Content-Type: multipart/mixed; boundary="-------------0007050920223" This is a multi-part message in MIME format. ---------------0007050920223 Content-Type: text/plain; name="00-283.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-283.keywords" Hartree theory, bosons, magnetic fields, atoms ---------------0007050920223 Content-Type: application/x-tex; name="baum_seir.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="baum_seir.tex" %%%%%%%%%%%%%%%%% %%FINAL VERSION%% %% 05/07/00 %% %%%%%%%%%%%%%%%%% \newcommand{\version}{July 5, 2000} \documentclass[12pt]{article} \usepackage{a4,amsthm,amsfonts,latexsym,amssymb} {\catcode `\@=11 \global\let\AddToReset=\@addtoreset} \AddToReset{equation}{section} \renewcommand{\theequation}{\thesection.\arabic{equation}} %Numbers in front \swapnumbers \pagestyle{myheadings} %Theoremstyles: \theoremstyle{plain} \newtheorem{thm}{THEOREM}[section] \newtheorem{cl}[thm]{COROLLARY} \newtheorem{lem}[thm]{LEMMA} \newtheorem{proposition}[thm]{PROPOSITION} \theoremstyle{definition} \newtheorem{rem}[thm]{Remark} %Macros \newcommand{\beq}{\begin{equation}} \newcommand{\eeq}{\end{equation}} \def\beqa{\begin{eqnarray}} \def\eeqa{\end{eqnarray}} \newcommand{\infspec}{{\rm inf\ spec\ }} \newcommand{\nab}{\left(-i\nabla+\beta\A(\x)\right)} \newcommand{\R}{{\mathbb R}} \newcommand{\C}{{\mathbb C}} \newcommand{\N}{{\mathbb N}} \newcommand{\D}{{\mathcal D}} \newcommand{\Ll}{{\mathcal L}} \newcommand{\Hh}{{\mathcal H}} \newcommand{\Cc}{{\mathcal C}} \newcommand{\eps}{\varepsilon} \newcommand{\A}{{\bf a}} \newcommand{\abf}{{\bf a}} \newcommand{\B}{{\bf b}} \newcommand{\Aa}{{\bf A}} \newcommand{\x}{{\bf x}} \newcommand{\y}{{\bf y}} \newcommand{\X}{{\bf X}} \newcommand{\0}{{\bf 0}} \newcommand{\rr}{{\bf r}} \newcommand{\bfeta}{{\bf z}} \newcommand{\xperp}{\x^\perp} \newcommand{\yperp}{\y^\perp} \newcommand{\rperp}{{\bf r}^\perp} \newcommand{\xpar}{x^\parallel} \newcommand{\ypar}{y^\parallel} \newcommand{\ppar}{p^\parallel} \newcommand{\Lpar}{L^\parallel} \newcommand{\Tr}{{\rm Tr}} \newcommand{\half}{\mbox{$\frac{1}{2}$}} \newcommand{\rg}{\rho_\Gamma} \newcommand{\rH}{\rho^{\rm H}} \newcommand{\rh}{\rho_{\Gamma^{\rm H}}} \newcommand{\dx}{\frac\partial{\partial x}} \newcommand{\dy}{\frac\partial{\partial y}} \newcommand{\E}{{\mathcal E}^{\rm MH}} \newcommand{\Ec}{E_{\rm conf}} \newcommand{\tE}{{\mathcal E}^{\rm MH}_{\beta , \rm lin}} \newcommand{\Edens}{{\hat{\mathcal E}}^{\rm MH}} \newcommand{\uw}{\underline w} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \date{\small\version} \begin{document} \markboth{\scriptsize{BS \version}}{\scriptsize{BS \version}} \title{\bf{Atoms with bosonic ``electrons'' in \\ strong magnetic fields}} \author{\vspace{5pt} Bernhard Baumgartner$^1$ and Robert Seiringer$^2$\\ \vspace{-4pt}\small{Institut f\"ur Theoretische Physik, Universit\"at Wien}\\ \small{Boltzmanngasse 5, A-1090 Vienna, Austria}} \maketitle \begin{abstract} We study the ground state properties of an atom with nuclear charge $Z$ and $N$ bosonic \lq\lq electrons\rq\rq\ in the presence of a homogeneous magnetic field $B$. We investigate the mean field limit $N\to\infty$ with $N/Z$ fixed, and identify three different asymptotic regions, according to $B\ll Z^2$, $B\sim Z^2$, and $B\gg Z^2$. In Region 1 standard Hartree theory is applicable. Region 3 is described by a one-dimensional functional, which is identical to the so-called Hyper-Strong functional introduced by Lieb, Solovej and Yngvason for atoms with fermionic electrons in the region $B\gg Z^3$; i.e., for very strong magnetic fields the ground state properties of atoms are independent of statistics. For Region 2 we introduce a general {\it magnetic Hartree functional}, which is studied in detail. It is shown that in the special case of an atom it can be restricted to the subspace of zero angular momentum parallel to the magnetic field, which simplifies the theory considerably. The functional reproduces the energy and the one-particle reduced density matrix for the full $N$-particle ground state to leading order in $N$, and it implies the description of the other regions as limiting cases. \end{abstract} \footnotetext[1]{E-Mail: \texttt{baumgart@ap.univie.ac.at}} \footnotetext[2]{E-Mail: \texttt{rseiring@ap.univie.ac.at}} \newpage \tableofcontents \section{Introduction} The ground states of atoms with many electrons in magnetic fields have been studied in \cite{LSY94a, LSY94b, BSY00}, and their energies have been evaluated, exactly to leading order, as some of the physical parameters tend to infinity. The atoms have been modeled by the nonrelativistic quantum mechanics of $N$ fermionic electrons, with an unmovable pointlike nucleus of charge $Z$ in a homogeneous magnetic field of strength $B$. In order to shed some more light onto the interplay of the involved laws of physics, we investigate the effects of changing one of them: What would happen, if the electrons were bosons? So we study the ground state of the Hamiltonian - written in appropriate units - \beq\label{ham} \underline H_{N,Z,B}=\sum_{i=1}^N\left( H_{B,i}-B- \frac{Z}{|\x_i|}\right)+ \sum_{i0$ and $\alpha >0$ \beqa E^{\rm MH}_{\rm ext}(\lambda,\beta,\zeta ,\alpha)&>& \lambda E^{\rm hyd}(\beta,\zeta ) \\ E^{\rm MH}_{\rm ext}(\lambda,\beta,\zeta ,\alpha)&<&\lambda E^{\rm hyd}(\beta,\zeta -\lambda \alpha /2). \label{hydupbd} \eeqa \end{proposition} \begin{proof} The first inequality follows from the strict positivity of $D[\rho,\rho]$. The upper bound can be given by choosing the projection onto the ground state of the Hamiltonian $H_{\beta}-(\zeta -\lambda \alpha /2)/|\x|$, as a test density matrix $\Gamma$. Then we apply the bound to the repulsion by attraction (\ref{aboundr}), together with the observation, that the nucleus has to be at the point of the maximum of the potential $A_y[\rho]$. (Otherwise the energy could be lowered by shifting $\rho$). \end{proof} We remark that the bound (\ref{hydupbd}) is of no use for $\lambda \alpha \sim 2\zeta $ or larger. In this case one can use the monotone decrease of the energy $E^{\rm MH}_{\rm ext}$ in $\lambda$, and bound $E^{\rm MH}_{\rm ext}$ by $\min_{\lambda} \{\lambda E^{\rm hyd}(\beta,\zeta -\lambda \alpha /2)\}$. % We will use these bounds in Subsection \ref{lowestlandauhartree}. As an obvious consequence of these bounds, using also the continuity of the hydrogen energy in $\zeta $, we add \begin{rem}[The limit $\lambda \to 0$]\label{ll0} In the limit $\lambda \to 0$ the energy per unit charge, $E^{\rm MH}_{\rm ext}(\lambda,\beta,\zeta ,\alpha)/\lambda$, converges to the energy of the hydrogen atom, $E^{\rm hyd}(\beta,\zeta )$. \end{rem} The bounds of Proposition \ref{hydroprop} will actually be used in a simplified form: \begin{lem}[Simple bounds]\label{simple} \beqa E^{\rm MH}(\lambda,\beta)&\geq&-(1/4+\beta)\lambda, \\ E^{\rm MH}(\lambda,\beta)&\leq&-(1/4)\lambda (1-\lambda /2)^2, \qquad {\rm for} \quad \lambda \leq 2. \eeqa \end{lem} \begin{proof} Applying the diamagnetic inequality \cite{S76} to the lower bound in proposition (\ref{hydroprop}), we get \beq\label{214} H_{\beta}-\beta-\frac{1}{|\x|} \geq\infspec (H_0-\beta-\frac{1}{|\x|}) =-\frac{1}{4}-\beta. \eeq To get the upper bound, we apply Lieb's inequality (Theorem A.1. in \cite{AHS78}) \beq \label{lieb} \infspec (H_{\beta}-\beta +V) \leq \infspec (H_0 +V) \eeq to the upper bound in proposition (\ref{hydroprop}). \end{proof} \subsection{Minimizers} \begin{thm}[Existence