof of the following lemma is taken up in Appendix. \bl{form} (1) Suppose \rIl. Then we have $$E(q)=\half\jjj\frac{q_j^2}{m_j} -\half \eI\left|\jjj\frac{\al}{\mj}q_j\right|^2 +\gI(m)$$ where $$\eI:= \frac{\dd \|\la/\so\|^2}{1+ \epsilon\rho \|\la/\so\|^2},$$ $$\gI(m):=\frac{d }{2\pi}\int_{-\infty}^\infty \frac{\epsilon \rho s^2\|\so\la/(s^2+\omega^2)\|^2}{1+\epsilon\rho \|\so\la/\sqrt{s^2+\omega^2}\|^2}ds,$$ $$\rho:=\dd \jjj\atmm j.$$ (2) Suppose \rIIl. Then we have $$E(q)=\half\jjj\frac{q_j^2}{m_j} -\half\jjj\frac{q_j^2}{\mj} \eII^j +\gII(m),$$ where $$\eII^j:= \frac{\rj\|\laj/\so\|^2}{1+\e \rj \|\laj/\so\|^2},$$ $$\gII(m):= \jjj\frac{d}{2\pi}\int_{-\ii}^\ii\frac{\epsilon\rj s^2\|\so\laj/(s^2+\omega^2)\|^2} {1+\epsilon \rj \|\so\laj/\sqrt{s^2+\omega^2}\|^2}ds,$$ $$\rj:=\dd\atmm j.$$ \el \proof See Appendix. \qed It is an interesting case where $\epsilon=1$. \begin{corollary} (1) Suppose \rIIl and $\epsilon=1$. Then $$E(q) =\jjj \frac{1}{2\mass}q_j^2 + \jjj\frac{d}{2\pi}\int_{-\ii}^\ii\frac{\rj s^2\|\so\laj/(s^2+\omega^2)\|^2} {1+ \rj \|\so\laj/\sqrt{s^2+\omega^2}\|^2}ds,$$ where $$\mass:=m_j+ \at \dd \|\laj/\so\|^2.$$ (2) Suppose $N=1$ and $\epsilon=1$. Then $$E(q)=\frac{1}{2\masss}q^2+ \frac{d}{2\pi}\int_{-\ii}^\ii \frac{ \dd \alpha_1^2 s^2\|\so\la_1/(s^2+\omega^2)\|^2}{m+ \dd \alpha_1^2 \|\so\la_1 /(s^2+\omega^2)\|^2}ds,$$ where $$\masss:=m_1+\alpha_1^2 \dd \|\la_1/\so\|^2.$$ \end{corollary} \bl{ii} (1) Suppose $\ie$. Then for all $q\in\BRN$ $$ \jjjjq-\half \eI\left|\jjj \frac{\al}{\mj}q_j\right|^2>0.$$ (2) Suppose $\iie$. Then for all $q\in\BRN$ $$ \jjjjq-\half \jjj \frac{q_j^2}{\mj} \eII^j>0.$$ \el \proof By the Schwartz inequality we have $$\half \eI\left|\jjj \frac{\al}{\mj}q_j\right|^2\leq \frac{\rho \oo}{1+\e \rho \oo}\jjjjq.$$ Hence $$ \jjjjq-\half \eI\left|\jjj \frac{\al}{\mj}q_j\right|^2\geq \half\lk 1-\frac{\rho\oo}{1+\e\rho\oo}\rk \jjjjq.$$ Since Hypothesis $\ie$ implies that $$1-\frac{\rho\oo}{1+\e\rho\oo}>0,$$ (1) follows. (2) is similarly proven. \qed \subsection{Observed mass} We define the observed mass by $$\ms(\mu,\nu,j):=\left. \lk \frac{\partial^2}{\partial q_{j\mu}\partial q_{j\nu}}E(q)\right \lceil_{q=0}\rk\f.$$ Put $$\ms(\mu,\mu,j):=\ms({\mu,j}).$$ By Lemma \ref{form} we have the following corollaries. \bc{bc} (1) Suppose \rIl. Then $$m^\ast(\mu,\nu, j)=\delta_{\mu\nu} {\mj}\lk\frac{1+\dd (\e \sum_{i=1}^N\atmm i-\atmm j)\|\la/\so\|^2} {1+\e \dd\jjj \atmm j\|\la/\so\|^2}\rk\f. $$ In particular for the case of $\e=1$, $$m^\ast(\mu,\nu, j)=\delta_{\mu\nu} {\mj}\lk\frac{1+\dd\sum_{i\not=j}\atmm i\|\la/\so\|^2} {1+\dd\jjj \atmm j\|\la/\so\|^2}\rk\f. $$ (2) Suppose \rIIl. Then $$ m^\ast(\mu,\nu, j)=\delta_{\mu\nu} \frac{ m_j+\e \dd \al^2\|\laj/\so\|^2}{ 1+(\e-1)\dd \atmm j\|\laj/\so\|^2}. $$ In particular for the case of $\e=1$, $$ m^\ast(\mu,\nu, j)=\delta_{\mu\nu} \lk m_j+\dd \al^2\|\laj/\so\|^2\rk.$$ \ec \proof It is a direct calculation. \qed It is interesting to compare the sizes of observed masses for \rIl and \rIIl. \bc{bbc} Suppose $\e=1$. Let $\ms_{\rm I}(\mu,j)$ be the observed mass for \rIl\ and ${\ms_{\rm II}(\mu,j)}$ for \rIIl. Suppose that $\|\la/\sqrt{\omega}\|\leq \|\laj/\sqrt{\omega}\|$ for $j$. Then $${\ms_{\rm I}(\mu,j)}< {\ms_{\rm II}(\mu,j)}.$$ \end{corollary} \proof Note that $$f(x)=\frac{x}{a+x},\ \ \ a\geq 0,$$ is a monotonously increasing function of $x$. Directly we have $$\frac{1}{{\ms_{\rm I}(\mu,j)}}= \frac{1}{\mj}\lk\frac{1+\dd\sum_{i\not=j}\atmm i\|\la/\so\|^2} {1+\dd\jjj \atmm j\|\la/\so\|^2}\rk $$ $$> \frac{1}{m_j}\lk \frac{1}{1+\dd\atmm j \|\la/\so\|^2}\rk \geq \frac{1}{m_j}\lk \frac{1}{1+\dd\atmm j \|\la_j/\so\|^2}\rk$$ $$= \frac{1}{\ms_{\rm II}(\mu,j)}. $$ Hence the corollary follows. \qed \subsection{Diagonalization of $\hz$ and its self-adjointness} We define $$\theta:= \exp\lk \jjj\rrr\mmm \frac{\al }{\sqrt{2}} p_{j\mu} \otimes \lkk a^r\lk\ov{ \frac{\emr R_j}{\omega^{3/2}}}\rk -\add{r} \lk \frac{\emr R_j}{\omega^{3/2}}\rk\rkk\rk$$ and $$\cu:= \theta U(\W).$$ Note that $\cu$ maps $\ddd$ onto itself: $$\cu:\ddd\rightarrow \ddd.