Content-Type: multipart/mixed; boundary="-------------0111050516119" This is a multi-part message in MIME format. ---------------0111050516119 Content-Type: text/plain; name="01-407.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-407.keywords" Pauli-Fierz model, ground states, degeneracy, polaron model, spin, total momentum ---------------0111050516119 Content-Type: application/x-tex; name="2spin.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="2spin.tex" \documentstyle[12pt]{article} \topmargin=0in \oddsidemargin=0truein \evensidemargin=0truein \textheight=8.5in \textwidth=15.5cm \newcommand{\nt}[3]{\newtheorem{#1}[#2]{\rm #3}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \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}r \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{conjecture}[theorem]{Conjecture} \newtheorem{assumption}[theorem]{Assumption} %-------------------------------------------------------------- \newcommand{\proof}{{\noindent \it Proof:\ }} \newcommand{\qed}{\hfill $\Box$\par\medskip} \newcommand{\BR}{{{\Bbb R}^3}} \newcommand{\ct}{{\Bbb C}^2} \newcommand{\BC}{{\Bbb C}} \newcommand{\RR}{{{\Bbb R}}} \newcommand{\RRR}{[0,\infty)} \newcommand{\LL}{Suppose that $|\aa|$ is sufficiently small} \newcommand{\xo}{x\otimes\Omega} \newcommand{\cmp}[5]{{#1}, {#2}, {\it Commun. 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We assume $\omega:\BR\rightarrow \RR$ to be continuous with the properties \bi \item[(A.1)] $\infk \omega(k)\geq \dm>0,$ \item[(A.2)] $\omega(k_1)+\omega(k_2)\geq \omega(k_1+k_2),$ \item[(A.3)] $\omega(k)=\omega(Rk)$ for an arbitrary rotation $R$. \ei The quantized transverse vector potential is defined through $$\Av(x)= \jj \int \frac{\vp(k)}{\sqrt{2\omega(k)}} e_j(k) \lk e^{-i\kx} \add(k,j)+e^{i\kx} a(k,j)\rk dk.$$ Here $e_1$ and $e_2$ are polarization vectors which together with $\hk=k/|k|$ form a standard basis in $\BR$. $\varphi:\BR\rightarrow \RR$ is the form factor which ensures an ultraviolet cutoff. It is assumed to be rotational invariant, $\varphi(Rx)=\varphi(x)$ for an arbitrary rotation $R$, continuous, bounded with some decay at infinity, and normalized as $\int \varphi(x) dx=1$. We will mostly work with the Fourier transform $\vp(k)=(2\pi)^{-3/2}\int\varphi(x) e^{-i\kx} dx$. It satisfies (1) $\vp(Rk)=\vp(k)$ for an arbitrary rotation $R$, (2) $\overline{\vp}=\vp$ for notational simplicity, (3) the normalization $\vp(0)=(2\pi)^{-3/2}$, and (4) the decay $$\int\lk \omega(k)^{-2}+\omega(k)\f+1+\omega(k)\rk |\vp(k)|^2 dk<\infty.