Content-Type: multipart/mixed; boundary="-------------0012130443589" This is a multi-part message in MIME format. ---------------0012130443589 Content-Type: text/plain; name="00-499.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-499.comments" AMS CODE: 81V45,81V70 PACS: 31.10.+z, 31.15.Bs,,31.15.Gy ---------------0012130443589 Content-Type: text/plain; name="00-499.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-499.keywords" Hartree-Fock, ionization, semiclassics, Thomas-Fermi, atoms ---------------0012130443589 Content-Type: application/x-tex; name="final-dec-13.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="final-dec-13.tex" \documentclass[12pt]{article} \usepackage{a4,amsthm,amsfonts, } \swapnumbers \font\notefont=cmsl8 \font\smallcal=cmsy8 %% %HEADING: \pagestyle{myheadings} \newcommand\heading[1]{\markright{\notefont JPS/13-Dec-2000---Hartree-Fock Ionization, Sect.~\thesection, #1\hfill}} %% %%FORMATTING MACRO: \renewcommand{\baselinestretch}{1.3} %% %% %%THEOREMSTYLES: \theoremstyle{plain} \newtheorem{thm}{THEOREM}[section] \newtheorem{lm}[thm]{LEMMA} \newtheorem{cl}[thm]{COROLLARY} \newtheorem{prop}[thm]{PROPOSITION} \theoremstyle{definition} \newtheorem{defi}[thm]{DEFINITION} \theoremstyle{remark} \newtheorem{remark}[thm]{REMARK} %% %%MATH MACROS: \renewcommand{\phi}{\varphi} \newcommand{\const}{\mbox{\rm \small (Const.)}} \newcommand{\iint}{\mathop{\displaystyle\int\!\!\int}} \newcommand{\set}[2]{{\left\{#1 : #2\right\}}} \newcommand{\downskip}[1]{\rule{0pt}{#1pt}} \newcommand{\upskip}[1]{\rule[-#1pt]{0pt}{#1pt}} \newcommand{\upchi}[1]{\raise1pt\hbox{$\chi$}_{#1}} \newcommand{\thetar}{\theta_r} \newcommand{\thetam}{\theta_{\scriptscriptstyle(0)}} \newcommand{\thetaplus}{\theta_{\downskip{5}_+}} \newcommand{\chiplus}[1]{\upchi{#1}^{\upskip{3}+}} \newcommand{\chiminus}[1]{\upchi{#1}} \newcommand{\uprho}{\raise1pt\hbox{$\rho$}} \newcommand{\upgamma}{\raise1pt\hbox{$\gamma$}} \newcommand{\I}{{\mathord{I}}} \newcommand{\ea}{{{\cal E}^{\upskip{3}\hbox{\tiny A}}}} \newcommand{\acn}{A_{\rm cn}} \newcommand{\etf}[1]{{{\cal E}^{\upskip{3}\TF}_{#1}}} \newcommand{\ehf}{{{\cal E}^{\upskip{3}\HF}}} \newcommand{\HF}{{\hbox{\rm \tiny HF}}} \newcommand{\TF}{{\hbox{\rm \tiny TF}}} \newcommand{\OTF}{{\hbox{\rm \tiny OTF}}} \newcommand{\Ex}{{\mathord{{\cal E}\!\hbox{\smallcal X}}}} \newcommand{\D}{{\mathord{\cal D}}} \newcommand{\B}[1]{{B(#1)}} \newcommand{\A}[2]{{A(#1,#2)}} \newcommand{\CK}{\mathord{C}_K} \newcommand{\CM}{\mathord{C}_M} \newcommand{\CPhi}{{\mathord{C_{\Phi}}}} \newcommand{\Om}[1]{{\Omega(#1)}} \newcommand{\Tr}{{\mathop{\rm Tr}}} \newcommand{\R}{{\mathord{\mathbb R}}} \newcommand{\C}{{\mathord{\mathbb C}}} \newcommand{\rC}{{\rm C}} \newcommand{\supp}{{\mathop{\rm supp\ }}} \renewcommand{\epsilon}{{\varepsilon}} \newcommand{\phihf}{\phi^{\upskip{3}\HF}} \newcommand{\Phihf}[1]{\Phi^{\upskip2 \HF}_{\downskip{6}{#1}}} \newcommand{\Phitf}[1]{\Phi^{\upskip2 \TF}_{\downskip{6}{#1}}} \newcommand{\rhohf}{\uprho^{\upskip{3}\HF}} \newcommand{\phitf}{\phi^{\upskip{3}\TF}} \newcommand{\mutf}{\mu^{\upskip{3}\TF}} \newcommand{\mutfr}{\mu^{\OTF} _{\downskip{6}{r}}} \newcommand{\rhotf}{\uprho^{\upskip{3}\TF}} \newcommand{\rhotfv}{{\displaystyle\uprho^{\TF}_{\downskip{8}V}}} \newcommand{\mutfv}{{\displaystyle\mu^{\TF}_{\downskip{8}V}}} \newcommand{\phitfv}{{\displaystyle\phi^{\TF}_{\downskip{8}V}}} \newcommand{\rhotfr}{\uprho^{\OTF}_{\!\downskip{6}r}} \newcommand{\etfr}{{{\cal E}^{\upskip{3}\OTF}_{r}}} \newcommand{\phitfr}{\phi^{\OTF}_{\!\downskip{6}r}} \newcommand{\rhohfr}{\uprho^{\HF}_{\!\downskip{6}r}} \newcommand{\phihfr}{\phi^{\HF}_{\!\downskip{6}r}} \newcommand{\gammar}{\upgamma^{\HF}_{\!\downskip{6}r}} \newcommand{\gammahf}{\upgamma^{\upskip{3}\HF}} \newcommand{\gammam}{\upgamma^{\HF}_{\!\downskip{6}\scriptscriptstyle(0)}} \newcommand{\gammai}{\upgamma^{\HF}_{\!\downskip{6}-}} \newcommand{\omegap}[1]{\omega^+_{#1}} \newcommand{\omegam}[1]{\omega^-_{#1}} \newcommand{\V}[1]{V_{#1}} \newcommand{\W}[1]{W_{#1}} \newcommand{\spec}{{\mathop{\rm spec}}} \newcommand{\HN}{{H_{\rho}^{(N)}}} \newcommand{\HMF}{H_{\gammahf}} \newcommand{\KMF}{{\cal K}_{\gamma^\HF}} \newcommand{\mfr}[2]{{\textstyle\frac{#1}{#2}}} \renewcommand{\Re}{{\mathop{\rm Re}}} %% %% \begin{document} \title{THE IONIZATION CONJECTURE IN HARTREE-FOCK THEORY} \author{Jan Philip Solovej\\ \normalsize Department of Mathematics, University of Copenhagen\\ \normalsize Universitetsparken 5, 2100 Copenhagen, Denmark\\ \normalsize {\it e-mail\/}: solovej@math.ku.dk} \date{Dec. 13, 2000} \maketitle \begin{abstract} We prove the ionization conjecture within the Hartree-Fock theory of atoms. More precisely, we prove that, if the nuclear charge is allowed to tend to infinity, the maximal negative ionization charge and the ionization energy of atoms nevertheless remain bounded. Moreover, we show that in Hartree-Fock theory the radius of an atom (properly defined) is bounded independently of its nuclear charge. \end{abstract} \vfill\eject \tableofcontents \vfil\eject \section{Introduction and main results} \heading{Introduction} One of the great triumphs of quantum mechanics is that it explains the order in the periodic table qualitatively as well as quantitatively. In elementary chemistry it is discussed how quantum mechanics implies the shell structure of atoms which gives a qualitative understanding of the periodic table. In computational quantum chemistry it is found that quantum mechanics gives excellent agreement with the quantitative aspects of the periodic table. It is a very striking fact, however, that the periodic table is much more ``periodic'' than can be explained by the simple shell structure picture. As an examples it can be mentioned that e.g., the radii of different atoms belonging to the {\it same} group in the periodic table do not vary very much, although the number of electrons in the atoms can vary by a factor of 10. Another related example is the fact that the maximal negative ionization (the number of extra electrons that a neutral atom can bind) remains small (possibly no bigger than 2) even for atoms with large atomic number (nuclear charge). These experimental facts can to some extent be understood numerically, but there is no good qualitative explanation for them. In the mathematical physics literature the problem has been formulated as follows (see e.g., problems 9 and 10 in \cite{Simon21}). Imagine that we consider `the infinitely large periodic table', i.e., atoms with arbitrarily large nuclear charge $Z$, is it then still true that the radius and maximal negative ionization remain bounded? This question often referred to as the {\it ionization} conjecture is the subject of this paper. To be completely honest neither the qualitative nor the quantitative explanations of the periodic table use the full quantum mechanical description. On one hand the simple qualitative shell structure picture ignores the interactions between the electrons in the atoms. On the other hand even in computational quantum chemistry one most often uses approximations to the full many body quantum mechanical description. There are in fact a hierarchy of models for the structure of atoms. The one which is usually consider most complete is the Schr\"odinger many-particle model. There are, however, even more complicated models, which take relativistic and/or quantum field theoretic corrections into account. A description which is somewhat simpler than the Schr\"odinger model is the Hartree-Fock (HF) model. Because of its greater simplicity it has been more widely used in computational quantum chemistry than the full Schr\"odinger model. Although, chemists over the years have developed numerous generalizations of the Hartree-Fock model, it is still remarkable how tremendously successful the original (HF) model has been in describing the structure of atoms and molecules. A model which is again much simpler than the Hartree-Fock model is the Thomas-Fermi (TF) model. In this model the problem of finding the structure of an atom is essentially reduced to solving an ODE. The TF model has some features, which are qualitatively wrong. Most notably it predicts that atoms do not bind to form molecules (Teller's no binding Theorem, see \cite{LiebSimon:TF}). In this work we shall show that the TF model is, indeed, a much better approximation to the more complicated HF model than generally believed. In fact, we shall show that it is only the outermost region of the atom which is not well described by the TF model. As a simple corollary of this improved TF approximation we shall prove the {\it ionization conjecture} within HF theory. The corresponding results for the full Schr\"odinger theory are still open and only much simpler results are known (see e.g., \cite{Fef-Sec,LSST,Ruskai,Sigal,SSS}). In \cite{BenguriaLieb} the ionization conjecture was solved in the Thomas-Fermi-von Weizs{\"a}cker generalization of the Thomas-Fermi model. In \cite{Solovej:RHF} the ionization conjecture was solved in a simplified Hartree-Fock mean field model by a method very similar to the one presented here. In the simplified model the atoms are entirely spherically symmetric. In the full HF model, however, the atoms need not be spherically symmetric. This lack of spherical symmetry in the HF model is one of the main reasons for many of the difficulties that have to be overcome in the present paper, although this may not always be apparent from the presentation. We shall now describe more precisely the results of this paper. In common for all the atomic models is that, given the number of electrons $N$ and the nuclear charge $Z$, they describe how to find the electronic ground state density $\rho\in L^1(\R^3)$, with $\int\rho=N$. Or more precisely how to find {\it one} ground state density, since it may not be unique. In the TF model the ground state is described only by the density, whereas in the Schr\"odinger and HF models the density is derived from more detailed descriptions of the ground state. For all models we shall use the following definitions. We distinguish quantities in the different models by adding superscripts TF, HF. (In this work we shall not be concerned with the Schr\"odinger model at all.) Througout the paper we use units in which $\hbar=m=e=1$, i.e., {\it atomic units}. \begin{defi}[Mean field potentials] \label{defi:phiPhi} Let $\rhohf$ and $\rhotf$ be the densities of atomic ground states in the HF and TF models respectively. We define the corresponding {\it mean field potentials} \begin{eqnarray} \phihf(x)&:=&Z|x|^{-1}-\rhohf*|x|^{-1}= Z|x|^{-1}-\int \rhohf(y)|x-y|^{-1}dy\label{eq:phihf}\\ \phitf(x)&:=&Z|x|^{-1}-\rhotf*|x|^{-1}= Z|x|^{-1}-\int \rhotf(y)|x-y|^{-1}dy\label{eq:phitf} \end{eqnarray} and for all $R\geq0$ the {\it screened nuclear potentials at radius} $R$ \begin{eqnarray} \Phihf{R}(x)&:=&Z|x|^{-1}-\int_{|y|0$ are universal constants. \end{thm} This theorem is proved in Section~\ref{sec:provingmain} on page \pageref{proof:potentialestimate}. The significance of the power $|x|^{-4}$ is that for $N\geq Z$ we have $\lim_{Z\to\infty}\phitf(x)=3^42^{-3}\pi^2|x|^{-4}$. The existence of this limit known as the Sommerfeld law \cite{Sommerfeld} follows from Theorem~2.10 in \cite{Lieb:tf}, but we shall also prove it in Theorems~\ref{thm:sommerfeldup} and \ref{thm:sommerfeldlow} below. Note that the bound in Theorem~\ref{thm:potentialestimate} is uniform in $N$ and $Z$. The second main theorem is the universal bound on the atomic radius mentioned in the beginning of the introduction. In fact, not only do we prove uniform bounds but we also establish a certain exact asymptotic formula for the radius of an ``infinite atom''. \begin{thm}[The radius of an infinite neutral HF atom]\hfil \label{thm:radius} Both $\liminf\limits_{Z\to\infty}R_{Z,Z}^{\HF}(\nu)$ and $\limsup\limits_{Z\to\infty}R_{Z,Z}^{\HF}(\nu)$ are bounded and have the asymptotic behavior $$ 2^{-1/3}3^{4/3}\pi^{2/3}\nu^{-1/3}+ o(\nu^{-1/3}) $$ as $\nu\to\infty$. \end{thm} The proof of this theorem can be found in Sect.~\ref{sec:provingmain} on page~\pageref{proof:radius}. The universal bound on the maximal ionization is given in Theorem~\ref{thm:maxN}. The proof is given in Sect.~\ref{sec:provingmain} on page~\pageref{proof:maxN}. A universal bound on the ionization energy (the energy it takes to remove one electron) is formulated in Theorem~\ref{thm:ie}. The proof is given in Sect.~\ref{sec:provingmain} on page~\pageref{proof:ie}. The paper is organized as follows. In Sect.~\ref{sec:notations} we fix our notational conventions and give some basic prerequisites. In Sect.~\ref{sec:hf} we discuss Hartree-Fock theory. In Sects.~\ref{sec:tf} and \ref{sec:atomictf} we discuss Thomas-Fermi theory. In particular we show that the TF model, indeed, has the universal behavior for large $Z$ that we want to establish for the HF model. In the TF model the universality can be expressed very precisely through the Sommerfeld asymptotics. In Sect.~\ref{sec:io} we begin the more technical work. We show in this section that the HF atom in the region $\set{x}{|x|>R}$ is determined to a good approximation, in terms of energy, from knowledge of the screened nuclear potential $\Phihf{R}$. It is this crucial step in the whole argument that I do not know how to generalize to the Schr\"odinger model or even to the case of molecules in HF theory. For the outermost region of the atom one cannot use the energy to control the density. In fact, changing the density of the atom far from the nucleus will not affect the energy very much. Far away from the nucleus one must use the exact energy minimizing property of the ground state, i.e., that it satisfies a variational equation. This is done in Sect.~\ref{sec:l1} to estimate the $L^1$ norm of the density in a region of the form $\set{x}{|x|>R}$. In Sect.~\ref{sec:sc} we establish the semiclassical estimates that allow one to compare the HF model with the TF model. To be more precise, there is no semiclassical parameter in our setup, but we derive bounds that in a semiclassical limit would be asymptotically exact. It turns out to be useful to use the electrostatic energy (or rather its square root) as a norm in which to control the difference between the densities in TF and HF theory. The properties of this norm, which we call the Coulomb norm, are discussed in Sect.~\ref{sec:cn}. Sects.~\ref{sec:tf}-\ref{sec:cn} can be read almost independently. In Sect.~\ref{sec:me} we state and prove the main technical tool in the work. It is a comparison of the screened nuclear potentials in HF and TF theory. Using a comparison between the screened nuclear potentials at radius $R$ one may use the result of the separation of the outside from the inside given in Sect.~\ref{sec:io} to get good control on the outside region $\set{x}{|x|>R}$. Using an iterative scheme one establishes the main estimate for all $R$. The two main technical lemmas are proved in Sect.~\ref{sec:smallx} and Sect.~\ref{sec:iteration} respectively. Finally the main theorems are proved in Sect.~\ref{sec:provingmain}. The main results of this paper were announced in \cite{Solovej:announcement} and a sketch of the proof was given there. \section{Notational conventions and basic prerequisites} \heading{Notations and prerequisites}\label{sec:notations} We shall throughout the paper use the definitions \begin{eqnarray} \B{r}&:=&\set{y\in\R^3}{|y|\leq r}\label{def:b}\\ \B{x,r}&:=&\set{y\in\R^3}{|y-x|\leq r}\label{def:bx}\\ \A{r_1}{r_2}&:=&\set{x\in\R^3}{r_1\leq |x|\leq r_2}.