$$ Moreover we define $E(p)$ by the pseudo differential operator with the kernel $E(q)$ in $\LR$. \bl{dia1} Suppose \rIl or \rIIl. Then for $\Phi\in \ccc\widehat{\otimes} D(\hf)$, \eq{lll} \cu\f\hz\cu \Phi =(E(p)\otimes \I+\I\otimes \hf)\Phi. \en \el \proof Let $\Phi=g\otimes\psi$ and $\Psi=f\otimes \phi$, $g,f\in\ccc$, $\psi,\phi\in D(\hf)$. We note that $\cu \Phi\in \ddd$. Then the left hand side of \kak{lll} is well defined. We have by Proposition \ref{dia} $$(\Psi, \cu\f \hz \cu \Phi)_\hhh$$ $$=\int_{\BRN} dq (\widehat{\cu \Psi}(q), \widehat{\hz \cu \Phi}(q))_\fff= \int_{\BRN} dq (\cu(q)\widehat{\Psi}(q), \heq \cu(q) \widehat{\Phi}(q))_\fff$$ $$ \int_{\BRN} dq \ov{\widehat{f}}(q) \widehat{g}(q) (\cu(q){\psi}, \heq \cu(q) {\phi})_\fff = \int_{\BRN} dq \ov{\widehat{f}}(q) \widehat{g}(q) ({\psi}, (E(q)+\hf){\phi})_\fff$$ $$ =(\Psi, (E(q)\otimes\I+\I\otimes \hf)\Phi)_\hhh.$$ Thus the lemma follows. \qed \bl{self2} Suppose $\ie$ or $\iie$. Then \eq{sp} \cu\f\hz\cu=E(p)\otimes \I+\I\otimes \hf \en holds on $\ddd$. In particular $\he$ is self-adjoint on $\ddd$ and bounded from below. \el \proof By the assumption on $\aa$, $E(p)$ is nonnegative. Then $E(p)+\hf$ is self-adjoint on $\ddd$ and bounded from below. For $\Phi\in\ddd$, we can take a sequence $\Phi_n\in \ccc\widehat{\otimes}D(\hf)$ such that $\Phi_n\rightarrow\Phi$ and $(E(p)\otimes\I+\I\otimes \hf) \Phi_n\rightarrow (E(p)\otimes\I+\I\otimes \hf)\Phi$ strongly as $n\rightarrow \infty$. Since $\he$ is closed, $\he\cu\Phi_n\rightarrow \he\cu\Phi$ as $n\rightarrow\infty$. Thus \kak{sp} holds. Since $\cu$ maps $\ddd$ onto itself, $\he$ is self-adjoint on $\ddd$ and bounded from below. \qed \begin{remark} \label{999} Let $\e<1$. Then $E(p)$ is unbounded from below for sufficiently large coupling constants. Actually in the case of $N=1$, for $$\alpha_1^2\geq m_1 \lk \|\la_1/\so\|^2(1-\e)\rk\f,$$ $E(p)$ is unbounded from below. Hence $\hz$ is unbounded from below for sufficiently large coupling constants. We should not overlook that the approximation of neglecting the self-interaction $A^2$ in $H$ is not reasonable for sufficiently large coupling constants. \end{remark} \bt{self4} Suppose $\ie$ or $\iie$. Then $\he$ is self-adjoint on $\ddd$ and bounded from below. \et \proof By Lemma \ref{self2} and the closed graph theorem there exists a constant $c$ such that $$\|(\jjjj+\hf)\Psi\|\leq c(\|\he \Psi\|+\|\Psi\|),\ \ \ \Psi\in\ddd.$$ Thus $$\|V\Psi\|\leq ac (\|\he \Psi\|+\|\Psi\|)+b\|\Psi\|$$ with some sufficiently small $a$, and some $b$, the Kato-Rellich theorem leads to the lemma. \qed \bc{gs} Suppose \rIl. Then $$\inf \sigma(\hz) =\gi(m)=\frac{d }{2\pi}\int_{-\infty}^\infty \frac{\epsilon \dd \jjj \frac{\al^2}{\mj} s^2\|\so\la/(s^2+\omega^2)\|^2}{1+\epsilon \dd \jjj \frac{\al^2}{\mj} \|\so\la/\sqrt{s^2+\omega^2}\|^2}ds.$$ Suppose \rIIl. Then $$\inf \sigma(\hz)=\gii(m)= \jjj\frac{d}{2\pi}\int_{-\ii}^\ii\frac{\epsilon \dd \frac{\al^2}{\mj} s^2\|\so\laj/(s^2+\omega^2)\|^2} {1+\epsilon \dd \frac{\al^2}{\mj}\|\so\laj/\sqrt{s^2+\omega^2}\|^2}ds. $$ \ec \proof It follows from Lemmas \ref{form} and \ref{dia1}. \qed \section{Effective Hamiltonian} \subsection{Scaled Hamiltonians} Here we recall the scaled Hamiltonian $\hk$; $$\hk:=\ele\otimes\I -\k^l\jjj \frac{\al }{\mj}\vp\otimes \vA+\k^2 \I\otimes \lk \jjj \frac{\e \at }{2\mj} A^2_j+\hf\rk.$$ $\hk$ is derived from $\he$ scaled as \eq{1}\omega\rightarrow \k^2\omega, \en \eq{2}\al\rightarrow \k^l\al, \en \eq{3}\epsilon\rightarrow \k^{2-2l}\epsilon. \en Replacing $\omega,\al,\epsilon$ as in \kak{1}--\kak{3}, we see that \eq{i2} \theta\longrightarrow \exp\lk \k^{l-2}\jjj \mmm \frac{\al}{\sqrt{2}} p_{j\mu}\otimes \lkk a^r\lk\ov{\frac{\emr R_j}{\omega}}\rk-\add{r}\lk\frac{\emr R_j}{\omega}\rk\rkk\rk, \en $$ -\half \eI\left|\jjj\frac{\al}{\mj}q_j\right|^2 \longrightarrow -\k^{2l-2} \half \eI\left|\jjj\frac{\al}{\mj}q_j\right|^2, $$ $$ -\half\jjj\frac{q_j^2}{\mj} \eII^j \longrightarrow -\k^{2l-2} \half\jjj\frac{q_j^2}{\mj} \eII^j,$$ $$ \gI(m)\rightarrow \gI(m/k^2),$$ $$ \gII(m)\rightarrow \gII(m/k^2).$$ \bl{i1} $U(\W)$ leaves invariant under the scaling \kak{1},\kak{2} and \kak{3}. \el \proof See Lemma \ref{bog} in Appendix. \qed Let \eq{we} \er(p):=\lkk\bea{ll} -\half \eI\left|\jjj\frac{\al}{\mj}p_j\right|^2,&\ {\rm for} \ {\rm I_\lambda},\\ \ & \\ -\half\jjj\frac{p_j^2}{\mj} \eII^j, &\ {\rm for}\ {\rm II_\lambda}. \ena\right. \en Moreover we set $$g(m):=\lkk \bea{ll} \gi(m),& {\rm for}\ {\rm I_\lambda},\\ \ & \\ \gii(m),& {\rm for}\ {\rm II_\lambda}. \ena\right.