Content-Type: multipart/mixed; boundary="-------------0010030518351" This is a multi-part message in MIME format. ---------------0010030518351 Content-Type: text/plain; name="00-391.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-391.keywords" atoms, magnetic fields, lowest Landau band, ground state ---------------0010030518351 Content-Type: application/x-tex; name="ddm.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="ddm.tex" %%%%%%%%%%%%%%%%%%%% %%% FINAL VERSION %% %%%%%%%%%%%%%%%%%%%% \newcommand{\version}{October 3, 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+\eta\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{\xpp}{\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{\rddm}{\rho^{\rm DDM}} \newcommand{\rdm}{\rho^{\rm DM}} \newcommand{\gddm}{\Gamma^{\rm DDM}} \newcommand{\hddm}{h^{\rm DDM}} \newcommand{\Eddm}{E^{\rm DDM}} \newcommand{\Pddm}{\Phi^{\rm DDM}} \newcommand{\dx}{\frac\partial{\partial x}} \newcommand{\dy}{\frac\partial{\partial y}} \newcommand{\ED}{{\mathcal E}^{\rm DDM}_{B,Z}} \newcommand{\Elin}{{\mathcal E}^{\rm DDM}_{\rm lin}} \newcommand{\Econf}{E_{\rm conf}} \newcommand{\tE}{{\mathcal E}^{\rm MH}_{\eta , \rm lin}} \newcommand{\Edens}{{\hat{\mathcal E}}^{\rm MH}} \newcommand{\uw}{\underline w} \newcommand{\rhs}{\rho^{\rm HS}} \newcommand{\bsigma}{\mathord{\hbox{\boldmath $\sigma$}}} \newcommand{\rmd}{{\rm d}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \date{\small\version} \begin{document} \markboth{\scriptsize{HS \version}}{\scriptsize{HS \version}} \title{\bf{A discrete density matrix theory for atoms in strong magnetic fields}} \author{\vspace{5pt} Christian Hainzl$^1$ 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} This paper concerns the asymptotic ground state properties of heavy atoms in strong, homogeneous magnetic fields. In the limit when the nuclear charge $Z$ tends to $\infty$ with the magnetic field $B$ satisfying $B \gg Z^{4/3}$ all the electrons are confined to the lowest Landau band. We consider here an energy functional, whose variable is a sequence of one-dimensional density matrices corresponding to different angular momentum functions in the lowest Landau band. We study this functional in detail and derive various interesting properties, which are compared with the density matrix (DM) theory introduced by Lieb, Solovej and Yngvason. In contrast to the DM theory the variable perpendicular to the field is replaced by the discrete angular momentum quantum numbers. Hence we call the new functional a {\it discrete density matrix (DDM) functional}. We relate this DDM theory to the lowest Landau band quantum mechanics and show that it reproduces correctly the ground state energy apart from errors due to the indirect part of the Coulomb interaction energy. \end{abstract} \footnotetext[1]{E-Mail: \texttt{hainzl@doppler.thp.univie.ac.at}} \footnotetext[2]{E-Mail: \texttt{rseiring@ap.univie.ac.at}} \section{Introduction} The ground state properties