\label{def:a} \end{eqnarray} For any $r>0$ we shall denote by $\chiminus{r}$ the characteristic function of the ball $\B{r}$ and by $\chiplus{r}=1-\chiminus{r}$. We shall as in the introduction use the notation $[t]_\pm=(t)_\pm:=\max\{\pm t,0\}$. Our convention for the Fourier transform is \begin{equation} \hat{f}(p):=(2\pi)^{-3/2}\int e^{ipx} f(x)\, dx. \end{equation} Then \begin{equation} \widehat{f*g}=(2\pi)^{3/2}\hat{f}\hat{g},\quad \|f\|_2=\|\hat f\|_2,\quad |\hat f(p)|\leq (2\pi)^{-3/2}\|f\|_1 \end{equation} and \begin{equation} \iint f(x)|x-y|^{-1}\overline{g(y)}dx\,dy =2(2\pi)\int \hat{f}(p)\overline{\hat{g}(p)}|p|^{-2}\, dp. \end{equation} \begin{defi}[Density Matrix] Here we shall use the definition that a {\it density matrix}, on a Hilbert space $\cal H$, is a positive trace class operator satisfying the operator inequality $0\leq\gamma\leq I$. When $\cal H$ is either $L^2(\R^3)$ or $L^2(\R^3;\C^2)$ we write $\gamma(x,y)$ for the integral kernel for $\gamma$. It is $2\times 2$ matrix valued in the case $L^2(\R^3;\C^2)$. We define the {\it density} $0\leq\uprho_\gamma\in L^1(\R^3)$ corresponding to $\gamma$ by \begin{equation}\label{eq:den} \uprho_\gamma:=\sum_j\nu_j|u_j(x)|^2, \end{equation} where $\nu_j$ and $u_j$ are the eigenvalues and corresponding eigenfunctions of $\gamma$. Then $\int\uprho_\gamma=\Tr[\gamma]$. \end{defi} \begin{remark}\label{remark:deltagamma} Whenever $\gamma$ is a density matrix with eigenfunctions $u_j$ and corresponding eigenvalues $\nu_j$ on either $L^2(\R^3)$ or $L^2(\R^3;\C^2)$ we shall write \begin{equation}\label{eq:trdeltagamma} \Tr\left[-\Delta\gamma\right] :=\sum_j\nu_j\int |\nabla u_j(x)|^2\,dx. \end{equation} If we allow the value $+\infty$ then the right side is defined for all density matrices. The expression $-\Delta\gamma$ may of course define a trace class operator for some $\gamma$, i.e., if the eigenfunctions $u_j$ are in the Sobolev space $H^2$ and the right side above is finite. In this case the left side is well defined and is equal to the right side. On the other hand, the right side may be finite even though $-\Delta\gamma$ does not even define a bounded operator, i.e., if an eigenfunction is in $H^1$, but not in $H^2$. Then the sum on the right is really $$ \Tr\left[(-\Delta)^{1/2}\gamma(-\Delta)^{1/2}\right] =\Tr\left[\nabla\cdot\gamma\nabla\right]. $$ It is therefore easy to see that (\ref{eq:trdeltagamma}) holds not only for the spectral decomposition, but more generally, whenever $\gamma$ can be written as $\gamma f=\sum_j\nu_j (u_j,f)u_j$, with $0\leq\nu_j$ (the $u_j$ need not be orthonormal). The same is also true for the expression (\ref{eq:den}) for the density. \begin{prop} The map $\gamma\mapsto\Tr[-\Delta\gamma]$ as defined above on all density matrices is affine and weakly lower semicontinuous \end{prop} \begin{proof} Choose a basis $f_1,f_2,\ldots$ for $L^2$ consisting of functions from $H^1$. Then $$\Tr[-\Delta\gamma]=\sum_m (\nabla f_m,\gamma \nabla f_m).$$ The affinity is trivial and the lower semicontinuity follows from Fatou's Lemma. \end{proof} We are of course abusing notation when we define $\Tr[-\Delta\gamma]$ for all density matrices. This is, however, very convenient and should hopefuly not cause any confusion. If $V$ is a positive measurable function. We always identify $V$ with a multiplication operator on $L^2$. If $V\uprho_\gamma\in L^1(\R^3)$ we abuse notation and write $$ \Tr\left[V\gamma \right]:=\int V\uprho_\gamma. $$ As before if $V\gamma $ happens to be trace class then the left side is well defined and finite and is equal to the right side. Otherwise, we really have $\int V\uprho_\gamma=\Tr\left[[V]_+^{1/2}\gamma[V]_+^{1/2}\right] -\Tr\left[[V]_-^{1/2}\gamma[V]_-^{1/2}\right]$. \end{remark} \begin{lm}[The IMS formulas]\label{lm:IMS} If $u$ is in the Sobolev space $H^1(\R^3;\C^2)$ or $H^1(\R^3)$ and if $\Xi\in C^1(\R^3)$ is real, bounded, and has bounded derivative then\footnote{We denote by $u^*$ the complex conjugate of $u$. In the case when $u$ takes values in $\C^2$ this refers to the complex conjugate matrix.} \begin{equation}\label{eq:IMSformula} \Re\int\nabla\left(\Xi^2{u}^*\right)\cdot \nabla u =\int \left|\nabla(\Xi u)\right|^2-\int\left|\nabla \Xi\right|^2 |u|^2. \end{equation} If $\gamma$ is a density matrix on $L^2(\R^3;\C^2)$ or $L^2(\R^3)$ and if $\Xi_1,\ldots,\Xi_m\in C^1(\R^3)$ are real, bounded, have bounded derivatives, and satisfy $\Xi_1^2+\ldots+\Xi_m^2=1$ then \begin{eqnarray}\label{eq:IMS} \Tr\left[-\Delta\gamma\right]&=& \Tr\left[-\Delta(\Xi_1\gamma\Xi_1)\right] -\Tr\left[\left(\nabla\Xi_1\right)^2\gamma\right] +\ldots\nonumber\\ &&{}+\Tr\left[-\Delta(\Xi_m\gamma\Xi_m)\right] -\Tr\left[\left(\nabla\Xi_m\right)^2\gamma\right]. \end{eqnarray} Note that $\Xi_j\gamma\Xi_j$ again defines a density matrix (where we identified $\Xi_j$ with a multiplication operator). \end{lm} \begin{proof} The identity (\ref{eq:IMSformula}) follows from a simple computation. If we sum this identity and use $\Xi_1^2+\ldots+\Xi_m^2=1$ we obtain $$ \int|\nabla u|^2= \int \left|\nabla(\Xi_1 u)\right|^2-\int\left|\nabla \Xi_1\right|^2 |u|^2+\ldots+ \int \left|\nabla(\Xi_m u)\right|^2-\int\left|\nabla \Xi_m\right|^2 |u|^2. $$ If we allow the value $+\infty$ this identity holds for all functions $u$ in $L^2$. Thus (\ref{eq:IMS}) is a simple consequence of the definition (\ref{eq:trdeltagamma}). \end{proof} \begin{thm}[Lieb-Thirring inequality] We have the Lieb-Thirring inequality \begin{equation}\label{eq:LTdensity} \Tr\left[-\mfr{1}{2}\Delta \gamma\right] \geq K_1\int\uprho_\gamma^{5/3}, \end{equation} where $K_1:=20.49$. Equivalently, If $V\in L^{5/2}(\R^3)$ and if $\gamma$ is any density matrix such that $\Tr[-\Delta\gamma]<\infty$ we have \begin{equation}\label{eq:LT} \Tr\left[-\mfr{1}{2}\Delta\gamma\right] -\Tr\left[V\gamma\right] \geq -L_1\int [V]_+^{5/2}, \end{equation} where $L_1:=\mfr{2}{5}\left(\mfr{3}{5K_1}\right)^{2/3} =0.038$. \end{thm} The original proofs of these inequalities can be found in \cite{LiebThirring}. The constants here are taken from \cite{HundertmarkLaptevWeidl}. {F}rom the min-max principle it is clear that the right side of (\ref{eq:LT}) is in fact a lower bound on the sum of the negative eigenvalues of the operator $-\mfr{1}{2}\Delta -V$. \begin{thm}[Cwikel-Lieb-Rozenblum inequality] If $V\in L^{3/2}(\R^3)$ then the number of non-positive eigenvalues of $-\mfr{1}{2}\Delta -V$, i.e., $\Tr\left[\chi_{(-\infty,0]}\left( -\mfr{1}{2}\Delta -V\right)\right]$, where $\chi_{(-\infty,0]}$ is the characteristic function of the interval $(-\infty,0]$, satisfies the bound \begin{equation}\label{eq:CLR} \Tr\left[\chi_{(-\infty,0]}\left( -\mfr{1}{2}\Delta -V\right)\right] \leq L_0\int[V]_+^{3/2}, \end{equation} where $L_0:=2^{3/2}0.1156=0.3270$. \end{thm} The original (independent) proofs can be found in Cwikel~\cite{Cwikel}, Rozenblum~\cite{Rozenblum}, and Lieb~\cite{Lieb:CLR}. The constant is from Lieb~\cite{Lieb:CLR}. \section{Hartree-Fock Theory}\heading{HF theory}\label{sec:hf} In Hartree-Fock theory, as opposed to Schr\"odinger theory, one does not consider the full $N$-body Hilbert space $\bigwedge^NL^2(\R^3;\C^2)$. One rather restricts attention to the pure wedge products (Slater determinants) \begin{equation}\label{eq:Slater} \Psi=(N!)^{-1/2}u_1\wedge\ldots\wedge u_N, \end{equation} where $u_1,\ldots,u_N\in L^2(\R^3;\C^2)$. Then one minimizes the energy expectation $$ \frac{\left(\Psi, H_{N,Z}\Psi\right)}{\left(\Psi, \Psi\right)} $$ of the Hamiltonian \begin{equation}\label{eq:NHamiltonian} H_{N,Z}:=\sum_{i=1}^N \left(-\mfr{1}{2}\Delta -\frac{Z}{|x|}\right)+\sum_{1\leq i0$ be a real number and $N\geq 0$ be an integer. The Hartree-Fock ground state energy is $$ E^\HF(N,Z):=\inf\set{\ehf(\gamma)}{\gamma^*=\gamma,\ \gamma=\gamma^2,\ \Tr[\gamma]=N}. $$ If a minimizer $\gammahf$ exists we say that the atom has an HF ground state described by $\gammahf$. In particular, its density is $\rhohf(x)=\uprho_{\gammahf}(x)$. \end{defi} \begin{thm}[Bound on the Hartree-Fock energy] \label{thm:HFenergy} For $Z>0$ and any integer $N>0$ we have $$ E^\HF(N,Z)\geq -3(4\pi L_1)^{2/3}Z^{2}N^{1/3}, $$ where $L_1$ is the constant in the Lieb-Thirring inequality (\ref{eq:LT}). \end{thm} \begin{proof} Let $\gamma$ be an $N$ dimensional projection. Since the last term in $H_{N,Z}$ is positive we see that $\ehf(\gamma)\geq \Tr\left[\left(-\mfr{1}{2}\Delta-Z|x|^{-1}\right)\gamma\right]$. It the follows from the Lieb-Thirring inequality (\ref{eq:LT}) that for all $R>0$ we have $$ \ehf(\gamma)\geq -L_1\int_{|x|2Z+1$ there are no minimizers for the Hartree-Fock functional among $N$-dimensional projections, i.e., there does not exist an $N$-dimensional projection $\gamma$ such that $\ehf(\gamma)=E^\HF(N,Z)$. \end{thm} This Theorem will, in fact, follow from the proof of Lemma~\ref{lm:extl1} below (see page~\pageref{proof:extl1}). Although this result is very good for $Z=1$ it is far from optimal for large $Z$. In particular the factor $2$ should rather be $1$. This fact known as the ionization conjecture is one of the of the main results of the present work. \begin{thm}[Optimal bound on maximal ionization charge] \label{thm:maxN} There exists a universal constant $Q>0$ such that for all positive integers satisfying $N\geq Z+Q$ there are no minimizers for the Hartree-Fock functional among $N$-dimensional projections. \end{thm} \begin{remark} Although, it is possible to calculate an exact value for the constant $Q$ above it is quite tedious to do so. Moreover, the present work does not attempt to optimize this constant. The result of this work is mainly to establish that such a {\it finite} constant exists. This of course raises the very interesting question of finding a {\it good} estimate on the constant, but we shall not address this here. \end{remark} The proof Theorem~\ref{thm:maxN} is given in Sect.~\ref{sec:provingmain} on page~\pageref{proof:maxN}. \begin{thm}[Bound on the ionization energy] \label{thm:ie} The ionization energy of a neutral atom $E^\HF(Z-1,Z)-E^{\HF}(Z,Z)$ is bounded by a universal constant (in particular, independent of $Z$). \end{thm} This theorem is proved in Sect.~\ref{sec:provingmain} on page \pageref{proof:ie}. The variational equations (Euler-Lagrange equations) for the minimizer was also given in \cite{LiebSimon:HF}. Since the Hartree-Fock variational equations shall be used later in this work, we shall derive them in Theorem~\ref{thm:HFequations} below. We first note that the Hartree-Fock functional $\ehf$ may be extended from projections (i.e., density matrices with $\gamma^2=\gamma$) to all density matrices. If $\Tr\left[-\Delta\gamma\right]<\infty$ all the terms of $\ehf$ are finite. In fact, $\Tr\left[Z|x|^{-1}\gamma\right]$ is finite by the Lieb-Thirring inequality (\ref{eq:LTdensity}) since $Z|x|^{-1}\in L^\infty(\R^3)+ L^{5/2}(\R^3)$. The term $\D(\gamma)$ is finite by the Hardy-Littlewood-Sobolev inequality since $\uprho_\gamma\in L^1(\R^3)\cap L^{5/3}(\R^3)\subset L^{6/5}(\R^3)$. Finally, $\Ex(\gamma)\leq D(\gamma)$ since $$ D(\gamma)-\Ex(\gamma) =\mfr{1}{4}\sum_{i,j}\nu_i\nu_j\iint \frac{\left\|u_i(x)\otimes u_j(y) -u_j(x)\otimes u_i(y)\right\|_{\C^2\otimes\C^2}^2} {|x-y|}dx\,dy\geq 0, $$ when $\nu_i$ are the eigenvalues of $\gamma$ with $u_i$ being the corresponding eigenfunctions. If $\Tr\left[-\Delta\gamma\right]=\infty$ we set $\ehf(\gamma):=\infty$. It is clear that $\lim_n\ehf(\gamma_n)=\infty$ if $\lim_n\Tr\left[-\Delta\gamma_n\right]\to\infty$. \begin{remark} It is important to realize that although $D(\gamma)-\Ex(\gamma)$ is positive it is {\it not} a convex functional on the set of density matrices. In particular, the Hartree-Fock minimizer need not be unique. (A simple example of non-uniqueness occurs for the case $N=1$. For a one-dimensional projection $\gamma$, it is clear that $D(\gamma)-\Ex(\gamma)=0$, hence the minimizer in this case is simply the projection onto a ground state of the operator $-\mfr{1}{2}\Delta-Z|x|^{-1}$ on the space $L^2(\R^3;\C^2)$. There are many ground states since the spin can point in any direction.) \end{remark} Another fact related to the non-convexity of the Hartree-Fock functional is the important observation first made by Lieb in \cite{Lieb:Var} that the infimum of the Hartree-Fock functional is not lowered by extending the functional to all density matrices. For a simple proof of this see \cite{Bach}. \begin{thm}[Lieb's variational principle]\label{thm:liebvar} For all non-negative integers $N$ we have $$ \inf\set{\ehf(\gamma)}{\gamma^*=\gamma,\ \gamma=\gamma^2,\ \Tr[\gamma]=N}= \inf\set{\ehf(\gamma)}{0\leq\gamma\leq \I,\ \Tr[\gamma]=N} $$ and if the infimum over all density matrices (the inf on the right) is attained then so is the infimum over projections (the inf on the left). \end{thm} We now come to the properties of the Hartree-Fock minimizers, especially that they satisfy the Hartree-Fock equations. These equations state that a minimizing $N$-dimensional projection $\gammahf$ is the projection onto the $N$-dimensional space spanned by eigenfunctions with lowest possible eigenvalues for the {\it HF mean field operator} \begin{equation}\label{eq:HMF} \HMF:=-\mfr{1}{2}\Delta-Z|x|^{-1}+\rhohf*|x|^{-1} -\KMF. \end{equation} Here $\KMF$ is the {\it exchange operator} defined by having the $2\times2$-matrix valued integral kernel $$ \KMF(x,y):=|x-y|^{-1}\gammahf(x,y). $$ Thus $\gammahf(x,y)=\sum_{i=1}^Nu_i(x)u_i(y)^*$, where $\HMF u_i=\epsilon_i u_i$, and $\epsilon_1,\epsilon_2,\ldots,\epsilon_N\leq0$ are the $N$ lowest eigenvalues of $\HMF$ counted with multiplicities. This self-consistent property of a minimizer $\gammahf$ may equivalently be stated as in the theorem below. \begin{thm}[Properties of HF minimizers]\label{thm:HFequations} If $\gammahf$ with density $\rhohf$ is a projection minimizing the HF functional $\ehf$ under the constraint $\Tr\left[\gammahf\right]=N$ then $\rhohf\in L^{5/3}(\R^3)\cap L^1(\R^3)$ and $\HMF$ defines a semibounded self-adjoint operator with form domain $H^1(\R^3;\C^2)$ having at least $N$ non-positive eigenvalues. Moreover, $\gammahf$ is the $N$-dimensional projection minimizing the map $\gamma\mapsto\Tr\left[\HMF\gamma\right]$. \end{thm} \begin{remark} The reader may worry that, because of degenerate eigenvalues of $\HMF$, the $N$-dimensional projection $\gamma$ minimizing $\Tr\left[\HMF\gamma\right]$ may not be unique. That it is, indeed, unique was proved in \cite{BLLS}. \end{remark} \begin{proof}[Proof of Theorem~\ref{thm:HFequations}] We note that $\Tr\left[\gammahf\right]=N$, $\Tr\left[-\Delta\gammahf\right]<\infty$, and the Lieb-Thirring inequality (\ref{eq:LTdensity}) implies that $\rhohf\in L^{5/3}(\R^3)\cap L^1(\R^3)$. {F}rom this it is easy to see that $\rhohf*|x|^{-1}$ is a bounded function (in fact, it is continuous and tends to $0$ as $|x|\to\infty$). Moreover, in the operator sense $\KMF\leq \rhohf*|x|^{-1}$. This follows, since for $f\in L^2(\R^3;\C^2)$ we have \begin{eqnarray*} \int \rhohf*|x|^{-1}|f(x)|^2dx- \iint f(x)^*\KMF(x,y) f(y) dx\,dy&=&\\ \sum_{i=1}^N\mfr{1}{2}\iint \frac{\left\|u_i(x)\otimes f(y) -f(x)\otimes u_i(y)\right\|_{\C^2\otimes\C^2}^2}{|x-y|}dx\,dy,&& \end{eqnarray*} where $u_1,\ldots,u_N$ is a complete set of eigenfunctions of $\gammahf$. It is therefore clear that $\HMF$ defines a semibounded operator with form domain $H^1(\R^3;\C^2)$. Thus it makes sense to compute $\Tr\left[\HMF\gamma\right]$ if and only if $\Tr\left[-\Delta\gamma\right]<\infty$. Let $\gamma'$ be an $N$-dimensional projection with $\Tr\left[-\Delta\gamma'\right]<\infty$. We shall prove that $$ \Tr\left[\HMF\gamma'\right]\geq \Tr\left[ \HMF\gammahf\right]. $$ For $0\leq t\leq1$, consider the density matrix $\gamma_t= (1-t)\gammahf+t\gamma'$. It satisfies $\Tr[\gamma_t]=N$. By the Lieb variational principle, Theorem~\ref{thm:liebvar}, we have that $\ehf(\gammahf)=\ehf(\gamma_0)\leq\ehf(\gamma_t)$, for all $0\leq t\leq 1$. Hence $$ 0\leq\left.