$$ We define $\cu_\k$ by $\cu$ with $\omega$,$\al$ and $\e$ replaced as in \kak{1},\kak{2}, and \kak{3}, respectively. \bl{dia2} Suppose ${\rm I}_\lambda^\alpha(\k^{2l-2}\e)$ or ${\rm II}_\lambda^\alpha(\k^{2l-2}\e)$. Then $$\cu_\k\f \hk\cu_\k= \half\jjj\frac{p_j^2}{m_j} +\k^{2l-2}\er(p)+\kt \hf + g(m/\kt)+\cu_\k\f V\cu_k.$$ In particular $\hk$ is self-adjoint on $\ddd$ and bounded from below. \el \proof It follows from Lemma \ref{self2}. \qed \bl{gro} Suppose \rIl or \rIIl. Then there exists the unique ground state $\Omega_\e$ of $\xx+\hf$. Moreover \eq{gp} P_\e:=U(\W)P_\Omega U(\W)\f \en is the projection to the eigenspace spanned by $\Omega_\e$. \el \proof Set $q=0$ in Lemma \ref{dia} (2). Then $$\cu(0)\f\lk \epsilon\jjj \frac{\al^2}{2\mj}A^2_j +\hf\rk\cu(0)= \hf+g(m).$$ The right hand side has the unique ground state $\Omega$. Thus $\cu(0)\Omega=U(\W)\Omega=\Omega_\e$ is the unique ground state of $\xx+\hf$. Hence the lemma follows. \qed \bl{scaling} (1) Suppose $l=2$. Then $\cu_\k$ is independent of $\k$. (2) Suppose $l<2$. Then $s-\limk \cu_\k=U(\W).$ \el \proof It follows from \kak{i1} and \kak{i2}. \qed We write as $$\cu_\ii:=\lkk\bea{ll} \cu,&l=2,\\ \ & \\ U(\W),&l<2. \ena \right. $$ \subsection{Effective potentials} Define $\veff$ acting on $\LR$ by $$(f, \veff g)_{\LR}: =(f\otimes \Omega, \lk\cu_\infty\f V \cu_\infty\rk g\otimes \Omega)_\hhh.$$ Note $$D(\veff)\supset D(\jjjj).$$ We define the $N\times N$ correlation matrix $\tlm=\lk\tlm_{ij}\rk_{1\leq i,j\leq N}$ by $$\tlm_{ij}:=\half \aij\dd \lk \frac {Q_i}{\omega^{3/2}}, \frac {{Q_j}}{\omega^{3/2}}\rk.$$ Note that $$ \lk {L}_i^\mu,{{L}_j^\nu}\rk=\frac{\aij}{2}\dd \delta_{\mu\nu} \lk \frac {Q_i}{\omega^{3/2}}, \frac {{Q_j}}{\omega^{3/2}}\rk=\delta_{\mu\nu}\tlm_{ij}. $$ Suppose that ${\rm Rank}\tlm=M$ and $T=T_\Delta $ is an $N\times N$ unitary matrix such that $$T\tlm T\f={\rm diag}\{\mu_1,\cdots,\mu_M,0,\cdots,0\}, \ \mu=\mu_1\cdots\mu_M\not=0.$$ Let $$\TT:=\lk\bea{ccc} T_{11}\I_d&\cdots&T_{1N}\I_d\\ \vdots&\ &\vdots\\ T_{N1}\I_d&\cdots&T_{NN}\I_d \ena\rk.$$ Here $\I_d$ denotes the $d\times d$ identity matrix. \bl{eff2} Let $l=2$. Suppose $$\int_{\BRM} |V_{\TT\f}|(y_1,\cdots,y_M, (\TT x)_{M+1},\cdots,(\TT x)_N) e^{-\sum_{j=1}^K|(\TT x)_j-y_j|^2/(2\mu)}dy \in L_{\rm loc}^1(\BRN).$$ \bi \item[(1)] Suppose that ${\rm Rank}\tlm=N.$ Then we have $$\veff(x)= (2\pi)^{-dN/2} ({\rm det}\tlm )^{-d/2} \int_{\BRN} e^{-|x-y|^2/(2{\rm det}\tlm)}V(y) dy.$$ \item[(2)] Suppose that ${\rm Rank}\tlm=M0$ independent of large $\k$. Then \eq{vv} \|V_\k\Psi\|\leq \frac{a}{\e(\all)}\|(\jjjj+\er(p)) \Psi\|+b\|\Psi\|. \en Hence $\effk$ for $l\not=1$ is self-adjoint on $\ddd$ and bounded from below. By \kak{dia1}, \kak{p} holds on $\ccc\widehat{\otimes}D(\hf)$, and $\cu_\k$ maps $\ddd$ onto itself. Then \kak{p} holds on $\ddd$ by a limiting argument. Thus (1) follows. For $l=1$, $\xxx+\er(p)$ is a nonnegative self-adjoint operator. Thus $\effk$ is self-adjoint on $\ddd$ and bounded from below. Thus (2) follows similarly. \qed \bl{self5} (1) Suppose $\ie$ or $\iie$, and $l=1$. Then $\eff$ is self-adjoint on $D(\xxx)$ and bounded from below. (2) Let $l\not=1$. Then $\eff$ is self-adjoint on $D(\xxx)$ and bounded from below. \el \proof It is clear. We omit it.\qed Let $P_\e$ denotes the projection to the subspace spanned by the unique ground state of $\jjj\frac{\e \al^2}{2\mj}A_j^2+\hf$. Note $P_0:=P_\Omega$ is the projection to $\Omega$. \bt{m21} Let $l=2$. Suppose \rIl or \rIIl. Moreover suppose $|\veff|\in L_{\rm loc}^1(\BRN)$. Then for all $\aa$ and $\zcr$, $$s-\!\limk (\ren-z)\f = \theta\lkk \lk \eff -z\rk\f\otimes \P\rkk \theta\f.$$ \et \bt{m22} Let $12.$$ \ec \proof The self-adjointness follows from Proposition \ref{self6}. Since $\sigma_j$ and $\cu_k$ are commutative, by Theorem \ref{m23}, we have $$s-\limk (H(\k)-g(m/\kt)-z)\f =\lk\frac{1}{2(m+\dm)}p^2-\frac{g}{4m}\sigma\cdot \curl a+V-z\rk\f\otimes P_1.$$ Hence the corollary follows. \qed \section{Appendix} \subsection {A proof of Proposition \ref{dia}} \subsubsection{Construction of canonical transformations} We recall $$ D_{ij}:= \mij-\epsilon \aij \frac{ \dd}{2} S_{d-1}\lk \lime \int_{|\omega(k)^2-x|>\eta, x>0} \frac{\tvp(x)}{\omega(k)^2-x}dx-2\pi i \tvp(\omega(k)^2)\rk.