$$ With these preparations the Pauli-Fierz Hamiltonian, including spin, is defined by \eq{11} H=\frac{1}{2m}\lkk\si\cdot\lk-i\nabla_x-e\Av(x)\rk\rkk^2+\hf, \en where $m$ is the bare mass and $e$ the charge of the electron. Translations for the electron are generated by $-i\nabla_x$ and translations for the photon field by the field momentum \eq{02} \pf=\jj \int k\add(k,j)\aaa(k,j)dk. \en By translation invariance of $H$ the total momentum $$p=-i\nabla_x+\pf$$ is thus conserved, $$[H,p]=0,$$ as can be checked also directly. We unitarily transform $H$ such that the fibering with respect to $p$ becomes transparent. In momentum representation, momentum multiplication by $k$, position $i\nabla_k$, an element $\psi\in\hhh$ is written as $\psi^{(n)}(k,k_1,j_1,...,k_n,j_n)$ with values in $\BC^2$. For each $n$ let \eq{12} U \psi^{(n)}(k,k_1,j_1,...,k_n,j_n)= \psi^{(n)}(k-\sum_{j=1}^n k_j, k_1,j_1,...,k_n,j_n) \en with inverse \eq{13} U\f \psi^{(n)}(k,k_1,j_1,...,k_n,j_n)= \psi^{(n)}(k+\sum_{j=1}^n k_j, k_1,j_1,...,k_n,j_n). \en Clearly $U$ is unitary. We set $$\Av=\Av(0)$$ as an operator on $\fff$. Similarly for the quantized magnetic field, $$\Bv(x) = i \jj \int\ \frac{\vp(k)}{\sqrt{2\omega(k)}} (k \wedge e_j(k)) \lk e^{-i\kx}\add(k,j)- e^{i\kx} a(k,j)\rk dk,$$ and we denote $$\Bv=\Bv(0)$$ as an operator on $\fff$. Then, first working out the square in \kak{11}, one obtains \eq{14} UHU\f =\frac{1}{2m}(p-\pf-e\Av)^2-\frac{\aa}{2m}\si\cdot \Bv+\hf. \en In \kak{14} $p$ is multiplication by $k$ in the representation from \kak{12} and \kak{13}. Thus the fibering is $$\hhh=\int_\BR^\oplus \hhh_p dp$$ with $\hhh_p$ isomorphic to $\BC^2\otimes\fff$ and $$UHU\f= \int_\BR^\oplus \hp dp.$$ In the following we will regard $p\in\BR$ simply as a parameter. We also choose units such that $m=1$. Then the operator under study is \eq{hp} \hp=\half\lk \p- \pf-e \Av\rk^2-\frac{e}{2}\si \cdot \Bv+ \hf \en acting on $$\hhh_p=\BC^2\otimes \fff.$$ For simplicity the index $p$ of $\hhh_p$ will be omitted. For $\aa=0$ \kak{hp} reduces to the noninteracting Hamiltonian \eq{16} \hpz=\half(p-\pf)^2+\hf. \en Clearly, with the definitions \kak{01} and \kak{02}, $\hpz$ is self-adjoint with the domain $D(\hf+\pf^2)=D(\hf)\cap D(\pf^2)$. The rest of \kak{hp} is regarded as the interaction part of the Hamiltonian, \eq{17} \hi=\hp-\hpz =-{\aa}(\p-\pf)\cdot \Av+\frac{\at}{2}\Av^2-\frac{\aa}{2}\sigma\cdot \Bv. \en For sufficiently small $\aa$, $\mm0$ implies $\cco<1$. The goal of our paper is to study some ground state properties of $\hp$. The ground state energy of $\hp$ is $$E(p)=\inf\s(\hp)=\inf_{\psi\in D(\hp), \|\psi\|=1} (\psi, \hp\psi).$$ It is easily seen that resolvent $(\hp-z)\f$ with $z\not\in \s(\hp)$ is continuous in both $\p$ and $\aa$ in the operator norm. Thus $\epr$ is continuous in both $\p$ and $\aa$. If $E(p)$ is an eigenvalue, the corresponding spectral projection is denoted by $\pg $. $\tr \pg $ is the degeneracy of the ground state. The bottom of the continuous spectrum is denoted by $E_{\rm c}(p)$. Under the assumptions (A.1)--(A.3) one knows that $$E_{\rm c}(p)=\inf_{k\in\BR}\lkk E(p-k)+\omega(k)\rkk,$$ see \cite{fr2,sp3}. Thus it is natural to set $$\D(p)=E_{\rm c}(p)-E(p)=\inf_{k\in\BR}\lkk E(p-k)+\omega(k)-E(p)\rkk.