of a minimizer]\label{exist} For each $\beta\geq 0$ and $\lambda>0$ there is a minimizer $\Gamma^{\rm H}$ for $\E _\beta $ under the condition $\Tr[\Gamma^{\rm H}]\leq\lambda $, i.e. \beq E^{\rm MH}(\lambda,\beta)=\E _\beta [\Gamma^{\rm H}]. \eeq \end{thm} \begin{proof} We follow closely the proof of the analogous theorems 2.2 and 4.3 in \cite{LSY94a}. Let $\Gamma_n$ be a minimizing sequence for $\E _\beta $ with $\Tr[\Gamma_n]\leq\lambda$. First note that $\Tr[H_\beta\Gamma_n]$ is bounded above, because $\E _\beta [\Gamma_n]$ is bounded above, and since the other contributions to the energy are bounded relative to $H_\beta $. Now we show that the magnetic-kinetic energy is bounded below by the 3-norm of $\rho$: Using the diamagnetic inequality (the prerequisite for the final result in \cite{S76}) \beq \langle \psi ,(H_{\beta} +V)\psi \rangle \geq \langle |\psi |,(H_0 +V)|\psi |\rangle , \eeq and the decomposition of $\Gamma$ as in (\ref{gdia}) , we get \beq \Tr[\nab ^2\Gamma]\geq \sum_nw_n\int\left|\nabla| \psi _n|\right|^2. \eeq Moreover, using the Cauchy-Schwarz inequality, \begin{eqnarray}\nonumber |\nabla\rg|^2&=&\left|\sum_n2w_n| \psi _n|\nabla| \psi _n|\right|^2\\ &\leq&4\left(\sum_nw_n| \psi _n|^2\right)\left(\sum_nw_n \left|\nabla| \psi _n|\right|^2\right). \end{eqnarray} Because $\nabla\rg=2\rg^{1/2}\nabla\rg^{1/2}$ this gives \beq\label{l3} \Tr[\nab^2\Gamma]\geq \int\left|\nabla\rg^{1/2}\right|^2\geq 3\left(\frac\pi 2\right)^{4/3}\left(\int\rg^3\right)^{1/3}, \eeq where we have used the Sobolev inequality in the last step. We can then conclude that the corresponding sequence $\rho_n\equiv\rho_{\Gamma_n}$ is bounded in $L^3\cap\Ll^1$ and $\rho_n^{1/2}$ is bounded in $\Hh^1$. Therefore, for each $p \in (1, 3]$, there exists a subsequence, again denoted by $\rho_n$, that converges to some $\rho_\infty$ weakly in $\Ll^3\cap\Ll^p$, and pointwise almost everywhere. It follows from weak convergence that $\rho_\infty\geq 0$ and $\int\rho_\infty\leq\lambda$. From Fatou's lemma we infer that \beq \label{dconv} \liminf_{n\to\infty}D[\rho_n,\rho_n]\geq D[\rho_\infty,\rho_\infty]. \eeq Moreover, since $|\x|^{-1}\in\Ll^{3/2}+\Ll^{3+\eps}$ for every $\eps>0$, and choosing $p$ the dual of $3+\eps$, we see that \beq \label{aconv} \lim_{n\to\infty}\int\frac{1}{|\x|}\rho_n =\int\frac{1}{|\x|}\rho_\infty. \eeq Since the $\Gamma_n$'s are trace class operators on $\Ll^2$, we can pass to a subsequence such that for some $\Gamma^{\rm H}$ \beq \lim_{n\to\infty}\Tr[\Gamma_n A]=\Tr[\Gamma^{\rm H} A] \eeq for every compact operator $A$. (Here we used the Banach-Alaoglu Theorem and the fact that the dual of the compact operators is the trace class operators). In particular, we have \beq\label{weak} \Gamma_n\rightharpoonup\Gamma^{\rm H} \eeq in the weak operator sense. It is clear that $\Gamma^{\rm H}\geq 0$. Now let $\phi_j\in \Cc_0^\infty(\R^3)$ be an orthonormal basis for $\Ll^2$. Again by Fatou's Lemma \beq \Tr[\Gamma^{\rm H}]=\sum_j\langle\phi_j|\Gamma^{\rm H}|\phi_j\rangle\leq \liminf_{n\to\infty}\Tr[\Gamma_n]\leq\lambda. \eeq In the same way one shows that \begin{eqnarray}\nonumber \Tr[\left(H_\beta-\beta\right)\Gamma^{\rm H}]&=&\sum_j\langle \left(H_\beta-\beta\right)^{1/2} \phi_j|\Gamma^{\rm H}|\left(H_\beta-\beta\right)^{1/2}\phi_j\rangle\\ &\leq& \liminf_{n\to\infty}\Tr[\left(H_\beta-\beta\right)\Gamma_n]. \end{eqnarray} It remains to show that $\rho_{\Gamma^{\rm H}}=\rho_\infty$. We already mentioned that for some constant $C$ \beq\label{bound} \Tr[H_\beta\Gamma_n]\bar \zeta $, \beq -\int\frac{1}{|\x|}\rho < \frac{E^{\rm MH}_{\rm ext}(\lambda,\beta,\zeta ,\alpha)-% E^{\rm MH}_{\rm ext}(\lambda,\beta,\bar \zeta ,\alpha)}{\zeta -\bar \zeta }< -\int\frac{1}{|\x|}\bar \rho. \eeq We conclude, using the continuity (\ref{aconv}), where the $\rho_n$ are now the unique densities of the minimizers for $\bar \zeta _n \to \zeta $, that the Hartree energy is differentiable in $\zeta $, and \beq \left.\frac{\partial E^{\rm MH}_{\rm ext}}{\partial \zeta } \right|_{\zeta =1}=-A. \eeq In the same way, using the inequality (\ref{dconv}) instead of (\ref{aconv}), we show that \beq \left.\frac{\partial E^{\rm MH}_{\rm ext}}{\partial \alpha} \right|_{\alpha =1}=R. \eeq Having established the differentiability in $\lambda$, $\zeta $ and in $\alpha$, the scaling relation (\ref{rel}) implies differentiability in $\beta$. So, the {\bf magnetic moment $\theta $}, \beq \theta =\frac{\partial E^{\rm MH}}{\partial \beta}, \eeq is well defined as a function of $\beta$. Taking the partial derivatives of (\ref{rel}) in $\alpha$ and $\zeta $ at $\alpha =1$ and $\zeta =1$ gives \beqa R&=&-E+\lambda \mu , \label{mu}\\ -A&=&3E-\lambda \mu -2\beta \theta . \label{mutheta} \eeqa The relation (\ref{mu}) is, because of the negativity of the chemical potential $\mu$ for $\lambda < \lambda_c$ , a {\bf virial inequality}, \beq \label{virin} R<|E|, \eeq turning into an equality at $\lambda = \lambda_c$. Inserting (\ref{mu}) into (\ref{mutheta}), we get \beq \label{theta} \beta \theta=\frac{K-|E|}{2}. \eeq The equation (\ref{mu}) and the inequality (\ref{virin}) are also valid in the Regions 1 and 3, where there is a scaling relation analogous to (\ref{rel}). But this relation is without a $\beta$, so instead of (\ref{theta}), in Regions 1 and 3 the well known {\bf virial equality} \beq \label{virial} K=|E|=\frac{A-R}{2} \eeq holds (see also \cite{LSY94a, B84}). Moreover, for $\lambda =\lambda _c$ we have $|E|:K:A:R=1:1:3:1$. \subsection{The linearized theory} Since the density $\rho^{\rm H}$ of a minimizer is unique and integrable, we can define a linearized Hartree functional by \beq \tE[\Gamma]=\Tr[H^{\rm H}\Gamma], \eeq with the one-particle Hartree Hamiltonian \beq\label{hartop} H^{\rm H} \equiv H_\beta-\beta-\Phi^{\rm H}(\x). \eeq Here the Hartree potential $\Phi^{\rm H}(\x)$ is given by \beq \Phi^{\rm H}(\x)=\frac{1}{|\x|}-\rho^{\rm H}*\frac{1}{|\x|}. \eeq Note that $H^{\rm H}$ depends on $\lambda$ only via $\rho^{\rm H}$. \newline Since $\Phi^{\rm H}$ is in $\Ll^2+\Ll^\infty_\eps$, it is relative compact with respect to $H_\beta$ \cite{AHS78}, so we know that the essential spectrum of $H^{\rm H}$ is $[0,\infty)$. \begin{lem}[Linear Hartree functional]\label{linear} Let $\Gamma^{\rm H}$ be a minimizer of $\E_{\beta}$ under the constraint $\Tr[\Gamma]\leq\lambda$. Then $\Gamma^{\rm H}$ also minimizes $\tE$ (under the same constraint). \end{lem} \begin{proof} (We proceed essentially as in \cite{LSY94a}). For any $\Gamma$ \beq \E_{\beta}[\Gamma]=\tE[\Gamma]+D[\rg-\rho^{\rm H},\rg-\rho^{\rm H}]-D[\rho^{\rm H},\rho^{\rm H}]. \eeq Especially, for all $\delta>0$, \begin{eqnarray}\nonumber \E_{\beta}[(1-\delta)\Gamma^{\rm H}+\delta\Gamma]&=& (1-\delta)\tE[\Gamma^{\rm H}]+\delta\tE [\Gamma]-D[\rho^{\rm H},\rho^{\rm H}]\\ &&+\delta^2D[\rg-\rho^{\rm H},\rg-\rho^{\rm H}]. \end{eqnarray} Now if there exists a $\Gamma_0$ with $\Tr[\Gamma_0]\leq\lambda$ and $\tE[\Gamma_0]<\tE[\Gamma^{\rm H}]$ we can choose $\delta$ small enough to conclude that \beq \E_{\beta}[(1-\delta)\Gamma^{\rm H}+\delta\Gamma_0]<\tE [\Gamma^{\rm H}]-D[\rho^{\rm H},\rho^{\rm H}]=\E_{\beta}[\Gamma^{\rm H}], \eeq which contradicts the fact that $\Gamma^{\rm H}$ minimizes $\E_{\beta}$. \end{proof} % The {\bf Hartree equation}, the Euler-Lagrange equation corresponding to the minimization of the functional $\E_{\beta}$, is \beq H^{\rm H} \Gamma = \mu \Gamma . \eeq % % The ground state energy of $H^{\rm H}$ is given by the chemical potential $\mu (\lambda ,\beta )=(E+R)/\lambda$. For the overcritical values of $\lambda \geq \lambda_c$ the ground state energy $\mu$ of $H^{\rm H}$ is $0$, and there is just one density corresponding to the minimizer. We can now ensure that $\lambda_c=\lambda_c(\beta)$ is not too small. In fact we state \begin{lem}[Lower bound on the critical charge]\label{critlow} \beq \lambda_c(\beta)>1\qquad{\rm for\ all\ \beta\geq 0}. \eeq \end{lem} \begin{proof} For $\beta=0$ this was shown in \cite{BL83}, and $\lambda_c(0)$ was computed numerically to be $1.21$ in \cite{B84}. Fix $\beta>0$. We assume that $\lambda_{\rho}\equiv \int\rho^{\rm H}\leq 1$, and will show that $H^{\rm H}$ has an eigenvalue strictly below zero. Using $\psi(r,z)=\exp (-\beta r^2/4-a|z|)$ with some $a>0$ as a (not normalized) test function we compute \begin{eqnarray}\nonumber \langle\psi|H^{\rm H}\psi\rangle&=&\frac{2\pi}\beta a-2\pi\int\Phi^{\rm H}(r,z) e^{-\half \beta r^2-2a|z|}rdrdz\\ &\leq&\frac{2\pi}\beta a-\int\rho^{\rm H}(\x) \left(\phi({\bf 0})-\phi(\x)\right)d^3\x, \end{eqnarray} where $\phi(\x)$ is the potential generated by the charge distribution $\exp(-\half \beta r^2-2a|z|)$, i.e. \beq \phi(\y)=\int\frac{e^{-\half \beta r^2-2a|z|}}{|\x-\y|}d^3\x. \eeq %strictly decreasing Note that $\phi (\0) >\phi (\y)$ for $\y\not= \0$, so we can choose $a$ small enough to conclude that $H^{\rm H}$ has an eigenvalue strictly below zero and binds more charge than $\lambda_{\rho}$. \end{proof} Lemma \ref{critlow} shows that in Hartree theory there are always negative ions. The following lemma gives an upper bound on the maximal negative ionization. \begin{lem}[Upper bound on the critical charge]\label{critup} For some con\-stant $C$ independent of $\beta$ \beq \lambda_c(\beta)\leq 2+\frac 12\min\left\{1+\beta, C\left(1+[\ln\beta]^2\right)\right\}. \eeq \end{lem} \begin{proof} This follows from Lemma \ref{critlem} and the bound on $N_c$ given in \cite{S00} (see Remark \ref{critrem} below). \end{proof} \subsection{The minimizer has $\Lpar =0$} \label{min0} % Let $H$ be the operator \beq\label{defhr} H=H_\beta-\beta-\Phi(\x), \eeq where the potential is given by \beq \Phi(\x)=\frac 1{|\x|}-\frac 1{|\x|}*\rho. \eeq The function $\rho$ is assumed to be {\it axially symmetric}, non-negative, and $\rho\in\Ll^1\cap\Ll^q$, where $q>3/2$. This implies that $|\x|^{-1}*\rho$ is a bounded, continuous function going to zero at infinity. The operator $H$ is essentially self-adjoint on $\Cc_0^\infty(\R^3)$. Since $\Phi\in\Ll^2+\Ll_\eps^\infty$, it is a relatively compact perturbation of $H_\beta$ \cite{AHS78}, so we know that the essential spectrum of $H$ is given by $[0,\infty)$. The symmetry of $\rho$ implies that $H$ is axially symmetric, i.e. it commutes with the rotations generated by the parallel component of the angular momentum, \beq \Lpar=-i\left(x\dy -y\dx\right). \eeq For $m\geq 0$ let now $\Psi_m$ be a ground state of $H$, if there is one, with angular momentum $\Lpar\Psi_m=-m\Psi_m$. Note that we can restrict ourselves to considering non-negative $m$'s, since $(H \upharpoonright \Lpar=-m)$ is antiunitarily equivalent to $(H+2\beta m \upharpoonright \Lpar=m)$ by complex conjugation, so in the ground state $m$ is certainly non-negative. Writing \beq\label{polar} \Psi_m(\x)=e^{-im\varphi}r^m f(r,z), \eeq where $(r,\varphi)$ denote polar coordinates for $(x,y)$, we see that $f$ is a ground state for \beq \widetilde H_m=-\frac{\partial^2}{\partial r^2}-\frac{2m+1}r \frac{\partial}{\partial r}-\frac{\partial^2}{\partial z^2}+\frac{\beta^2}4 r^2-(m+1)\beta-\Phi(r,z) \eeq on $\Ll^2(\R^3,r^{2m}d\x)$. If $\widetilde H_m$ has a ground state, it is unique and strictly positive (\cite{RS78}; to apply the theorems therein note that the first three terms in $\widetilde H_m$ are just the radial part of the Laplacian in $2m+3$ dimensions). So $\Psi_m$ is the only ground state of $H$ with $\Lpar=-m$. Moreover, $f$ is a bounded, continuous function \cite{LL97}, and hence \beq \label{smallr} |\Psi_m (r,\phi ,z)|\leq {\rm const.}r^m, \eeq for small $r$ and some constant independent of $z$. In particular, $\Psi_m({\bf 0})=0$ if $m>0$. We now are able to prove the main result of this subsection: \begin{thm}[$m=0$ in the ground state of $H$]\label{m0} Let $H$ be a Hamiltonian as in (\ref{defhr}), with $\rho$ axially symmetric and non-negative. If \ $\infspec H$ is an eigenvalue, the corresponding eigenvector is unique and has zero angular momentum, for all values of $\beta\geq 0$. \end{thm} \begin{proof} For $\rho=0$, i.e. the hydrogen atom, this was shown in \cite{AHS81} (and also in \cite{GS95}). So we can restrict ourselves to considering the case $\int \rho >0$. Let $\Psi_m$ be a normalized ground state for $H$, with angular momentum $-m$. Define \beq f(\B)=-\int |\Psi_m(\x)|^2 \Phi(\x-\B)d^3\x. \eeq The function $f$ is continuous and bounded. It achieves its minimal value at $\B=\0$, because otherwise one could lower the energy by translating $|\Psi_m|^2$. Strictly speaking, \beq f(\B)-f(\0)=\langle\widetilde\Psi_m|H|\widetilde\Psi_m\rangle -\langle\Psi_m|H|\Psi_m\rangle\geq 0, \eeq with $\B=(b_1,b_2,b_3)$ and \beq\label{phase} \widetilde\Psi_m(\x)=e^{-\mbox{$\frac i2$} \beta (b_2 x-b_1 y)}\Psi_m(\x+\B). \eeq The phase in (\ref{phase}) is chosen such that the kinetic energy remains invariant. (Note that translating $H_\beta$ is equivalent to changing the gauge of the magnetic potential.) In the sense of distributions, \beq\label{dist} \Delta f(\B)=4\pi\left(|\Psi_m(\B)|^2-|\Psi_m|^2*\rho_-(\B)\right), \eeq with $\rho_-(\x)\equiv\rho(-\x)$. The function $|\Psi_m|^2*\rho_-$ is pointwise strictly positive and continuous. Assume now that $m>0$. Since, by (\ref{smallr}), $|\Psi_m(\x)|\leq C|\x|^m$ for some $C>0$, there is an $R>0$ such that \beq \Delta f(\B)<0 \quad {\rm for}\quad |\B|< R, \eeq i.e. $f$ is superharmonic in some open region containing $\0$. This contradicts the fact that $f$ achieves its minimum at $\B=\0$. As a consequence, a ground state of $H$ must have angular momentum $m=0$. \end{proof} An analogous result holds also for the restriction of $H$ to the lowest Landau band. This fact will be used in the proof of Theorem \ref{nrglarge}. \begin{cl}[$m=0$ in the ground state of $\Pi_0 H \Pi_0$]\label{conf0} Let $\Pi_0$ be \newline the projector onto the lowest Landau band. If\, $\Pi_0H\Pi_0$ has a ground state, the ground state wave function $\Psi_0$ is given by \beq\label{psi0} \Psi_0(\x)=\sqrt\frac\beta{2\pi}\exp\left(-\frac\beta 4r^2\right)\psi(z) \eeq for some $\psi(z)$ with $\int|\psi(z)|^2dz=\|\Psi_0\|_2^2$. \end{cl} \begin{proof} Mimicking the proof of the last theorem, we see that if\, $\Pi_0H\Pi_0$ has a ground state, the corresponding wave function $\Psi_0=\Pi_0\Psi_0$ has $\Lpar=0$. Hence it is given by (\ref{psi0}). \end{proof} Let now $\rho$ be the Hartree density $\rho^{\rm H}$, the unique density of the minimizer of $\E_\beta $ (depending on $\lambda$ and $\beta$). Applying Lemma (\ref{linear}) and the theorem above, we immediately get \begin{cl}[Uniqueness of $\Gamma^{\rm H}$]\label{unique} The functional $\E _\beta $ has a unique minimizer $\Gamma^{\rm H}$ under the condition $\Tr[\Gamma]\leq \lambda$, which is proportional to the projection onto the positive function $\sqrt{\rho^{\rm H}}$. In particular, $\Gamma^{\rm H}$ has rank $1$. Moreover, $\rho^{\rm H}$ minimizes the is density functional (\ref{densfunc}) under the condition $\int\rho\leq\lambda$. It is $\Cc^\infty$ away from the origin, continuous at $\x=\0$, and strictly positive. \end{cl} The properties of $\rho^{\rm H}$ stated above follow from the fact that $\rho^{\rm H}$ minimizes $\Edens_\beta$, and therefore satisfies the variational equation \beq\label{varrho} -\Delta\sqrt{\rho^{\rm H}}+\left(\frac{\beta^2}4 r^2- \beta-\Phi^{\rm H}\right)\sqrt{ \rho^{\rm H}}=\mu\sqrt{\rho^{\rm H}}, \eeq where $\mu$ is the chemical potential of the MH theory. Note that Corollary \ref{unique} proves the assertion made in the Introduction that both $\E$ and $\Edens$ have the same ground state energy and density. \section{The limits of Region 2} \label{betalimits} \subsection{The limit of very weak magnetic fields} \label{lowfieldlimit} The limit $\beta \to 0$ is rather easy to handle: \begin{thm}[Hartree energy for small $\beta$]\label{nrgsmall} In the limit $\beta \to 0$, \beq\label{ob} E^{\rm MH}(\lambda,\beta)=E^{\rm H}(\lambda)-\beta\lambda+O(\beta ^2). \eeq \end{thm} \begin{proof} Fix $\lambda>0$, and let $\rho_\beta$ be the minimizer of the density functional $\Edens_\beta$ under the constraint $\int\rho\leq\lambda$. Using $\rho_0$ as a trial function we get \beq E^{\rm MH}(\lambda,\beta)\leq \Edens_\beta[\rho_0]=E^{\rm MH}(\lambda,0)-\beta\lambda+\frac{\beta^2}4\int r^2\rho_0. \eeq It is well known that $\rho_0$ falls off more quickly that the inverse of any polynomial, so $\int r^2\rho_0$ is finite \cite{vw2}. For the converse, we estimate \beq\label{converse} E^{\rm MH}(\lambda,\beta)=\Edens_\beta[\rho _\beta] \geq \Edens_0[\rho _\beta]-\beta\lambda \geq E^{\rm MH}(\lambda,0)-\beta\lambda. \eeq The observation $E^{\rm MH}(\lambda,0)=E^{\rm H}(\lambda)$ then leads to (\ref{ob}). \end{proof} \subsection{Lowest Landau band confinement in Hartree theory}\label{lowestlandauhartree} We now show that for large $\beta$ most of the charge is confined to the lowest Landau band. To do this, we write the Hartree functional as \beq \E_\beta[\Gamma]=\Tr\left[\left(H_\beta-\beta-\frac 1{|\x|}\right)\Gamma\right] +\half\Tr_2\left[\Gamma\otimes\Gamma \frac 1{|\x-\y|}\right], \eeq where $\Tr_2$ means the trace over the doubled space $\Ll^2(\R^3,d^3\x)\otimes \Ll^2(\R^3,d^3\y)$. Let $\Pi_0$ be the projector onto the lowest Landau band, and let $\Pi_>=1-\Pi_0$. We will use the decomposition \beq\label{35} H_\beta-\beta=\Pi_0(H_\beta-\beta)\Pi_0+\Pi_>(H_\beta-\beta)\Pi_> \eeq and \beq \frac 1{|\x|}=\Pi_0\frac 1{|\x|}\Pi_0+\Pi_>\frac 1{|\x|}\Pi_>+ \Pi_0\frac 1{|\x|}\Pi_>+\Pi_>\frac 1{|\x|}\Pi_0. \eeq The off-diagonal terms can be bound using \beq \left(\sqrt\eps\Pi_0-\frac 1{\sqrt\eps}\Pi_>\right)\frac 1{|\x|} \left(\sqrt\eps\Pi_0-\frac 1{\sqrt\eps}\Pi_>\right)\geq 0 \eeq for some $0<\eps<1$, with the result that \beq\label{38} \frac 1{|\x|}\leq(1+\eps)\Pi_0\frac 1{|\x|}\Pi_0+\left(1+\frac 1\eps\right) \Pi_>\frac 1{|\x|}\Pi_>. \eeq In the same way one shows that, see \cite{LSY94a}, \begin{eqnarray}\nonumber \frac 1{|\x-\y|}&\geq& (1-3\eps)\Pi_0\otimes\Pi_0\frac 1{|\x-\y|}\Pi_0\otimes\Pi_0\\ \nonumber &+&\left(1-\frac 3\eps\right)\Pi_>\otimes\Pi_>\frac 1{|\x-\y|}\Pi_>\otimes\Pi_>\\ \nonumber &+&\left(1-2\eps-\frac 1\eps\right)\Pi_0\otimes\Pi_>\frac 1{|\x-\y|}\Pi_0\otimes\Pi_>\\ &+&\left(1-\eps-\frac 2\eps\right)\Pi_>\otimes\Pi_0\frac 1{|\x-\y|}\Pi_>\otimes\Pi_0. \end{eqnarray} Moreover, we will use \beq \Tr_2\left[\Pi_{\#}\otimes\Pi_>\frac 1{|\x-\y|}\Pi_\#\otimes\Pi_> \Gamma\otimes\Gamma\right] \leq\Tr[\Pi_\#\Gamma]\sup_{\abf}\Tr\left[\Pi_>\Gamma\Pi_>\frac 1{|\x-\abf|}\right], \eeq where $\#$ stands for either $0$ or $>$. If we restrict ourselves to considering $\Gamma$'s with $\Tr[\Gamma]\leq\lambda$ we therefore have \begin{eqnarray}\nonumber \E_\beta[\Gamma]&\geq& (1-3\eps)\E_\beta[\Pi_0\Gamma\Pi_0]+3\eps\lambda\left(\frac 83\right)^2 \infspec\left(H_\beta-\beta-\frac 1{2|\x|}\right)\\ \label{abc} &+&\Tr[\Pi_>\Gamma]\left(\frac \beta 2+\inf_\abf\infspec\left(\half H_\beta- \frac 2\eps\frac 1{|\x|}-\frac{2\lambda}{\eps}\frac 1{|\x-\abf|}\right) \right), \end{eqnarray} where we have used that \beq\label{311a} \Pi_>(H_\beta-\beta)\Pi_>\geq\half\Pi_>(H_\beta+\beta)\Pi_> \eeq and that \beq\label{39} \Pi_0\left(H_\beta-\beta-\frac\delta {2|\x|}\right)\Pi_0\geq \delta^2\Pi_0\left(H_\beta-\beta-\frac1 {2|\x|}\right)\Pi_0 \eeq for $\delta\geq 1$, which can easily be seen by scaling $z\to\delta^{-1}z$. Now we use the comparison with the hydrogen atom, Proposition (\ref{hydroprop}): \beq\label{310} \infspec\left(H_\beta-\beta-\frac 1{2|\x|}\right)\geq \max\{1,\lambda^{-1}\} E^{\rm MH}(\lambda,\beta). \eeq By the same argument as in the proof of this inequality, we can set $\abf={\bf 0}$ in (\ref{abc}). Using the diamagnetic inequality we see that $\half H_\beta-c|\x|^{-1}\geq -\half c^2$, so finally \begin{eqnarray}\nonumber \E_\beta[\Gamma]&\geq& (1-3\eps)\E_\beta[\Pi_0\Gamma\Pi_0]+3\eps\left(\frac 83\right)^2 \max\{1,\lambda\}E^{\rm MH}(\lambda,\beta)\\ \label{314} &+&\frac 12\left(\beta - \frac{4(1+\lambda)^2}{\eps^2}\right)\Tr[\Pi_>\Gamma]. \end{eqnarray} Now if $\beta$ is large enough, we can set $\eps^2=4(1+\lambda)^2/\beta$ to conclude \begin{lem}[Comparison with the confined Hartree theory]\label{compconf} \beq\label{confconst} E_{\rm conf}^{\rm MH}(\lambda,\beta)\leq E^{\rm MH}(\lambda,\beta)(1-{\rm const.}(1+\lambda)^2\beta^{-1/2}), \eeq where \beq E_{\rm conf}^{\rm MH}(\lambda,\beta)=\inf_{\Tr[\Pi_0\Gamma]\leq\lambda}\E_\beta[\Pi_0\Gamma \Pi_0]. \eeq \end{lem} Note that the constant in (\ref{confconst}) can be chosen such that (\ref{confconst}) is valid for all values of $\beta>0$ and $\lambda>0$. \begin{rem} Equation (\ref{314}) can be used to estimate the part of the Hartree state $\Gamma^{\rm H}$ not confined to the lowest Landau band. In fact, if $\eps^2>4(1+\lambda)^2/\beta$ in (\ref{314}), and with $\Gamma=\Gamma^{\rm H}$, we get \beq \Tr[\Pi_>\Gamma^{\rm H}]\leq -{\rm const.}(1+\lambda)\frac {\eps E^{\rm MH}(\lambda,\beta)} {\beta -4(1+\lambda)^2\eps^{-2}}. \eeq Using the simple lower bound on $E^{\rm MH}$, given in Lemma \ref{simple}, and optimizing over $\eps$ gives \beq \Tr[\Pi_>\Gamma^{\rm H}]\leq {\rm const.}(1+\lambda)^3\beta^{-1/2}. \eeq \end{rem} \subsection{Confinement for the mean field Hamiltonian} An analogous result as in the previous subsection holds also for the linearized theory. The following estimate will be used in Subsection \ref{reg3}: \begin{lem}[Confinement for the mean field Hamiltonian]\label{confmean} Let $H$ be given as in (\ref{defhr}), with $\rho\geq 0$ and $\int\rho\leq\lambda$. Then for all $\beta>0$ \beq\label{320} \infspec H \geq \infspec \Pi_0 H \Pi_0 +(2+\lambda)\beta^{-1/2} E^{\rm hyd}(\beta,2). \eeq \end{lem} \begin{proof} For $0<\eps<1$ we use again (\ref{35}) and (\ref{38}), and the analogous inequality in the other direction for $|\x|^{-1}*\rho$, to conclude that \begin{eqnarray}\nonumber H &\geq& \Pi_0\left(H_\beta-\beta-(1+\eps)\frac 1{|\x|}+(1-\eps) \frac 1{|\x|}*\rho \right)\Pi_0 \\ \nonumber & & + \Pi_>\left(H_\beta-\beta- (1+\eps^{-1})\frac 1{|\x|}+(1-\eps^{-1}) \frac 1{|\x|}*\rho\right)\Pi_>\\ \nonumber &\geq& (1-\eps)\Pi_0 H \Pi_0+\eps \Pi_0\left(H_\beta-\beta- \frac 2{|\x|}\right)\Pi_0 \\ & & + \half \Pi_>\left(H_\beta+\beta-\frac {4\eps^{-1}}{|\x|}- \frac {2\eps^{-1}}{|\x|}*\rho\right)\Pi_>, \end{eqnarray} where we