of atoms in strong magnetic fields have been the subject of intensive mathematical studies during the last decade. This paper is based on the comprehensive work of Lieb, Solovej and Yngvason \cite{LSY94a,LSY94b}, which we refer to for an extensive list of references concerning the history of this subject. The starting point of our investigation is the Pauli Hamiltonian for an atom with $N$ electrons and nuclear charge $Z$ in a homogeneous magnetic field ${\bf {B}} = (0,0,B)$ with vector potential ${\bf A}(\x) = \half {\bf B}\times\x$, \begin{equation} \label{pho} H = \sum_{1 \leq j \leq N} \left\{ \left[\bsigma_j\cdot\left(-i\nabla_j + {\bf A}(\x_j)\right)\right]^{2} - \frac{Z}{|\x_j|} \right\} + \sum_{1 \leq i0$, $B>0$ and $Z>0$ there exists a minimizer $\gddm$ for $\ED$ under the condition $\sum_m \Tr[\Gamma_m]\leq N$. \end{thm} \begin{proof} Let $\Gamma^{(i)}$ be a minimizing sequence for $\ED$ under the normalization condition $\sum_m \Tr[\Gamma_m]\leq N$, with corresponding densities $\rho^{(i)}=(\rho_m^{(i)})_m$. Using the one-dimensional Lieb-Thirring inequality \cite{LT76} we have \beq\label{3norm} \int\left|\rho_m^{(i)}\right|^3\leq {\rm const.} \Tr[-\partial_z^2\Gamma_m^{(i)}]. \eeq Moreover (cf. \cite{LSY94a}, Eq. (4.5b)), \beq\label{sqrt} \int\left(\frac{\rmd\sqrt{\rho_m^{(i)}(z)}}{\rmd z}\right)^2\rmd z \leq \Tr[-\partial_z^2 \Gamma_m^{(i)}]. \eeq Since the potential energy is relatively bounded with respect to the kinetic energy (compare Thm. 2.2 in \cite{LSY94a}), the right hand sides of (\ref{3norm}) and (\ref{sqrt}) are uniformly bounded. Hence the sequence $\rho_m^{(i)}$ is bounded in $\Ll^3(\R,\rmd z)\cap \Ll^1(\R,\rmd z)$, and $\sqrt{\rho_m^{(i)}}$ is bounded in $\Hh^1(\R,\rmd z)$. By the \lq\lq diagonal sequence trick\rq\rq, there is a subsequence, again denoted by $\rho^{(i)}$, such that $\rho_m^{(i)}$ converges to some $\rho_m^{(\infty)}$ for each $m$, weakly in $\Ll^3(\R,\rmd z)\cap \Ll^p(\R,\rmd z)$ for some $10$ and for all $m$. Moreover, $V_m(z)\leq 1/|z|$, so $V_m\in\Ll^p$ for all $p>1$. By the weak convergence we can conclude that \beq \lim_{i\to\infty} \int \rmd z V_m(z)\rho_m^{(i)}(z)=\int \rmd z V_m(z)\rho_m^{(\infty)}(z) \eeq for each $m$. Moreover, by the dominated convergence theorem, we get \beq \lim_{i\to\infty} \sum_m \int \rmd z V_m(z)\rho_m^{(i)}(z)= \sum_m \int \rmd z V_m(z)\rho_m^{(\infty)}(z). \eeq Using the \lq\lq diagonal sequence trick\rq\rq\ once more, we can use the trace class property of the $\Gamma_m^{(i)}$'s to conclude that there exists a subsequence of $\Gamma^{(i)}$ and a $\gddm=(\gddm_m)_m$ such that \beq \Gamma_m^{(i)}\rightharpoonup \gddm_m \eeq in weak operator sense, for each $m$. It follows from weak convergence that $0\leq\gddm_m\leq 1$. Using Fatou's lemma twice, we have \beq \sum_m \Tr[\gddm_m]\leq \liminf_{i\to\infty} \sum_m\Tr[\Gamma_m^{(i)}]\leq N. \eeq By the same argument \beq \sum_m \Tr[-\partial_z^2 \gddm_z]\leq \liminf_{i\to\infty} \sum_m \Tr[-\partial_z^2\Gamma_m^{(i)}]. \eeq Let now $\rddm_m$ denote the density of $\gddm_m$. It remains to show that $\rho_m^{(\infty)}=\rddm_m$ for each $m$. From weak convergence it follows that \beq\label{WEAK} (1-\partial_z^2)^{1/2}\Gamma_m^{(i)}(1-\partial_z^2)^{1/2}\rightharpoonup (1-\partial_z^2)^{1/2}\gddm_m(1-\partial_z^2)^{1/2} \eeq weakly on the dense set $\Cc_0^\infty(\R)$. Since the operators are bounded by $\Tr[(1-\partial_z^2)\Gamma_m^{(i)}]\leq C$, we see that (\ref{WEAK}) holds weakly in $\Ll^2(\R,\rmd z)$. With $\eta\in\Cc_0^\infty(\R)$ considered as a multiplication operator, it is easy to see that $(1-\partial_z^2)^{-1/2}\eta(1-\partial_z^2)^{-1/2}$ is a compact operator (it is even Hilbert-Schmidt). Thus it can be approximated in norm by finite rank operators. Using (\ref{WEAK}) we can therefore conclude that \beq \lim_{i\to\infty} \Tr[\Gamma_m^{(i)}\eta]=\Tr[\gddm_m\eta], \eeq i.e. $\rho_m^{(i)}\to\rddm_m$ in the sense of distributions. Since we already know that $\rho_m^{(i)}$ converges to $\rho_m^{(\infty)}$ pointwise almost everywhere, we conclude that $\rho_m^{(\infty)}=\rddm_m$ for each $m$. We have thus shown that there exists a $\gddm$ with $\sum_m \Tr[\gddm_m]\leq N$ and $\ED[\gddm]\leq \liminf_{i\to\infty}\ED[\Gamma^{(i)}]=E^{\rm DDM}$. \end{proof} \begin{lem}[Uniqueness of the density] \, The density corresponding to\linebreak the minimizer is unique, i.e., if there are two minimizers $\Gamma^{(1)}$ and $\Gamma^{(2)}$, their densities $\rho_m^{(1)}$ and $\rho_m^{(2)}$ are equal, for all $m$. \end{lem} \begin{proof} Observe that \beq \widetilde D(\rho,\rho)=D( \tilde\rho,\tilde\rho), \eeq where we set $\tilde\rho(\x)=\sum_m |\phi_m(\xpp)|^2 \rho_m(z)$. Using the positive definiteness of the Coulomb kernel and the fact that $\tilde\rho^{(1)}=\tilde\rho^{(2)}$ implies $\rho_m^{(1)}=\rho_m^{(2)}$ for all $m$, we immediately get the desired result. \end{proof} Having established the uniqueness of the density, we can now define a {\it linearized DDM functional} by \beq \Elin[\Gamma]=\sum_m \Tr[\hddm_m \Gamma_m], \eeq with the one-particle operators \beq\label{defhm} \hddm_m=-\partial_z^2-\Pddm_m(z), \eeq where the potentials are given by \beq\label{defphim} \Pddm_m(z)=ZV_m(z)-\sum_n \int \rddm_n(z')V_{m,n}(z-z')\rmd z'. \eeq \begin{lem}[Equivalence of linearized theory] A minimizer $\gddm$ for $\ED$ under the constraint $\sum_m \Tr[\Gamma_m]\leq N$ is also a minimizer for the linearized functional $\Elin$ (with the same constraint). \end{lem} \begin{proof} We proceed essentially as in \cite{LSY94a}. For any $\Gamma$ \beq\label{essent} \ED[\Gamma]=\Elin[\Gamma]+\widetilde D(\rho_\Gamma-\rddm,\rho_\Gamma-\rddm)-\widetilde D(\rddm,\rddm). \eeq In particular, for all $\delta>0$, \begin{eqnarray}\nonumber \ED[(1-\delta)\gddm+\delta\Gamma]&=&(1-\delta)\Elin[\gddm]+\delta\Elin [\Gamma]-\widetilde D(\rddm,\rddm)\\ &&+\delta^2\widetilde D(\rho_\Gamma-\rddm,\rho_\Gamma-\rddm). \end{eqnarray} Now if there exists a $\Gamma^{(0)}$ with $\sum_m\Tr[\Gamma^{(0)}_m] \leq N$ and $\Elin[\Gamma^{(0)}]<\Elin[\gddm]$ we can choose $\delta$ small enough to conclude that \beq \ED[(1-\delta)\gddm+\delta\Gamma^{(0)}]<\Elin [\gddm]-\widetilde D(\rddm,\rddm)=\ED[\gddm], \eeq which contradicts the fact that $\gddm$ minimizes $\ED$. \end{proof} We now are able to construct the minimizer of $\ED$ by means of the eigenfunctions $e^i_m$ of the one-dimensional operators $\hddm_m$. If $\mu_m^1<\mu_m^2<\dots$ denote the corresponding eigenvalues, there is a $\mu\leq 0$ such that \beq\label{gddm} \gddm_m=\sum_{i=1}^{I_m-1} |e^i_m\rangle\langle e^i_m| + \lambda_m |e^{I_m}_m\rangle\langle e^{I_m}_m|, \eeq where $I_m=\max\{i\, :\, \mu_m^i\leq\mu\}$, and $0\leq \lambda_m\leq 1$ is the filling of the last level. Since $\lambda_m$ is determined by the unique density $\rddm_m$, we immediately get the following corollary: \begin{cl}[Uniqueness of $\gddm$] For any $B>0$ and $Z>0$ the minimizer of the functional $\ED$ under the condition $\sum_m \Tr[\Gamma_m]\leq N$ is unique. \end{cl} Note that $\gddm$ really depends on the three parameters $N$, $Z$ and $B$, but we suppress this dependence for the simplicity of the notation. Of course, the choice of $\mu$ is not unique, but $I_m$ is unique for every $m$. In the following, we will choose for $\mu$ the smallest possible, i.e. \beq \mu \equiv \max\{\mu_m^{I_m} \}. \eeq The energy $E^{\rm DDM}(N,Z,B)$ is a convex, non-increasing function in $N$. Moreover, it has the following property. \begin{thm}[Differentiability of $\Eddm$]\label{diffthm} For $N\not\in \N$ the energy $E^{\rm DDM}$ is differentiable in $N$, and the derivative is given by $\partial E^{\rm DDM}/\partial N=\mu$ with $\mu$ given above. For $N\in\N$, the right and left derivatives, $\partial\Eddm/\partial N_{\pm}$, are given by \beq \frac{\partial\Eddm}{\partial N_-}=\mu,\quad \frac{\partial\Eddm}{\partial N_+}=\min\{\mu_m^{I_m+1}\}. \eeq \end{thm} \begin{proof} For fixed $N$, $Z$ and $B$ define $\Elin$ as above. Let $E_{\rm lin}(N')$ be the infimum of $ \Elin$ under the constraint $\sum_m\Tr[\Gamma_m]\leq N'$. For $00$ and $\eta=B/Z^3$, \beqa\nonumber &&\left| |\psi(0)|^2-\frac 1{Z L^2}\int V_m(z/LZ) |\psi(z)|^2 \rmd z \right| \\ &&\qquad\qquad\qquad\leq \frac 1L \left(\lambda \sqrt{\frac\pi 2} \frac {Z^{1/2}}{(m+1)^{1/2}}+16 \lambda^{1/4} T^{3/4} \frac{(m+1)^{1/4}}{Z^{1/4}}\right). \eeqa \end{lem} \begin{proof} After an appropriate scaling, this is a direct consequence of \cite{BSY00}, Lemma 2.1, using the estimates \beq \int|\phi_m(\xpp)|^2 |\xpp|^{-1}\rmd\xpp\leq \sqrt{\frac\pi 2}\frac {B^{1/2}}{(m+1)^{1/2}} \eeq and \beq \int|\phi_m(\xpp)|^2 |\xpp|^{1/2}\rmd\xpp\leq 2 \frac {(m+1)^{1/4}}{B^{1/4}}. \eeq \end{proof} \begin{thm}[The limit $B\gg Z^3$]\label{largeb} For all $\lambda>0$ \beq\label{ddmhs} \lim_{\eta\to\infty}\frac{E^{\rm DDM}(\lambda Z,Z,\eta Z^3)}{Z^3 (\ln\eta)^2}=E^{\rm HS}(\lambda), \eeq uniformly in $Z$. \end{thm} \noindent{\it Remark:} The uniformity