$$ Let $\aa=0$. Then $\D(p)>0$ for $|p|0\}\not=\emptyset.$ As our main result we state \bt{main3} Suppose $|e|0$. Then $\tr \pg =2.$ \et We briefly comment on our assumptions. $\D(p)>0$ means that we assume, rather than prove, a spectral gap. If in addition one would impose the implicit but rather natural condition that \eq{nat} E(p)\geq E(0), \en then, following the arguments in \cite{fr2}, one concludes in case of the dispersion relation \kak{10} that $\D(p)>0$ for $|p|<\sqrt{3}-1$ and arbitrary $\aa\in\RR$. For a general dispersion relation satisfying (A.1) to (A.3) a corresponding bound can be established. For the spinless Pauli-Fierz Hamiltonian \kak{nat} is proven through a suitable variant of a diamagnetic inequality. We did not succeed to establish \kak{nat} for the Hamiltonian \kak{hp}. Note that on physical grounds one expects $\D(p)>0$ for $|p|< \pc$ and $\D(p)=0$ for $|p|>\pc$ with a simultaneous loss of the ground state. For the Hamiltonian \kak{hp} with $\omega$ given by \kak{10} one has $\pc <\infty$ at $e=0$ and expects $\pc<\infty$ to persist for all $e\not=0$ provided the spatial dimension $d\geq 3$. The constant $\aa_0$ does {\it not} depend on $m_{\rm ph}$. Thus, together with \kak{nat}, the domain of validity of Theorem \ref{main3} is independent of $m_{\rm ph}$ and it may seem that one could take the limit $m_{\rm ph}\rightarrow 0$. Of course, thereby $\D(p)\rightarrow 0$. Unfortunately, this will not work, since the Pauli-Fierz Hamiltonian is infrared divergent and the number of photons increases without bound as $m_{\rm ph}\rightarrow 0$ \cite{ch}. The physical ground states are no longer in Fock space. The only exception is $\p=0$, and one might hope to apply our method directly to the case $\p=0$ and $m_{\rm ph}=0$. This problem will be taken up in Section 3. Let us indicate the strategy for the proof of Theorem \ref{main3}. From a pull-through formula one estimates the overlap of a ground state with the subspace $P_0\hhh=\BC^2\otimes\{\BC \Omega\}$. If $\mm$ is sufficiently small, the overlap is large which implies $\tr \pg <3$. For a lower bound we will derive the algebraic relation $P_0\pg P_0=aP_0$, $a>0$, which implies $\tr \pg\geq 2$. The Hamiltonian \kak{hp} is invariant under rotations with axis $\widehat{p}=p/|p|$. To understand the implications let us define the field angular momentum relative to the origin by $$\jf=-i \jj\int \add(k,j) (k\wedge \nabla_k) a(k,j)dk$$ and the helicity by $$\ssf=i\int \hk \lkk \add(k,2)a(k,1)-\add(k,1)a(k,2) \rkk dk. $$ Let $\n\in\BR$ be an arbitrary unit vector, $\t\in\RR$, and let $R=R(\n,\t)$ be the rotation around $\n$ through the angle $\t$. We note that $$\ejt \hf \emjt=\hf,$$ $$\ejt \pf \emjt=R \pf,$$ $$\ejt \Av \emjt=R\Av .