have also used (\ref{311a}). With the aid of the inequality (\ref{aboundr}) one easily sees that \beq \infspec \left(H_\beta-\frac {4\eps^{-1}}{|\x|}- \frac {2\eps^{-1}}{|\x|}*\rho\right) \geq \infspec \left(H_\beta-\frac {(4+2\lambda)\eps^{-1}}{|\x|}\right). \eeq Using again $\half H_\beta-c|\x|^{-1}\geq -\half c^2$ we finally get \beq H \geq (1-\eps) \Pi_0 H \Pi_0 + \eps E^{\rm hyd}(\beta,2)+\half\Pi_>\left(\beta-(2+\lambda)^2 \eps^{-2}\right). \eeq In particular, if we choose $\eps=(2+\lambda)\beta^{-1/2}$, we arrive at the desired result, as long as $\beta$ is large enough to ensure $\eps<1$. But if $\beta\leq (2+\lambda)^2$, (\ref{320}) holds trivially because of the positivity of $\rho$. \end{proof} \subsection{The limit of very strong magnetic fields} \label{highfieldlimit} In the investigation of the limit $\beta \to \infty$ some special concepts from earlier studies are involved: The right scaling, the restriction to the lowest Landau band, the effects of the superharmonicity on the distribution of the density in the perpendicular directions, and the high field limit of the Coulomb interaction. We consider pure states $\Lambda =|\chi \rangle \langle \chi |$ defined by the wave functions with $\Lpar =0$ in the lowest Landau band, \beq \label{chi} \chi (\x )=\sqrt{\frac{\beta}{2\pi}}% e^{-\beta (\xperp)^2 /4}% \sqrt{L}\psi (L\xpar ). \eeq We denote here the perpendicular and parallel components of the coordinates as $\x = (\xperp , \xpar )$. The scaling factor $L=L(\beta )$ has been defined in \cite{BSY00} as the solution of the equation \beq \beta ^{1/2}=L(\beta )\sinh[L(\beta )/2]. \eeq Note that \beq L(\beta )=\ln \beta +O(\ln\ln \beta )\eeq as $\beta \to \infty$. In the following we use the scaled coordinates \beq \label{betascal} z=L(\beta)\xpar ,\qquad \rr = \sqrt{\beta}\xperp , \qquad r=|\rr |. \eeq % With the density matrix $\Lambda =|\chi \rangle \langle \chi |$, we calculate the contributions to $\E_\beta [\Lambda]$. The normalization of $\Lambda$, we denote it as $\lambda _\psi$, is equal to $\|\psi\|_2^2$. We restrict the set of $\psi$ to real valued wave functions, since they minimize the kinetic energy, if the density $|\psi|^2$ is kept fixed. \begin{lem}[The strong field limits]\label{highblim} Given the state $\Lambda$ defined by real valued $\psi \in \mathcal{H}^{1}(\R)$ as in (\ref{chi}), the contributions to the energy in the magnetic Hartree theory, magnetic-kinetic energy $K_{\beta ,\psi}$, attraction $A_{\beta ,\psi}$, and repulsion $R_{\beta ,\psi}$, are in the following way related to the kinetic energy, $K_\psi = \int (d\psi / dz)^2 dz$, attraction-energy $A_\psi =\psi (0)^2$, and repulsion-energy $R_\psi = \frac{1}{2} \int \psi (z)^4 dz$ of the density $\psi (z)^2$ in the HS-theory: \beqa K_{\beta ,\psi}&=& L^2 K_\psi , \label{khighblim} \\ \left|\frac{1}{L^2}A_{\beta ,\psi}-A_\psi \right| &\leq& \frac{C}{L}\left( \lambda _\psi + \lambda _\psi ^{1/4}K_\psi ^{3/4} \right), \label{ahighblim} \\ \left|\frac{1}{L^2}R_{\beta ,\psi}-R_\psi \right| &\leq& \frac{C\lambda _\psi}{L}\left( \lambda _\psi + \lambda _\psi ^{1/4}K_\psi ^{3/4} \right). \label{rhighblim} \eeqa \end{lem} \begin{proof} The equation for the kinetic energy is obvious by definition of $\Lambda$. We calculate the energy of attraction as \beq A_{\beta ,\psi}= L^2\int_0^{\infty}e^{-r^2/2} \left[\int_{-\infty}^{\infty}V_{\beta ,r}(z)\psi ^2(z)dz\right]rdr, \eeq where \beq V_{\beta ,r}(z)=\frac{1}{L\sqrt{L^2r^2/\beta +z^2}}. \eeq For the term in square brackets we use Lemma 2.1 of \cite{BSY00}, \beq \left| [...] - \psi (0)^2 \right| \leq \lambda _\psi /r+8\lambda _\psi^{1/4}K_\psi ^{3/4}r^{1/2}, \eeq to estimate the difference to the expected limit in the HS-theory: \beq \left|\frac{1}{L^2}A_{\beta ,\psi}-A_\psi \right| \leq \frac{1}{L}\left( \lambda _\psi \sqrt{\pi /2}+ 8\cdot 2^{1/4}\Gamma (5/4)\lambda _\psi^{1/4}K_\psi^{3/4}\right). \eeq % For the repulsion $R_{\beta ,\psi}$ we calculate \beq \label{rhighb} \left|\frac{1}{L^2}R_{\beta ,\psi}-R_\psi \right| \leq \frac{1}{2} \int_{\R^4}^{}d^2\rr d^2\rr '(\frac{1}{2\pi})^2 e^{-(\rr ^2+\rr '^2)/2}[...], \eeq where \beqa [...] = \left| \int_{\R^2}dz dz' \left( V_{\beta ,|\rr -\rr '|}(z-z')-\delta (z-z') \right) \psi^2 (z) \psi^2 (z') \right| \label{rpsi} \\ \leq \frac{1}{L}\left(\frac{\lambda _\psi ^2}{|\rr -\rr '|}+ 8\lambda _\psi^{5/4}K_\psi^{3/4}|\rr -\rr '|^{1/2}\right) . \label{rhighc} \eeqa The inequality is supplied by Lemma 2.2 of \cite{BSY00}, with an adaptation due to the different notation concerning the normalization. After inserting (\ref{rhighc}) in (\ref{rhighb}), the integrals can be evaluated as \beq \left|\frac{1}{L^2}R_{\beta ,\psi}-R_\psi \right|\leq \frac{1}{L}\left( (\sqrt{\pi}/4)\lambda _\psi^2 + 4\sqrt{2}\Gamma (5/4)\cdot \lambda _\psi^{5/4}K_\psi^{3/4} \right) . \eeq \end{proof} \begin{thm}[Magnetic Hartree energy for large $\beta$]\label{nrglarge} In the limit \newline $\beta \to \infty$ with $\lambda$ fixed, \beq \lim_{\beta \to\infty}\frac{E^{\rm MH}(\lambda,\beta)}{(\ln \beta )^2}= E^{\rm HS}(\lambda). \eeq \end{thm} % \begin{proof} Combining all the bounds of the Lemma, and using $\lambda _\psi ^{1/4}K_\psi ^{3/4}\leq \frac{1}{4}\lambda _\psi +\frac{3}{4}K_\psi$ to simplify, gives \beq \label{comparison} \left|\frac{1}{L^2}\E_{\beta}[\Lambda] -{\mathcal E}^{\rm HS}[|\psi |^2]\right| \leq \frac{C}{L(\beta)}(1+\lambda _\psi )(\lambda _\psi+ K_\psi). \eeq For the {\bf upper bound} on $E^{\rm MH}(\lambda,\beta)$ we specify $\psi$ as $\psi _\lambda ^{\rm HS}$, the minimizing wave function for the HS-theory. See equation (3.6) in \cite{LSY94a} or the following (\ref{rhohs}). It remains $\psi _2 ^{\rm HS}$, for all $\lambda > 2$, so these wave functions are normalized to $\|\psi \|_2^2 = \lambda _\psi = \min \{\lambda ,2\}$. Since the variational principle with fixed norm can be replaced by a variational principle with bounded norm, as we have remarked in the Subsection \ref{hartreedef}, the wave function $\psi _2 ^{\rm HS}$ is good for the upper bounds for all $\lambda \geq 2$. $K_\psi$ is the kinetic energy in the hyperstrong theory. Using the virial equation (\ref{virial}), we may replace $K_\psi $ on the right side of (\ref{comparison}) by $|E^{\rm HS}(\lambda )|$. With this choice of $\psi$ the equation (\ref{comparison}) gives the upper bound \beq \label{comparisonupper} \frac{1}{L^2}E^{\rm MH}(\lambda,\beta) \leq E^{\rm HS}(\lambda) + \frac{C}{L(\beta)} \left( \lambda+ \left| E^{\rm HS}(\lambda)\right| \right) . \eeq To derive a {\bf lower bound}, we use the error bound, when confining the theory to the lowest Landau band, which we have estimated in the Subsection \ref{lowestlandauhartree}. By Lemma \ref{compconf} we have, for $\beta$ large enough, \beq \label{compconf2} E^{\rm MH}(\lambda,\beta) \geq (1-{\rm const.