$$ \bl {ij} Suppose \rIl or \rIIl. Then $\|D_{ij}\|_\ii<\infty$. \el \proof Set $$ \ttvp(s):=\lime\int_{|s-x|>\eta, x>0} \frac{\tvp(x)}{s-x}dx.$$ Then by (A.3) and \cite{ti} we see that for $s,t\geq 0$ $$|\ttvp(s)-\ttvp(t)|\leq C_{ij}|s-t|.$$ Hence $\ttvp$ is continuous. Since $$\lim_{|s|\rightarrow \infty}\ttvp(s)=0,$$ it follows that $\|D_{ij}\|_\infty<\ii.$ \qed \bl{lambda} Suppose \rIl. Then there exists $$Q=\lk \bea{c}Q_1\\\vdots\\ Q_N\ena\rk\in\oplus^N\LL$$ such that (1) $DQ=\so\lk \bea{c} \la_1\\ \vdots\\ \la_N \ena\rk,$ (2) $\|\omega^{(d-1)/2}Q_j\|_\infty<\infty$ and $\|\omega^{(d-3)/2}Q_j\|_\infty<\infty$, (3) $Q_j/\omega^{3/2}\in \LL.$ \el \proof We set $$D(k):=M-\e\dd g(k) A(\alpha_1,\cdots,\alpha_N),$$ where $$M:= {\rm diag}\{m_1,\cdots,m_N\},$$ $$A(\alpha_1,\cdots,\alpha_N):= \lk\begin{array}{ccc} \alpha_1\alpha_1&\cdots&\alpha_1\alpha_N\\ \vdots& \ &\vdots \\ \alpha_N\alpha_1 &\cdots & \alpha_N\alpha_N \end{array} \rk,$$ and $$g(k):=\lim_{\eta\downarrow 0}\int_\BR\frac{\omega(k')|\la(k')|^2}{\omega(k)^2-\omega(k')^2+i\eta}dk'.$$ $A$ is a positive definite matrix and we denote its eigenvalues by \eq{aaaa} \{a_1,\cdots,a_M,0,\cdots,0\},\ \ \ a_j>0,\ \ \ j=1,...,M, \en with some $M\leq N$. $A_M$ denotes the $M$-dimensional space spanned by eigenvectors with eigenvalues $a_1,...,a_M$, and its orthogonal complement by $A_M^\perp$. Set $m=\min\{m_1,\cdots,m_N\}$, $a=\min \{a_1,\cdots,a_M\}$. For $z\in A_M^\perp$ we have $$(z,D(k)z)=(z,Mz)\geq m|z|^2.$$ For $z\in A_M$, we have $$|(z,D(k)z)|\geq \e \dd |\Im g(k)||(z,Az)|\geq \e a \dd |\Im g(k)| |z|^2.$$ Hence for $\e_0<|k|<\e_1$ with $0<\e_0$ and $0<\e_1$ there exists $d'=d'(\e_0,\e_1)>0$ such that $$ |(z,D(k)z)|\geq d'|z|^2.$$ For $k=0$ it follows that \eq{44} (z,D(0)z)=(z,Mz)-\e \dd g(0) (z, Az)\geq m |z|^2, \en and as $|k|\rightarrow \infty$ \eq{45} (z,D(k)z)\rightarrow (z,Mz)\geq m|z|^2.\en Combining \kak{44} with \kak{45} we see that for $|k|\leq \e_0$ or $|k|\geq\e_1$ with some sufficiently small $\e_0$ and sufficiently large $\e_1$ there exists $d''$ such that $$|(z,D(k)z)|>d''|z|^2.$$ Hence for all $k\in\BR$ $$|(z,D(k)z)|>\min\{d',d''\}|z|^2.$$ In particular the inverse $D\f$ of $D$ exists as a bounded operator of $\oplus^N\LL$. Putting $Q:=\sqrt{\omega} D\f \lk \bea{c} \la_1\\ \vdots\\ \la_N \ena\rk,$ we see that $Q$ satisfies (1). It is seen that $$\|\omega^n Q_j\|_\infty \leq \|D\f\|\|\omega^{n+(\han)}\laj\|_\infty.$$ From $\|\laj\omega^{(d-2)/2}\|_\infty<\infty$ and $\|\laj \omega^{d/2}\|_\infty<\infty$ (see (A.4)), (2) follows. Moreover form $$\|\omega^n Q_j\|_\LL \leq \|D\f\|\|\omega^{n+(\han)}\laj\|_\LL,$$ (3) follows, since $\la/\omega\in\LL$ (see (A.2)). Thus the lemma follows. \qed \bl{lambda2} Suppose \rIIl. Then there exists $Q$ such that (1)--(3) in Lemma \ref{lambda} are satisfied. \el \proof Note that $$D_{jj}(k)\not =0,\ \mbox{for}\ k\in \{k\in\BR|\laj(k)\not =0\}.$$ Hence we can define $Q=(Q_1,\cdots,Q_N)$ by $$Q_j(k):=\lkk \bea{ll} \frac{\sqrt{\omega(k)} \laj(k)}{D_{jj}(k)},& k\in \{k\in\BR|\laj(k)\not =0\},\\ 0,& k\not \in {\rm supp }\laj. \ena\right.$$ Since $\left|\frac{\laj(k)}{D_{jj}(k)}\right|\rightarrow 0$ as $k$ converges to the boundary of ${\rm supp}\laj$, $Q(k)$ is well defined for all $k\in\BR$, and we can easily check that $Q$ satisfies (1)--(3). Thus the lemma follows. \qed Let $$R_j:=\so D(0)\f\lk\bea{c} \la_1\\ \vdots\\ \la_N \ena\rk.$$ Define $$Gf(k):=\lime \int_\BR\frac{f(k')}{(\omega(k)^2-\omega(k')^2+i\eta) (\omega(k)\omega(k'))^{(d-2)/2}}dk'.$$ $G$ is a skew symmetric bounded operator of $\LR$. Let $$\tmn f:= \delta_{\mu\nu}f + \e \jjj \at Q_j \omega^{(d-2)/2}G\omega^{(d-2)/2}\dmn \sqrt{\omega}\laj f.$$ \bl{qs} Suppose \rIl or \rIIl. Then the following algebraic relations hold: \bi \item[(1)] $\sum_{\beta=1}^d T_{\mu\beta}^\ast d_{\beta \nu}(1/\omega^2) Q_j= d_{\mu\nu}{R_j}/{\omega^{3/2}}.$ \item[(2)] $[\omega^2,\tmn^\ast]f= -\jjj \al^2\lk Q_j, f \rk_\LL \dmn\so\laj.$ \item[(3)]$\tmn\so\laj=\delta_{\mu\nu}m_jQ_j.$ \item[(4)]$\tmn^\ast d_{\nu\beta}T_{\beta \gamma}=d_{\mu\gamma}.$ \item[(5)]$e_\mu^r \tmn d_{\nu\beta}T_{\beta\gamma}e_\gamma^s=\delta_{rs}.