$$ Define the total angular momentum through $$J=\jf+\ssf+\half\si.$$ Then it follows that $$\eejt \hp \eemjt =H_{R\f \p}.$$ In particular, the ground state energy is rotation invariant, \eq{ep} E(p)=E(R\p). \en If one sets $\n=\vP$, then \eq{ang} e^{i\t \pj} \hp e^{-i\t \pj}=\hp, \en which expresses the rotation invariance relative to $\vP$. Since $\s(\vP\cdot (\jf+\ssf))={\Bbb Z}$ and $ \s(\vP\cdot \si ) =\{-1,+1\}$, it follows that $$\s(\pj)=\hz,$$ where $\hz$ is the set of half integers $\{\pm\han,\pm3/2,\pm5/2,...\}$. By virtue of \kak{ang}, $\hhh$ and $\hp$ are decomposable as $$\hhh=\bigoplus_{z\in \hz} \hhh(z),$$ $$\hp=\bigoplus_{z\in \hz} \hp(z).$$ Therefore, if $\tr \pg=2$, the ground states $\griii\in\pg\hhh$ can be chosen such that $\gri\in\hhh(z)$ and $\grii\in\hhh(z')$ for some $z,z'\in\hz$. \bt{main2} Suppose $|e|0$. Then $\hp$ has two orthogonal ground states, $\griii$, with the property $\griii\in\hhh(\pm\han)$. \et \section{Spectral properties} \subsection{Upper bound} Let us denote the number operator by $$\nf=\jj\int\add(k,j)a(k,j)dk.$$ In what follows $\gr$ denotes an {\it arbitrary} normalized ground state of $\hp$. We note that $a(k,j)\psi$, $\psi\in D(\nf^\han)$, is well defined and $$(a(k,j)\psi)^{(n)}(\kj)=\sqrt{n+1}\psi^{(n+1)}(k,j,\kj).$$ Moreover it follows that $\|a(k,j)\psi\|\leq \|\nf^\han \psi\|$ and $$(\psi, \nf \phi)=\jj\int(a(k,j)\psi, a(k,j)\phi) dk.$$ \bl{fp} Suppose $\D(p)>0$. Then $$(\gr, \nf\gr)\leq 2 \at\int \frac{(|k|^2/4)+6E(p)}{(E(\p-k)+\omega(k)-E(p))^2}\frac{|\vp(k)|^2}{\omega(k)}dk. $$ \el \proof By the pull-through formula $$a(k,j)\hp \gr=\lk \hpk +\omega(k)\rk a(k,j)\gr$$ $$-\aa\sa \lkk\PP \cdot e_j(k)+\half \si\cdot (ik\wedge e_j(k))\rkk\gr . $$ Hence we have $$ a(k,j)\gr=\aa \sa \lk \hpk+\omega(k)-E(p) \rk\f $$ $$\times \lkk \PP \cdot e_j(k)+\half\si\cdot (ik\wedge e_j(k))\rkk\gr.$$ Thus $$(\gr, \nf \gr)=\jj \int \|a(k,j)\gr\|^2dk $$ $$\leq \jj \int\left|\sa\frac{\|\PP \cdot e_j(k)\gr+(\han) \si\cdot (ik\wedge e_j(k))\gr\|}{E(\p-k)+\omega(k)-E(p)}\right|^2dk.$$ We note that $$\|\PP \cdot e_j(k)\gr\|^2=\|\mmm \si_\mu \PP_\mu \si_\mu e_j^\mu(k)\gr\|^2$$ $$\leq 3 \mmm \|\si_\mu\PP_\mu \gr\|^2\leq 6 E(p), $$ and $$\|\si\cdot (ik\wedge e_j(k))\gr\|^2\leq |k|^2 \mmm \|(i\hk\wedge e_j(k))_\mu \gr\|^2\leq |k|^2. $$ Thus $$ \|\PP \cdot e_j(k)\gr+(\han )\si\cdot (ik\wedge e_j(k))\gr\|^2\leq 2\lkk (|k|^2/4)+6E(p)\rkk, $$ which leads to $$(\gr, \nf\gr)\leq 2 \at\int \frac{(|k|^2/4)+6E(p)}{(E(\p-k)+\omega(k)-E(p))^2}\frac{|\vp(k)|^2}{\omega(k)}dk$$ and the lemma follows. \qed We set $$\t(p)=\t(p,\aa)=2 \int \frac{(|k|^2/4)+6E(p)}{(E(\p-k)+\omega(k)-E(p))^2}\frac{|\vp(k)|^2}{\omega(k)}dk.