}(1+\lambda)^2\beta^{-1/2})^{-1} E_{\rm conf}^{\rm MH}(\lambda,\beta). \eeq Also this confined Hartree theory has rotation invariant minimizers, since the lowest Landau band is mapped onto itself by rotations around the z-axis. Moreover, since the potential is superharmonic, also the minimizer of this confined theory has $\Lpar =0$. See Corollary \ref{conf0} of Subsection \ref{min0}. The variational principle can therefore be restricted to states $\Lambda =|\chi \rangle \langle \chi |$, where the wave functions $\chi$ are specified as in equation (\ref{chi}), with general $\psi \in \mathcal{H}^{1}(\R)$, normalized to $\|\psi\|_2^2 =\lambda$: \beq E_{\rm conf}^{\rm MH}(\lambda,\beta) = \inf_{\Lambda , \hspace{4pt} \Tr[\Lambda]=\lambda , \hspace{4pt} \Lambda =|\chi \rangle \langle \chi |} \E_{\beta}[\Lambda]. \eeq % The comparison with HS-theory in (\ref{comparison}) is now used as a lower bound \beq \frac{1}{L(\beta)^2}\E_{\beta}[\Lambda]\geq {\mathcal E}^{\rm HS}[|\psi |^2] -\frac{C}{L(\beta)}(1+\lambda )(\lambda + K_\psi ). \eeq % The right side of this inequality can be considered as a functional similar to the HS-functional, but with the constant $1-C(1+\lambda )/L(\beta)$ in front of the kinetic energy. With an appropriate scale transformation, this is equivalent to a HS-functional multiplied with $\left( 1-(1+\lambda )C/L(\beta) \right) ^{-1}$, as long as this parameter is positive. Taking the infima of both sides, we conclude that \beq \frac{1}{L(\beta)^2} E_{\rm conf}^{\rm MH}(\lambda,\beta) \geq \left( 1-C\frac{(1+\lambda )}{L(\beta)} \right)^{-1} E^{\rm HS}(\lambda) -C\frac{(1+\lambda )\lambda}{L(\beta)} \eeq for $\beta$ large enough. Considering the limit $\beta \to \infty$ at constant $\lambda$ we infer \beq \liminf_{\beta\to\infty} \frac{1}{L(\beta)^2} E_{\rm conf}^{\rm MH}(\lambda,\beta) \geq E^{\rm HS}(\lambda). \eeq Combining this with (\ref{compconf2}) proves, in union with (\ref{comparisonupper}), the theorem. \end{proof} \begin{rem}\label{specify} To be precise, we specify the asymptotics: \beqa \frac{E^{\rm MH}(\lambda,\beta)}{L( \beta )^2} &\leq& \left( 1-C\frac{\lambda}{L(\beta)} \right) E^{\rm HS}(\lambda) + C\frac{\lambda}{L(\beta)}, \\ \nonumber \frac{E^{\rm MH}(\lambda,\beta)}{L( \beta)^2} &\geq& \left(1-C\frac{(1+\lambda)^2}{\beta^{1/2}}\right)^{-1}\left( 1-C\frac{1+\lambda}{L(\beta)}\right)^{-1} E^{\rm HS}(\lambda) - C\frac{\lambda+\lambda^2}{L(\beta)},\\ \eeqa as long as the terms in braces are positive. \end{rem} \begin{rem} The convergence both of the energy of the atom, and of the energy per unit charge, when divided by $(\ln \beta)^2$, is uniform in $\lambda$ on bounded sets of $\lambda$. \end{rem} Analyzing the limit $N \to \infty$ of many particle quantum mechanics, we will be led to the linearized mean field theory, where we need the \begin{lem}[Generalized strong field limits] \label{highblim2} Given two states which are defined by real valued $\psi \in \mathcal{H}^{1}(\R)$ and real valued $\varphi \in \mathcal{H}^{1}(\R)$ as in (\ref{chi}), with densities $\rho_{\beta,\psi} = \beta /2\pi e^{-\beta (\xperp)^2 /2} L\psi (L\xpar )^2$, the following estimate for the Coulomb repulsion holds: \beq \left|\frac{1}{L^2} D[\rho_{\beta,\varphi},\rho_{\beta,\psi}] - \frac{1}{2}\int_{-\infty}^{+\infty}\varphi (z)^2\psi(z)^2dz\right| \leq \frac{C\lambda_\varphi}{L}\left(\lambda_\psi+\lambda_\psi^{1/4} K_\psi^{3/4}\right). \eeq \end{lem} \begin{proof} As for $R_{\beta,\psi}$ in the proof of Lemma \ref{highblim}, with $\psi(z)^2$ in (\ref{rpsi}) changed to $\varphi(z)^2$. \end{proof} \section{The mean field limit} \label{highnlimit} We now prove the theorems that we have stated in the introduction. We will derive appropriate upper and lower bounds to the quantum mechanical energy of symmetric states obeying Bose statistics. The atom with many interacting particles will be compared to models with independent particles in an effective field. In this comparison the upper bound is rather easy: One uses symmetric wave functions of product form as trial wave functions. Then one adds the ``self energies'' of the one-particle densities. The production of lower bounds is not so easy. For the Regions $1$ and $2$ we will use the Lieb-Oxford bound, \cite{LO81}, and then we will borrow a part of the kinetic energy to estimate the correction in comparison to the mean field model. But this method breaks down in the presence of magnetic fields, when they are too strong. In Region $3$ its use is restricted to the subregion $\beta \lesssim N^{4/3}$. This situation is similar to the case of fermionic electrons, discussed in Sect.~$7$ of \cite{LSY94a}. For $\beta \gtrsim N^{4/3}$ we have to borrow some kinetic energy at the level of quantum mechanics for $N$ particles. Here we develop a new way of producing bounds, extending the method of \cite{BSY00}. \subsection{Upper bounds} \label{nparticleupperbounds} \begin{lem}[Upper bounds for Regions $1$ and $2$]\label{up1-2} The Hartree energies provide upper bounds to the quantum mechanical energies for each $N$, $\lambda$ and $\beta$: \beq \frac{E(N,\lambda,\beta)}{N}\leq \frac{E^{\rm MH}(\lambda,\beta)}{\lambda}. \eeq \end{lem} \begin{proof} We apply the variational principle, using $N$-fold products of $1$-particle states $\Gamma$ as $N$-particle states \beq E(N,\lambda,\beta)\leq N \Tr[(H_\beta-\beta)\Gamma]-N\int\frac{1}{|\x|}\rg +\lambda \frac{N(N-1)}{N^2} D[\rg,\rg] \eeq for all $\Gamma$ with $\Tr[\Gamma]\leq 1$. Setting $\Gamma=\Gamma^{\rm H}/\lambda$ we see that the inequality holds, for each $\beta$ and $\lambda$. \end{proof} \begin{lem}[Upper bound for Region $3$]\label{up3} In the limit $\beta \to \infty$, the energies of the hyperstrong theory are asymptotic upper bounds to the quantum mechanical energies for each $N$: \beq \limsup_{\beta \to \infty} \frac{E(N,\lambda,\beta)}{N (\ln \beta)^2}\leq \frac{E^{\rm HS}(\lambda)}{\lambda}. \eeq To be precise: \beq \frac{E(N,\lambda,\beta)}{N\,L( \beta )^2} \leq \left( 1-C\frac{\lambda}{L(\beta)} \right) \frac{E^{\rm HS}(\lambda)}{\lambda} + \frac{C}{L(\beta)}. \eeq \end{lem} \begin{proof} This results from combining Lemma \ref{up1-2} with Theorem \ref{nrglarge}. \end{proof} \subsection{Lower bounds for Regions $1$ and $2$} \label{nparticlelowerbounds} \begin{lem}[Lower bounds for Regions $1$ and $2$]\label{down1-2} In the limit $N \to \infty$, the Hartree energies are asymptotic lower bounds to the quantum mechanical energies, \beq \liminf_{N \to \infty} \frac{E(N,\lambda,\beta)}{N }\geq \frac{E^{\rm MH}(\lambda ,\beta )}{\lambda}, \eeq uniformly in $\beta$ for bounded $\beta$. \end{lem} \begin{proof} We use the Lieb-Oxford inequality \cite{LO81} \beq \langle\Psi|\sum_{i1$) into its effective part $V_\mu$ and the long range tail $V_{\rm long}$, \beqa V_\mu (\x) &=& \frac{e^{-|\x|}-e^{-\mu |\x|}}{|\x|},\qquad \mu >1, \\ V_{\rm long}(\x) &=& \frac{1-e^{-|\x|}}{|\x|}, \eeqa and borrow some kinetic energy. The {\bf construction of ${\bf W}$} proceeds in several steps. We begin with the observation that $V_\mu$ is integrable in $\xpar$ for any $\xperp$. It has moreover the property of being decreasing in $|\x|$. So the integrals along lines with fixed $\xperp$ can be bounded by the integral along the line, where $\xperp = 0$: \beq \int_{-\infty}^\infty V_\mu (\x) d\xpar \leq 2 \ln \mu. \eeq We now use Lemma $6.3$ of \cite{BSY00}, the {\bf operator inequality}, which holds for $\alpha>0$, $F>0$, and for each $b>0$, $b$ independent of $\alpha$ and $F$, \beq \label{deltiq} \alpha p_x^2 +F\delta (x) \geq F\cdot w_{F/\alpha , b}(x), \eeq where \beq \label{deltappr} w_{D, b}(x) = D\frac{b^2}{2b+1}e^{-bD|x|} \eeq and $p_x^2=-(\partial/\partial x)^2$. These functions have the properties of being positive definite and producing delta-function sequences in the limit $bD \to \infty$ coupled with $b \to \infty$. (The variable $x$ will in the application be replaced by $\xpar $.) We extend this lemma to the \begin{proposition}[Operator inequalities] Let V(x) define a positive Borel measure $V(x)dx$, with $\int_{-\infty }^{+\infty }V(x)dx \leq F$, $F>0$, $\alpha>0$. For each $b>0$ the operator inequality \beq \alpha p_x^2 +V(x)\geq (V * w_{F/\alpha , b})(x) \label{opiq4} \eeq holds, in the sense of an inequality for quadratic forms, with the Sobolev space ${\mathcal H}^1 (\R)$ as the form domain. \end{proposition} \begin{proof} For $\psi \in {\mathcal H}^1 (\R)$ consider the wave functions $\psi_y(x) = \psi (x+y)$. By (\ref{deltiq}) there holds for each $y$ the inequality \beq \langle \psi_y | ( \alpha p^2 +F\delta ) |\psi_y \rangle% \geq F\langle \psi_y | w_{F/\alpha , b} |\psi_y \rangle. \label{functiq} \eeq Both sides of this inequality are Borel measurable functions of $y$. Observe $\langle \psi_y | p^2 |\psi_y \rangle =% \langle \psi | p^2 |\psi \rangle$, $\langle \psi_y | w(x) |\psi_y \rangle =% \langle \psi | w(x-y) |\psi \rangle$, and integrate both sides of (\ref{functiq}) with $V(y)dy$. Divide by $F$, and use $\int_{-\infty }^{+\infty }V(x)dx / F \leq 1$ in front of the kinetic energy term. \end{proof} We may consider $\alpha (\ppar_i)^2 + V_{\mu }(\x_{i}-\x_{j})$ as a 1-particle Hamiltonian acting in $\mathcal{L}^{2}(\R ,d\xpar_{i})$, which is parametrized by the perpendicular coordinates and by $\x_{j}$. However, the operator inequality (\ref{opiq4}) is also valid if the operators act in the extended space $\mathcal{L}^{2}(\R ^{3N},d^3 \x_{1}...d^3 \x_{N})$, so we can apply this proposition, with $F=2\ln \mu$, to \beq \alpha (\ppar _i)^2 + V_{\mu }(\x_i-\x_j). \eeq We add the missing long range tail and define \beq \label{wdef} W(\x)=\int V_{\mu }(\x-\y) \delta ^2 (\yperp ) w_{2\ln \mu /\alpha ,b}(\ypar)d^3\y +\frac{1-e^{-|\x|}}{|\x|}, \eeq where $w_{...}(\xpar)$ is defined in (\ref{deltappr}). The essential effect of the convolution of $V_\mu$ with the distribution \beq \uw (\x)=\delta(\xperp)w(\xpar) \eeq is the lowering of the value of the interactions at the points of coincidences of particles, \beq W(\0) \leq \left( \int V_\mu d\xpar \right) w_{2\ln \mu /\alpha ,b}(0) +1 < \frac{2b}{\alpha}(\ln \mu)^2 +1. \eeq Thus we have shown, that $W(\x)$ is a lower bound to the Coulomb potential, when some borrowed part of the kinetic energy is added, \beq \label{wleqc} W=V_\mu *\uw + V_{\rm long} \leq \alpha (\ppar)^2 + V_\mu +V_{\rm long} \leq \alpha (\ppar)^2 +1/|\x|. \eeq Now the function $W = V_\mu *\uw +V_{\rm long}$ is also positive definite, since the distribution $\uw$ and all the involved functions are positive definite. So we can refer to the technique (\ref{pdef}) of reduction to one-particle models. \begin{proposition}[Bounds by independent particles]\label{bdbyind} Considered as quadratic forms, the N-particle Hamiltonians $H_{N,\lambda ,\beta }$ are bounded from below by sums of one-particle Hamiltonians and constants: \beq H_{N,\lambda ,\beta } \geq \sum_{i=1}^{N}h_{i} - \frac12 \frac{\lambda}{N} \int\!\!\!\int \sigma (\x )W(\x -\y)\sigma (\y)d^3\x \,d^3\y -\lambda \frac{b}{\alpha}(\ln \mu )^{2}-\frac{\lambda}{2}, \eeq % where \beq \label{oneham} h_{i}= H_{\beta , i} - \beta -\frac{\alpha \lambda}{2}(\ppar _i)^2 - \frac{1}{|\x_{i}|}+\frac{\lambda}{N}\int W(\x _{i}-\y )\sigma (\y )d^3\y, \eeq % and where $W(\x)$ is defined in (\ref{wdef}). The parameters $\alpha ,b$ have to be positive, $\mu>1$, $\sigma (\x)$ should be an element of $\Ll^1 (\R^3)$. \end{proposition} \begin{proof} From $H_{\beta ,i}$ we borrow a part of $(\ppar_i)^2 = - (\partial /\partial \xpar_i)^2$ to use (\ref{wleqc}), \beq \alpha (\ppar _i)^2 + \frac{1}{|\x_i-\x_j|} \geq W(\x_i-\x_j), \eeq sum over all pairs $i\neq j$ , and multiply with $\lambda /2N$. This gives \beq \label{opiq3} \alpha \lambda \frac{N-1}{2N}\sum_{i=1}^{N}(\ppar _i)^2 +\frac{\lambda }{N} \sum_{iC_\beta$ for some $C_\beta>1$ in the following. We choose the parameters as \beq \mu=\beta^{1/2+\delta}, \alpha=N^{-\eps}, b=N^\eta , \eeq with $\delta, \eps, \eta$ all greater than zero, $\eps+\eta<1$. This guarantees first of all the relative vanishing of the constant which stems from $W(\0)$: \beq \frac{\lambda}{N} \left(\frac{b}{\alpha}(\ln \mu )^{2}+\frac 12\right) /(\ln \beta )^2 \to 0, \eeq as $\beta,N \to \infty$. It remains to study the Hamiltonian $h$ and the self energy of $\rho$ due to the interaction $W=V_\mu *\uw + V_{\rm long}$. The inequality \beq \int \!\!\!\int \rho(\x)(V_\mu *\uw )(\x -\y) \rho (\y)d^3\x \,d^3\y \leq \int \!\!\!\int \rho(\x)V_\mu(\x -\y) \rho (\y)d^3\x \,d^3\y \eeq can be observed in Fourier space, where $\widetilde V_\mu$ and $\widetilde \uw$ are positive, and $\widetilde \uw \leq 1$. We add $V_{\rm long}$, and use the pointwise inequality in $\x$-space $(V_\mu +V_{\rm long})(\x) \leq V^C(\x)\equiv 1/|\x|$. So the self-energy of $\rho$ due to $W$ is smaller than the self-energy $D[\rho,\rho]$ due to the interaction by the Coulomb potential $V^C$. The potential $(V_\mu *\uw +V_{\rm long})*\rho$ in the Hamiltonian $h$ is now changed to $V^C*\rho$. First we give a bound to the difference of $V_{\mu}*\uw *\rho$ to $V_{\mu}*\rho$ in the form of an operator inequality, which follows from \beqa |\,\langle \psi| V_{\mu}*(\uw *\rho -\rho )|\psi\rangle\,| &\leq& \|\,(|\psi|^2*V_{\mu})\cdot(\uw *\rho -\rho)\|_1 \nonumber \\ &\leq& \|\uw *\rho -\rho\|_1\,\,\|\,|\psi|^2*V_{\mu}\|_\infty \nonumber \\ &=& \|\uw *\rho -\rho\|_1\,\sup_{\y}\langle\psi|V_{\mu,\y}|\psi\rangle, \label{vy} \eeqa where $V_{\mu,\y}=V_{\mu}(\x -\y)$. Since $V_{\mu}\leq V^C$, and, with $E^{\rm hyd}(\beta)\equiv E^{\rm hyd}(\beta,\zeta=1)$, \beq H_\beta -\beta - V_{\y}^C \geq E^{\rm hyd}(\beta), \eeq the last term in (\ref{vy}) is bounded by \beq \langle\psi|V_{\y}^C|\psi\rangle \leq \langle\psi| H_\beta -\beta - E^{\rm hyd}(\beta)|\psi\rangle, \eeq multiplied with $\|\uw *\rho -\rho\|_1$. We observe, see Remark \ref{ll0}, \beq E^{\rm hyd}(\beta)=\lim_{\lambda\to 0}\frac{E^{\rm MH}(\lambda,\beta)}{\lambda}, \qquad \lim_{\lambda\to 0}\frac{E^{\rm HS}(\lambda)}{\lambda}=-\frac{1}{4}. \eeq The precise bound in the Remark \ref{specify} gives thus (see also \cite{AHS81}) \beq E^{\rm hyd}(\beta) \geq \left( -\frac{1}{4}-C\frac{1}{L(\beta)}\right) L^2(\beta)\geq -C(\ln\beta)^2 \eeq for $\beta>C_\beta$. An analogous bound holds also for $E^{\rm hyd}(\beta,2)$. Note that $L(\beta)$ can be replaced by $(\ln\beta)$ (and vice versa), since for $\beta>C_\beta>1$ there are positive constants $C_1$ and $C_2$ such that $C_1\leq\ln(\beta)/L(\beta)\leq C_2$. To complete the estimate (\ref{vy}), we need a bound to $\|\uw *\rho -\rho\|_1$. There, the integrals in perpendicular coordinates can be done explicitly. Note that for all $\lambda$ we have $|d\rho^{\rm HS}_\lambda/d z|\leq \rho^{\rm HS}_\lambda$. Using this and the monotonicity of $\rho^{\rm HS}_\lambda$ in $|z|$ we can estimate \begin{eqnarray}\nonumber \left|\rho^{\rm HS}_\lambda(z-y)-\rho^{\rm HS}_\lambda(z)\right|&\leq& |y| \sup_{x\in[z-y,z]}\left|\frac{d \rho^{\rm HS}_\lambda}{dz}(x)\right| \\ \nonumber &\leq& |y|\left( \rho^{\rm HS}_\lambda(z-y)+\rho^{\rm HS}_\lambda(z)+\Theta(|y|-|z|) \rho^{\rm HS}_\lambda(0) \right).\\ \label{Theta} \end{eqnarray} Since for $|y|\geq 1$ (\ref{Theta}) is obviously true even without the last term, we see that (\ref{Theta}) holds with $\Theta(|y|-|z|)$ replaced by $\Theta(1-|z|)$. This implies \begin{eqnarray}\nonumber \|\rho*\uw -\rho\|_1&\leq& \frac 1\lambda\int d\xpar d\eta L w(\eta)\\ \nonumber & &\times\left( \left|\rho^{\rm HS}_\lambda(L\xpar -L\eta)-\rho^{\rm HS}_\lambda(L\xpar )\right| +\left((\mbox{$\int w$})^{-1}-1\right)\rho^{\rm HS}_\lambda(L\xpar )\right) \\ \nonumber &\leq& \lambda^{-1}\mbox{$\int\rho^{\rm HS}_\lambda$}(1-\mbox{$\int w$})+ 2\lambda^{-1}\left(\mbox{$\int\rho^{\rm HS}_\lambda$}+ \rho^{\rm HS}_\lambda(0)\right) L \int |\eta|w(\eta)d\eta.