\frac{d\ehf(\gamma_t)}{dt}\right|_{t=0}= \Tr\left[\HMF\gamma'\right]-\Tr\left[\HMF\gammahf\right]. $$ The fact that $\Tr\left[\HMF\gamma\right]$ is minimized among $N$-dimensional projections implies in particular that $\HMF$ has at least $N$ non-positive eigenvalues. \end{proof} \section{Thomas-Fermi Theory} \heading{Thomas-Fermi theory}\label{sec:tf} In this section we discuss the facts needed from Thomas-Fermi theory. We focus only on the results that we shall use in our study of Hartree-Fock theory. \begin{defi}[Thomas-Fermi functional] \label{defi:TFfunctional} Let $V\in L^{5/2}(\R^3)+L^\infty(\R^3)$ with $$ \inf\set{\|W\|_{L^\infty(\R^3)}}{V-W\in L^{5/2}(\R^3)}=0. $$ Corresponding to $V$ we define the {\it Thomas-Fermi (TF) energy functional} $$ \etf{V}(\rho) =\mfr{3}{10}(3\pi^2)^{2/3} \int\rho^{5/3}-\int V\rho +\mfr{1}{2}\iint\rho(x)|x-y|^{-1}\rho(y)dx\,dy, $$ on functions $\rho$ with $0\leq\rho\in L^{5/3}(\R^3)\cap L^1(\R^3)$. \end{defi} Note that the Hardy-Littlewood-Sobolev inequality implies that $D(\rho,\rho)= \mfr{1}{2}\iint\rho(x)|x-y|^{-1}\rho(y)dx\,dy$ is finite for functions $\rho\in L^{5/3}\cap L^1\subset L^{6/5}$. Hence $\etf{V}$ is finite on these functions. The proof of existence and uniqueness of minimizers to the TF functional and the characterization of their properties can be found in the work of Lieb and Simon \cite{LiebSimon:TF} (see also \cite{Lieb:tf}). We state the properties that we need in the following theorem. \begin{thm}[The TF minimizer]\label{thm:tf} Let $V$ be as in Definition~\ref{defi:TFfunctional}. For all $N'\geq0$ there exists a unique non-negative $\rhotfv\in L^{5/3}(\R^3)$ such that $\int\rhotfv\leq N'$ and \begin{equation}\label{eq:tfvariation} \etf{V}(\rhotfv) =\inf\set{\etf{V}(\uprho)}{\uprho\in L^{5/3}(\R^3),\ \int\rho\leq N'}. \end{equation} On the other hand there exists a (unique) {\bf chemical potential} (Lagrange multiplier) $\mutfv(N')$, with $0\leq\mutfv(N')\leq\sup V$, such that $\rhotfv$ is uniquely characterized by \begin{eqnarray} \lefteqn{\etf{V}(\rhotfv)+\mutfv(N')\int\rhotfv}&&\nonumber\\ &=&\inf\set{\etf{V}(\uprho)+\mutfv(N')\int\uprho} {0\leq \uprho\in L^{5/3}(\R^3)\cap L^1(\R^3)}.\label{eq:tfmuvariation} \end{eqnarray} Moreover, $\rhotfv$ is the unique solution in $L^{5/3}\cap L^1$ to the {\bf Thomas-Fermi equation} (the Euler-Lagrange equation for the variational problem (\ref{eq:tfmuvariation})) \begin{equation}\label{eq:tfeqgeneral} \mfr{1}{2}(3\pi^2)^{2/3}\left(\rhotfv(x)\right)^{2/3} =\left[V(x)-\rhotfv*|x|^{-1}-\mutfv(N')\right]_+. \end{equation} If $\mutfv(N')>0$ then $\int \rhotfv=N'$. Therefore $ \mutfv(N')\int\rhotfv = \mutfv(N') N'. $ For all $0<\mu$ there is a unique minimizer $0\leq \rho\in L^{5/3}\cap L^1$ to $\etf{V}(\rho)+\mu\int\rho$. (If $\mu\geq \sup V$ then $\rho$ is simply zero). \end{thm} We shall be interested in properties of the {\it Thomas-Fermi potential} \begin{equation}\label{eq:tfpotgeneral} \phitfv:=V(x)-\rhotfv*|x|^{-1}. \end{equation} The Thomas-Fermi equation (\ref{eq:tfeqgeneral}) can be turned into the {\it Thomas-Fermi differential equation} \begin{equation}\label{eq:tfdiffeqgeneral} \Delta \phitfv={2}^{7/2}(3\pi)^{-1}\left[\phitfv -\mutfv(N')\right]_+^{3/2} +\Delta V, \end{equation} which holds in distribution sense. \begin{thm}[Maximal ionization]\label{thm:maxion} There exists a non-negative real number $N_c$, possibly equal to $+\infty$, such that $\mutfv(N')>0$ if and only if $N'0$ we must have $\mutfv(N')=\mutfv(N_c)=0$ for $N'\geq N_c$. On the other hand since $\int\rhotfv=N'$ if $N'c r^{-3/2}$ for some positive constant $c$ and all large enough $r$. Hence $N_c =\int\rhotfv=\infty$ in contradiction with our assumption. \end{proof} Proving a bound on $N_c$ in the opposite direction is in general more difficult. We shall return to a partial converse to (\ref{eq:ncupper}) in Corollary~\ref{cl:uppermaxion} below. Usually the Thomas-Fermi model is studied for the potential $V$ being the Coulomb potential, i.e., $Z|x|^{-1}$. In this case we denote $\rhotfv$, $\phitfv$, and $\mutfv$ simply by $\rhotf$, $\phitf$, and $\mutf$. These are the functions discussed in the introduction. In fact, the equations (\ref{eq:tfsystemphi}) and (\ref{eq:tfsystemrho}) correspond to (\ref{eq:tfeqgeneral}) and (\ref{eq:tfdiffeqgeneral}). {F}rom Theorem~\ref{thm:maxion} we see that in this case $N_c\geq Z$. We shall see below after Corollary~\ref{cl:uppermaxion} that indeed $N_c=Z$. The first mathematical study of the atomic TF equation was done by Hille~\cite{Hille}; a much more complete analysis can be found in \cite{LiebSimon:TF}. The function $\phitf$ satisfies the asymptotics $\phitf(x)\approx 3^42^{-3}\pi^2|x|^{-4}$ for large $x$. The important thing to note about this asymptotics, first discovered by Sommerfeld~\cite{Sommerfeld}, is that it is independent of $Z$. The Sommerfeld asymptotics is central to the present work and we shall prove a strong version of it in Theorems~\ref{thm:sommerfeldup} and \ref{thm:sommerfeldlow} below. Similar asymptotic estimates may be derived for the density using the TF equation (\ref{eq:tfeqgeneral}). We shall more generally prove asymptotic bounds for $\phitfv$, in the case when the potential $V$ is harmonic in certain regions of space. We now come to the main technical lemma in this section, which is a version of the Sommerfeld estimate.\footnote{A version of this Sommerfeld estimate was stated in the announcement~\cite{Solovej:announcement}. The result stated was weaker than here in the sense that the exponents in the error terms were different for the upper and lower bounds. The result in the announcement also contained a minor error because the lower bound had been stated incorrectly. The better and correct version is the one stated and proved here.} \begin{lm}[Sommerfeld estimate] \label{lm:sommerfeld} Assume that $\phi\geq0$ is a smooth function on $|x|>R$ and satisfies the differential equation $$ \Delta\phi(x)=2^{7/2}(3\pi)^{-1}\phi(x)^{3/2},\quad\hbox{for $|x|>R$}, $$ for some $R\geq0$. Let $\zeta:=(-7+\sqrt{73})/2\approx0.77$. Define \begin{eqnarray*} a(R)&:=&\liminf_{r\searrow R}\sup_{|x|=r} \left[\left(\frac{\phi(x)}{3^42^{-3}\pi^2r^{-4}}\right)^{-1/2}-1\right] r^\zeta \\ \noalign{\hbox{and}} A(R)&:=& \liminf_{r\searrow R}\sup_{|x|=r} \left[\frac{\phi(x)}{3^42^{-3}\pi^2r^{-4}}-1\right] r^\zeta. \end{eqnarray*} Then for $|x|>R$ we have \begin{equation}\label{eq:tfestimate} \left(1+a(R)|x|^{-\zeta}\right)^{-2}\leq \frac{\phi(x)}{3^42^{-3}\pi^2|x|^{-4}}\leq \left(1+A(R)|x|^{-\zeta}\right). \end{equation} \end{lm} \begin{remark} It is important to realize that we are not assuming that $\phi$ is spherically symmetric. The lemma above can therefore not be proved by ODE techniques. By elliptic regularity the smoothness of $\phi$ would of course be a consequence of a much weaker assumption. \end{remark} \begin{proof}[Proof of Lemma~\ref{lm:sommerfeld}] We first prove that $\phi(x)\to0$ as $|x|\to\infty$. For this purpose consider $L>4R$ and for $L/4<|x|f(x)}$. This is an open set on which $\Delta(\phi-f)\geq 2^{7/2}(3\pi)^{-1}(\phi^{3/2}-f^{3/2})>0$, i.e., $\phi-f$ is subharmonic on the set and is zero on its boundary. Hence $\phi(x)\leq f(x)$ on the set which is a contradiction unless the set is empty. Thus for all $L>4R$ we have $ \phi(L/2)\leq C\left((1/4)^{-4}+(1/2)^{-4}\right) L^{-4}. $ Hence, $\phi(x)|x|^4$ is bounded. Next we turn to the proof of the main estimate. Let $R'>R$ and set $A'=A(R')$ and $a'=a(R')$. Then $a'$ and $A'$ are finite. We consider the two functions $$ \omegap{A'}(x):=3^42^{-3}\pi^2|x|^{-4}(1+A'|x|^{-\zeta}) \quad\hbox{and}\quad \omegam{a'}(x):=3^42^{-3}\pi^2|x|^{-4}(1+a'|x|^{-\zeta})^{-2}. $$ Note that by the definition of $a'$ and $A'$ both functions are well-defined and positive for $|x|>R'$. We claim that \begin{equation}\label{eq:omegainequality} \Delta \omegap{A'}(x)\leq 2^{7/2}(3\pi)^{-1}\omegap{A'}(x)^{3/2} \quad\hbox{and}\quad \Delta \omegam{a'}(x)\geq 2^{7/2}(3\pi)^{-1}\omegam{a'}(x)^{3/2}. \end{equation} As we shall first show the lemma is a simple consequence of the estimates in (\ref{eq:omegainequality}). We give the proof for the upper bound. The lower bound is similar. Let $$ \Omega_+:=\set{|x|>R'}{\phi(x)>\omegap{A'}(x)}. $$ On $\Omega_+$, $\phi-\omegap{A'}$ is subharmonic. On the boundary of $\Omega_+$, $\phi-\omegap{A'}$ vanishes. For the subset $\partial\Omega_+\cap\set{x}{|x|=R'}$ this follows from the choice of $A'$. Since $\phi(x)$ and $\omegap{A'}(x)$ both tend to zero as $|x|$ tends to infinity we conclude that $\Omega_+=\emptyset$. Therefore $\phi(x)\leq \omegap{A(R')}(x)$ for $|x|>R'$. For $|x|>R$ we get $\phi(x)\leq\liminf_{R'\searrow R}\omegap{A(R')}(x)=\omegap{A(R)}(x)$. It remains to check (\ref{eq:omegainequality}). For $\omegam{a'}$ we get \begin{eqnarray*} \Delta \omegam{a'}(x) =2^{7/2}(3\pi)^{-1}\omegam{a'}(x)^{3/2} &\biggl(&1+\left(1-\mfr{1}{6}\zeta(\zeta+7)\right)a'|x|^{-\zeta}\\ &&+6(1+a'|x|^{-\zeta})^{-1}(\zeta a' |x|^{-\zeta})^2\biggr). \end{eqnarray*} Since $\zeta(\zeta+7)=6$ and $1+a'|x|^{-\zeta}>0$ we see that $\Delta \omegam{a'}(x)\geq 2^{7/2}(3\pi)^{-1}\omegam{a'}(x)^{3/2}$. For $\omegap{A'}$ we have \begin{eqnarray*} \Delta \omegap{A'}(x) &=&2^{7/2}(3\pi)^{-1}\omegap{A'}(x)^{3/2} (1+A'|x|^{-\zeta})^{-3/2}\left(1+\left(1+\frac{\zeta(\zeta+7)}{12} \right)A'|x|^{-\zeta}\right)\\ &\leq&2^{7/2}(3\pi)^{-1}\omegap{A'}(x)^{3/2}, \end{eqnarray*} where we have used that $$(1+A'|x|^{-\zeta})^{3/2}\geq 1+\mfr{3}{2} A'|x|^{-\zeta}=1+(1+\mfr{1}{12}\zeta(\zeta+7))A'|x|^{-\zeta}. $$ \end{proof} We can immediately use this lemma to get estimates on $\phitfv$ when $\mutfv=0$. For general $\mutfv$ the result can be generalized as follows. \begin{thm}[Sommerfeld estimate for general $\mutfv$]\label{thm:sommerfeldmu} Assume that $V$ is continuous and harmonic for $|x|>R$ and satisfies $\lim_{|x|\to\infty}V(x)=0$. Consider the corresponding Thomas-Fermi potential $\phitfv$, which satisfies the TF differential equation (\ref{eq:tfdiffeqgeneral}). Assume that $\mutfv< \liminf\limits_{r\searrow R}\inf\limits_{|x|=r}\phitfv(x) $. Define \begin{eqnarray} a(R)&:=&\liminf_{r\searrow R}\sup_{|x|=r} \left[\left(\frac{\phitfv(x)}{3^42^{-3}\pi^2r^{-4}}\right)^{-1/2}-1 \right] r^\zeta \label{eq:armu}\\ \noalign{\hbox{and}} A(R,\mutfv)&:=& \liminf_{r\searrow R}\sup_{|x|=r} \left[\frac{\phitfv(x)-\mutfv}{3^42^{-3}\pi^2r^{-4}}-1\right] r^\zeta.\label{eq:Armu} \end{eqnarray} Then again, with $\zeta=(-7+\sqrt{73})/2\approx0.77$, we find for all $|x|>R$ \begin{eqnarray} \phitfv(x)&\leq& 3^42^{-3}\pi^2|x|^{-4} \left(1+A(R,\mutfv)|x|^{-\zeta}\right)+\mutfv\label{eq:tfupestimate}\\ \noalign{\hbox{and}} \phitfv(x)&\geq&\max\left\{ 3^42^{-3}\pi^2|x|^{-4}\left(1+a(R)|x|^{-\zeta}\right)^{-2},\, \nu(\mutfv)|x|^{-1}\right\}, \label{eq:tflowestimate} \end{eqnarray} where \begin{equation} \nu(\mutfv):=\inf_{|x|\geq R}\max\Bigl\{ 3^42^{-3}\pi^2|x|^{-3}\left(1+a(R)|x|^{-\zeta}\right)^{-2}, \mutfv|x|\Bigr\}.\label{eq:nudef} \end{equation} \end{thm} \begin{proof} Since $\rhotfv\in L^{5/3}\cap L^1$ it is easy to see that $\rhotfv*|x|^{-1}$ is continuous and tends to zero as $x$ tends to infinity. Thus from the assumption on $V$ it follows that $\phitfv$ is continuous on $|x|>R$ and satisfies $\phitfv(x)\to0$ as $|x|\to\infty$. Let $R'>R$ and set $A'=A(R',\mutfv)$ and $a'=a(R')$. Then $a'$ is well-defined if $R'$ is close enough to $R$ since then we may assume that $\phitfv(x)>\mutfv\geq 0$ for all $|x|=R'$. Both $a'$ and $A'$ are finite. Using the notation from the proof of Lemma~\ref{lm:sommerfeld} we define $$ \omegap{\mutfv,A'}(x):=\omegap{A'}(x)+\mutfv \quad\hbox{and}\quad \omegam{\mutfv,a'}(x):= \max\left\{\omegam{a'}(x),\ \nu'|x|^{-1}\right\}, $$ where $$ \nu':= \min_{|x|\geq R'}\max\left\{ |x|\omegam{a'}(x),\, |x|\mutfv\right\}.. $$ Note that, since we assume that $\phitfv(x)>\mutfv$ for $|x|=R'$, we have that both $\omegap{A'}(x)$ and $\omegam{a'}(x)$ are positive for all $|x|>R'$. We also have that $\omegam{a'}(x)>\mutfv$ for $|x|=R$ and hence that $\omegam{a'}(x_0)=\mutfv $ at points $x_0$ where the minimum, defining $\nu'$, is attained. (Note that $|x|\omegam{a'}(x)$ is a radially decreasing function for $|x|>R'$.) The proof of the present lemma is now similar to that of Lemma~\ref{lm:sommerfeld} if we can show that for $|x|>R'$ \begin{equation}\label{eq:omega+muineq} \Delta\omegap{\mutfv,A'}(x) \leq 2^{7/2}(3\pi)^{-1}\left[\omegap{\mutfv,A'}(x)-\mutfv \right]_+^{3/2} \end{equation} and \begin{equation}\label{eq:omega-muineq} \Delta\omegam{\mutfv,a'}(x) \geq 2^{7/2}(3\pi)^{-1} \left[\omegam{\mutfv,a'}(x)-\mutfv\right]_+^{3/2} \end{equation} (in distribution sense). The inequality (\ref{eq:omega+muineq}) follows immediately from the first inequality in (\ref{eq:omegainequality}). The inequality (\ref{eq:omega-muineq}) is slightly more complicated. Note that the definitions of $\omegam{\mutfv,a'}$ and of $\nu'$ imply that $\omegam{\mutfv,a'}(x)=\nu'|x|^{-1}$ if $\omegam{\mutfv,a'}(x)<\mutfv$ and $\omegam{\mutfv,a'}(x)=\omegam{a'}(x)$ if $\omegam{\mutfv,a'}(x)>\mutfv$. Thus if $\omegam{\mutfv,a'}(x)<\mutfv$ we have that $\omegam{\mutfv,a'}$ is harmonic. Hence (\ref{eq:omega-muineq}) holds in this region. If $\omegam{\mutfv,a'}(x)>\mutfv$ then $\omegam{\mutfv,a'}(x)=\omegam{a'}(x)$ and (\ref{eq:omega-muineq}) follows in this region from the second inequality in (\ref{eq:omegainequality}). Finally, since the maximum of two subharmonic functions is also subharmonic, it is clear that the distribution $\Delta\omegam{\mutfv,a'}$ is a positive measure and in particular positive on the set (of Lebesgue measure) zero where $\omegam{\mutfv,a'}(x)=\mutfv$. Hence, (\ref{eq:omega-muineq}) holds in distribution sense for all $|x|>R$. \end{proof} As an application of the lower bound on $\phitfv$ in (\ref{eq:tflowestimate}) we can get an estimate on the chemical potential $\mutfv$. \begin{cl}[Chemical potential estimate]\label{cl:muestimate} With the assumptions and definitions in Theorem~\ref{thm:sommerfeldmu}, in particular, if $\mutfv<\liminf_{r\searrow R}\inf_{|x|=r}\phitfv(x)$ we have \begin{equation}\label{eq:muestimate} (\mutfv)^{3/4}\leq \frac{2^{3/4}}{3\pi^{1/2}}(1+|a(R)|R^{-\zeta})^{1/2} \left(\lim_{r\to\infty} (4\pi)^{-1}\int_{{\cal S}^2}rV(r\omega)d\omega -\int_{\R^3}\rhotfv(y)dy\right). \end{equation} \end{cl} \begin{proof} According to (\ref{eq:tflowestimate}) we have $ \nu(\mutfv)\leq \liminf_{|x|\to\infty}|x|\phitfv(x). $ Using that $V$ is harmonic and tends to zero at infinity we have that for all $r>R$ $$ \liminf_{|x|\to\infty}|x|\phitfv(x) \leq(4\pi)^{-1}\int_{{\cal S}^2}rV(r\omega)d\omega -\int_{\R^3}\rhotfv(y)dy. $$ Moreover since, $\mutfv\geq0$ the assumption $\mutfv<\liminf_{r\searrow R}\inf_{|x|=r}\phitfv(x)$ implies that the spherical average of $V$ is non-negative. On the other hand, since $\left(1+|a(R)|R^{-\zeta}\right)^{-2}\leq\left(1+a(R)|x|^{-\zeta}\right)^{-2}$ for $|x|\geq R$, we have from (\ref{eq:nudef}) that $\nu(\mutfv)\geq\nu'$, where \begin{eqnarray*} \nu'&=&\min_{|x|\geq R} \max\Bigl\{ 3^42^{-3}\pi^2|x|^{-3}\left(1+|a(R)|R^{-\zeta}\right)^{-2}, \mutfv|x|\Bigr\}\\&=& 3\cdot 2^{-3/4}\pi^{1/2}\left(1+|a(R)|R^{-\zeta}\right)^{-1/2} (\mutfv)^{3/4}. \end{eqnarray*} \end{proof} This corollary immediately gives a partial converse to Theorem~\ref{thm:maxion}. \begin{cl}[Upper bound on maximal ionization] \label{cl:uppermaxion} If $V$ is harmonic and continuous for $|x|>R$ and satisfies $V(x)\to0$ as $|x|\to\infty$ and if moreover $\mutfv< \liminf_{r\searrow R}\inf_{|x|=r}\phitfv(x)$ then $$ \int\rhotfv\leq \lim_{r\to\infty} (4\pi)^{-1}\int_{{\cal S}^2} rV(r\omega)d\omega. $$ In particular, if $\liminf_{r\searrow R}\inf_{|x|=r}\phitfv(x)>0$ (which may not necessarily be true) we have $$ N_c\leq \lim_{r\to\infty}(4\pi)^{-1}\int_{{\cal S}^2} rV(r\omega)d\omega. $$ \end{cl} \begin{remark} The limit above of course exists since by the harmonicity of $V$ and since $V$ tends to zero at infinity we have that $\int_{{\cal S}^2} rV(r\omega)d\omega$ is independent of $r$. The difficulty in using Corollaries \ref{cl:muestimate} and \ref{cl:uppermaxion} in concrete examples lies in establishing the condition \begin{equation}\label{eq:chemcondition} \mutfv<\liminf_{r\searrow R}\inf_{|x|=r}\phitfv(x). \end{equation} \end{remark} \section{Estimates on the standard atomic TF theory} \heading{Standard TF theory} \label{sec:atomictf} In the usual atomic case the Coulomb potential $V(x)=Z|x|^{-1}$ is harmonic away from $x=0$ and we can use Corollary~\ref{cl:uppermaxion} for all $R>0$. Since $\rhotf*|x|^{-1}$ is a bounded function it follows that $\phitf(x)\to\infty$ as $x\to0$. The condition (\ref{eq:chemcondition}) is therefore satisfied if $R$ is chosen small enough. It therefore follows from Theorem~\ref{thm:maxion} and Corollary~\ref{cl:uppermaxion} that $N_c=Z$. Thus the neutral atom corresponds to $\mutf=0$. \begin{lm}\label{lm:ltf3.8} Let $\phitf_0$ be the TF potential for the neutral atom then if $\phitf$ is the potential for a general $\mutf\geq0$ we have $$ \phitf_0(x)\leq \phitf(x)\leq\phitf_0(x)+\mutf. $$ \end{lm} \begin{proof} See Corollary~3.8 (i) and (iii) in \cite{Lieb:tf}. \end{proof} We now easily get an upper bound agreeing with the atomic Sommerfeld asymptotics. \begin{thm}[Atomic Sommerfeld upper bound]\label{thm:sommerfeldup} The atomic TF potential satisfies the bound $$ \phitf(x)\leq \min\{3^42^{-3}\pi^2|x|^{-4}+\mutf,\ Z|x|^{-1}\} $$ \end{thm} \begin{proof} This follows immediately from and (\ref{eq:tfpotgeneral}) and (\ref{eq:tfupestimate}) together with the fact that $\rhotf$ is non-negative. Simply note that since $\phitf(x)|x|\to Z$ as $x\to0$ we have that $A(0,\mutf)=0$ in (\ref{eq:tfupestimate}). \end{proof} \begin{lm}[Lower bound on the TF potential]\label{lm:firstphitfbound} In the atomic case we have for all $N>0$ and $Z>0$ $$ \phitf(x)\geq Z|x|^{-1}-\min\left\{N|x|^{-1},\frac{22}{(9\pi)^{2/3}}Z^{4/3}\right\}. $$ \end{lm} \begin{proof} We have by Newton's Theorem \begin{eqnarray*} \rhotf*|x|^{-1}&=& |x|^{-1}\int_{|y|<|x|}\rhotf(y)dy+\int_{|y|>|x|}\rhotf(y)|y|^{-1}dy\\ &\leq& \min\left\{N|x|^{-1},\ \int\rhotf(y)|y|^{-1}dy\right\}. \end{eqnarray*} {F}rom the Sommerfeld upper bound Theorem~\ref{thm:sommerfeldup} and the TF equation (\ref{eq:tfeqgeneral}) we have $$ \rhotf(x)^{2/3}\leq 2(3\pi^2)^{-2/3}\min\{3^42^{-3}\pi^2|x|^{-4},\ Z|x|^{-1}\}. $$ Hence $$ \rhotf(x)\leq\min\left\{c_1Z^{3/2}|x|^{-3/2},\,c_2|x|^{-6}\right\}, $$ where $c_1:=2^{3/2}(3\pi^2)^{-1}$ and $c_2:=3^52^{-3}\pi$. Let $r_0:=(c_2/c_1)^{2/9}Z^{-1/3}$. When $|x|=r_0$ the two functions, in the minimum above, are equal. Thus \begin{eqnarray*} \int\rhotf(y)|y|^{-1}dy&\leq& 4\pi c_1Z^{3/2}\int_0^{r_0}t^{-1/2}dt +4\pi c_2\int_{r_0}^\infty t^{-5}dt=\frac{11\pi}{3}c_1^{8/9}c_2^{1/9} Z^{4/3}\\ &=&\frac{22}{(9\pi)^{2/3}}Z^{4/3}. \end{eqnarray*} The lemma follows from the definition (\ref{eq:tfpotgeneral}) of the TF potential. \end{proof} \begin{thm}[Atomic Sommerfeld Lower bound] \label{thm:sommerfeldlow} The TF potential satisfies $$ \phitf(x)\geq\left\{ \begin{array}{ll} Z|x|^{-1}-22(9\pi)^{-2/3}Z^{4/3}, &\mbox{if } |x|\leq \beta_0Z^{-1/3}\\ \\ \begin{array}{rl} \max\Bigl\{&3^42^{-3}\pi^2\left(1+aZ^{-\zeta/3}|x|^{-\zeta}\right)^{-2}|x|^{-4},\\ &(Z-N)_+|x|^{-1}\Bigr\}, \end{array} &\mbox{if } |x|\geq \beta_0Z^{-1/3} \end{array} \right. , $$ where $\beta_0=\frac{(9\pi)^{2/3}}{44}$ and $\zeta=(-7+\sqrt{73})/2$ as in Theorem~\ref{thm:sommerfeldmu} and $ a= 43.7. $ \end{thm} \begin{proof}Let $R=(9\pi)^{2/3}Z^{-1/3}/44$. Note that for $|x|\leq R$ the bound we want to prove is identical to the bound in Lemma.~\ref{lm:firstphitfbound}. If $N\geq Z$, i.e., $\mutf=0$ the lower bound follows from Theorem~\ref{thm:sommerfeldmu} since $a$ is chosen so as to make the lower bound continuous at $|x|=R$ and at these points we clearly have $\phitf(x)>0=\mutf$. For general $N$ the lower bound follows from the case $N=Z$ because of Lemma~\ref{lm:ltf3.8} and Lemma~\ref{lm:firstphitfbound}. \end{proof} We end this section by giving a bound on the {\it screened nuclear potential} $\Phitf{R}$ at radius $R$ in the atomic TF theory. \begin{lm}[Bound on $\Phitf{R}$] \label{lm:Phitfrbound} We have $$ \Phitf{|x|}(x)\leq 3^42^{-1}\pi^2|x|^{-4}+\mutf $$ \end{lm} \begin{proof} We write $\Phitf{|x|}(x)=\phitf(x)+\int_{|y|>|x|}\rhotf(y)|x-y|^{-1}dy$. {F}rom Theorem~\ref{thm:sommerfeldup} and the TF equation (\ref{eq:tfeqgeneral}) we see that $$ \rhotf(y)\leq 2^{3/2}(3\pi^2)^{-1}[\phitf-\mutf]_+^{3/2} \leq 2^{-3}3^{5}\pi|y|^{-6} $$ and hence \begin{eqnarray*} \int_{|y|>|x|}\rhotf(y)|x-y|^{-1}dy&\leq& \int_{|y|>|x|}2^{-3}3^{5}\pi|y|^{-6}|x-y|^{-1}dy\\ &=&\int_{|y|>|x|}2^{-3}3^{5}\pi|y|^{-7}dy=2^{-3}3^{5}\pi^2|x|^{-4}. \end{eqnarray*} The lemma follows from Theorem~\ref{thm:sommerfeldup}. \end{proof} \section{Separating the outside from the inside} \heading{Outside-inside} \label{sec:io} We shall here control the energy coming from the regions far from the nucleus. Let $\gammahf$ be an HF minimizer with $\Tr[\gammahf]=N$. (We are thus assuming that $N$ is such that a minimizer exists.) \begin{defi}[The localization function]\label{defi:localization} Fix $0<\lambda<1$ and let $G:\R^3\to\R$ be given by $$ G(x)=\left\{\begin{array}{ll} 0&\mbox{if }|x|\leq1\\ {(\pi/2)(|x|-1)}\left[(1-\lambda)^{-1}-1\right]^{-1} &\mbox{if }1\leq|x|\leq(1-\lambda)^{-1}\\ {\pi}/{2}&\mbox{if }(1-\lambda)^{-1}\leq|x| \end{array}\right. $$ We introduce the cutoff radius $r>0$ and define the outside localization function $\thetar(x)=\sin G(|x|/r)\index{$\theta_r$}$. Then $$ 0\leq\thetar(x) \left\{\begin{array}{ll}=0&\mbox{if $|x|\leq r$}\\ \leq1&\mbox{if $r\leq|x|\leq(1-\lambda)^{-1}r$}\\ =1&\mbox{if $(1-\lambda)^{-1}r\leq|x|$} \end{array} \right. $$ and $|\nabla\thetar(x)|^2+|\nabla (1-\thetar(x)^2)^{1/2}|^2\leq ({\pi}/{(2\lambda r)})^2$ (since $(1-\lambda)^{-1}-1\geq\lambda$). \end{defi} We shall consider the HF minimizer restricted to the region $\set{x}{|x|>r}$. We therefore define the exterior part of the minimizer \begin{equation}\label{eq:gammar} \gammar=\thetar\gammahf\thetar \end{equation} and its density $\rhohfr(x)=\thetar(x)^2\rhohf(x)$. In order to control $\gammar$ we introduce an auxilliary functional defined on all density matrices with $\Tr\left[-\Delta\gamma\right]<\infty$ (See Remark~\ref{remark:deltagamma}) by \begin{equation}\label{eq:eadef} \ea(\gamma)=\Tr\left[(-\mfr{1}{2}\Delta-\Phihf{r}) \gamma\right]+ \mfr{1}{2}\iint\limits_{|x|\geq r\atop|y|\geq r} \uprho_\gamma(x)|x-y|^{-1}\uprho_{\gamma}(y)dx\,dy, \end{equation} where the screened nuclear potential $\Phihf{r}$ is defined in (\ref{eq:Phihf}) in Definition~\ref{defi:phiPhi}. Note that the functional $\ea$, in contrast to the HF functional $\ehf$ in (\ref{eq:ehf}), does not contain an exchange term. The main result in this section is that $\gammar$ almost minimizes $\ea$. More precisely, we shall prove the following theorem. \begin{thm}[The outside energy]\label{thm:outsideenergy} For all $0<\lambda<1$ and all $r>0$ we have \begin{equation}\label{eq:outsideenergy} \ea\left[\gammar\right]\leq \inf\set{\ea(\upgamma)}{ \supp\uprho_{\gamma}\subset\set{y}{|y|\geq r},\ \int\uprho_{\gamma}\leq \int\chiplus{r}\rhohf} +{\cal R}, \end{equation} where the error is \begin{equation}\label{eq:ioerror} {\cal R}=C_\lambda(r)\!\! \int\limits_{|x|\geq(1-\lambda)r}\!\!\rhohf(x)dx\nonumber +2L_1\hspace{-5ex}\int\limits_{{(1-\lambda)r\leq|x|\leq (1-\lambda)^{-1}r}} \hspace{-5ex}\left[\Phihf{(1-\lambda)r}(x)\right]_+^{5/2} dx+ \Ex\left(\gammar\right) \nonumber \end{equation} with $$ C_\lambda(r)=\left(\frac{\pi^2}{8\left(\lambda(1-\lambda)r\right)^{2}} +\frac{\pi}{r\lambda}\right). $$ Here $L_1$ is the constant in the Lieb-Thirring inequality (\ref{eq:LT}). \end{thm} \begin{proof} Besides $\thetar$ we introduce two other localization functions $$ \theta_-=(1-\theta_{r(1-\lambda)}^2)^{1/2}\qquad\hbox{and}\qquad \thetam=\left(\theta_{r(1-\lambda)}^2-\thetar^2\right)^{1/2}. $$ Note that $\theta_-^2+\thetam^2+\thetar^2=1$ and that $(\nabla\theta_-)^2+(\nabla\thetam)^2+(\nabla\thetar)^2\leq \left(\pi/(2\lambda(1-\lambda)r)\right)^2$. We introduce the inside part of the HF minimizer \begin{equation}\label{eq:gammai} \gammai=\theta_-\gammahf\theta_-. \end{equation} We shall prove (\ref{eq:outsideenergy}) by showing that for all density matrices $\upgamma$ with $\supp\uprho_\gamma\subset\set{x}{|x|\geq r}$ and $\int\uprho_{\gamma}\leq\int\chiplus{r}\rhohf$ we have \begin{equation}\label{eq:outsideenergy1} \ea\left[\gammar\right]+\ehf\left[\gammai\right]-{\cal R} \leq\ehf\left[\gammahf\right]\leq\ea[\upgamma]+ \ehf\left[\gammai\right], \end{equation} with ${\cal R}$ given by (\ref{eq:ioerror}). The estimate (\ref{eq:outsideenergy}) follows immediately from (\ref{eq:outsideenergy1}). \medskip \noindent{\it Proof of the upper bound in (\ref{eq:outsideenergy1})\/}: Since $\gammahf$ is a minimizer for $\ehf$ under the condition $\Tr[\gammahf]\leq N$ (see (\ref{eq:rel})) we have for any density matrix $\widetilde{\upgamma}$ with $\Tr\left[\widetilde{\upgamma}\right]\leq N$ that $\ehf\left(\gammahf\right)\leq\ehf\left(\widetilde{\upgamma}\right)$. We take $$ \widetilde{\upgamma}=\gammai+\upgamma. $$ Since the support of $\uprho_\gamma$ is disjoint from the support of $\theta_-$ we see that $\gammai\upgamma=0$ and hence $\widetilde{\upgamma}$ is a density matrix. Note that $$ \Tr\left[\widetilde{\upgamma}\right]=\Tr\left[\gammai\right] +\int\uprho_{\gamma} \leq\Tr\left[\gammai\right]+\int\chiplus{r}\rhohf= \int(\theta_-^2+\chiplus{r})\rhohf\leq\int\rhohf\leq N $$ We shall compute $\ehf\left(\widetilde{\upgamma}\right)$. The only terms in $\ehf$ that are not linear in the density matrix (and thus do not simply split into a sum of terms for $\gammai$ and $\upgamma$) are the exchange and direct Coulomb energies. Because of the support properties of $\gammai$ and $\upgamma$ we have that $\widetilde{\upgamma}^2=\left(\gammai\right)^2+\upgamma^2$ and therefore even the exchange term satisfies $$ \Ex\left(\widetilde{\upgamma}\right)=\Ex\left(\gammai\right) +\Ex\left(\upgamma\right)\geq\Ex\left(\gammai\right). $$ We are thus left with the direct Coulomb energy. For this we find that $$ \D\left(\widetilde{\upgamma}\right)= \D\left(\gammai\right)+\D\left(\upgamma\right) +\int \theta_-^2(y)\rhohf(y)|x-y|^{-1}\uprho_{\gamma}(y)dy. $$ By the choice of the support of $\theta_-$ we have that $$ \int \theta_-^2(y)\rhohf(y)|x-y|^{-1}\uprho_{\gamma}(x)dx\,dy \leq \int\left( \frac{Z}{|x|} -\Phihf{r}(x)\right)\uprho_{\gamma}(x)dx. $$ We have thus proved the upper bound in (\ref{eq:outsideenergy1}). \noindent{\it Proof of the lower bound in (\ref{eq:outsideenergy1})\/}: Let again the inside part of the HF minimizer be $\gammai$ defined by (\ref{eq:gammai}) and introduce also the middle part $\gammam=\thetam\gammahf\thetam$. Since $\theta_-^2+\thetam^2+\thetar^2=1$ we have from the IMS formula (\ref{eq:IMS}) that \begin{eqnarray} \Tr\left[-\mfr{1}{2}\Delta\gammahf\right]&=&\Tr\left[-\mfr{1}{2}\Delta \left(\gammai+\gammam +\gammar \right)\right]\nonumber\\ \nonumber\\ &&{}-\mfr{1}{2}\Tr\left[\gammahf\left((\nabla\theta_-)^2+(\nabla\thetam)^2 +(\nabla\thetar)^2\right) \right]\nonumber\\ \nonumber \\ &\geq&\Tr\left[-\mfr{1}{2}\Delta\left(\gammai+\gammam +\gammar \right)\right]\nonumber\\ \nonumber \\ &&{}-({\pi}^2/8)\left(\lambda(1-\lambda)r\right)^{-2} \int\limits_{(1-\lambda)r<|x|r$ $$ \thetar(x)^2\leq \thetar(x)\leq (\pi/2)(|x|/r-1)\left((1-\lambda)^{-1}-1\right)^{-1}. $$ Thus if $|y|r\}$ and tends to zero at infinity we get for all $|y|>r$ that $\Phihf{r}(y)\leq |y|^{-1}r\sup_{|x|=r}\Phihf{r}(x)$. Hence \begin{eqnarray*} \ea\left[\gammar\right]&\geq& K_1\int\rhohfr(y)^{5/3}dy -\left[r\sup_{|x|=r}\Phihf{r}(x)\right]\int|y|^{-1}\rhohfr(y)dy\\ &&{} +\mfr{1}{2}\iint\rhohfr(y)|y-y'|^{-1}\rhohfr{\gamma'}(y') dy\, dy' \end{eqnarray*} It follows from standard atomic Thomas-Fermi theory that the right hand side is bounded below by the energy of a neutral Thomas-Fermi atom with nuclear charge $\left[r\sup_{|x|=r}\Phihf{r}(x)\right]_+$ and with the constant $K_1$ in front of the first term. A simple scaling argument shows that this is $-\frac{3}{10}(3\pi^2)^{2/3}K_1^{-1} \left[r\sup_{|x|=r}\Phihf{r}(x)\right]_+^{7/3}e_0$. By repeating this argument with only a fraction of the term $\int (\rhohfr)^{5/3}$ we conclude that for all $01$ (\cite{SteinWeiss}, p.~58) we obtain $$ \Ex\left[\gamma\right]\leq \frac{384}{\pi}\left(\frac{8\pi}{3}\right)^{1/3}\int \uprho_{\gamma}^{4/3}=248.3\int \uprho_{\gamma}^{4/3}. $$ \end{proof} \section{Exterior $L^1$ estimate} \heading{Exterior $\scriptstyle L^1$ estimate} \label{sec:l1} The aim of this section is to control $\int_{|x|>r}\rhohf(x)dx$, for all $r>0$. As before $\rhohf$ is the density of a Hartree-Fock minimizer $\gammahf$ with $\Tr[\gammahf]=N$. Thus $\int\rhohf=N$. The difficulty in estimating $\int_{|x|>r}\rhohf(x)dx$ is that this quantity cannot be controlled in terms of the energy $\ehf(\gammahf)$. More precisely, $\int_{|x|>r}\uprho_\gamma(x)dx$ can be arbitrarily large even when $\ehf(\gamma)$ is arbitrarily close to the absolute minimum. The simple reason is that ``adding electrons at infinity'' will not raise the energy. Therefore, in order to control $\int_{|x|>r}\rhohf(x)dx$, we must use the minimizing property of $\gammahf$. In contrast, it follows from the Lieb-Thirring inequality that $\int_{|x|>r}\rhohf(x)^{5/3}dx$ can be controlled in terms of the energy. By H\"older's inequality it then also follows that the integral of $\rhohf$ over any bounded set can be controlled by the energy. The philosophy here will be, to use the minimizing property of $\gammahf$, to control the integral of $\rhohf$ over an unbounded set, in terms of the integral over a bounded set. Our main result in this section is stated in the next lemma. The proof of the lemma uses an idea of Lieb~\cite{Lieb:2z+1}. \begin{lm}[Exterior $L^1$ estimate]\label{lm:extl1} For all $r>0$ and all $0<\lambda<1$ the density $\rhohf$ of an HF minimizer $\gammahf$ satisfies the bound \begin{eqnarray*} \int_{|x|>(1-\lambda)^{-1}r}\rhohf(x)dx&\leq & 1+2\lambda^{-1}+2\left[\sup_{|x|=(1-\lambda)r} |x|\Phihf{(1-\lambda)r}(x)\right]_+\\ &&{}+\left(K_\lambda r^{-1}\int_{r<|x|<(1-\lambda)^{-1}r} \rhohf(x)dx\right)^{1/2}, \end{eqnarray*} where $K_\lambda:=(\frac{2\lambda}{\pi}+(1-\lambda)^{-1}) \left(\frac{\pi}{2\lambda}\right)^2$. Here $\Phihf{(1-\lambda)r}$ is the screened nuclear potential introduced in Definition~\ref{defi:phiPhi}. \end{lm} \begin{proof}\label{proof:extl1} Since $\gammahf$ is a minimizer we know that it satisfies the Hartree-Fock equations. I.e, according to Theorem~\ref{thm:HFequations}, $\gammahf$ is a projection onto a space spanned by functions $u_1,\ldots,u_N\in L^2(\R^3;\C^2)$ satisfying $\HMF u_i =\epsilon_i u_i$, were $\epsilon_i\leq0$. Let $\Xi\in C^1(\R^3)$ have compact support away from $x=0$, be real and satisfy $\Xi(x)^2\leq 1$. Then $$ 0\geq\sum_{i=1}^N\epsilon_i\int\left|u_i(x)\right|^2|x|\Xi(x)^2\,dx =\sum_{i=1}^N\int u_i(x)^*|x|\Xi(x)^2 \HMF u_i(x) dx. $$ {F}rom the definition (\ref{eq:HMF}) of the mean field operator $\HMF$ we obtain \begin{eqnarray} 0&\geq&\sum_{i=1}^N\mfr{1}{2}\int \nabla\left(u_i(x)^*|x|\Xi(x)^2\right) \cdot\nabla u_i(x)dx -Z\int\rhohf\Xi^2\nonumber\\ \nonumber\\ &&+\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\Xi(y)^2}{|x-y|}dx\,dy.