$$ Note that $\t(p)$ is rotation invariant, i.e. $\t(Rp)=\t(p)$ for an arbitrary rotation $R$. Let $P_0=1\otimes P_\Omega$ be the projection onto $\BC^2\otimes \{\BC \Omega\}$. \bl{35} Let $\D(p)>0$. Suppose $\mm <1/\sqrt{3 \t(p)} $. Then $\tr \pg \leq 2$. \el \proof By Lemma \ref{fp} we have $\tr(\P\nf)\leq \at \t(p)\tr \P.$ Therefore $$\tr \P-\tr(\P P_0)=\tr\P(I-P_0)\leq \tr(\P \nf)\leq \at \t(p) \tr \P,$$ and $$(1-\at \t(p))\tr \P\leq \tr (\P P_0)\leq 2,$$ which implies $$\tr\P\leq \frac{2}{1-\at \t(p)}<3.$$ Thus the lemma follows. \qed \subsection{Lower bound} We say that $\psi\in\fff$ is {\it real}, if for all $n\geq 0$, $\psi^{(n)}$ is a real-valued function on $L^2(\RR^{3n}\times\{1,2\})$. The set of real $\psi$ is denoted by $\fffr$. We define the set of reality-preserving operators $\rpof$ as follows: $$\rpof=\left\{A| A:\fffr\cap D(A)\longrightarrow \fffr\right\}.$$ It is seen that $\hf$ and $\pf$ are in $\rpof$. Since for all $k\in \RR$ and $z\in\RR$ with $z\not\in\s(\hpz)$, $$((\hpz-z)^k\psi)^{(n)}\nm$$ $$= \lk \half\lk \p-\sumi k_i\rk^2+\sumi \omega(k_i) +z\rk^k \psi^{(n)}\nm,$$ $(\hpz-z)^k$ is also in $\rpof$. Moreover $\aaa(f)$ and $\add(f)$ are in $\rpof$ for real $f$'s. In particular $\Av$ and $i\Bv$ are in $\rpof$. Note that, if $\psi\in D(\Bv)\cap \fffr$, then $(\psi,\Bv \psi)=0.$ Let $\ffff=\bigcup_{N=0}^\infty \oplus_{n=0}^N \lk L^2(\BR\times\{1,2\}) \rk_{\rm sym}^n$ denote the finite particle subspace of $\fff$. Note that $\Av$, $\Bv$, $\zes$ and $\zess$ leave $\ffff$ invariant. \bl{real} Suppose $\mm<\az$. Let $x\in\ct$. Then there exists $a(t)\in\RR$ independent of $x$ such that for $t\geq 0$ \eq{eht} (\xo, e^{-t (\hp-\epr) } \xo)_\hhh=a(t)(x,x)_{\ct}. \en \el \proof Note that $\|\hi(1+\hpz)\f\|<1$ for $\mm<\az$ by \kak{co}. Then, by spectral theory, one has $$ \ehtt=\ln\left(1+\frac{t}{n} (\hp-\epr) \right)^{-n}$$ $$ =\lnk \left\{ \sum_{k=0}^m\ress \left\{\left(-\frac{t}{n}\hib\right)\ress\right\}^k \right\}^n $$ $$=\lnk \left\{ \res\left( \sum_{k=0}^m \lk-\frac{t}{n} \hiw\rk ^k \right) \res\right\}^n.$$ Here $$\hiw=\hiiw+i\si\cdot\BB,$$ $$\hiiw=\res(\hii-\epr) \res,$$ $$\BB=\res (i \Bv) \res,$$ $$\hii=-\aa(p-\pf)\cdot\Av+\frac{\at}{2}\Av^2.$$ It is seen that $$\hiw^2= \hiiw \hiiw- \BB\cdot\BB +i\si\cdot (\hiiw \BB+\BB\hiiw-\BB\wedge\BB)=M+i\si\cdot L.$$ Here both of $M=\hiiw \hiiw- \BB\cdot\BB$ and $L=\hiiw \BB+\BB\hiiw-\BB\wedge \BB$ are in $\rpof$. Moreover $$\hiw^3= \hiiw M-\BB L+i\si\cdot(\BB M+\hiiw L-\BB\wedge L),$$ where both of $\hiiw M-\BB L$ and $\BB M+\hiiw L-\BB\wedge L$ are also in $\rpof$. Thus, repeating above procedure, one obtains $$\sum_{k=0}^m\lk-\frac{t}{n} \hiw\rk ^k=a_m+i\si\cdot b_m,$$ where $a_m$ and $b_m$ are in $\rpof$. Hence there exist $a_{nm}\in \rpof$ and $b_{nm}\in\rpof$ such that $$\left\{\res\left(\sum_{k=0}^m \left(-\frac{t}{n} \hiw\rk ^k\right)\res\right\}^n=a_{nm}+ i\si\cdot b_{nm}.