\\ \end{eqnarray} Now $\int\rho^{\rm HS}_\lambda\leq \lambda$, $\rho^{\rm HS}_\lambda(0)\leq \half\lambda$, $(1-\int w)\leq (2b)^{-1}$ and $\int |\eta|w\leq \alpha /(2b\ln \mu)$, so we get \beq \|\rho*\uw -\rho\|_1\leq \frac1{2b}+ \frac{3}{2}\frac{\alpha L(\beta)}{b\ln \mu} \leq C(N^{-\eta} + N^{-\eta-\eps} L(\beta)/\ln \beta )\leq C' N^{-\eta}. \eeq \bigskip Finally, we add to $V_\mu$ the short-range term \beq V_{\rm short}(\x)=\frac{e^{-\mu |\x|}}{|\x|}, \eeq and we find the pointwise bound to $(V^C-V_\mu)*\rho$ as \beq \|V_{\rm short}\|_1 \|\rho\|_\infty \leq C\mu^{-2} \beta L(\beta)= C\beta^{-2\delta}L(\beta). \eeq \bigskip We collect the bounds we have obtained so far: \beq \label{collbounds} \frac{E(N,\lambda,\beta)}{N} \geq \infspec \{h_b\} -\lambda D[\rho,\rho]-C\lambda(\ln \beta)^2\left(N^{\eps+\eta -1}+N^{-\eta}+ \beta^{-2\delta}L(\beta)^{-1}\right), \eeq with the one-particle Hamiltonian \beq \label{cllh} h_b = \left( H_{\beta} -\beta \right)\left( 1-C\lambda N^{-\eta}-2\lambda N^{-\eps}\right) - V^C +\lambda V^C*\rho, \eeq where we have used $(\ppar)^2\leq H_\beta-\beta$. We now use (\ref{eps}) to estimate the error terms in the kinetic energy. Then we apply the statements of the subsection on the confinement in the mean field theory, Lemma \ref{confmean}, to estimate \begin{eqnarray}\nonumber \frac{E(N,\lambda,\beta)}{N} &\geq& \infspec\left\{\Pi_0\left( H_\beta-\beta-V^C+\lambda V^C*\rho\right)\Pi_0\right\} -\lambda D[\rho,\rho] \\ \nonumber &&-C\lambda(\ln\beta)^2\left(N^{-\eta}+N^{-\eps}+N^{\eps+\eta-1} +\frac{\beta^{-2\delta}}{L(\beta)}+\frac{2+\lambda}\lambda \beta^{-1/2}\right).\\ \end{eqnarray} Corollary \ref{conf0} states that the minimizer of the operator in question has $\Lpar =0$. The search for the infimum of the spectrum can therefore be restricted to the use of the same type of wave-functions as in the investigations on the limit of very strong magnetic fields in Subsection \ref{highfieldlimit}, Lemma \ref{highblim} and \ref{highblim2}, and these investigations can be applied, too. We choose $\eps=\eta=1/3$ and get the lower bound \beqa\nonumber &&\frac{E(N,\lambda,\beta)}{N\,L^2(\beta)} \\ \nonumber &&\geq \left( \infspec \left\{p_z^2-\delta (z) + \rho^{\rm HS}_\lambda (z) \right\} -\frac1{2\lambda} \int (\rho^{\rm HS}_\lambda (z))^2 dz \right) \left( 1-C\frac{1+\lambda}{L(\beta)}\right)^{-1} \\ \label{bbbound} &&\quad -\ C\left(\lambda N^{-1/3}+L(\beta)^{-1}\left(\lambda \beta^{-2\delta}+1+\lambda\right) +(2+\lambda)\beta^{-1/2}\right). \eeqa The operator in curly brackets is the Hamiltonian for the linearized HS theory, with ground state energy $(E^{\rm HS}+\half\int(\rho^{\rm HS}_\lambda)^2)/\lambda$ (cf. eq. (\ref{mu})). So our final result can be stated as \begin{lem}[Lower bound for Region 3] If $L(\beta)>C(1+\lambda)$, then \beq \label{bbbbound} \frac{E(N,\lambda,\beta)}{N\,L^2(\beta)} \geq \frac{E^{\rm HS}(\lambda)}{\lambda}\left( 1-C\frac{1+\lambda}{L(\beta)}\right)^{-1}-C \left(\lambda N^{-1/3}+\frac{1+\lambda}{L(\beta)}\right). \eeq \end{lem} Hence the convergence of the lower bound, in the limit $N\to \infty$, $\beta \to \infty$, is proven. \begin{rem} The limits $N\to \infty$ and $\beta \to \infty$ can be considered independently of each other, in any succession or combination. \end{rem} \subsection{Critical particle number}\label{critpart} Recall that $\underline E(N,Z,B)$ denotes the ground state energy of the unscaled Hamiltonian (\ref{ham}). \begin{lem}[Limit of the critical charge]\label{critlem} Define \beq N_c(Z,B)=\max\{N|\underline E(N,Z,B)<\underline E(N-1,Z,B)\}. \eeq We have \beq\label{limnc} \liminf_{Z\to\infty}\frac{N_c(Z,\beta Z^2)}{Z}\geq \lambda_c(\beta). \eeq \end{lem} \begin{proof} This follows from the convergence of the energies, by analogous arguments as in \cite{BL83} or \cite{LSY94a}. \end{proof} \begin{rem}\label{critrem} In \cite{S00}, Theorem 3, the following upper bound on $N_c$ is proven: \beq\label{nc} N_c<2Z+1+\frac Z2 \min\left\{\left(1+\frac B{Z^2}\right), C\left(1+\left[\ln\left(\frac B{Z^2}\right)\right]^2\right)\right\} \eeq for some constant $C$ independent of $B$ and $Z$. Inserting (\ref{nc}) into (\ref{limnc}) this proves Lemma \ref{critup}. \end{rem} \subsection{Convergence of the density matrices}\label{convdens} \begin{lem}[Ground states are Hartree minimizers]\label{statesmin} For fixed $\lambda$ and $\beta$, let $\Psi_N$ be an $\eps$-approximate ground state of $H_{N,\lambda,\beta}$ defined in (\ref{ham2}), i.e. \beq \langle\Psi_N,H_{N,\lambda,\beta}\Psi_N\rangle\leq E(N,\lambda,\beta)+\eps(N) \eeq for all $N$, with $0<\eps=o(N)$. Let $\Gamma_N$ be the corresponding one-particle reduced density matrix. Then $\lambda\Gamma_N/N$ is a minimizing sequence for $\E _\beta $, i.e. \beq \lim_{N\to\infty}\E _\beta [\lambda\Gamma_N/N]=E^{\rm MH}(\lambda,\beta). \eeq \end{lem} \begin{proof} Let $\rho_N$ be the density of $\Gamma_N$. We have \begin{eqnarray}\nonumber E^{\rm MH}&\leq&\E _\beta [\lambda\Gamma_N/N]=\frac\lambda N \Tr[(H_\beta-\beta-|\x|^{-1})\Gamma_N]+\frac{\lambda^2}{N^2} D[\rho_N,\rho_N]\\ \nonumber &=&\frac\lambda N\langle\Psi_N|H_{N,\lambda,\beta}\Psi_N\rangle-\frac{\lambda^2}{N^2} \langle\Psi_N|\sum_{i0$. This is shown to be the case for our model. \begin{thm}[Bose condensation in the mean field limit]\label{bec} Let $\Gamma_N$ be as in Lemma \ref{statesmin}. Then for each fixed $\beta$ and $\lambda$ \beq \lim_{N\to\infty}\frac{\|\Gamma_N\|}{\Tr[\Gamma_N]}= \frac{\min\{\lambda,\lambda_c\}}{\lambda}. \eeq In particular, if $\lambda\leq \lambda_c$, and if $\varphi_N$ denotes the normalized eigenvector corresponding to the largest eigenvalue of $\Gamma_N$ with appropriate phase-factor, we have \beq \lim_{N\to\infty}\frac{\|\Gamma_N\|}{\Tr[\Gamma_N]}=1,\quad \lim_{N\to\infty}\| \varphi_N-\varphi_{\rm H}\|_2=0, \eeq where $\varphi_{\rm H}=\lambda^{-1/2}\sqrt{\rho^{\rm H}}$ is the normalized ground state of $H^{\rm H}$. \end{thm} \begin{proof} We use the notations of the proof of Theorem \ref{densconv}. The first assertion is easily proved, using that \beq \left|\|a_N\|-\|a\|\right|\leq \|a_N-a\|\leq \|a_N-a\|_1 \eeq and $\|a\|=\min\{\lambda,\lambda_c\}$. To prove the second we denote $P=|\varphi_{\rm H}\rangle\langle\varphi_{\rm H}|$ and $P_N=|\varphi_N\rangle\langle\varphi_N|$, and compute \begin{eqnarray}\nonumber \Tr[a_NP]&=&\|a_N\|\langle\varphi_N|P\varphi_N\rangle+ \Tr[a_N^{1/2}Pa_N^{1 /2}(1-P_N)]\\ &\leq&\|a_N\|\langle\varphi_N|P\varphi_N\rangle+\Tr[a_N]-\|a_N\|, \end{eqnarray} where we have used $P\leq 1$ in the last step. Therefore $\langle\varphi_N|P\varphi_N\rangle\to 1$ as $N\to\infty$, which gives immediately the desired result. \end{proof} \section*{Acknowledgements} We thank Jakob Yngvason for continuing interest and fruitful discussions. \begin{thebibliography}{abbrv} \bibitem[LSY94a]{LSY94a} Elliott H. Lieb, Jan Philip Solovej, and Jakob Yngvason: {\it Asymptotics of Heavy Atoms in High Magnetic Fields: I. Lowest Landau Band Regions}, Commun. Pure Appl. Math. {\bf XLVII}, 513--591 (1994) \bibitem[LSY94b]{LSY94b} Elliott H. 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