$ \ei \el \proof By (1) of Lemmas \ref{lambda} and \ref{lambda2}, it is proven. See \cite[Lemma 2.6]{hi2} for details. \qed We define the bounded operator $W_\pm=(\wpmrs)_{1\leq r,s\leq d-1}:W\rightarrow W$ by $$\wprs f:=\half \emr\lk \omega^{-\han}\tmn^\ast\omega^\han+\omega^\han\tmn^\ast\omega^{-\han}\rk\ens f,$$ $$\wmrs f:=\half \emr\lk \omega^{-\han}\tmn^\ast\omega^\han-\omega^\han\tmn^\ast\omega^{-\han}\rk\widetilde{\ens f}.$$ It is checked by (A.4), Lemmas \ref{lambda} and \ref{lambda2} that $$\|\wpmrs f\|\leq c\|f\|,$$ where $$ c= 1+\half\lk \| \omega^{(d-3)/2} Q_j\|_\ii\|\omega^{d/2}\laj\|_\ii +\|\omega^{(d-1)/2}Q_j\|_\ii \|\omega^{(d-2)/2}\|_\ii\rk \|G\|.$$ We define $$\W:=\mat{W_+}{\ov{W_-}}{W_-}{\ov{W_+}}.$$ \bl{bog} Suppose \rIl or \rIIl. Then $\W\in\bog.$ Moreover $\W$ is invariant under the scaling $\omega\rightarrow \kt \omega$, $\al\rightarrow \k^l\al$, and $\e\rightarrow \k^{2l-2}\e$. \el \proof It follows from Lemma \ref{qs} (4) and (5). \qed We define $$\v{b(f)}{\bdd(f)}:= { }^T\W\v{a(f)}{\add {}(f)}.$$ Let $$\vl:=\jjj \mmm q_{j\mu}{L}_j^\mu\in W$$ with $${L}_j^\mu :=\frac{\al }{\sqrt{2}} \oplus_{r=1}^{d-1} \frac{e_\mu^r Q_j}{\omega^{3/2}}.$$ $L_j^\mu$ is well defined by Lemmas \ref{lambda} and \ref{lambda2}. We define $$\v{B_q(f)}{\Bdd_q(f)}:= \v{b(f)}{\bdd {}(f)}+ \v{(\vl,f)_W}{(\ov{\vl},f)_W}. $$ By \kak{aa} we have $$\v{a(f)}{\add{}(f)}={ }^T(\W\f) \v{B_q(f)}{\Bdd_q(f)} -\v{ \lk W_+{\vl}-\ov{W_-}\ov{\vl}, f\rk_W}{ \lk \ov{W_+{\vl}-\ov{W_-}\ov{\vl}}, f\rk_W}. $$ By Proposition \ref{sym} there exists a unitary operator $U(\W)$ such that $$U(\W)\f \bss(f)U(\W)=\ass{}(f).$$ Recall $$\theta(q):=\exp\lk \jjj\rrr\mmm \frac{\al q_{j\mu}}{\sqrt{2}} \lkk a^r\lk\ov{ \frac{\emr R_j}{\omega}}\rk -\add{r} \lk \frac{\emr R_j}{\omega}\rk\rkk\rk$$ and $$\cu(q):=\theta(q)U(\W).$$ {\it A proof of Proposition \ref{dia} (1)} By Lemma \ref{qs} (1), we have $$W_-\vl-\ov{W_+\vl}=-\jjj \mmm q_{j\mu}\frac{\al}{\sqrt{2}}\oplus_{r=1}^{d-1} \frac{\emr R_j}{\omega}.$$ Hence $\theta(q,\W,L)=\theta(q)$ follows. \qed {\it A proof of Proposition \ref{dia} (2)}\\ It is seen that by Lemma \ref{qs} (2) and (3), for $f\in W$, $$[\heq, B_q(f)]=-B_q((\oplus^{d-1}\omega) f),$$ $$[\heq, \Bdd_q(f)]=\Bdd_q ((\oplus^{d-1}\omega) f)$$ on some dense domain. Using these commutation relations we can prove that \eq{h11} e^{it\heq}B^\dagger (f)e^{-it\heq}=B^\dagger (e^{it\oplus^{d-1}\omega} f), \en \eq{h2} e^{it\heq}B (f)e^{-it\heq}=-B (e^{it\oplus^{d-1}\omega} f). \en Then we have $$B(f)e^{it\heq}\cu(q)\Omega=0,$$ for all $f\in W$. It implies that $e^{it\heq}\cu(q)\Omega=\mbox{constant}\times \cu(q)\Omega$. Since $e^{it\heq}$ is one parameter unitary group, \eq{h3} e^{it\heq}\cu(q)\Omega=e^{itE(q)}\cu(q)\Omega \en with some constant $E(q)$. Hence by \kak{h11},\kak{h2} and \kak{h3} we obtain that $$\cu(q)\f e^{it \heq} \cu(q)\add {}(f_1)\cdots \add{}(f_n)\Omega= e^{it(\hf-E(q))}\add {}(f_1)\cdots \add{}(f_n)\Omega.$$ Then $$\cu(q)\f e^{it \heq}\cu(q)=e^{it (\hf-E(q))}.$$ Since $\heq$ is self-adjoint on $D(\hf)$, we see that $\cu(q)$ maps $D(\hf)$ onto itself and $\cu(q)\f \heq \cu(q)=\hf-E(q)$ holds by Stone's theorem. \qed \subsubsection{Analytic continuation} \bl{ana1} Suppose \rIl. Let $f\in\LL$. Then $Q_jf$ can be strongly analytically continued to a neighborhood ${\cal O}$ of $\RR$ in each $\alpha_i$'s. \el \proof We fix $j$. We can find a neighborhood ${\cal O}$ of $\RR$ such that, for $z_j\in{\cal O}$, $$D(z_j)=D(z_j,k):=M-\e \dd g(k) A(\alpha_1,\cdots,z_j,\cdots,\alpha_N)$$ is a strongly analytic bounded operator on $\oplus^N\LL$, and $$\inf_{k\in\BR}|(w, D(z_j,k)w)|>\delta|w|^2,\ \ \ \mbox{for}\ w\in\BC^N,$$ with some positive $\delta$. Hence $D(z_j)\f$ exists as a bounded operator for $z_j\in{\cal O}$ and $$\|D(z_j)\f\|\leq\frac{1}{\delta}.$$ We can take a sufficiently small neighborhood ${\cal O}(z_j)\subset{\cal O}$ of $z_j$ such that \eq{iff} \inf_{z_j\in{\cal O}(z_j)}\|D(z_j)\f\|<\infty. \en Since $$\frac{D(z_j)\f-D(z_j')\f}{z_j-z_j'}=D(z_j)\f\lk \frac{D(z_j')-D(z_j)}{z_j-z_j'}\rk D(z_j')\f,$$ by \kak{iff} we conclude that $D(z_j)\f$ is strongly differentiable in ${\cal O}$. Thus $D(z_j)\f$ is a strongly analytic bounded operator. Hence the lemma follows. \qed \bl{ana2} Suppose \rIIl. Let $f\in\LL$. Then $Q_jf$ can be strongly analytically continued to a neighborhood ${\cal O}$ of $\RR$ in each $\alpha_i$'s. \el \proof Note $$Q_j(k)=\frac{1}{\e\dd\alpha_j^2}\frac{1}{\frac{m_j}{\e \dd \alpha_j^2}-g_j(k)}.