$$ Finally $$(\xo,\ehtt\xo)=\lnk(x,x)(\Omega,a_{nm}\Omega)+ i\lnk(x,\si x) (\Omega, b_{nm}\Omega).$$ Since the left-hand side is real, the second term of the right-hand side vanishes and $a(t)=\lnk(\Omega, a_{nm}\Omega)$ exists, which establishes the desired result. \qed \bl{po} Suppose that $\D(p)>0$ and $\mm< 1/\sqrt{\t(p)}$. Then $(\gr, P_0 \gr)\not =0.$ \el \proof Since $P_\Omega+\nf\geq 1$, we have from Lemma \ref{fp} $$(\gr, P_0 \gr)\geq \|\gr\|^2-\| (1\otimes \nf^\han) \gr\|^2>1-\at \t(p)>0.$$ Thus the lemma follows. \qed \bl{projection} Suppose $\mm<\az$ and $\mm<1/\sqrt{\t(p)}$. Then there exists $a>0$ such that \eq{contr} P_0\pg P_0=aP_0. \en \el \proof Note that $$\pg =s-\lim_{t\rightarrow\infty}\ehtt.$$ Thus by Lemma \ref{real}, $$(\xo,\pg \xo)=\lim_{t\rightarrow\infty}(\xo,\ehtt\xo) =\lim_{t\rightarrow\infty}a(t)(x,x)$$ for all $x\in\ct$. Since by Lemma \ref{po}, $(\xo,\pg \xo)\not=0$ for some $x\in\BC^2$, $\lim_{t\rightarrow\infty}a(t)$ exists and it does not vanish. For arbitrary $\phi_1,\phi_2\in\hhh$, the polarization identity leads to $(\phi_1,P_0\P P_0\phi_2)=a(\phi_1,P_0\phi_2).$ The lemma follows. \qed \bl{low} Suppose $\mm<\az$ and $\mm<1/\sqrt{\t(p)}$. Then $\tr \pg \geq 2$. \el \proof Suppose $\tr \pg =1$. Then $P_0\pg P_0/\tr P_0\pg P_0$ is a one-dimensional projection which contradicts \kak{contr}. Thus the lemma follows. \qed \subsection{Proofs of Theorems \ref{main3} and \ref{main2}} We define $$e_0= \inf \lkk \mm \, \left| \mm<1/\sqrt{3\t(p)}, \mm<\aa^\ast\right.\rkk.$$ \noindent {\it Proof of Theorem \ref{main3}:} \\ Since $\D(p)>0$ and $\mm0, \en \eq{aaa} (\mmo,\pg\mmo)=a>0. \en Then $Q_\han \pg \ppo\not=0$ and $Q_{-\han} \pg\mmo \not=0$. The alternative $Q_\han\fri\not=0$ or $Q_{\han}\frii\not=0$ holds by \kak{aa}, the alternative $Q_{-\han}\fri\not=0$ or $Q_{-\han}\frii\not=0$ by \kak{aaa}. We may set $Q_\han \fri\not=0$. Then $\fri\in\hhh(\han)$ and $\frii\in\hhh(-\han)$. \qed \section{Zero total momentum} \subsection{Spinless Hamiltonian} In the spinless case $\hp$ simplifies to $$\hp=\half\PP^2+\hf$$ acting on $\fff$. The bound \kak{nat} is available and $\hp$ has at least one ground state for $|p|0$ and arbitrary $\aa$. To show the uniqueness of the ground state only the pull-through argument seems to be available. The details of Section 2.1 remain unchanged and one concludes that if $$\mm^2\leq \half \lkk \int\frac{E(p)}{(E(p-k)+\omega(k)-E(p))^2}\frac{|\vp(k)|^2}{\omega(k)}dk\rkk\f,$$ then $\tr \pg\leq 1$, which implies that the ground state of $\hp$ is unique. \subsection{Confining