\label{eq:|x|HMF} \end{eqnarray} We consider separately the different terms above. By the IMS formula (\ref{eq:IMSformula}) \begin{eqnarray} \Re\lefteqn{\mfr{1}{2}\int \nabla\left(u_i(x)^*|x|\Xi(x)^2\right) \cdot\nabla u_i(x)dx}\nonumber\\ &=&\mfr{1}{2}\int \left|\nabla\left( \Xi(x)|x|^{1/2}u_i(x)\right)\right|^2 -\mfr{1}{4}|x|^{-1}\Xi(x)^2 \left|u_i(x)\right|^2 dx\nonumber\\ &&-\mfr{1}{2}\int \left(\mfr{1}{2}\nabla\left(\Xi(x)^2\right) +|x|\left(\nabla\Xi(x)\right)^2\right) \left|u_i(x)\right|^2 dx\nonumber\\ &\geq&-\mfr{1}{2}\int\left(\mfr{1}{2}\nabla\left(\Xi(x)^2\right)+ |x|\left(\nabla\Xi(x)\right)^2\right) \left|u_i(x)\right|^2 dx,\label{eq:hardy} \end{eqnarray} where we have used Hardy's inequality $\int|\nabla f(x)|^2dx \geq\mfr{1}{4}\int|x|^{-2}|f(x)|^2dx$. For the Coulomb terms we estimate it using that $$ \rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right] =\mfr{1}{2}\sum_{i,j=1}^N\left\|u_i(x)\otimes u_j(y) -u_j(x)\otimes u_i(y)\right\|_{\C^2\otimes\C^2}^2 $$ is non-negative. Hence \begin{eqnarray*} \lefteqn{\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\Xi(y)^2}{|x-y|}dx\,dy}\\ \nonumber\\ &=&\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\left(1-\Xi(x)^2\right)\Xi(y)^2}{|x-y|}dx\,dy\\ \nonumber\\ &&{}+\mfr{1}{2}\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|x|+|y|}{|x-y|}\Xi(x)^2\Xi(y)^2dx\,dy, \end{eqnarray*} where we expressed the last term symmetrically in $x$ and $y$. If we now use the triangle inequality and the fact $\int\Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\Xi(y)^2dy \leq \rhohf(x)$, which follows from $\Xi(x)^2\leq 1$ and $(\gammahf)^2=\gammahf$, we arrive at \begin{eqnarray} \lefteqn{\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\Xi(y)^2}{|x-y|}dx\,dy}\nonumber\\ \nonumber\\ &\geq&\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\left(1-\Xi(x)^2\right)\Xi(y)^2}{|x-y|}dx\,dy\nonumber\\ \nonumber \\ &&{}+\mfr{1}{2}\left(\int\rhohf\Xi^2\right)^2- \mfr{1}{2}\int\rhohf\Xi^2.\label{eq:triangle} \end{eqnarray} Inserting the inequalities (\ref{eq:hardy}) and (\ref{eq:triangle}) into (\ref{eq:|x|HMF}) gives \begin{eqnarray} 0&\geq& -\mfr{1}{2}\int\left(\mfr{1}{2}\nabla\left(\Xi(x)^2\right)+ |x|\left(\nabla\Xi(x)\right)^2\right) \rhohf(x) dx -Z\int\rhohf\Xi^2\nonumber\\ \nonumber\\ &&{}+\iint \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\left(1-\Xi(x)^2\right)\Xi(y)^2}{|x-y|}dx\,dy\nonumber\\ \nonumber\\ &&{}+\mfr{1}{2}\left(\int\rhohf\Xi^2\right)^2-\mfr{1}{2}\int\rhohf\Xi^2. \label{eq:Xifinal} \end{eqnarray} By an approximation argument it is clear that we can use (\ref{eq:Xifinal}) for any real function $\Xi$ for which $\Xi^2\leq1$ and the function $\left(\nabla\left(\Xi(x)^2\right)+ |x|\left(\nabla\Xi(x)\right)^2\right)$ is bounded. In particular we can choose $\Xi$ identically equal to $1$, which will recover Lieb's result from \cite{Lieb:2z+1}, i.e., $\int\rhohf\leq 2Z+1$. We shall now choose $\Xi:=\thetar$, where $\thetar$ is the localization function given in Definition~\ref{defi:localization}. Then \begin{equation} \mfr{1}{2}\nabla\left(\Xi(x)^2\right)+ |x|\left(\nabla\Xi(x)\right)^2 \leq \frac{\pi}{2\lambda r} +(1-\lambda)^{-1}r\frac{\pi^2}{\left(2\lambda r\right)^2} =K_\lambda r^{-1} \label{eq:locerror} \end{equation} and \begin{eqnarray*} \iint\left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\left(1-\Xi(x)^2\right)\Xi(y)^2}{|x-y|}dx\,dy\\ \nonumber\\ \geq \iint_{|x|<(1-\lambda)r} \left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\thetar(y)^2}{|x-y|}dx\,dy. \end{eqnarray*} If we now use that $|x|<(1-\lambda)r$ and $y\in\supp\thetar$ imply that $|y||x-y|^{-1}\leq \lambda^{-1}$ we obtain \begin{eqnarray} \iint\left[\rhohf(x)\rhohf(y)- \Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]\right] \frac{|y|\left(1-\Xi(x)^2\right)\Xi(y)^2}{|x-y|}dx\,dy\nonumber\\ \nonumber\\ \geq \iint_{|x|<(1-\lambda)r} \rhohf(x)\rhohf(y)\frac{|y|\thetar(y)^2}{|x-y|}dx\,dy -\lambda^{-1}\int\thetar(y)^2\rhohf(y)\,dy,\label{eq:effpotential} \end{eqnarray} where we have also used that $\int\Tr_{\C^2}\left[\left|\gammahf(x,y)\right|^2\right]dx =\rhohf(y)$. If we insert (\ref{eq:locerror}) and (\ref{eq:effpotential}) into (\ref{eq:Xifinal}) we arrive at \begin{eqnarray*} 0&\geq& -\frac{K_\lambda}{2r}\int_{r<|x|<(1-\lambda)^{-1}r}\rhohf(x)dx -\int\thetar(y)^2|y|\Phihf{(1-\lambda)r}(y)\rhohf(y)dy\\ &&{}+\mfr{1}{2}\left(\int\rhohf\thetar^2\right)^2 -\left(\mfr{1}{2}+\lambda^{-1}\right)\int\rhohf\thetar^2. \end{eqnarray*} Using now that $\Phihf{(1-\lambda)r}(y)$ tends to zero at infinity and is harmonic for $|y|>(1-\lambda)r$, which contains the support of $\thetar$, we see by a simple comparison argument that $$ \thetar(y)^2|y|\Phihf{(1-\lambda)r}(y) \leq \thetar(y)^2 \left[\sup_{|x|=(1-\lambda)r} |x|\Phihf{(1-\lambda)r}(x)\right]_+. $$ Thus \begin{eqnarray*} 0&\geq&\left(\int\rhohf\thetar^2\right)^2 -\left(1+2\lambda^{-1}+2\left[\sup_{|x|=(1-\lambda)r} |x|\Phihf{(1-\lambda)r}(x)\right]_+\right)\int\rhohf\thetar^2\\ && -K_\lambda r^{-1}\int_{r<|x|<(1-\lambda)^{-1}r}\rhohf(x)dx. \end{eqnarray*} Finally, in order to arrive at the result of the lemma we simply use that $0\geq X^2-BX -C$ for $B,C>0$ implies $X\leq B+\sqrt{C}$. \end{proof} \section{The semiclassical estimates} \heading{Semiclassical estimates} \label{sec:sc} In this section we derive the relevant semiclassical estimates. We do not attempt to give optimal results. We shall be satisfied with what is needed for the application we have in mind. In a certain sense it is misleading to refer to the estimates in this section as {\it semiclassical}. Usually, semiclassics refers to the limit as Planck's constant $\hbar$ tends to zero. One then expands the relevant physical quantities like energy and density in powers of $\hbar$. In our setting there is, however, no semiclassical parameter which could play the role of Planck's constant. It is rather that we consider potentials for which the semiclassical expressions for the energy and density are approximately valid. We must then estimate the errors directly in terms of certain norms of the potential. The estimates are semiclassical in the sense that if one introduces a semiclassical parameter then the errors are of smaller order than the leading semiclassical expression. We are interested in a semiclassical approximation to the sum of the negative eigenvalues of a Schr\"odinger operator $$\index{$h_r$} h:=-\mfr{1}{2}\Delta-V, $$ on $\R^3$. We shall in this section always assume that the potential $V:\R^3\to\R$ is locally in $L^1$ and that its positive part satisfies that $[V]_+\in L^{5/2}(\R^3)$. This ensures (by the Lieb-Thirring inequality or even by Sobolev's inequality) that $h$ is bounded below and can be defined as a Friedrichs' extension from the domain $C^\infty_0(\R^3)$. \footnote{Note that we are not including spin. The operator $h$ is acting in the space $L^2(\R^3)$ and {\it not} in the space $L^2(\R^3;\C^2)$.} The semiclassical approximation to the sum of the negative eigenvalues of $h$ is given by \begin{equation}\label{eq:scenergy} (2\pi)^{-3}\iint_{\mfr{1}{2}p^2-V(x)\leq0}\mfr{1}{2}p^2-V(x)\ dp\, dx =-2^{3/2}(15\pi^2)^{-1}\int \left[V(x)\right]_+^{5/2} dx. \end{equation} Moreover, the semiclassical approximation to the density, i.e., the sum of the absolute square of the eigenfunctions corresponding to the negative eigenvalues, is \begin{equation}\label{eq:scdensity} (2\pi)^{-3}\int_{\mfr{1}{2}p^2-V(x)\leq0}1dp =2^{3/2}(6\pi^2)^{-1}\left[V(x)\right]_+^{3/2}. \end{equation} \begin{defi}\label{defi:g} For $s>0$, let $g:\R^3\to\R\index{$g$}$ be the ground state of the Dirichlet Laplacian for a ball of radius $s$, i.e., the function with $g(x)=0$ if $|x|>s$ and $g(x)=(2\pi s)^{-1/2} |x|^{-1}\sin(\pi|x|/s)$ if $|x|\leq s$. Then \begin{equation}\label{eq:gproperties} 0\leq g\leq1,\quad\int g^2=1\quad\int|\nabla g|^2=(\pi/s)^2. \end{equation} \end{defi} \begin{lm}[Semiclassical approximation]\label{lm:sc} We assume about the potential that $[V]_+,[V-V*g^2]_+\in L^{5/2}(\R^3)$ with $g$ as in Definition~\ref{defi:g} above . Let $e^{(1)}\leq e^{(2)}\leq\ldots<0$, denote the negative eigenvalues of $h=-\mfr{1}{2}\Delta-V$ as an operator on $L^2(\R^3)$. Then for all $0<\delta<1$, all integers $N>0$, and all $s>0$ we have \begin{eqnarray} \sum_{j=1}^Ne^{(j)}&\geq& -2^{3/2}(15\pi^2)^{-1}(1-\delta)^{-3/2}\int \left[V\right]_+^{5/2} \nonumber\\ &&-\mfr{1}{2}\pi^2s^{-2}N -L_1\delta^{-3/2}\left\|\left[V -V*g^2\right]_+\right\|_{5/2}^{5/2} \label{eq:sclower1} \end{eqnarray} where $L_1$ is the constant in the Lieb-Thirring estimate (\ref{eq:LT}). If there are fewer than $N$ negative eigenvalues the sum on the left refers simply to the sum over all the negative eigenvalues. On the other hand, if also $[V]_+\in L^{3/2}(\R^3)$, we can, for all $s>0$, find a density matrix $\gamma$ with $\uprho_\gamma(x)=2^{3/2}(6\pi^2)^{-1}\left[V\right]_+^{3/2}*g^2(x)$ such that \begin{eqnarray} \Tr[-\mfr{1}{2}\Delta\gamma]=2^{1/2}(5\pi^2)^{-1} \int \left[V\right]_+^{5/2}+\mfr{1}{2}\pi^2s^{-2}\int 2^{3/2}(6\pi^2)^{-1} \left[V\right]_+^{3/2}. \label{eq:scupper1}\end{eqnarray} \end{lm} \begin{remark} We are not proving that the true density of the projection onto the negative eigenvalues of $h$ is approximated by the semiclassical expression (\ref{eq:scdensity}). We only claim that there is a `good' trial density matrix. In the context where we shall use the semiclassics we shall infer the approximation of the true density by other means. \end{remark} \begin{proof}[Proof of Lemma~\ref{lm:sc}] We prove the result using the method of coherent states (see Thirring~\cite{Thirring81} and Lieb~\cite{Lieb:tf}). For $u,p\in\R^3$ let $\Pi_{u,p}\index{\Pi_{u,p}}$ be the one-dimensional projection in $L^2(\R^3)$, projecting onto the space spanned by the function $f_{u,p}(x):=\exp(ipx)g(x-u)\index{f_{u,p}}$. We then have the {\it coherent states identities} \begin{eqnarray} \Tr[\Pi_{u,p}]&=&1,\ \mbox{for all }p,u\label{eq:coherenttr}\\ (2\pi)^{-3}\iint\Pi_{u,p}dp\,du&=&\I,\ \mbox{on } L^2(\R^3). \label{eq:coherentint} \end{eqnarray} We also have the identity \begin{eqnarray} \Tr[-\mfr{1}{2}\Delta\Pi_{u,p}]&=&\mfr{1}{2}p^2+\mfr{1}{2}\int |\nabla g|^2 =\mfr{1}{2}p^2+\pi^2/(2s^2),\label{eq:PiDelta}\\ \noalign{\noindent and for all density matrices $\gamma$} \Tr[-\mfr{1}{2}\Delta\gamma]&=&(2\pi)^{-3}\iint \mfr{1}{2}p^2 \Tr[\Pi_{u,p}\gamma]dp\, du -\pi^2/(2s^2)\Tr[\gamma]\label{eq:gammaDelta}\\ \Tr[(V*g^2)\gamma]&=&(2\pi)^{-3}\iint V(u)\Tr[\Pi_{u,p}\gamma]dp\, du\,. \label{eq:gammaV} \end{eqnarray} {\it Proof of the lower bound (\ref{eq:sclower1}):\/} Let $f_1,f_2,\ldots\in L^2(\R^3)$ denote the normalized eigenfunctions of $h$ corresponding to negative eigenvalues. It is clear that we may without loss of generality assume that there are $N$ negative eigenvalues We decompose the operator $h$ as $h=-\mfr{1}{2}(1-\delta)\Delta-V*g^2 +[-\mfr{1}{2}\delta\Delta -(V-V*g^2)]$. We may then write $\sum_{j=1}^Ne^{(j)}=\sum_{j=1}^N(f_j,hf_j)={\cal A}+{\cal B}$, where $$ {\cal A}:=\sum_{j=1}^N(f_j,[-\mfr{1}{2}(1-\delta)\Delta-V*g^2]f_j), \quad {\cal B}:=\sum_{j=1}^N(f_j,[-\mfr{1}{2}\delta\Delta-(V -V*g^2)]f_j). $$ {F}rom (\ref{eq:gammaDelta}) and (\ref{eq:gammaV}) we hence obtain $$ {\cal A}=(2\pi)^{-3}\iint\left(\mfr{1}{2}(1-\delta)p^2-V(u)\right) \sum_{j=1}^N(f_j,\Pi_{u,p}f_j)dp\,du -N\pi^2(2s^2)^{-1}. $$ As a consequence of (\ref{eq:coherenttr}) we have that $0\leq\sum_{j=1}^N(f_j,\Pi_{u,p}f_j)\leq 1$. It is therefore clear that \begin{eqnarray*} {\cal A}&\geq& (2\pi)^{-3}\iint_{(1-\delta)\mfr{1}{2}p^2-V(u)\leq0} \left((1-\delta)\mfr{1}{2}p^2-V(u)\right)dp\,du-\pi^2(2s^{2})^{-1}N\\ &=&-2^{3/2}(15\pi^2)^{-1}(1-\delta)^{-3/2}\int \left[V\right]_+^{5/2} -\pi^2(2 s^2)^{-1}N. \end{eqnarray*} The estimate (\ref{eq:sclower1}) follows by applying the Lieb-Thirring estimate (\ref{eq:LT}) to conclude that $ {\cal B}\geq -L_1\delta^{-3/2}\|\left[V -V*g^2\right]_+\|_{5/2}^{5/2}. $ {\it Proof of the existence of $\gamma$.\/} We shall prove that $$ \gamma:= (2\pi)^{-3}\iint_{\mfr{1}{2}p^2-V(u)\leq 0} \Pi_{u,p} dp\,du $$ has the desired properties. {F}rom (\ref{eq:coherentint}) we see that $\gamma$ is a density matrix, i.e., $0\leq \gamma\leq \I$. The density corresponding to $\gamma$ is easily computable $$ \uprho_\gamma(x)=\gamma(x,x)=(2\pi)^{-3} \iint_{\mfr{1}{2}p^2-V(u)\leq 0}\Pi_{u,p}(x,x) dp\,du =2^{3/2}(6\pi^2)^{-1}\left[V\right]_+^{3/2}*g^2(x). $$ {F}rom (\ref{eq:PiDelta}) we immediately obtain (\ref{eq:scupper1}). \end{proof} Although we shall use the semiclassical approximation in the form given in the lemma we shall for completeness state a less technical semiclassical result which follows very easily from the lemma. \begin{thm}[Semiclassical approximation]\label{thm:sc} Assume that $0\leq V\in L^{5/2}(\R^3)\cap L^{3/2}(\R^3)$ and $|\nabla V|\in L^{5/2}(R^3)$. Let $e^{(1)}\leq e^{(2)}\leq\ldots<0$, denote the negative eigenvalues of $h=-\mfr{1}{2}\Delta-V$ as an operator on $L^2(\R^3)$. Then we have \begin{eqnarray} \sum_je^{(j)}&\geq& -2^{3/2}(15\pi^2)^{-1}\int V(x)^{5/2} dx \left\{1+A_{\rm L}\|V\|_{5/2}^{-3/2}\|\nabla V\|_{5/2}^{2/3} \|V\|_{3/2}^{1/2}\right\}^{5/3}\label{eq:scthmlower1}\\ \noalign{\hbox{and}} \sum_je^{(j)}&\leq& -2^{3/2}(15\pi^2)^{-1}\int V(x)^{5/2} dx+\mfr{3}{2}\pi^{2/3} \|V\|_{5/2}\|\nabla V\|_{5/2}^{2/3}\|V\|_{3/2}^{1/2} \label{eq:scthmupper1} \end{eqnarray} where $A_{\rm L}:=\frac{9}{4}2^{-9/10}(15\pi^2)^{3/5} \left(\frac{2\pi^2}{5}\right)^{1/3}L_0^{1/3}L_1^{4/15}$. Here $L_0$ and $L_1$ are the constants in the CLR and Lieb-Thirring estimates (\ref{eq:CLR}) and (\ref{eq:LT}) respectively. \end{thm} \begin{proof} We may estimate \begin{eqnarray*} \left|V(u)-V*g^2(u)\right|&\leq& \int_{\R^3}\int_0^1|\nabla V(u-ty)||y|g(y)^2dt\,dy\\ &=&\int_{\R^3}|\nabla V(u-y)||y|\int_0^1t^{-4}g(y/t)^2dtdy\\ &\leq& (4\pi)^{-1}\int_{|y|\leq s}|\nabla V(u-y)||y|^{-2} dy, \end{eqnarray*} where we have used that $\int_0^1t^{-4}g(y/t)^2dt \leq |y|^{-3}\int_0^\infty t^2 g(t)^2dt =(4\pi)^{-1}|y|^{-3}$ (identifying $g$ with a function on $\R_+$). Hence \begin{equation}\label{eq:gapproximation} \|V-V*g^2\|_{5/2}\leq (4\pi)^{-1}\|\nabla V\|_{5/2}\int_{|y|\leq s}|y|^{-2}dy =s\|\nabla V\|_{5/2}. \end{equation} For any density matrix $\gamma$, $\Tr[h\gamma]$ is an upper bound to the sum of the negative eigenvalues of $h$. {F}rom (\ref{eq:scupper1}) we find for the density matrix constructed in Lemma~\ref{lm:sc} that \begin{eqnarray*} \Tr[h\gamma]&=&-2^{3/2}(15\pi^2)^{-1}\int \left[V\right]_+^{5/2}\\ &&{}+2^{3/2}(6\pi^2)^{-1}\int \left[V(u)\right]_+^{3/2} \left[V(u)-V*g^2(u)+ \mfr{1}{2}\pi^2s^{-2}\right]du. \end{eqnarray*} The bound (\ref{eq:scthmupper1}) follows from applying H\"older's inequality, (\ref{eq:gapproximation}), and optimizing in $s$. By the CLR bound (\ref{eq:CLR}) we know that $h$ has only finitely many negative eigenvalues and that their number $N$ is bounded by $N\leq L_0\int \left[V\right]_+^{3/2}$. {F}rom (\ref{eq:sclower1}) and (\ref{eq:gapproximation}) we therefore obtain \begin{eqnarray*} \sum_je^{(j)} &\geq& -2^{3/2}(15\pi^2)^{-1}(1-\delta)^{-3/2}\int V^{5/2}\\ &&{} -L_0\pi^2(2 s^2)^{-1}\int V^{3/2} -L_1\delta^{-3/2}s^{5/2}\int|\nabla V|^{5/2}\\ &=& -2^{3/2}(15\pi^2)^{-1}(1-\delta)^{-3/2}\int V^{5/2}\\ &&{} -\mfr{9}{4}\left(\mfr{2}{5}\right)^{5/9} L_0^{5/9}L_1^{4/9}\pi^{10/9}\delta^{-2/3} \left(\int V^{3/2}\right)^{5/9}\left(\int|\nabla V|^{5/2}\right)^{4/9}, \end{eqnarray*} where we have optimized in the parameter $s$. We now optimize in the parameter $\delta$. Define $\delta'$ by $(1-\delta)^{-3/2}=(1-\delta')^{-2/3}$. (Note that $0<\delta<1$ if and only if $0<\delta'<1$.) Then $\delta^{-2/3}\leq (4\delta'/9)^{-2/3}$. Thus $$ \sum_je^{(j)} \geq-2^{3/2}(15\pi^2)^{-1}(1-\delta')^{-2/3}\int V^{5/2} -A_1\delta'^{-2/3}\left(\int