$$ Here $$g_j(k):= \lim_{\eta\downarrow 0}\int_\BR \frac{\omega(k')|\laj(k')|^2}{\omega(k)^2-\omega(k')^2+i\eta}dk'.$$ It is seen that $$(1)\ g_j(0)<0,\ \ \ (2)\ \Im g_j(k)<0\ {\rm for}\ k\in {\rm supp}\laj,$$ $$ (3)\ g(k)=0\ {\rm for}\ k\not \in {\rm supp}\laj, \ \ \ (4)\ g_j\ \mbox{is continuous}.$$ Hence, for arbitrary $\delta>0$, $${\rm dist}({\rm Ran}g_j,[\delta,\infty))>M$$ with some positive $M$. Thus $Q_j f$ can be analytically continued to $${\cal O}':=\{a+ib,a\in\RR,b\in\RR||a|>\delta,|b|0.$$ Let $1/a>0$ be the width of the momentum lattice, and some $L>0$ be fixed. For $$l=(l^1,\cdots,l^d) \in (2\pi {\Bbb Z}/a)^d$$ we define $$\Gamma(l):= \left[l^1,l^1+\frac{2\pi}{a}\rk \times\cdots\times \left[l^d,l^d+\frac{2\pi}{a}\rk.$$ Let $|l|:=\max_{r}|l^r|$. Suppose $$|l|\leq 2\pi L.$$ Set $$D:=(2[aL]+1)^d,$$ where $[z]$ denotes the integer part of $z$. We named the lattice points of the rectangle with width $2[aL]$ centered at the origin as $$l_1,l_2,\cdots, l_{D}.$$ We set $$\qq:=\sh\frac{1}{\sqrt{\od (l)}} \lkk \add{r}\lk{\chi_{\Gamma(l)}}\rk+ a^r\lk {\chi_{\Gamma(l)}}\rk \rkk_{1\leq r\leq d-1,|l|\leq 2\pi L},$$ $$\pp:=\frac{i}{\sqrt{2}} \sqrt{\od (l)} \lkk \add{r}\lk {\chi_{\Gamma(l)}}\rk - a^r\lk {\chi_{\Gamma(l)}}\rk \rkk_{1\leq r\leq d-1,|l|\leq 2\pi L},$$ where $\chi_I$ denotes the characteristic function of $I\subset \BR$. Define $$ \heqq:=\jjj \lkk \frac{1}{2\mj}q_j^2-\frac{\al}{\mj}\mmm (q_{j\mu}\vv^j,\qq)+\epsilon \mmm \frac{\al^2}{2\mj}(\qq,\vv^j)(\vv^j,\qq)\rkk $$ \eq{ap} +\half\lkk (\pp, \pp)+(\qq, A_0\qq)\rkk-\half \tr\sqrt{A_0} . \en Here $A_0$ is the $(d-1)D\times (d-1)D$ symmetric matrix defined by $$A_0:=\lk \I_{(d-1)} \delta_{ll'}\od (l)^2\rk_{|l|,|l'|\leq 2\pi L}$$ \eq{az} = \lk \begin{array}{cccc} \od (l_1)\I_{(d-1)}& \ &\ &\ \\ \ &\od (l_2)\I_{(d-1)}& \ &\ \\ \vdots &\ldots &\ddots &\vdots \\ \ & \ & \ & \od (l_{D})\I_{(d-1)} \end{array} \rk, \en and $$\vv^j:=\vv:=\lk \begin{array}{c} \sqrt{\od (l_1)}\la(l_1)e_\mu^1(l_1)\\ \vdots \\ \sqrt{\od (l_1)}\la(l_1)e_\mu^{(d-1)}(l_1)\\ \vdots\\ \vdots\\ \sqrt{\od (l_D)}\la(l_D)e_\mu^1(l_D)\\ \vdots \\ \sqrt{\od (l_D)}\la(l_D)e_\mu^{(d-1)}(l_D) \end{array} \rk\in \RR^{(d-1)D}.$$ Here $\I_{(d-1)}$ denotes the $(d-1)\times (d-1)$ identity matrix. It is seen that \eq{uni} (\heqq-z)\f \rightarrow (\he(q,\delta)-z)\f \en uniformly as $a\rightarrow \infty$ and then $L\rightarrow \infty$. Here $\he(q,\delta)$ denotes $\heq+\delta N$. We define the projection $P$ by $$P:=\epsilon \jjj \mmm \atmm j|\vv\rangle\langle \vv|.$$ Then $$\heqq= \jjj \lkk \mhalf q_j^2-\frac{\al}{\mj}\mmm (q_{j\mu}\vv,\qq)\rkk +\half (\qq, P\qq)$$ $$+ \half(\pp,\pp)+\half(\qq, A_0\qq)-\half\tr\sqrt{A_0}.$$ Set $$A:=A_0+P.$$ Since $\od (k)>\delta>0$, the inverse of $A_0$ exists, and hence $A$ also has the inverse, and it is symmetric. Then we have $$= \jjj \mhalf q_j^2-\jjj \mmm \frac{\al}{\mj}(AA\f q_{j\mu}\vv,\qq)+\half (\qq, A\qq)+ \half(\pp,\pp)-\half\tr\sqrt{A_0}$$ $$= \jjj \mhalf q_j^2+\half((\qq-\ff), A(\qq-\ff))-\half(\ff,A\ff)+ \half(\pp,\pp)-\half\tr\sqrt{A_0}.$$ Here we put $$f:=\jjj \mmm \frac{\al}{\mj}A\f q_{j\mu}\vv.$$ By the commutation relation on $e^{is\pp}$ and $e^{it\qq}$: $$e^{is\pp_{rl_n}}e^{it\qq_{r'l_n'}}= e^{its \delta_{rr'}\delta_{l_nl_n'}}e^{it\qq_{r'l_n'}}e^{is\pp_{rl_n}},$$ and the von Neumann uniqueness theorem, we can identify $\pp$ and $\qq$ with the momentum and position operators, respectively, $$\pp_{t,l_n}\cong-i\frac{\partial}{\partial x_{r,n}},\ \ \ \qq_{r,l_n}\cong x_{r,n},$$ for $r=1,...,d-1$ and $n=1,...,D.$ Hence we identify $\heqq$ as a harmonic oscillator in $L^2(\RR^{(d-1)D})$. Generally for the harmonic oscillator $$H_T:=\half (\pp,\pp) +\half (\qq, T\qq)$$ with nonnegative symmetric matrix $T$, we have $$\inf \sigma(H_T)=\tr \sqrt{T}.$$ Then it follows that $$E(q)=\inf\sigma(\heqq)$$ $$= \inf\sigma\lk \half(\pp,\pp) +\half((\qq-\ff), A(\qq-\ff))\rk +\jjj \mhalf q_j^2 -\half(\ff,A\ff) -\half\tr\sqrt{A_0}$$ $$= \jjj \mhalf q_j^2-\half (f,Af)+\half\tr(\sqrt{A}-\sqrt{A_0}).$$ Separately we calculate \bi \item[(1)] $\jjj \mhalf q_j^2-\half (f,Af)$, \item[(2)] $\half\tr(\sqrt{A}-\sqrt{A_0})$. \ei First we calculate (1). We give a remark. Let $$\dmn(k):=\rrr e_\mu^r(k) e_\nu^r(k)=\lk \I-\frac{k}{|k|} \otimes \frac{k}{|k|}\rk_{\mu\nu}= \delta_{\mu\nu}-\frac{k_\mu k_\nu}{|k|^2}.