in $Z$ will be important for the proof of Theorem \ref{rank1}. It is non-trivial in contrast to DM, where one has the scaling relation (\ref{scaldm}), which implies that the left hand side of (\ref{ddmhs}) (with DDM replaced by DM) is independent of $Z$. \begin{proof} The lower bound is quite easy, using the results of \cite{LSY94a}. As shown in Section \ref{compar}, we have \beq\label{ddmdm} \frac{E^{\rm DDM}(\lambda Z,Z,\eta Z^3)}{Z^3}\geq \frac{E^{\rm DM}(\lambda Z,Z,\eta Z^3)}{Z^3}= E^{\rm DM}(\lambda ,1,\eta), \eeq where we have used the scaling properties of $E^{\rm DM}$. It is shown in \cite{LSY94a} that the right hand side of (\ref{ddmdm}) divided by $(\ln\eta)^2$ converges to $E^{\rm HS}(\lambda)$. For the upper bound we assume $N\in\N$ for the moment. We use as trial density matrices \beqa\nonumber \Gamma_m(z,z')&=&\frac{Z^2}N L\sqrt{\rhs(L Z z)}\sqrt{\rhs(L Z z')}, \quad 0\leq m \leq N-1 \\ \label{trial} \Gamma_m(z,z')&=&0, \quad m\geq N, \eeqa where $\rhs$ is given in (\ref{RHS}) and $L=L(\eta)$ is defined in (\ref{lbeta}). The kinetic energy is easily computed to be \beq Z^3 L^2 \int |\rmd\sqrt{\rhs}/\rmd z|^2. \eeq For the attraction term we use Lemma \ref{scallem} to estimate \beqa\nonumber \frac 1{Z^3 L^2} \sum_m Z\int V_m \rho_m &\geq& \frac 1N \sum_{m=0}^{N-1}\left( \rhs(0)-\frac {C_\lambda}L\left(\frac {Z^{1/2}}{(m+1)^{1/2}}+ \frac{(m+1)^{1/4}} {Z^{1/4}}\right)\right) \\ &\geq& \rhs(0)- \frac {C'_\lambda}L \eeqa for some constant $C'_\lambda$ depending on $\lambda$. For the repulsion term we first estimate \beq \sum_{0\leq m,n\leq N-1}V_{m,n}(z)\leq \left(\frac B{2\pi}\right)^2 \int \frac{\Theta(\sqrt{2 N/B}-|\xpp|)\Theta(\sqrt{2 N/B}-|\xpp'|)}{\sqrt{|\xpp-\xpp'|^2+z^2}}\rmd\xpp \rmd\xpp', \eeq which follows from monotonicity of $1/|\x|$ in $|\xpp|$ and the fact that $\sum_m |\phi_m|^2\leq B/2\pi$. Therefore we have \beqa\nonumber &&\frac 1{Z^3 L^2} \sum_{0\leq m,n\leq N-1} \int V_{m,n}(z-z')\rho_m(z)\rho_m(z')\rmd z\rmd z'\\ &&\qquad\qquad\qquad\leq \frac 1L \xi \int f(\xi (z-z'))\rhs(z)\rhs(z')\rmd z \rmd z', \eeqa where we set $\xi=(\eta/\lambda)^{1/2}/L$, and the function $f$ is given by \beq f(z)=(2\pi)^{-2}\int \frac{\Theta(\sqrt{2}-|\xpp|)\Theta(\sqrt{2} -|\xpp'|)} {\sqrt{|\xpp-\xpp'|^2+z^2}}\rmd\xpp \rmd\xpp'. \eeq We now claim that $L^{-1}\xi f(\xi z)\to \delta(z)$ as $\eta\to\infty$. Since $f(z)\leq 1/|z|$ we have, for any $\chi\in\Hh^1\cap\Ll^1$, \beqa\nonumber &&\left| L^{-1}\xi\int f(\xi z)\chi(z)\rmd z-\chi(0) L^{-1}\int_{-\xi}^\xi f(z) \rmd z \right| \\ \nonumber &&= \left|L^{-1} \xi \int_{-1}^1 f(\xi z) (\chi(z)-\chi(0))\rmd z+L^{-1} \xi\int_{|z|\geq 1}f(\xi z)\chi(z)\rmd z\right|\\ && \leq {\rm const.} L^{-1} \xi \int_{-1}^1 f(\xi z) z^{1/2}\rmd z+L^{-1} \int_{|z|\geq 1}|\chi(z)|\rmd z\leq {\rm const.