V^{3/2}\right)^{5/9} \left(\int|\nabla V|^{5/2}\right)^{4/9}, $$ where $ A_1:=\left(\mfr{9}{4}\right)^{5/3}\left(\mfr{2\pi^2}{5}\right)^{5/9} L_0^{5/9}L_1^{4/9}. $ Using that $\min_{\delta'}\left[(1-\delta')^{-2/3}a+\delta'^{-2/3}b\right] = a[1+(b/a)^{3/5}]^{5/3}$ we arrive at (\ref{eq:scthmlower1}). \end{proof} We shall need the semiclassical estimates also for the operator $h$ restricted to functions on the set $\set{x}{|x|\geq r}$ satisfying Dirichlet boundary conditions. \begin{lm}[Dirichlet Boundary conditions]\label{lm:DBC} Let the assumptions be as in the beginning of Lemma~\ref{lm:sc}. For $r>0$ let $h_r$ denote the restriction of the operator $h=-\mfr{1}{2}\Delta-V$ to functions on the set $\set{x}{|x|\geq r}$ satisfying Dirichlet boundary conditions. Denote by $e^{(1)}\leq e^{(2)}\leq \ldots<0$ the negative eigenvalues of $h$ and by $e^{(1)}_r\leq e^{(2)}_r\leq \ldots<0$ the negative eigenvalues of $h_r$. Then $ \sum_je^{(j)}\leq \sum_je^{(j)}_r. $ Moreover, if $\gamma$ is a density matrix on $L^2(\R^3)$ we may, for all $0<\lambda<1$, find a density matrix $\widetilde\gamma$ such that $\uprho_{\widetilde\gamma}$ is supported in $\set{x}{|x|\geq r}$ and $\uprho_{\widetilde\gamma}\leq\uprho_\gamma$ and $$ \Tr[h_r\widetilde\gamma]\leq \Tr[h\gamma] +L_1\int_{|x|\leq(1-\lambda)^{-1}r} \left[V\right]_+^{5/2} +\mfr{1}{2}({\pi}/{(2\lambda r)})^2 \int_{|x|\leq(1-\lambda)^{-1}r}\uprho_\gamma. $$ \end{lm} \begin{proof} That the Dirichlet eigenvalues are upper bounds to the eigenvalues on $\R^3$, is a well known simple consequence of the variational principle. Let $\thetar$ be the localization function from Definition~\ref{defi:localization}. We shall choose $\widetilde\gamma =\thetar\gamma\thetar$. Then by the IMS formula (\ref{eq:IMS}) we have \begin{eqnarray*} \Tr[h\gamma]&=&\Tr[h_r\thetar\gamma\thetar] +\Tr[h(1-\thetar^2)^{1/2}\gamma(1-\thetar^2)^{1/2}]\\&&{} -\mfr{1}{2}\Tr[\left((\nabla\thetar)^2+ (\nabla(1-\thetar^2)^{1/2})^2\right)\gamma]. \end{eqnarray*} By the Lieb-Thirring inequality (\ref{eq:LT}) we have $\Tr[h(1-\thetar^2)^{1/2}\gamma(1-\thetar^2)^{1/2}] \geq -L_1\int_{|x|\leq(1-\lambda)^{-1}r} \left[V\right]_+^{5/2}$. Thus the lemma follows from the bound on the gradient of $\thetar$ and $(1-\thetar^2)^{1/2}$ given in Definition~\ref{defi:localization}. \end{proof} \section{The Coulomb norm estimates} \heading{Coulomb norm estimates} \label{sec:cn} In this section we introduce and study the Coulomb norm. \begin{defi}\label{defi:cn} For $f,g\in L^{6/5}(\R^3)$ we define the {\it Coulomb inner product} \begin{equation}\label{eq:Coulombinner} D(f,g):=\mfr{1}{2} \iint f(x) |x-y|^{-1} \overline{g(y)} dx\,dy \end{equation} and the corresponding {\it Coulomb norm}, $$ \|g\|_\rC:=D(g,g)^{1/2}. $$ \end{defi} By the Hardy-Littlewood-Sobolev estimate we have \begin{equation}\label{eq:HLS} \|g\|_\rC\leq \pi^{-1/6}2^{1/3}3^{-1/2}\|g\|_{6/5}. \end{equation} In this sharp form the inequality was proved by Lieb~\cite{Lieb:sob}. Using the Fourier transform we may write \begin{equation}\label{eq:Dfourier} D(f,g) =(2\pi)\int \hat{f}(p)\overline{\hat{g}(p)}|p|^{-2} dp, \end{equation} from which it follows that the Coulomb norm really is a norm on $L^{6/5}(\R^3)$. The following estimate was first used in the context of atomic problems by Fefferman and Seco~\cite{Fef-Sec}. \begin{lm}[Coulomb norm estimate]\label{lm:cn} If $f\in L^6(\R^3)$ with $|\nabla f|\in L^2(\R^3)$ and $g\in L^{6/5}(\R^3)$ then $$ \left|\int f\overline g\right|\leq (2\pi)^{-1/2}\|\nabla f\|_2\|g\|_\rC. $$ \end{lm} \begin{proof} Using Plancherel's identity and the representation (\ref{eq:Dfourier}) we have $$ \left|\int f\overline g\right| =\left|\int \hat{f}\overline{\hat{g}}\right| \leq \||p|\hat{f}(p)\|_2\||p|^{-1}\hat{g}(p)\|_2 =(2\pi)^{-1/2}\|\nabla f\|_2\|g\|_\rC. $$ \end{proof} We shall next give some simple but very useful consequences of this estimate. \begin{cl} \label{cl:cn} Consider $f\in L^{5/3}(\R^3)\cap L^{6/5}(\R^3)$. For all $x\in\R^3$ and $s>0$ we have \begin{equation}\label{eq:cn1} f*|x|^{-1} \leq (5/4)^{2/5}(\pi s)^{1/5} \|[f]_+\|_{L^{5/3}(\B{x,s})}+ (4\pi)^{1/2}s^{-1/2}\|f\|_\rC. \end{equation} For all $x\in\R^3$ and all $\kappa>0$ denote by $A(|x|,\kappa)$ the annulus $$ A(|x|,\kappa)= \set{y}{(1-2\kappa)|x|\leq|y|\leq|x|} $$ we then have \begin{eqnarray} \int_{|y|<|x|}f(y)|x-y|^{-1}dy&\leq& 2^{7/5}\pi^{2/5}(\kappa|x|)^{1/5} \left\|[f]_+\right\|_{L^{5/3}\left(A(|x|,\kappa)\right)} \nonumber\\&&{} +4\pi^{1/2}\kappa^{-1}|x|^{-1/2} \|f\|_\rC. \label{eq:cn2} \end{eqnarray} \end{cl} \begin{remark} Note that we do not restrict to $\kappa\leq1/2$. We do this to avoid having to check this condition in the applications of the corollary. \end{remark} \begin{proof}[Proof of Corollary~\ref{cl:cn}] Consider the function $\xi_s:\R^3\to\R$ defined by $$ \xi_s(z):=\left\{\begin{array}{cl} s^{-1},&\mbox{if }|z|\leq s\\ |z|^{-1},&\mbox{if }|z|\geq s\end{array}\right.. $$ It satisfies $\|\nabla \xi_s\|_2\leq(4\pi /s)^{1/2}$. Hence from Lemma~\ref{lm:cn} we obtain \begin{eqnarray*} f*|x|^{-1}&\leq& \int_{|y-x|\leq s}[f(y)]_+\left(|x-y|^{-1} -s^{-1}\right)dy +\int_{\R^3} f(y)\xi_s(x-y)dy\\ &\leq& (5/4)^{2/5}(\pi s)^{1/5}\|[f]_+\|_{L^{5/3}(\B{x,s})} +(4\pi /s)^{1/2}\|f\|_\rC, \end{eqnarray*} where we have used that $\int_{|y|<1}(|y|^{-1}-1)^{5/2}dy=\frac{5\pi^2}{4}$. In order to prove the second half of the corollary we introduce the function $\Xi_{x,\kappa}:\R^3\to\R^3$ given by $$ \Xi_{x,\kappa}(z):=\left\{\begin{array}{cl} 1,&\mbox{if }|z|\leq (1-2\kappa)|x|\\ 1-(|x|\kappa)^{-1}(|z|-|x|(1-2\kappa)), &\mbox{if }(1-2\kappa)|x|\leq|z|\leq(1-\kappa)|x|\\ 0,&\mbox{if }(1-\kappa)|x|\leq|z|\end{array}\right.. $$ Then $|\Xi_{x,\kappa}(z)|\leq1$ and we can estimate \begin{eqnarray*} \int_{|y|<|x|}f(y)|x-y|^{-1}dy&\leq& \int_{A(|x|,\kappa)}[f(y)]_+|x-y|^{-1}dy\\ &&{}+\int_{\R^3}f(y)|x-y|^{-1}\Xi_{x,\kappa}(y)dy\\ &\leq& \|[f]_+\|_{L^{5/3}\left(A(|x|,\kappa)\right)} \left(\int_{A(|x|,\kappa)}|x-y|^{-5/2}dy\right)^{2/5}\\ &&{}+\|f\|_\rC\left(\int\left|\nabla_y \left(|x-y|^{-1}\Xi_{x,\kappa}(y)\right)\right|^2dy\right)^{1/2}. \end{eqnarray*} It remains to estimate the two integrals above. For the first integral we find for $\kappa\leq 1/2$ \begin{eqnarray*} \int_{A(|x|,\kappa)}|x-y|^{-5/2}dy &=&2\pi|x|^{1/2}\int_{r=1-2\kappa}^1\int_{u=-1}^1 (1-2ru+r^2)^{-5/4}r^2du\,dr\\ &=&2\pi|x|^{1/2}\alpha_1(\kappa), \end{eqnarray*} where $$ \alpha_1(\kappa):=4(2\kappa)^{1/2}-(4/3)(2\kappa)^{3/2} +(4/3)\sqrt{2}[1-(1-\kappa)^{1/2}(1+2\kappa)] \leq 4(2\kappa)^{1/2}. $$ The last inequality follows from a straightforward careful analysis of $\alpha_1(\kappa)$. For the second integral we get \begin{eqnarray*} \left(\int\left|\nabla_y \left(|x-y|^{-1}\Xi_{x,\kappa}(y)\right)\right|^2dy\right)^{1/2} &\leq&\left(\int_{|x-y|>\kappa|x|}|x-y|^{-4}dy\right)^{1/2}\\ &&{} +(\kappa|x|)^{-1}\left(\int_{A(|x|,\kappa)}|x-y|^{-2} dy\right)^{1/2}\\ &=&(4\pi)^{1/2}(\kappa|x|)^{-1/2}(1+\alpha_2(\kappa)^{1/2}), \end{eqnarray*} with $$ \alpha_2(\kappa): =1+(1-\kappa)\ln\left(\frac{1-\kappa}{\kappa}\right)\leq 1-\ln\kappa \leq\kappa^{-1}, $$ where we have used that $$ \int_{A(|x|,\kappa)}|x-y|^{-2} dy =2\pi|x|\int_{r=1-2\kappa}^1\int_{u=-1}^1 (1-2ru+r^2)^{-1}r^2 du\,dr=4\pi\kappa |x|\alpha_2(\kappa). $$ Using $ (4\pi)^{1/2}\kappa^{-1/2}(1+\alpha_2(\kappa)^{1/2}) \leq 4\pi^{1/2}\kappa^{-1} $ we get (\ref{eq:cn2}). The estimate holds also for $\kappa>1/2$ since the last term in (\ref{eq:cn2}) can be ignored in this case. \end{proof} \section{Main Estimate} \heading{Main estimates}\label{sec:me} We now restrict attention to the case $N\geq Z$. Throughout the remaining part of this paper $\rhotf,\phitf,\Phitf{R},\rhohf, \phihf$ and $\Phihf{R}$ always refer to the problems with particle number $N$. In fact, since $N\geq Z$ the TF functions correspond to the neutral atom, i.e., $\mutf=0$. We shall suppress the dependence on $N$ everywhere since it is held fixed throughout the discussion. From now on we shall no longer explicitly compute the constants involved in the estimates. We shall use the notation $\const$ to refer to any universal (in principle explicitly computable) {\it positive} constant. Thus $\const$ does not mean the same constant in all equations or inequalities. Even within the same equation we shall use the notation $\const$ to refer to possibly different universal constants. Universal constants of particular importance will be given separate names. We begin by stating the main result of this section. \begin{thm}[Main Estimate]\label{thm:me} Assume $Z\geq 1$ and $N\geq Z$. There exist universal constants $0<\epsilon<4$ and $\CM,\CPhi>0$ such that for all $x\in\R^3$ we have \begin{equation}\label{eq:me1} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| \leq \CPhi |x|^{-4+\varepsilon}+\CM. \end{equation} \end{thm} We shall prove Theorem~\ref{thm:me} by an iterative procedure. The first step is to control ``small'' $x$. \begin{lm}[Control of region close to the nucleus]\label{lm:smallx} \hfill\\ Assume $Z\geq1$ and $N\geq Z$. For all $\beta>0$ and all $|x|\leq \beta Z^{-1/3}$ we have $$ \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| \leq A_\Phi \beta^{49/12-\varepsilon_1} |x|^{-4+\varepsilon_1}, $$ where $\varepsilon_1=1/66$ and $A_\Phi>0$ is a universal constant. \end{lm} \begin{lm}[Iterative step]\label{lm:iterate} Assume $N\geq Z$. For all $\delta>0$ smaller than some universal constant there exist constants $\epsilon_2,C'_\Phi>0 $ depending only on $\delta$ such that, given $\varepsilon',{\sigma}>0$, there exists a constant $D=D(\varepsilon',{\sigma})>0$ depending only on $\varepsilon',{\sigma}$ with the following property. For all $R_0R_0$ such that \begin{equation}\label{eq:itphi} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| \leq C'_\Phi |x|^{-4+\varepsilon_2} \end{equation} for all $x$ with $R_0<|x|0$ such that if (\ref{eq:phiasp}) holds for all $|x|\leq D$ then we have \begin{equation} \int_{|y|>|x|}\rhohf(y)^{5/3}\,dy\leq \const |x|^{-7},\label{eq:me5/3} \end{equation} for all $|x|\leq D$. \end{lm} \begin{lm}[Bound on $\int\rhohf$]\label{lm:meN} Assume that (\ref{eq:phiasp}) holds for all $|x|\leq R$ for some $R>0$ and some $\epsilon',\sigma>0$. Then for $0r}\rhohf(y)dy\leq\const(1+\sigma r^{\epsilon'}) \left(r^{-3}+1\right). $$ \end{lm} We shall prove Lemma~\ref{lm:smallx} in Sect.~\ref{sec:smallx} and Lemmas~\ref{lm:iterate} and \ref{lm:me5/3} in Sect.~\ref{sec:iteration}. We end this section with the proofs of Lemma~\ref{lm:meN} and the main estimate Theorem~\ref{thm:me}. \begin{proof}[Proof of Lemma~\ref{lm:meN}] First note that for $0r/2}\rhotf(y)\,dy\leq\const r^{-3}$$ and hence $$ \int_{r/2<|y|r}\rhohf(y)dy$. \end{proof} We finally show how to use Lemmas~\ref{lm:smallx}--\ref{lm:meN} to prove the main estimate Theorem~\ref{thm:me}. \begin{proof}[Proof of Theorem~\ref{thm:me}] We first show that we may choose $\delta>0$ small enough such that if we choose $\widetilde{R}^{1+\delta}=\beta_0 Z^{-1/3}$ we have for all $|x|<\widetilde{R}$ that \begin{equation}\label{eq:itphi0} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| \leq C''_\Phi |x|^{-4+\frac{\epsilon_1}{2}} \end{equation} for a universal constant $C''_\Phi>0$ and with $\epsilon_1$ given in Lemma~\ref{lm:smallx}. To see this let $\beta>0$ be such that $\left(\beta Z^{-1/3}\right)^{1+\delta}=\beta_0Z^{-1/3}$, i.e., $\beta^{1+\delta}=\beta_0Z^{\delta/3}$. We then see from Lemma~\ref{lm:smallx} that for all $|x|<\beta Z^{-1/3}$ we have \begin{eqnarray*} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| &\leq& A_\Phi \beta^{\frac{49}{12}-\frac{\epsilon_1}{2}} Z^{-\frac{\epsilon_1}{6}}|x|^{-4+\frac{\epsilon_1}{2}}\\ &=&A_\Phi (\beta_0Z^{\delta/3})^{\left(\frac{49}{12(1+\delta)} -\frac{\epsilon_1}{2(1+\delta)}\right)} Z^{-\frac{\epsilon_1}{6}}|x|^{-4+\frac{\epsilon_1}{2}}. \end{eqnarray*} Since $\widetilde{R}^{1+\delta}=\beta_0 Z^{-1/3}$, i.e., $\widetilde{R}=\beta Z^{-1/3}$ we see that if $\delta$ is small enough and $C''_\Phi$ is chosen appropriately then (\ref{eq:itphi0}) holds for all $|x|<\widetilde{R}$. We now assume that $\delta$ is also small enough that we may apply Lemma~\ref{lm:iterate}. This gives us constants $\epsilon_2,C'_\Phi>0$ and for all $\sigma,\epsilon'>0$ a $D>0$ with the properties stated in Lemmas~\ref{lm:iterate} and \ref{lm:me5/3}. We may without loss of generality assume that $D\leq1$. Now choose $\sigma=\max\{C'_\Phi,C''_\Phi\}$ and $\epsilon'=\min\{\epsilon_1/2,\epsilon_2\}$. Note that $\sigma,\epsilon$, and $D$ are now universal constants. We shall prove that for all $|x|\leq D$ we have \begin{equation} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| \leq \sigma |x|^{-4+\varepsilon'}\label{eq:me1-CM}. \end{equation} Since we are assuming that $D\leq 1$ it is sufficient to prove (\ref{eq:me1-CM}) with $\epsilon'$ replaced by $\epsilon_1/2$ or $\epsilon_2$. We have to prove that $D$ belongs to the set $$ {\cal M}=\set{0R_0:=\sup {\cal M}$. In order to reach a contradiction we therefore assume this and hence in particular that $R_0<1$. {F}rom (\ref{eq:itphi0}) it follows that either $\widetilde{R}>1$ or $\widetilde{R}\in{\cal M}$. If $\widetilde{R}>1$ then $R_0=\sup {\cal M}=1$ which contradicts our assumption. On the other hand if $\widetilde{R}\in{\cal M}$ then $R_0^{1+\delta}\geq \widetilde{R}^{1+\delta}=\beta_0 Z^{-1/3}$. It is then an immediate consequence of Lemma~\ref{lm:iterate} that there exists $R_0'\in{\cal M}$ with $R_0'>R_0$ and this is of course also a contradiction. This establishes an inequality of the form (\ref{eq:me1}) for all $|x|\leq D$. We shall now prove (\ref{eq:me1}) for $|x|>D$. We write $$ \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right|\leq \left|\Phihf{D}(x)-\Phitf{D}(x)\right|+ \left|\int_{D<|y|<|x|}\frac{\left(\rhotf(y)-\rhohf(y)\right)}{|x-y|}\, dy\right|. $$ We shall estimate the last term using Lemma~\ref{lm:me5/3} and the similar bound $$ \int_{|y|>|x|}\rhotf(y)^{5/3}\,dy\leq \const |x|^{-7}, $$ which holds for all $x$ by the Sommerfeld estimate Theorem~\ref{thm:sommerfeldup} and the TF equation (\ref{eq:tfsystemrho}). Hence using H\"older's inequality we have \begin{eqnarray*} \left|\,\,\int\limits_{D<|y|<|x|} \frac{\left(\rhotf(y)-\rhohf(y)\right)}{|x-y|}dy\right| &\leq& \const D^{-21/5} \left(\int_{|x-y|D}\left(\rhotf(y)+\rhohf(y)\right)dy. \end{eqnarray*} By Lemma~\ref{lm:meN} and the bound $\int_{|y|>D}\rhotf(y)\,dy\leq\const D^{-3}$, which is again a consequence of the Sommerfeld estimate Theorem~\ref{thm:sommerfeldup} and the TF equation (\ref{eq:tfsystemrho}), we see that this last expression is bounded by a universal constant. Since $\Phihf{D}(x)-\Phitf{D}(x)$ is harmonic for $|x|>D$ and tends to zero at infinity we have for all $|x|>D$ that $$ \left|\Phihf{D}(x)-\Phitf{D}(x)\right| \leq \sup_{|z|=D}\left|\Phihf{D}(z)-\Phitf{D}(z)\right| \leq \sigma D^{-4+\varepsilon'}, $$ which is also bounded by a universal constant. Thus (\ref{eq:me1}) holds for all $x$. \end{proof} \section{Control of region close to nucleus: proof of Lemma~\ref{lm:smallx} } \heading{Region close to nucleus}\label{sec:smallx} In order to prove Lemma~\ref{lm:smallx} we need some basic estimates. \begin{lm}[Global $L^{5/3}$ and Coulomb norm estimates] \label{lm:5/3cn}\hfill\\ For all $N$ and $Z$ we have the bound \begin{equation} \int_{\R^3}\rhohf(y)^{5/3}dy\leq\const Z^{7/3}. \label{eq:5/3} \end{equation} Moreover, if $Z\geq 1$ \begin{equation} \|\rhohf-\rhotf\|_\rC^2\leq \const Z^{7(1-\varepsilon_3)/3}\label{eq:||C}, \end{equation} with $\varepsilon_3:=2/77$. \end{lm} \begin{proof} Although we shall only use this result for $N\geq Z$ the proof is almost as easy without this restriction, so we treat the more general case here. We first estimate the $L^{5/3}$ norm of $\rhohf$. It is easy to see that $\ehf(\gammahf)\leq0$. Thus since $\D(\gammahf)-\Ex(\gammahf)\geq0$ we have from the Lieb-Thirring inequality (\ref{eq:LTdensity}) that $$ 0\geq\ehf(\gammahf)\geq \Tr\left[\left(-\mfr{1}{2}\Delta-Z|y|^{-1}\right)\gammahf\right] \geq \int \left(K_1\rhohf(y)^{5/3}-Z|y|^{-1}\rhohf(y)\right)dy. $$ If we use that $\int\rhohf=N$ and the inequality $ab\leq \mfr{3}{5}a^{5/3}+\mfr{2}{5}b^{5/2}$ we get for all $\delta>0$ and all $r>0$, $$ 0\geq \int (K_1-\mfr{3}{5}\delta^{5/3})\rhohf(y)^{5/3}dy- \mfr{2}{5}\int_{|y|0$. Recall that the function $g$ was given in Definition~\ref{defi:g}. Here we have used the semiclassical estimate for the space $L^2(\R^3;\C^2)$. The estimate above therefore has an extra factor of $2$ in the first and the last term compared to (\ref{eq:sclower1}). Thus \begin{eqnarray*} \lefteqn{\ehf(\gammahf)\geq-2^{5/2}(15\pi^2)^{-1}(1-\delta)^{-3/2}\int \left[\phitf-\mutf\right]_+^{5/2} -\mutf N -D(\rhotf,\rhotf)}&&\nonumber \\ &&{} +\|\rhotf-\rhohf\|_{\rC}^2-\mfr{1}{2}\pi^2s^{-2}N -2L_1\delta^{-3/2} \left\|\left[\phitf-\phitf*g^2\right]_+\right\|_{5/2}^{5/2} -\Ex(\gammahf). \end{eqnarray*} Since $|y|^{-1}-g^2*|y|^{-1}\geq0$ (because the function $|y|^{-1}$ is superharmonic) we have $$ \left\|\left[\phitf-\phitf*g^2\right]_+\right\|_{5/2}^{5/2}\leq Z\left\||y|^{-1}-g^2*|y|^{-1}\right\|_{5/2}^{5/2}\leq 8\pi Z^{5/2} s^{1/2}, $$ where we have used that $|y|^{-1}-g^2*|y|^{-1}$ is non-negative, bounded by $|y|^{-1}$, and vanishes for $|y|>s$. If we insert this above and optimize in $s$ we obtain \begin{eqnarray} \ehf(\gammahf)&\geq&-2^{5/2}(15\pi^2)^{-1}(1-\delta)^{-3/2}\int \left[\phitf-\mutf\right]_+^{5/2} -\mutf N -D(\rhotf,\rhotf)\nonumber \\ &&{} +\|\rhotf-\rhohf\|_{\rC}^2 -\const\delta^{-6/5}N^{1/5}Z^2 -\Ex(\gammahf). \end{eqnarray} We choose $\delta:=\mfr{1}{2}Z^{-2/33}$ (this is not optimal). Then for $Z\geq1$ we have $\delta\leq 1/2$ and thus $(1-\delta)^{-2/3}\leq 1+(2^{5/2}-2)\delta$. Hence \begin{eqnarray} \ehf(\gammahf)\geq-2^{5/2}(15\pi^2)^{-1}\int \left[\phitf-\mutf\right]_+^{5/2} -\mutf N -D(\rhotf,\rhotf)\nonumber \\ +\|\rhotf-\rhohf\|_{\rC}^2- \const Z^{7/3-2/33} -\Ex(\gammahf)\label{eq:hflower}, \end{eqnarray} where we have again used $N\leq 2Z+1$ and the fact that by (\ref{eq:5/3}) we have $$ 2^{5/2}(15\pi^2)^{-1}\int \left[\phitf-\mutf\right]_+^{5/2}=\mfr{2}{3}\left[ (3\pi^2)^{2/3}\mfr{3}{10}\int\left(\rhotf\right)^{5/3}\right] \leq\const Z^{7/3}. $$ On the other hand, since $\gammahf$ minimizes $\ehf$ among all density matrices $\gamma$ with $\Tr[\gamma]\leq N$, we can find an upper bound to $\ehf(\gammahf)$ by choosing an appropriate trial density matrix. We choose as trial matrix $\gamma$ an operator which acts identically on the two spin components. On each spin component we choose it to be the density matrix constructed in Lemma~\ref{lm:sc} satisfying (\ref{eq:scupper1}) with $V=\phitf-\mutf$. Note that the Thomas-Fermi equation (\ref{eq:tfsystemrho}) and the properties of $\gamma$ stated in Lemma~\ref{lm:sc} imply that $\uprho_\gamma=2^{5/2}(6\pi^2)^{-1}\left[\phitf-\mutf\right]_+^{3/2}*g^2=\rhotf*g^2$ (the extra factor of $2$ compared to Lemma~\ref{lm:sc} is of course due to the spin degeneracy). Thus $\Tr[\gamma]=\int \rhotf\leq N$. {F}rom (\ref{eq:ehf}) and (\ref{eq:scupper1}) we find, since $\Ex(\gamma)\geq0$, that $$ \ehf(\gamma)\leq 2^{3/2}(5\pi^2)^{-1} \int \left[\phitf-\mutf\right]_+^{5/2}+\mfr{1}{2}\pi^2s^{-2}N -\int Z |y|^{-1}\uprho_\gamma(y)dy +D(\uprho_\gamma,\uprho_\gamma) $$ Since $\iint g(x-z)^2|z-w|^{-1}g(y-w)^2dw\,dz\leq |x-y|^{-1}$ we see that $D(\uprho_\gamma,\uprho_\gamma)\leq D(\rhotf,\rhotf)$. Thus from the definition (\ref{eq:phitf}) of $\phitf$ we can write \begin{eqnarray*} \ehf(\gamma)&\leq& 2^{3/2}(5\pi^2)^{-1} \int \left[\phitf-\mutf\right]_+^{5/2} -\int \left[\phitf(y)-\mutf\right]\rhotf(y)dy- \mutf N\nonumber \\ &&{} -D(\rhotf,\rhotf) +\mfr{1}{2}\pi^2s^{-2}N+\int Z \left(|y|^{-1}-g^2*|y|^{-1}\right) \rhotf(y)dy \end{eqnarray*} If we use the TF equation (\ref{eq:tfsystemrho}), the estimate $\rhotf(y)\leq 2^{3/2}(3\pi^2)^{-1}Z^{3/2}|y|^{-3/2}$ which follows from the TF equation, and again the facts that $|y|^{-1}-g^2*|y|^{-1}$ is non-negative, bounded by $|y|^{-1}$, and vanishes for $|y|>s$, we obtain after optimizing in $s$ \begin{eqnarray} \ehf(\gamma)&\leq&-2^{5/2}(15\pi^2)^{-1}\int \left[\phitf-\mutf\right]_+^{5/2} -\mutf N -D(\rhotf,\rhotf)\nonumber \nonumber\\ &&{} +\const N^{1/5}Z^2.\label{eq:hfupper} \end{eqnarray} Comparing (\ref{eq:hflower}) and (\ref{eq:hfupper}) and recalling that $\ehf(\gamma)\geq\ehf(\gammahf)$ we get that $$ \|\rhotf-\rhohf\|_{\rC}^2\leq \const Z^{7/3-2/15} +\const Z^{7/3-2/33}+\Ex(\gammahf). $$ If we finally use the exchange inequality in Theorem ~\ref{thm:exchineq} and the estimate (\ref{eq:5/3}) we see that $$ \Ex(\gammahf)\leq 1.68\left(\int\left(\rhohf\right)^{5/3} \right)^{1/2} \left(\int\left(\rhohf\right)\right)^{1/2} \leq \const N^{1/2}Z^{7/6}. $$ Inserting this above and again using $N\leq 2Z+1$ we arrive at (\ref{eq:||C}). \end{proof} \begin{proof}[End of proof of Lemma~\ref{lm:smallx}] We write $$ \Phihf{|x|}(x)-\Phitf{|x|}(x) =\int_{|y|<|x|}\left[\rhotf(y)-\rhohf(y)\right]|x-y|^{-1}dy. $$ Using the Coulomb norm estimate (\ref{eq:cn2}) we find \begin{eqnarray*} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| &\leq& 2^{7/5}\pi^{2/5}(\kappa |x|)^{1/5} \max\left\{\|\rhotf\|_{L^{5/3}(\R^3)}, \,\|\rhohf\|_{L^{5/3}(\R^3)}\right\} \\ &&{} +4\pi^{1/2}\kappa^{-1}|x|^{-1/2}\|\rhohf-\rhotf\|_\rC. \end{eqnarray*} Thus from Lemma~\ref{lm:5/3cn} and the fact that $\int\left(\rhotf\right)^{5/3} \leq\const Z^{7/3}$ (which can be seen for instance from the Sommerfeld estimate Theorem~\ref{thm:sommerfeldup} together with the TF equation (\ref{eq:tfsystemrho})) we obtain \begin{eqnarray*} \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| &\leq& \const (\kappa |x|)^{1/5}Z^{7/5} \\&&{}+\const \kappa^{-1}|x|^{-1/2} Z^{7(1-\varepsilon_3)/36} =\const |x|^{1/12} Z^{7/6+7(1-\varepsilon_3)/36}, \end{eqnarray*} where the last equality above follows from choosing the optimal value for $\kappa$. Hence if $|x|\leq \beta Z^{-1/3}$ we have $$ \left|\Phihf{|x|}(x)-\Phitf{|x|}(x)\right| \leq A_{\Phi} \beta^{49/12-7\varepsilon_3/12} |x|^{-4+7\varepsilon_3/12}. $$ The lemma follows since $7\varepsilon_3/12=1/66$. \end{proof} \section{Proof of the iterative step Lemma~\ref{lm:iterate} and of Lemma~\ref{lm:me5/3}}\heading{Iteration} \label{sec:iteration} We begin by fixing some $0r$. Let $\rhotfr$ denote the minimizer for the TF functional, $\etfr(\rho)$, under the constraint $\int\rho\leq\int\rhohf \chiplus{r}$. Denote the corresponding TF potential by $$ \phitfr(y):=\V{r}(y)-\rhotfr*|y|^{-1} $$ and the corresponding chemical potential by $\mutfr$. We shall prove below (see Lemma~\ref{lm:sommerfeldotf}) that if $r$ is chosen appropriately then $\mutfr=0$. Note that according to the Thomas-Fermi equation (\ref{eq:tfeqgeneral}), $\rhotfr$ has support on the set $\set{y}{|y|\geq r}$. Since $\V{r}$ on the support of $\rhotfr$ is the potential coming from the true HF density for $|y|r$, of the atom. The notation OTF refers to {\it Outside} TF. \begin{lm}[Preliminary bounds on TF and OTF functions] \label{lm:tfotfprelims}\hfill\\ Assume that $N\geq Z$ then for all $y$ $$ \phitf(y)\leq 3^42^{-3}\pi^2|y|^{-4}\quad \hbox{and}\quad \rhotf(y)\leq 3^52^{-3}\pi|y|^{-6}. $$ For all $|y|\geq \beta_0 Z^{-1/3}$ we have $$ \phitf(y)\geq\const |y|^{-4}\quad \hbox{and}\quad \rhotf(y)\geq \const|y|^{-6}. $$ Given $\epsilon',\sigma>0$, and $r>0$ such that (\ref{eq:phiasp}) holds for all $|x|\leq r$ and $\sigma r^{\epsilon'}\leq 1$ then for all $|y|\geq r$ we have $$ \rhotfr(y)\leq \const r^{-6}\quad \hbox{and}\quad \phitfr(y)\leq |\V{r}(y)|=|\Phihf{r}(y)|\leq \const r^{-4}. $$ \end{lm} \begin{proof} The upper bounds on the TF functions follow immediately from the Atomic Sommerfeld estimate Theorem~\ref{thm:sommerfeldup}) and the TF equation (\ref{eq:tfsystemrho}) if we recall that $\mutf=0$. The lower bounds follow from Theorem~\ref{thm:sommerfeldlow}. Since $\Phihf{r}$ is harmonic on the set $\{|y|>r\}$ and tends to zero at infinity it follows from the assumptions on $\epsilon',\sigma,r$ that for all $|y|\geq r$ we have $$ |\Phihf{r}(y)|\leq \sup_{|z|=r}|\Phihf{r}(z)| \leq \const r^{-4}, $$ where in the last inequality we have used the iterative assumption (\ref{eq:phiasp}) and Lemma~\ref{lm:Phitfrbound} for the case $\mutf=0$ and the fact that $\Phitf{r}\geq\phitf\geq0$ (see e.g., Theorem~\ref{thm:sommerfeldlow}). The inequality $\phitfr(y)\leq |\V{r}(y)|$ is trivial from the definition of $\phitfr$. Finally, from the TF equation (\ref{eq:tfeqgeneral}) we conclude that for all $|y|\geq r$ $$ \rhotfr(y)\leq \const \V{r}(y)^{3/2}\leq \const r^{-6}. $$ \end{proof} \begin{lm}[Preliminary comparison of HF and TF]\label{lm:n(rtf)-n(rhf)} \hfill\\ Assume that $N\geq Z$. Given $\epsilon',\sigma>0$, and $r>0$ such that (\ref{eq:phiasp}) holds for all $|x|\leq r$ then \begin{equation} \int\chiplus{r}\left(\rhotf -\rhohf\right) \leq {\sigma} r^{-3+\varepsilon'}. \end{equation} \end{lm} \begin{proof} We have $$ \int_{|y|r$ we may write \begin{equation}\label{eq:AAA} \Phihf{|x|}(x)-\Phitf{|x|}(x) ={\cal A}_1(r,x) +{\cal A}_2(r,x)+{\cal A}_3(r,x), \end{equation} where \begin{eqnarray} {\cal A}_1(r,x)&=&\phitfr(x)-\phitf(x)\label{eq:A1defi}\\ {\cal A}_2(r,x) &=&\int_{|y|>|x|} \left[\rhotfr(y)-\rhotf(y)\right]|x-y|^{-1}dy\label{eq:A2defi}\\ {\cal A}_3(r,x)&=& \int_{r<|y|<|x|}\left[\rhotfr(y)-\rhohf(y)\right]|x-y|^{-1}dy. \label{eq:A3defi} \end{eqnarray} We turn first to estimating ${\cal A}_1$ and ${\cal A}_2$. Thus we need to control the difference between the full TF approximation and the TF approximation for the outside region. Our strategy is to first prove that $\phitfr(x)$ and $\phitf(x)$ are close on the set $\{|x|=r\}$. An application of the Sommerfeld estimates in Theorem~\ref{thm:sommerfeldmu} will then give excellent control on the difference $\phitfr(x)-\phitf(x)$ for all $|x|>r$. Controlling the behavior on the set $\{|x|=r\}$ is difficult and we begin with a weak estimate on the difference between $\rhotfr$ and $\rhotf$. In fact, we first estimate the difference in Coulomb norm. \begin{lm}[Coulomb norm comparison of TF and OTF]\label{lm:TFOTF} \hfill\\ Assume $N\geq Z$. Given constants $\varepsilon',{\sigma}>0$ there exists a constant $D>0$ depending only on $\varepsilon',{\sigma}$ such that for all $r$ with $\beta_0Z^{-1/3}\leq r\leq D$ for which (\ref{eq:phiasp}) holds for all $|x|\leq r$ we have \begin{equation}\label{eq:cnorm} \|\rhotfr-\chiplus{r}\rhotf\|_\rC^2\leq \const{\sigma} r^{-7+\varepsilon'}. \end{equation} Here again $\chiplus{r}$ denotes the characteristic function of the set $\set{y}{|y|\geq r}$. Moreover, \begin{equation}\label{eq:cmu} \mutfr\leq \const {\sigma}^{1/2}r^{-4+\frac{\varepsilon'}{2}}. \end{equation} \end{lm} \begin{proof} To prove this we make a perturbation analysis of the TF functional. We introduce the perturbation potential \begin{equation}\label{eq:perturbationpotential} W(x)=\Phihf{r}(x)-\Phitf{r}(x). \end{equation} Then for all $|x|>r$ $$ \Phitf{r}(x)=\V{r}(x)-W(x). $$ Note that $W$ is harmonic for $|x|>r$ and tends to zero at infinity. Hence, since we assume that the iterative assumption (\ref{eq:phiasp}) holds for $|x|=r$, we have \begin{equation}\label{eq:iterativeassumption} \sup_{|x|\geq r}|W(x)|= \sup_{|x|=r}|W(x)|\leq {\sigma} r^{-4+\epsilon'}. \end{equation} We claim that there exist two functions $W_1$, $W_2$ with $\supp W_1\subset \set{x}{|x|<3r}$ and $\supp W_2\subset\set{x}{|x|>2r}$ such that $W(x)=W_1(x)+W_2(x)$ and \begin{eqnarray} \sup_{|x|\geq r}|W_1(x)|&\leq& \sup_{|x|=r}|W(x)|\leq {\sigma} r^{-4+\epsilon'}\label{eq:W1}\\ \int|\nabla W_2|^2&\leq& \frac{76\pi}{3} r \sup_{|x|=r}|W(x)|^2 \leq \frac{76\pi}{3} {\sigma}^2 r^{-7+2\epsilon'}\label{eq:W2}. \end{eqnarray} In order to prove this we let $$ F(x)=\left\{\begin{array}{ll} 0&\mbox{if }|x|< 2r\\ (|x|-2r)r^{-1}, &\mbox{if }2r \leq|x|\leq 3r\\ 1&\mbox{if }|x|>3r \end{array}\right. . $$ Set $W_1(x)=(1-F(x))W(x)$ and $W_2(x)=F(x)W(x)$. The first estimate (\ref{eq:W1}) follows immediately from (\ref{eq:iterativeassumption}). By a simple integration by parts, similar to the one used to prove the IMS formula (\ref{eq:IMSformula}), we obtain (note that $W(x)$ behaves like $c|x|^{-1}$ and $|\nabla W(x)|$ behaves like $c|x|^{-2}$ at infinity so there are no contributions from infinity to the integration by parts) \begin{eqnarray*} \int|\nabla W_2|^2&=&\int|\nabla F|^2 |W|^2-\int|F|^2W\Delta W =\int|\nabla F|^2 |W|^2\\ &\leq& \sup_{|x|\geq r}|W(x)|^2\int_{2r\leq |x|\leq 3r}r^{-2}dx= \frac{76\pi}{3} r \sup_{|x|\geq r}|W(x)|^2, \end{eqnarray*} where the second equality follows since $W$ is harmonic on the support of $F$. The estimate (\ref{eq:W1}) now also follows from (\ref{eq:iterativeassumption}). We are now ready to estimate the TF densities. We shall use $\chiplus{r}\chiminus{R}\rhotf$ for some $R\geq r$ (possibly $R$ is infinity) as a trial density in $\etfr$. Since $\rhotfr$ minimizes $\etfr\left(\uprho\right)+\mutfr\int\uprho$ and $\mutfr=0$ unless $\int\rhotfr=\int\chiplus{r}\rhohf$ we have \begin{equation}\label{eq:etfrtrial} \mutfr\left(\int\chiplus{r}\rhohf-\int\chiplus{r}\chiminus{R}\rhotf\right) \leq \etfr\left(\chiplus{r}\chiminus{R}\rhotf\right)- \etfr\left(\rhotfr\right) \end{equation} We write the left side as \begin{eqnarray} \etfr\left(\chiplus{r}\chiminus{R}\rhotf\right)- \etfr\left(\rhotfr\right) &=&\etfr\left(\chiplus{r}\chiminus{R}\rhotf\right)- \etfr\left(\chiplus{r}\rhotf\right)\nonumber\\ &&{}+ \etfr\left(\chiplus{r}\rhotf\right) -\etfr\left(\rhotfr\right). \label{eq:etfrsplit} \end{eqnarray} We have for the first two terms \begin{eqnarray} \lefteqn{\etfr\left(\chiplus{r}\chiminus{R}\rhotf\right)- \etfr\left(\chiplus{r}\rhotf\right)}&& \nonumber\\ \nonumber\\ &=& \int\phitf\rhotf\chiplus{R}+ \|\chiplus{R}\rhotf\|_\rC^2 +\int\left(\Phihf{r}-\Phitf{r}\right)\chiplus{R}\rhotf -\mfr{3}{10}(3\pi^2)^{2/3}\int \chiplus{R}(\rhotf)^{5/3} \nonumber\\ \nonumber\\ &\leq&\int\phitf\rhotf\chiplus{R}+ \|\chiplus{R}\rhotf\|_\rC^2 +{\sigma}r^{-4+\varepsilon'} \int\chiplus{R}\rhotf,\label{eq:etfrfirst} \end{eqnarray} and since $\mutf=0$ \begin{eqnarray} \lefteqn{\etfr(\chiplus{r}\rhotf)-\etfr(\rhotfr) =\int W(\rhotfr-\chiplus{r}\rhotf)- \|\rhotfr-\chiplus{r}\rhotf\|_\rC^2}\qquad && \nonumber\\ \nonumber\\ &&\mbox{}+\int_{|y|\geq r}\biggl( \Bigl[\mfr{3}{10}(3\pi^2)^{2/3}\rhotf(y)^{5/3} -\phitf(y)\rhotf(y)\Bigr]\nonumber\\ \nonumber\\ &&\phantom{\mbox{}+\int_{|y|\geq r}}- \Bigl[\mfr{3}{10}(3\pi^2)^{2/3}\rhotfr(y)^{5/3}- \phitf(y)\rhotfr(y)\Bigr]\biggr)dy. \label{eq:etfrsecond} \end{eqnarray} Since $\rhotf$ satisfies the TF equation (\ref{eq:tfsystemrho}) we see that for fixed $y$ the expression $$ \mfr{3}{10}(3\pi^2)^{2/3}t^{5/3}-\phitf(y)t, \qquad t\geq0 $$ takes its minimal value for $t=\rhotf(y)$. Hence we conclude that the last integral above is negative. Thus combining (\ref{eq:etfrtrial})--(\ref{eq:etfrsecond}) we have \begin{eqnarray*} \lefteqn{\mutfr\left(\int\chiplus{r}\rhohf -\int\chiplus{r}\chiminus{R}\rhotf\right) \leq\int W(\rhotfr-\chiplus{r}\rhotf)- \|\rhotfr-\chiplus{r}\rhotf\|_\rC^2}\hspace{100pt}&&\\ \\ &&{} +\int\phitf\rhotf\chiplus{R} + \|\chiplus{R}\rhotf\|_\rC^2 +{\sigma}r^{-4+\varepsilon'} \int\chiplus{R}\rhotf. \end{eqnarray*} Lemma~\ref{lm:tfotfprelims} implies that $\|\chiplus{R}\rhotf\|_\rC^2\leq\const R^{-4}\int\chiplus{R}\rhotf$. If we now also use Lemma~\ref{lm:n(rtf)-n(rhf)} we arrive at \begin{eqnarray} \mutfr\left(\int\chiplus{r}\rhohf -\int\chiplus{r}\chiminus{R}\rhotf\right) &\leq&\int W(\rhotfr-\chiplus{r}\rhotf)- \|\rhotfr-\chiplus{r}\rhotf\|_\rC^2 \nonumber\\ &&{} \!