potentials} For $p=0$ no infrared divergence is expected and for the remainder of this section we set $$\omega(k)=|k|$$ as the physical dispersion relation. We did not succeed to apply the methods of Section 2 to this case. Therefore rather than considering $p=0$ directly we add to the Hamiltonian \kak{11} a confining potential, $V:\BR\rightarrow \RR$, which in spirit amounts to the same physical situation. The Hamiltonian under study is \eq{31} \hv=\half\lkk\si\cdot\lk-i\nabla_x-e\Av(x)\rk \rkk^2+V(x)+\hf. \en Let $\pa=-\half\Delta+V$ on $\LR\otimes\BC^2$, $\sp=\inf\s_{\rm ess}(\pa)$, $\epp=\inf\s(\pa)$ and $E=\inf\s(\hv)$. Let $V$ be relatively bounded with respect to $-\Delta$ with a bound less than 1. Then $\hv$ is self-adjoint on $D(\Delta)\cap D(\hf)$ as established in \cite{hi3}. For arbitrary $\aa$ the existence of ground states has been proven by Griesemer, Lieb, and Loss \cite{gll} under the condition that $\pa $ has a ground state separated by a gap from the continuous spectrum, i.e. \eq{gap} \gap>0. \en We also refer to \cite{bfs3} for prior results, where in particular it is proven that the charge density of an arbitrary ground state $\gr$ is localized, i.e. \eq{ex} \|e^{c|x|}\gr\|\leq c_1 \|\gr\| \en with some constant $c$. In the spinless case the uniqueness of the ground state, $\tr\pg=1$, would follow from a positivity argument \cite{hi9}. Let $\pel$ be the projection of the subspace spanned by ground states of $\pa$. Suppose \kak{gap}. Take $\aa$ such that \eq{33} \sp-E>0, \en which can be satisfied by the continuity of $E$ in $\aa$. Then a pull-through argument and \kak{ex} yield that \eq{pl} (\gr,(1\otimes\nf+\pel^\perp\otimes P_0)\gr)\leq \ee, \en where $\lime \ee=0$. Hence in the similar way as Lemma \ref{35} with $P_0$ and $\nf$ replaced by $\pel\otimes P_0$ and $1\otimes \nf+\pel^\perp\otimes P_0$, respectively, we see that if, in addition to \kak{33}, $\aa$ satisfies $ \ee<1/3$, then \eq{2} \tr \pg < 3. \en The realness argument of Section 2.2 requires some extra conditions on $V$. \bt{main4} Suppose $V(x)=V(-x)$, $\sp-\epp>0$, and $\tr\pel=2$. Then there exists a positive constant $e_{00}$ such that, if $\mm0. \en Let $F$ denote the Fourier transform of $\LR$ and define the unitary operator of $\hhh$ by $T=F e^{ix\cdot\pf}$. Then we have $$T\hv T\f=\half\lkk \si\cdot (x-\pf-\aa \Av(0))\rkk^2+FVF\f +\hf.$$ The assumption $V(x)=V(-x)$ implies that $F V F\f$ is a reality preserving operator on $\LR$. Let $\varphi_{\rm el}$ be the ground state of $\pa$. In the similar way as Lemmas \ref{real} and \ref{projection} with $\Omega$ and $P_0$ replaced by $\varphi_{\rm el}\otimes\Omega$ and $\pel\otimes P_0$, respectively, it is established that by \kak{th} there exists a positive constant $c^\ast$ such that if $\mm