} L^{-1}. \eeqa Now $L^{-1}\int_{|z|\leq \xi}f \to 1$ as $\eta\to\infty$, which proves our claim. And since $L(\eta)\approx \ln\eta$ for large $\eta$, this finishes the proof of Theorem \ref{largeb}, in the case where $N$ is an integer. The proof for $N\not\in \N$ is analogous, using (\ref{trial}) with the density matrix corresponding to $m=[N]$ multiplied by $N-[N]$ as trial density matrices. \end{proof} \begin{cl}[HS limit of the density]\label{hsdens} For fixed $\lambda=N/Z$ \beq\label{345} \lim_{\eta\to\infty} \frac 1 {Z^2 \ln\eta} \sum_m \rddm_m(z/Z\ln\eta)=\rhs(z) \eeq in the weak $\Ll^1$ sense, uniformly in $Z$. \end{cl} \begin{proof} The convergence of the densities in (\ref{345}) follows from the convergence of the energies in a standard way by considering perturbations of the external potential (cf. e.g. \cite{LSY94a}). Moreover, since the convergence in (\ref{ddmhs}) is uniform in $Z$, (\ref{345}) holds for any function $Z=Z(\beta)$, so we can conclude that (\ref{345}) holds uniformly in $Z$, too. \end{proof} Using the results above we can now prove the analogue of Theorem 4.6 in \cite{LSY94a}. \begin{thm}[$\gddm_m$ has rank at most $1$ for large $\eta$]\label{rank1} There exists a constant $C$ such that $\eta\geq C$ implies that $\gddm_m$ has rank at most $1$ for all $m$. \end{thm} \begin{proof} We first treat the case $\lambda<2$. From Theorems \ref{diffthm} and \ref{largeb} and the fact that $\Eddm(\lambda Z,Z,\eta Z^3)$ is convex in $\lambda$, we get \beq\label{241} \lim_{\eta\to\infty} \frac{\mu}{Z^2 (\ln\eta)^2}=\frac{d E^{\rm HS}(\lambda)}{d\lambda}<0. \eeq Suppose that $\mu$ is not the ground state energy of some $\hddm_m$. Then $\mu\geq -Z^2/4$, because the second lowest eigenvalue of $\hddm_m$ is equal to the ground state energy of the three-dimensional operator \beq\label{3d} -\Delta-\Pddm_m(|\x|). \eeq This follows because $\Pddm_m(z)$ is reflexion symmetric, so the eigenvector corresponding to the second lowest eigenvalue, $u_m(z)$, has a node at $z=0$. Therefore $u_m(|\x|)/|\x|$ is an eigenvector of (\ref{3d}), and because it does not change sign, it must be a ground state. Since $\Pddm_m(|\x|)\leq Z/|\x|$, the ground state energy of (\ref{3d}) is greater than $-Z^2/4$. So $\mu/Z^2(\ln\eta)^2$ would go to zero as $\eta\to\infty$, in contradiction to (\ref{241}). Therefore there exists a constant $C$ such that $\eta>C$ implies the assertion to the theorem. This constant can be chosen independent of $Z$, because the limit (\ref{241}) is uniform in $Z$ (by the same argument as in the proof of Corollary \ref{hsdens}). Now assume that $\lambda=1+\bar\delta$ for some $\bar\delta>0$. From Corollary \ref{hsdens} we infer that for large enough $\eta$ there is some $c_\delta>0$ such that \beq\label{cdelta} \sum_m \int_{|z|\leq c_\delta (Z\ln\eta)^{-1}} \rddm_m \geq (1+\half\delta)Z, \eeq where $\delta=\min\{1,\bar\delta\}$. We now will show that for $\eta$ large enough $\hddm_m$ has at most one eigenvalue, for all $0\leq m \leq N$. By the same argument as above, we need to show that the three-dimensional operator (\ref{3d}) has no eigenvalues. Using $V_m(z)\leq 1/|z|$ and (cf. the next section) \beq