$$ For rotation invariant functions $f$ and $g$ it follows that $$(\dmn f, g)=\delta_{\mu\nu}\dd(f,g).$$ We have $$(f,Af)=\sum_{i,j=1}^N\am i \am j q_{i\mu} q_{j\nu} (A\f v_\mu,v_\nu).$$ Note that we assume that $|\alpha_j|^2$ are suffieictly small. Then $$A\f=\sumn \ww A_0\f (PA_0\f)^{n-1}.$$ Hence $$(A\f v_\mu,v_\nu)= \sumn \ww (A_0\f (PA_0\f)^{n-1}v_\mu,v_\nu)$$ $$=\sumn\ww \sum_{j_1,...,j_{n-1}=1}^N\sum_{\mu_1,...,\mu_{n-1}=1}^{d} \epsilon^{n-1}\atm 1\cdots \atm {n-1}\times $$ $$ \times (v_\mu A_0\f\vvv 1)(\vvv 1,A_0\f \vvv 2) (\vvv 2,A_0\f \vvv 3)\cdots (\vvv {n-2},A_0\f \vvv {n-1}) (\vvv {n-1},A_0\f v_\nu)$$ $$=\delta_{\mu\nu} \sumn\ww \sum_{j_1,...,j_{n-1}=1}^N\sum_{\mu_1,...,\mu_{n-1}=1}^{d} \epsilon^{n-1} \dd^n \atm 1\cdots \atm{n-1} \xi^n, $$ where $$\xi:=\sum_{|l|\leq 2\pi L}\frac{|\la(l)|^2}{\od (l)}.$$ Then it follows that $$= \delta_{\mu\nu} \sumn\ww \dd \xi\lkk \epsilon \dd \xi \jjj \atmm j\rkk^{n-1}$$ $$=\delta_{\mu\nu} \frac{\dd\xi}{1+\epsilon \dd \xi \jjj \atmm j}.$$ We have $$(f, Af)= \sum_{i,j=1}^N\delta_{\mu\nu}\am i\am j q_{i\nu} q_{j\mu} \frac{\dd\xi}{1+\e \dd \xi \jjj \atmm j}$$ $$= \frac{\dd\xi}{1+\e \dd \xi \jjj \atmm j}\left| \jjj \am j q_j\right|^2.$$ Hence we have $$(1)=\half \jjj \frac{q_j^2}{\mj}-\half \frac{\dd\xi}{1+\e \dd \xi \jjj \atmm j}\left| \jjj \am j q_j\right|^2.$$ Next we calculate (2). Using the formula for nonnegative self-adjoint operator $B$, $$\sqrt{B}=\frac{1}{\pi}\int_{-\ii}^\ii B(s^2+B)\f ds,$$ we see that $$(2)=\frac{1}{2\pi}\int_{-\ii}^\ii\tr \lkk A(s^2+A)\f-A_0(s^2+A_0)\f\rkk ds.$$ Since $$(s^2+A)\f=(s^2+A_0)\f\sumn \ww \lkk P(s^2+A_0)\f\rkk^{n-1},$$ we have $$ A(s^2+A)\f-A_0(s^2+A_0)\f = \rI+\rII. $$ Here we put \eq{222}\rI:=A_0(s^2+A_0)\f\sumn\www\lkk P(s^2+A_0)\f\rkk^n, \en \eq{12} \rII:=P(s^2+A)\f. \en It follows that $$\tr \rI=\sumn\www \sum_{\varphi:\cons} (\varphi, A_0\sa \lkk P\sa\rkk^n\varphi),$$ where $\sum_{\varphi:\cons}$ means to sum up vectors in a complete orthonormal system. Then $$=\sumn \www \sum_{\varphi:\cons} (\varphi, A_0\sa \underbrace{P\sa P\sa\cdots P \sa}_{n\ \mbox{{\footnotesize times}}}\varphi)$$ $$=\sumn \www \sum_{\varphi:\cons} \sum_{j_1,...,j_n=1}^N\sum_{\mu_1,...,\mu_n=1}^d \epsilon^n \atm 1\cdots \atm n \times $$ $$\times (\varphi, A_0\sa \vvv 1)(\vvv 1,\sa \vvv 2)\times \cdots$$ $$\cdots \times (\vvv {n-1}, \sa \vvv n) (\vvv n, \sa \varphi) .$$ We take as a CONS $$\lkk \varphi_1:= \frac{\sa \vvv n}{\|\sa \vvv n\|},\varphi_2,\varphi_3,...,\rkk. $$ Then $$=\sumn \www \sum_{j_1,...,j_n=1}^N\sum_{\mu_1,...,\mu_n=1}^d \epsilon^n \atm 1\cdots \atm n \times $$ $$\times (\vvv n, A_0(s^2+A_0)^{-2} \vvv 1)(\vvv 1,\sa \vvv 2)\times \cdots$$ $$\cdots \times (\vvv {n-1}, \sa \vvv n) \frac{(\vvv n, (s^2+A_0)^{-2} \vvv n)}{ \|\sa \vvv n\|^2}$$ $$=d \sumn \www \lk\jjj\atmm j\rk^n \epsilon^n \dd^n \xi(s)^{n-1}\eta(s) ,$$ where $$\xi(s):=\sum_{|l|\leq 2\pi L}\frac{\od (l)|\la(l)|^2}{s^2+\od (l)^2},$$ and $$\eta(s):= \sum_{|l|\leq 2\pi L}\frac{\od (l)^3|\la(l)|^2}{(s^2+\od (l)^2)^2}.$$ Hence $$=d \sumn \www \lk \e \dd \xi(s) \jjj\atmm j\rk^n\frac{\eta(s)}{\xi(s)} $$ $$=d\frac{-\e\dd \xi(s) \jjj \atmm j } {1+\e \dd\xi(s) \jjj \atmm j } \frac{\eta(s)}{\xi(s)} =d \frac{-\epsilon\rho\eta(s)}{1+\epsilon \rho \xi(s) }.$$ Next we see that $$\tr \rII=\sum_{\varphi:\cons}(\varphi, P(s^2+A)\f \varphi)$$ $$=\sumn \ww \sum_{\varphi:\cons}(\varphi, P\sa \lk P\sa\rk^{n-1} \varphi)$$ $$=\sumn \ww \sum_{\varphi:\cons} \sum_{j_1,...,j_n=1}^N\sum_{\mu_1,...,\mu_n=1}^d \atm 1\cdots \atm n \epsilon^n\times $$ $$\times (\varphi, \vvv 1)(\vvv 1, \sa \vvv 2)\cdots (\vvv {n-1},\sa\vvv n) (\vvv n,\sa \varphi) .$$ Take as a CONS $$\lkk \varphi_1:=\frac{\sa \vvv n}{\|\sa\vvv n\|},\varphi_2, \varphi_3,...,\rkk.$$ Then we have $$\tr \rII =\sumn \ww \sum_{j_1,...,j_n=1}^N\sum_{\mu_1,...,\mu_n=1}^d \atm 1\cdots \atm n \epsilon^n\times $$ $$\times (\vvv n, \sa \vvv 1)(\vvv 1, \sa \vvv 2)\times \cdots $$ $$\cdots \times (\vvv {n-1},\sa\vvv n) \frac{(\vvv n,(s^2+A_0)^{-2}\vvv n)}{\|\sa\vvv n\|^2}$$ $$=\sumn \ww \sum_{j_1,...,j_n=1}^N\sum_{\mu_1,...,\mu_n=1}^d \atm 1\cdots \atm n \epsilon^n\dd^n d \xi(s)^n$$ $$=d \sumn \ww \lk \e \dd \xi(s) \jjj \atmm j \rk^n =d\frac{\epsilon \rho \xi(s) }{1+\epsilon \rho \xi(s) }.$$ Hence $$\tr (\rI+\rII)=\frac{d\epsilon \rho}{1+\epsilon \rho \xi(s)}(\xi(s)- \eta(s)).$$ From $$\xi(s)-\eta(s)=\sum_{|l|\leq 2\pi L}\lkk \frac{|\la(l)|^2\od (l)}{s^2+\od (l)^2}-\frac{\od (l)^3|\la(l)|^2}{(s^2+\od (l)^2)^2}\rkk =\sum_{|l|\leq 2\pi L}\frac{s^2|\la(l)|^2\od (l)}{(s^2+\od (l)^2)^2} := \eta'(s),$$ it follows that $$(2)=\frac{\epsilon }{2\pi}\int_{-\ii}^\ii\frac{\e \rho \eta'(s)} {1+\epsilon\rho \xi(s)}ds.$$ As was mentioned in \kak{uni}, as $a\rightarrow \infty$ and then $L\rightarrow \ii$, $\heqq$ uniformly converges to $\heq+\delta N$ in the resolvent sense. Thus as $a\rightarrow \ii$ and then $L\rightarrow \ii$, it follows that $$E(q, \delta, a, L)\rightarrow \inf\sigma (\he(q)+\delta N):=E(q,\delta).$$ Since $N\geq 0$, we have $$\liminf_{\delta\rightarrow 0}E(q,\delta)\geq E(q)$$ and the strong resolvent convergence of $\heq+\delta N$ to $H_\epsilon(q)$ as $\delta\rightarrow 0$ implies that $$E(q)\geq \limsup_{\delta\rightarrow 0} E(q,\delta).$$ Thus $$E(q,\delta)\rightarrow E(q)$$ as $\delta\rightarrow 0$. Thus the lemma follows for sufficiently small $|\alpha_j|$'s. By Proposition \ref{dia} (3), $\inf\sigma(\heq)$ can be analytically continued in each $\alpha_j$'s to ${\cal O}$, and clearly $E(q)$ also can be done. Hence the lemma follows for all $\alpha_j$'s in $\RR$. \qed {\it A proof of Lemma \ref{form} (2)}\\ The proof is similar to Lemma \ref{form}. Define the projection operator $$P:=\epsilon\jjj\mmm\atmm j |\vv^j\rangle\langle \vv^j|,$$ and $$A:=A_0+P,$$ where $A_0$ is defined in \kak{az}. Here $$\vv^j:=\lk \begin{array}{c} \sqrt{\od (l_1)}\laj (l_1)e_\mu^1(l_1)\\ \vdots \\ \sqrt{\od (l_1)}\laj (l_1)e_\mu^{(d-1)}(l_1)\\ \vdots\\ \vdots\\ \sqrt{\od (l_D)}\laj (l_D)e_\mu^1(l_D)\\ \vdots \\ \sqrt{\od (l_D)}\laj (l_D)e_\mu^{(d-1)}(l_D) \end{array} \rk\in \RR^{(d-1)D}.$$ Let $$f:=\jjj\mmm\frac{\al}{\mj}A\f q_{j\mu}\vv^j.$$ Similarly to Lemma \ref{form} it is enough to calculate \bi \item[(1)] $\half\jjj \frac{1}{\mj}q_j^2-\half (Af,f)$, \item[(2)] $\half\tr(\sqrt{A}-\sqrt{A_0})$. \ei First we calculate (1). We have $$(A\f\vv^i,{v}_\nu^j)=\delta_{\mu\nu}\delta_{ij} \sumn\ww \dd^n\epsilon^{n-1}\lk\atmm j\rk^{n-1}\xi_j^n$$ $$= \delta_{\mu\nu}\delta_{ij}\frac{\dd\xi_j}{1+ \e \dd \xi_j \atmm j },$$ where $$\xi_j:=\sum_{|l|\leq 2\pi L}\frac{|\laj(l)|^2}{\od (l)}.$$ Note that, for function $g$, $$(\laj, g \la_i)=\delta_{ij}(\laj, g\laj).$$ Then $$(Af,f)=\sum_{i,j=1}^N\sum_{\mu,\nu=1}^d (A\f\vv^i, {v}_\nu^j)$$ $$=\sum_{i,j=1}^N\sum_{\mu,\nu=1}^d \delta_{\mu\nu}\delta_{ij} \am i\am j q_{i\nu}q_{j\mu} \frac{\dd\xi_j}{1+\epsilon\dd \xi_j \atmm j}$$ $$=\jjj \lk\am j\rk^2q_j^2 \frac{\dd\xi_j}{1+ \e \dd \xi_j \atmm j}.$$ Then $$\half \jjj \frac{q_j^2}{\mj}-\half (Af,f)= \half \jjj\frac{q_j^2}{\mj}\lk \frac{\mj-(1-\epsilon)\dd \at \xi_j}{\mj+ \epsilon\dd \at \xi_j}\rk.$$ Next we calculate (2). Let (I) and (II) be as in \kak{222} and \kak{12}, respectively. We define $$\xi_j(s):=\sum_{|l|\leq 2\pi L}\frac{\od (l)|\laj(l)|^2}{s^2+\od (l)^2},$$ $$\eta_j(s):= \sum_{|l|\leq 2\pi L}\frac{\od (l)^3|\laj(l)|^2}{(s^2+\od (l)^2)^2},$$ and $$\eta_j'(s):=\sum_{|l|\leq 2\pi L}\frac{s^2|\laj(l)|^2\od (l)}{(s^2+\od (l)^2)^2}.$$ We have $$\tr \rI=d \jjj \sumn \www \lk\atmm j\rk^{n}\epsilon^n\dd^n \xi_j(s)^{n-1} \eta_j(s)$$ $$=-d\jjj \eta_j(s)\epsilon \atmm j \dd \frac{1}{1+\epsilon \xi_j(s) \dd \atmm j} =-d\jjj\frac{\epsilon \rj \eta_j(s)}{1+\epsilon \rj \xi_j(s)}.$$ Moreover $$\tr \rII=\jjj\sumn \ww \atmm j \epsilon^n\dd^n\xi_j(s)^n$$ $$=d \jjj\frac{\epsilon\rj \xi_j(s)}{1+\epsilon \rj\xi_j(s)}.$$ Hence $$\rI+\rII=d \jjj\frac{\epsilon \rj}{1+\epsilon \rj \xi_j(s)}(\xi_j(s)-\eta_j(s)) = d \jjj\frac{\epsilon \rj}{1+\epsilon \rj \xi_j(s)}\eta_j'(s).$$ Then $$ \half\tr(\sqrt{A}-\sqrt{A_0})= \half \tr(\rI+\rII)=\jjj\frac{d}{2\pi} \int_{-\ii}^\ii\frac{\epsilon \rj \eta'_j(s)}{1+\epsilon \rj\xi_j(s) }ds.$$ The same limiting argument on $a,L,\delta$ as in Lemma \ref{form} leads to the desired results. \qed {\bf Acknowledgements}\\ This work is supported by Grant-in-Aid 13740106 for Encouragement of Young Scientists from the Ministry of Education, Science, Sports and Culture. \begin{thebibliography}{99} \bibitem{bet} H. 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