V_{m,n}(z)\geq \frac 1 {\sqrt 2} V_{m+n}(z/\sqrt 2)\geq \frac 1{\sqrt{z^2+ 2(m+n+1)B^{-1}}} \geq \frac 1{\sqrt{z^2+ 2(2N+1)B^{-1}}} \eeq we can use (\ref{cdelta}) to estimate \beq \Pddm_m(|\x|)\leq \frac Z {|\x|} - \frac{ Z (1+\half\delta) } {\sqrt{\left(|\x|+c_\delta (Z\ln\eta)^{-1} \right)^2+ 2(2N+1)B^{-1}}} . \eeq Therefore, for $\eta$ large enough, \beq \Pddm_m(|\x|)\leq \left\{\begin{array}{cl} Z/|\x|\quad & {\rm for}\quad |\x|\leq 3 \delta^{-1} c_\delta (Z\ln\eta)^{-1} \\ 0\quad & {\rm otherwise}. \\ \end{array}\right. \eeq By means of the Cwikel-Lieb-Rosenbljum bound \cite{RS78} we can estimate the number of negative eigenvalues of (\ref{3d}) as \beq {\rm const.} \int|\Pddm_m(|\x|)|_+^{3/2}\rmd\x\leq c_\delta' (\ln\eta)^{-3/2}, \eeq which is less than 1 for $\eta$ large enough. \end{proof} \begin{rem}[Chemical potential for large $B/Z^3$] The theorem above,\linebreak together with Corollary \ref{monmu}, shows that for $B/Z^3$ large enough, the chemical potential is given by the ground state energy of $\hddm_{}$, i.e. $\mu=\mu_{}^1$, where $$ denotes the smallest integer $\geq N$. \end{rem} \section{Maximal negative ionization}\label{mi} The DDM energy is convex and monotonously decreasing in $N$. Because $\widetilde D(\rho,\rho)$ is strictly convex in $\rho$, $\Eddm$ is strictly convex up to some $N=N_c(Z,B)$, and constant for $N\geq N_c$. By uniqueness of $\gddm$ the minimizer for $N>N_c$ is equal to the one with $N=N_c$. In particular, $\sum_m\Tr[\gddm_m]=\min\{N,N_c\}$. The \lq\lq critical\rq\rq\ $N_c$ measures the maximal particle number that can be bound to the nucleus. We will proceed essentially as in \cite{S00} and use Lieb's strategy \cite{L84} to get an upper bound on $N_c$. In addition, the following Lemma is needed. Throughout, we use various properties of $V_m$ stated in \cite{BRW99}, Sect. 4., and proven in \cite{RW00}. \begin{lem}[Comparison of $V_m$ and $V_{m,n}$]\label{vmvmn} \beq\label{assert} \left(\frac 1{V_m(z)}+\frac 1{V_n(z)}+\frac 1{V_m(z')}+\frac 1{V_n(z')}\right) V_{m,n}(z-z')\geq 1 \eeq \end{lem} \begin{proof} Using the definition of $V_{m,n}$ it can be shown (cf. \cite{P82}) that \beq \sqrt 2 V_{m,n}(\sqrt 2 z)=\sum_{i=0}^{m+n} c_i V_i(z) \eeq for some coefficients $c_i\geq 0$ that fulfill $\sum_i c_i=1$. Since $V_m\geq V_{m+1}$ for all $m$ (\cite{BRW99}, 4b) we get \beq V_{m,n}(z)\geq \frac 1{\sqrt 2} V_{m+n}(z/\sqrt 2)\geq \frac 12 V_{m+n}(z/2), \eeq where we used the fact that $a V_m(az)\leq V_m(z)$ if $a\leq 1$ (\cite{BRW99}, 4g). Moreover, using convexity of $1/V_{m+n}$ (\cite{BRW99}, 4i), we arrive at \beq V_{m,n}(z-z')\geq \left(V_{m+n}(z)^{-1}+V_{m+n}(z')^{-1}\right)^{-1}. \eeq The assertion (\ref{assert}) follows if we can show that \beq \frac 1 {V_{m+n}(z)}\leq \frac 1 {V_{m}(z)}+\frac 1 {V_{n}(z)}. \eeq This is of course trivial if $n$ or $m$ equals zero. If $n,m\geq 1$ we use $\sqrt{z^2+m}^{-1}\geq V_m(z)\geq \sqrt{z^2+m+1}^{-1}$ (\cite{BRW99}, 4a) to estimate \beqa \nonumber \frac 1 {V_{m+n}(z)}&\leq& \sqrt{z^2+m+n+1}\leq 2\sqrt{z^2+(\sqrt m+\sqrt n)^2/4} \\ &\leq& \sqrt{z^2+m}+\sqrt{z^2+n}\leq \frac 1{V_m(z)}+\frac 1{V_n(z)}, \eeqa which finishes the proof. \end{proof} \begin{thm}[Critical particle number] \beq\label{NC} Z\leq N_c \leq 4Z-\frac 1{N_c}\frac {\partial \Eddm(N_c,Z,B)}{\partial Z} \eeq \end{thm} \noindent {\it Remark:} The factor $4$ stems from the symmetrization of (\ref{assert}) in $m$ and $n$. Due to this symmetrization one could expect that Lemma \ref{vmvmn} holds with 1 replaced by 2 on the right hand side. This would imply that $4Z$ could be replaced by $2Z$ in (\ref{NC}). \begin{proof} Let $e_m^i$ denote the eigenvectors of $\hddm_m$, i.e. \beq\label{heme} \hddm_m e_m^i = \mu_m^i e_m^i. \eeq Multiplying (\ref{heme}) with $e_m^i /V_m$ and integrating we get \beq\label{48} Z\geq \sum_n\int \frac 1 {V_m(z)} e_m^i(z)^2 V_{m,n}(z-z')\rddm_n(z')\rmd z\rmd z' +\langle e_m^i| \frac 1{V_m}(-\partial_z^2)| e_m^i\rangle, \eeq where we used that $\mu_m^i\leq 0$. Since $1/V_m$ is convex and $|z|V_m(z)\to 1$ as $|z|\to\infty$ we have $|(1/V_m)'|\leq 1$. Using this and partial integration we can estimate the last term in (\ref{48}) by \beqa\nonumber \langle e_m^i|\frac 1{V_m}(-\partial_z^2)|e_m^i\rangle &=& \langle e_m^i V_m^{-1/2}|-\partial_z^2-\frac 14 \left|\frac{V_m'}{V_m}\right|^2 |e_m^i V_m^{-1/2}\rangle \\ &\geq& -\frac 14\int V_m(z) e_m^i(z)^2 \rmd z . \eeqa Summing over all $m$ and $i$ (and, according to Eq. (\ref{gddm}), multiplying the factor corresponding to the largest $\mu_m^i$ by $\lambda_m$), we arrive at \beq\label{49} NZ\geq \sum_{m,n}\int \frac 1 {V_m(z)} \rddm_m(z) V_{m,n}(z-z')\rddm_n(z')\rmd z\rmd z' -\frac 14 \sum_{m}\int V_m \rddm_m. \eeq Note that the last term in (\ref{49}) is equal to $\partial\Eddm/\partial Z$. To treat the first term in (\ref{49}) we use symmetry and Lemma \ref{vmvmn} to get \beqa\nonumber &&\sum_{m,n}\int \frac 1{V_m(z)}\rddm_m(z) V_{m,n}(z-z') \rddm_n(z') \rmd z\rmd z' \\ \nonumber &&=\frac 14 \sum_{m,n}\int \left(\frac 1{V_m(z)}+\frac 1{V_n(z)}+\frac 1{V_m(z')} +\frac 1{V_n(z')}\right)\\ &&\qquad \qquad \times \rddm_m(z)V_{m,n}(z-z')\rddm_n(z') \rmd z\rmd z' \geq \frac 14 N^2. \eeqa Inserting this into (\ref{49}) and dividing by $N/4$ we arrive at \beq N\leq 4Z-\frac 1{N}\frac {\partial \Eddm}{\partial Z}. \eeq The lower bound on $N_c$ is quite easy. We just have to show that $\hddm_{[N+1]}$ has a bound state if $N0$ as a trial vector we compute \beqa\nonumber &&\langle\psi|\hddm_{[N+1]}|\psi\rangle\\ \nonumber &&= a-Z\int V_{[N+1]}(z)e^{-2a|z|}\rmd z+\sum_n \int V_{n,[N+1]}(z-z')\rddm_n(z')e^{-2a|z|}\rmd z\rmd z'\\ &&\leq a -Z\int V_{[N+1]}(z)e^{-2a|z|}\rmd z+N \max_{n\leq N}\int V_{n,[N+1]}(z)e^{-2a |z|}\rmd z. \eeqa Since \beq \lim_{a\to 0} \frac 1 {\ln(1/a)} \int V_{[N+1]}e^{-2a|z|}=\lim_{a\to 0} \frac 1{\ln(1/a)} \int V_{n,[N+1]} e^{-2a|z|}=1, \eeq $\langle\psi|\hddm_{[N+1]}|\psi\rangle$ will be negative for small enough $a$, if $N