\!\!\!+\left(\const R^{-4} +{\sigma}r^{-4+\varepsilon'}\right) \int\rhotf\chiplus{R}.\label{eq:cmain} \end{eqnarray} This is the main estimate from which we shall derive the estimates of the lemma. We shall do this by choosing different values for $R$. One for the estimate (\ref{eq:cnorm}) another for (\ref{eq:cmu}). We shall choose $R$ such that \begin{equation}\label{eq:Rchoice} \int\chiplus{r}\chiminus{R}\rhotf\leq\int\chiplus{r}\rhohf. \end{equation} Let $R_{\max}$ denote the largest possible $R$ for which this holds. Then \begin{equation}\label{eq:Rmax} \int\chiplus{R_{\max}}\rhotf= \left(\int\chiplus{r}\rhotf-\int\chiplus{r}\rhohf\right)_+ \leq {\sigma}r^{-3+\varepsilon'}, \end{equation} where the last inequality follows from Lemma~\ref{lm:n(rtf)-n(rhf)}. By Lemma~\ref{lm:tfotfprelims} we have for all $R\geq r\geq\beta_0Z^{-1/3}$ that \begin{equation}\label{eq:Rbound} \int\chiplus{R}\rhotf \geq \const R^{-3}. \end{equation} We shall now make the assumption that $D$ is chosen so small that if $r\leq D$ then ${\sigma}r^{\epsilon'}\leq1$. Thus from (\ref{eq:Rmax}) and (\ref{eq:Rbound}) we conclude that \begin{equation}\label{eq:Rmaxbound} R_{\max}^{-4}\leq \const {\sigma}^{4/3} r^{-4 +\frac{4}{3}\varepsilon'}\leq \const r^{-4}. \end{equation} {F}rom (\ref{eq:Rmax}) and (\ref{eq:cmain}) with $R=R_{\max}$ we get, again using the above assumption on $D$, that \begin{equation}\label{eq:coulombw} \|\rhotfr-\chiplus{r}\rhotf\|_\rC^2\leq\int W(\rhotfr-\chiplus{r}\rhotf) +\const {\sigma}r^{-7+\varepsilon'}. \end{equation} We estimate the integral on the right by dividing it in two parts \begin{eqnarray*} \lefteqn{\int W(\rhotfr-\chiplus{r}\rhotf) \leq \int|W_1|(\rhotfr+\chiplus{r}\rhotf) +\int W_2(\rhotfr-\chiplus{r}\rhotf)}&&\\ &\leq& {\sigma}r^{-4+\epsilon'}\int_{r<|x|<3r}\rhotfr(x) +\chiplus{r}(x)\rhotf(x)dx+\int W_2(\rhotfr-\chiplus{r}\rhotf), \end{eqnarray*} where we have also used (\ref{eq:W1}). {F}rom Lemma~\ref{lm:tfotfprelims} we arrive at \begin{equation}\label{eq:intW} \int W(\rhotfr-\chiplus{r}\rhotf) \leq \const {\sigma}r^{-7+\epsilon'} +\int W_2(\rhotfr-\chiplus{r}\rhotf). \end{equation} The last term in this estimate we now control using the Coulomb norm estimate Lemma~\ref{lm:cn}. Note that $\rhotf$ and $\rhotfr$ both belong to $L^{6/5}$ since they are in $L^{5/3}\cap L^{1}$. We find from (\ref{eq:W2}) that \begin{equation}\label{eq:W22} \int W_2(\rhotfr-\chiplus{r}\rhotf)\leq \const{\sigma}r^{-\frac{7}{2}+\epsilon'} \|\rhotfr-\chiplus{r}\rhotf\|_\rC. \end{equation} Inserting the last two estimates into (\ref{eq:coulombw}) gives (recall that ${\sigma}r^{\epsilon'}\leq1$) $$ \|\rhotfr-\chiplus{r}\rhotf\|_\rC^2 \leq \const {\sigma}r^{-7+\epsilon'} +\const \left({\sigma}r^{-7+\epsilon'}\right)^{1/2} \|\rhotfr-\chiplus{r}\rhotf\|_\rC $$ and (\ref{eq:cnorm}) follows immediately from this. We now return to (\ref{eq:cmain}) and make a new choice for $R$. It follows from the first inequality in (\ref{eq:Rmaxbound}) that that we can find $R$ with $R\leq R_{\max}$ satisfying $$ R^{-4}=\const{\sigma}^{1/2}r^{-4+\epsilon'/2}. $$ Note that, since $$ R^{-4}\leq \const{\sigma}^{1/2}D^{\epsilon'/2}r^{-4}, $$ we can choose $D$ small enough to ensure that $r\leq R$. {F}rom (\ref{eq:Rbound}) we have \begin{equation}\label{eq:intchiplusR} \int\chiplus{R}\rhotf\geq \const {\sigma}^{3/8}r^{-3+\frac{3}{8}\epsilon'}. \end{equation} It follows from (\ref{eq:Rmax}) and (\ref{eq:Rbound}) that we may assume that the constant in the definition of $R$ is chosen such as to ensure that $ \int\chiplus{R}\rhotf\geq2\int\chiplus{R_{\max}}\rhotf. $ Thus since (see (\ref{eq:Rmax})) $$ \int\chiplus{r}\rhohf -\int\chiplus{r}\chiminus{R}\rhotf\geq -\int\chiplus{R_{\max}}\rhotf +\int\chiplus{R}\rhotf $$ we have $$ \mfr{1}{2}\int\chiplus{R}\rhotf \leq\int\chiplus{r}\rhohf -\int\chiplus{r}\chiminus{R}\rhotf. $$ Thus from (\ref{eq:intchiplusR}), (\ref{eq:intW}), (\ref{eq:W22}), (\ref{eq:cnorm}), and (\ref{eq:cmain}) we conclude (\ref{eq:cmu}). \end{proof} Using these fairly weak estimates we shall now show that the outside TF potential and density satisfy Sommerfeld type estimates. \begin{lm}[Sommerfeld estimates for OTF]\label{lm:sommerfeldotf} Assume $N\geq Z$. Given constants $\varepsilon',{\sigma}>0$ there exists a constant $D>0$ depending only on $\varepsilon',{\sigma}$ such that for all $r$ with $\beta_0Z^{-1/3}\leq r\leq D$ for which (\ref{eq:phiasp}) holds for all $|x|\leq r$ then $\mutfr=0$ and for all $|x|\geq r$ we have \begin{eqnarray} \phitfr(x)&\leq& 3^42^{-3}\pi^2|x|^{-4} \left(1+Ar^\zeta|x|^{-\zeta}\right)\label{eq:phitfrup}\\ \noalign{\hbox{and}} \phitfr(x)&\geq& 3^42^{-3}\pi^2|x|^{-4}\left(1+ar^\zeta|x|^{-\zeta}\right)^{-2}, \label{eq:phitfrlow} \end{eqnarray} where $a$ and $A$ are universal constants (but not necessarily positive) and $a>-1$. Here as before $\zeta=(-7+\sqrt{73})/2\approx0.77$. \end{lm} \begin{proof} Note first that the potential $\V{r}$ satisfies the assumptions in Theorem~\ref{thm:sommerfeldmu} with $R=r$. Hence the potential $\phitfr$ satisfies the Sommerfeld estimates described in the theorem. In order to get some useful estimates we however need to control the behavior of $\phitfr(x)$ for $|x|=r$. In particular, we need to establish that $\mutfr< \inf_{|x|=r} \phitfr(x)$. In order to control $\phitfr(x)$ for $|x|=r$ we note that for all $|x|\geq r$ we have \begin{equation}\label{eq:phitfrbound} \phitfr(x)=\phitf(x)+\left(\chiplus{r}\rhotf-\rhotfr\right)*|x|^{-1} +W(x), \end{equation} where as in (\ref{eq:perturbationpotential}), $W=\Phihf{r}-\Phitf{r}$. According to the Coulomb norm Corollary~\ref{cl:cn} we get for all $s>0$ that \begin{eqnarray*} \lefteqn{\left|\left(\chiplus{r}\rhotf-\rhotfr\right)*|x|^{-1}\right|}&& \\ \\ &\leq&\const s^{1/5}\max\{\|\chiplus{r}\rhotf\|_{L^{5/3}(\B{x,s})}, \|\rhotfr\|_{L^{5/3}(\B{x,s})}\}\\ \\ &&{} +\const s^{-1/2}\|\chiplus{r}\rhotf-\rhotfr\|_\rC. \end{eqnarray*} {F}rom Lemma~\ref{lm:tfotfprelims} we see that $$ \max\{\|\chiplus{r}\rhotf\|_{L^{5/3}(\B{x,s})}, \|\rhotfr\|_{L^{5/3}(\B{x,s})}\}\leq \const r^{-6} s^{9/5}, $$ where we have assumed that $D$ is such that ${\sigma}r^{\epsilon'}\leq1$. Inserting this and the estimate (\ref{eq:cnorm}) from Lemma~\ref{lm:TFOTF} above we obtain that for all $|x|\geq r$ \begin{eqnarray} \left|\left(\chiplus{r}\rhotf-\rhotfr\right)*|x|^{-1}\right| &\leq& \const r^{-6}s^2 +\const s^{-1/2}\left( {\sigma}r^{-7+\epsilon}\right)^{1/2}\nonumber\\ &=&\const \sigma^{2/5} r^{-4+2\epsilon'/5} \label{eq:(rhotf-rhotfr)bound} \end{eqnarray} where we have optimized in $s$ in order to get the last expression. Thus from (\ref{eq:phitfrbound}), the iterative assumption (\ref{eq:phiasp}), and Lemma~\ref{lm:tfotfprelims} we obtain \begin{eqnarray*} \inf_{|x|=r}\phitfr(x)&\geq&\inf_{|x|=r}\phitf(x) -\const \sigma^{2/5} r^{-4+2\epsilon'/5}\\ &\geq&\const r^{-4}-\const\sigma^{2/5} r^{-4+2\epsilon'/5}. \end{eqnarray*} We have used that since $\sigma r^{\epsilon'}\leq 1$ the error from (\ref{eq:(rhotf-rhotfr)bound}) is worse than the error from (\ref{eq:phiasp}). Note that the constant in front of $r^{-4}$ above is positive. {F}rom Lemma~\ref{lm:TFOTF} we know that $ \mutfr\leq \const {\sigma}^{1/2}r^{-4+{\epsilon'}/{2}} $ Hence we may choose $D$ such that $\mutfr<\inf_{|x|=r}\phitfr(x)$. As above we of course also have $$ \sup_{|x|=r}\phitfr(x)\leq \const r^{-4}+\const\sigma^{2/5} r^{-4+2\epsilon'/5}. $$ Thus $\const r^{-4}\leq \inf_{|x|=r}\phitfr(x) \leq\sup_{|x|=r}\phitfr(x)\leq\const r^{-4}$. That $\mutfr=0$ follows from Corollary~\ref{cl:muestimate} as follows. By harmonicity of $\V{r}$ we have $$ \lim_{r'\to\infty}(4\pi)^{-1}\int_{{\cal S}^2}r'\V{r}(r'\omega)d\omega =\int_{{\cal S}^2}r\Phihf{r}(r\omega)d\omega =Z-\int\chiminus{r}\rhohf. $$ Thus from Corollary~\ref{cl:muestimate} we have if $\mutfr\ne0$, i.e., if $\int\rhotfr=\int\chiplus{r}\rhohf$ that $$ 0<(\mutfr)^{3/4}\leq \const\left( Z-\int\chiminus{r}\rhohf-\int\rhotfr\right) =\const (Z-\int\rhohf), $$ which is a contradiction since $\int\rhohf=N\geq Z$. The estimates (\ref{eq:phitfrup}) and (\ref{eq:phitfrlow}) now follow from Theorem~\ref{thm:sommerfeldmu}. \end{proof} We are now ready to give the bounds on ${\cal A}_1$ and ${\cal A}_2$ defined in (\ref{eq:A1defi}) and (\ref{eq:A2defi}). \begin{lm}[Control of ${\cal A}_1$ and ${\cal A}_2$]\label{lm:A1A2} \hfill\\ Assume $N\geq Z$. Given constants $\varepsilon',{\sigma}>0$ there exists a constant $D>0$ depending only on $\varepsilon',{\sigma}$ such that for all $r$ with $\beta_0Z^{-1/3}\leq r\leq D$ for which (\ref{eq:phiasp}) holds for all $|x|\leq r$ we have for all $|x|\geq r$ that \begin{equation}\label{eq:A1} \left|{\cal A}_1(r,x)\right| \leq \const r^{\zeta}|x|^{-4-\zeta} \end{equation} and \begin{equation}\label{eq:A2} \left|{\cal A}_2(r,x)\right| \leq \const r^{\zeta}|x|^{-4-\zeta}, \end{equation} with $\zeta=(-7+\sqrt{73})/2\approx0.77$. \end{lm} \begin{proof} Combining Theorems~\ref{thm:sommerfeldup}, \ref{thm:sommerfeldlow}, with Lemma~\ref{lm:sommerfeldotf} and recalling that $\mutf=0$ immediately gives the bound on ${\cal A}_1$. If we use the TF equation (\ref{eq:tfeqgeneral}) we obtain from Lemma~\ref{lm:sommerfeldotf} that for all $|y|\geq r$ $$ \left|\rhotfr(y)-3^52^{-3}\pi |y|^{-6}\right|\leq \const r^{\zeta}|y|^{-6-\zeta}. $$ Of course we similarly have from Theorems~\ref{thm:sommerfeldup}, \ref{thm:sommerfeldlow}, and the TF equation (\ref{eq:tfsystemrho}) that \begin{eqnarray*} \rhotf(y)&\leq& 3^52^{-3}\pi |y|^{-6} \\ \noalign{\hbox{and}} \rhotf(y)&\geq& 3^52^{-3}\pi |y|^{-6}-\const Z^{-\zeta/3}|y|^{-6-\zeta}. \end{eqnarray*} Since $r\geq \beta_0 Z^{-1/3}$ we conclude that for all $|y|>r$ \begin{equation}\label{eq:|rhorhotf|1} \left|\rhotf(y)-\rhotfr(y)\right| \leq \const r^{\zeta}|y|^{-6-\zeta}. \end{equation} Thus $$ |{\cal A}_2|\leq\const \int_{|y|>|x|} r^{\zeta}|y|^{-7-\zeta}\,dy $$ which gives the bound in (\ref{eq:A2}). \end{proof} We turn now to estimating ${\cal A}_3$. This requires estimating the difference between $\rhotfr$ and $\rhohf$. We again begin by estimating this difference in the Coulomb norm. More precisely, we estimate in Coulomb norm the difference between the ``outside'' TF density $\rhotfr$ and the ``outside'' HF density $\chiplus{r}\rhohfr$. This is done through a semiclassical analysis of the exterior region $\{|x|>r\}$. \begin{lm}[Coulomb norm comparison of HF and OTF]\label{lm:cnhfotf} \hfill\\ Assume $N\geq Z$. Given constants $\varepsilon',{\sigma}>0$ there exists a constant $D>0$ depending only on $\varepsilon',{\sigma}$ such that for all $r$ with $\beta_0Z^{-1/3}\leq r\leq D$ for which (\ref{eq:phiasp}) holds for all $|x|\leq r$ we have \begin{equation}\label{eq:cnhfotf} \|\rhotfr-\chiplus{r}\rhohf\|_{\rC}\leq \const r^{-\frac{7}{2}+\frac{1}{6}} \end{equation} and \begin{equation}\label{eq:me5/3hf} \int\left(\chiplus{r}\rhohf\right)^{5/3}\leq \const r^{-7}. \end{equation} \end{lm} \begin{proof} Let $\gamma$ be the density matrix on $L^2(\R^3;\C^2)$, but diagonal in spin, constructed in the semiclassical approximation Lemma~\ref{lm:sc} for the potential $V=\phitfr$. Note that from Lemma~\ref{lm:sommerfeldotf} we have $\phitfr(y)\geq0$ for $|y|\geq r$ and from its definition $\phitfr(y)\leq 0$ for $|y|r$ we conclude that $\V{r}*g^2(y)=\V{r}(y)$ for $|y|>r+s$. Hence we get from Lemma~\ref{lm:tfotfprelims} that \begin{equation}\label{eq:vrs} \int \V{r}\left(\rhotfr-\rhotfr*g^2\right)\leq \const r^{-4}\int_{|y|(1-\lambda)r}\rhohf\leq \const r^{-3}, \end{equation} where we have used that $r, {\sigma}r^{\epsilon'}\leq1$. Moreover, from Lemma~\ref{lm:tfotfprelims} with $r$ replaced by $(1-\lambda)r$ we have $$ \int\limits_{(1-\lambda)r<|y|<(1-\lambda)^{-1}r} \left[\Phihf{(1-\lambda)r}(y)\right]_+^{5/2}dy\leq \const r^{-7}\lambda. $$ Hence we have $$ {\cal R}\leq\const \lambda^{-2} r^{-5} + \const r^{-7}\lambda+\Ex\left[\gammar\right]. $$ If we now use the exchange inequality in Theorem~\ref{thm:exchineq} and (\ref{eq:l1rhohfr}) we get (recall that $\rhohfr=\thetar^2\rhohf$ is the density corresponding to $\gammar$) \begin{eqnarray*} \Ex\left[\gammar\right] &\leq& \const\int\left(\rhohfr\right)^{4/3}\leq \const\left(\int\rhohfr\right)^{1/2} \left(\int\left(\rhohfr\right)^{5/3}\right)^{1/2}\\ &\leq& \const r^{-3/2} \left({\cal R}+r^{-7} \right)^{1/2}, \end{eqnarray*} where we have also used that according to (\ref{eq:l5/3rhohfr}) and Lemma~\ref{lm:tfotfprelims} we have \begin{equation}\label{eq:me5/3rhohfr} \int\left(\rhohfr\right)^{5/3}\leq \const {\cal R}+\const r^{-7}. \end{equation} We may therefore conclude that \begin{equation}\label{eq:Restimate} {\cal R}\leq \const r^{-7}(r^2\lambda^{-2}+\lambda)+\const r^{-5}. \end{equation} We may use (\ref{eq:me5/3rhohfr}) and (\ref{eq:Restimate}) to prove (\ref{eq:me5/3hf}). Recall that $\rhohfr(y)=\rhohf(y)$ if $|y|>(1-\lambda)^{-1}r$. Now (\ref{eq:me5/3hf}) follows if we simply observe that (\ref{eq:me5/3rhohfr}) and (\ref{eq:Restimate}) hold with $r$ replaced by $r/2$ and $\lambda=1/2$. We shall make a possible different choice of $\lambda$ below. We shall now prove a lower bound on $\ea\left[{\gammar}\right]$. We write \begin{eqnarray*} \ea\left[\gammar\right]&=& \Tr\left[\left(-\mfr{1}{2}\Delta-\Phihf{r}\right)\gammar\right] +\mfr{1}{2}\iint\rhohfr(x)|x-y|^{-1}\rhohfr(y)dx\,dy\\ &=&\Tr\left[\left(-\mfr{1}{2}\Delta-\Phihf{r} +\rhotfr*|x|^{-1}\right)\gammar\right] +\|\rhohfr-\rhotfr\|_{\rC}^2\\{}&& -\mfr{1}{2}\iint\rhotfr(x)|x-y|^{-1}\rhotfr(y)dx\,dy. \end{eqnarray*} If we use that on the support of $\rhohfr$ we have $\Phihf{r}=\V{r}$ we may write this as \begin{eqnarray} \ea\left[\gammar\right]&=&\Tr\left[\left(-\mfr{1}{2}\Delta-\phitfr\right) \gammar\right] +\|\rhohfr-\rhotfr\|_{\rC}^2\nonumber\\{}&& -\mfr{1}{2}\iint\rhotfr(x)|x-y|^{-1}\rhotfr(y)\,dx\,dy.\label{eq:meealower} \end{eqnarray} The trace may be bounded below by the sum of the first $N'$ negative eigenvalues of the operator $-\mfr{1}{2}\Delta-\phitfr $, where $N'$ is the smallest integer larger than $\Tr[\gammar]=\int\rhohfr$. {F}rom Lemma~\ref{lm:sc} (again with an extra factor of $2$ due to spin) we therefore have that for all $s>0$ and all $0<\delta<1$ \begin{eqnarray} \Tr\left[\left(-\mfr{1}{2}\Delta-\phitfr \right)\gammar\right] &\geq& -\frac{2^{5/2}}{(15\pi^2)}(1-\delta)^{-3/2} \int \left(\phitfr\right)_+^{5/2}\nonumber\\ &&{}-\pi^2s^{-2}\left(\int\rhohfr+1\right)\label{eq:hfotfsc}\\ &&{} -2L_1\delta^{-3/2}\left\|\left[\phitfr -\phitfr*g^2\right]_+\right\|_{5/2}^{5/2}.\nonumber \end{eqnarray} We first estimate the last term. Since $\rhotfr*|x|^{-1}$ is superharmonic we have by the mean value property that $\rhotfr*|x|^{-1}\geq \rhotfr*|x|^{-1}*g^2$. Thus we have $$ \phitfr-\phitfr*g^2 =\V{r}-\V{r}*g^2+\rhotfr*|x|^{-1}*g^2- \rhotfr*|x|^{-1}\leq\V{r}-\V{r}*g^2. $$ The same argument which led to (\ref{eq:vrs}) gives that $\V{r}(y)-\V{r}*g^2(y)=0$ unless $r-s\leq |y|\leq r+s$. Since by Lemma~\ref{lm:tfotfprelims} we have $|\V{r}(y)|\leq \const r^{-4}$ (recall that $\V{r}$ is supported on $\{|y|\geq r\}$) we obtain $$ \left\|\left[\phitfr -\phitfr*g^2\right]_+\right\|_{5/2}^{5/2} \leq \const r^{-8}s, $$ if we assume that $s\leq r$. (Note that $s$ here does not have to be chosen as in the upper bound). {F}rom (\ref{eq:l1rhohfr}) we also get that $\int\rhohfr \leq\int\rhohf\chiplus{r}\leq \const r^{-3}$. Finally from the Sommerfeld estimate (\ref{eq:phitfrup}) and the fact that $\phitfr(x)$ is positive only if $\V{r}(x)>0$, i.e., only if $|x|\geq r$, we find that $$ \int \left(\phitfr\right)_+^{5/2}\leq \const r^{-7}. $$ We therefore see from (\ref{eq:hfotfsc}) and the TF equation (\ref{eq:tfeqgeneral}) (recall that $\mutfr=0$, by Lemma~\ref{lm:sommerfeldotf}) that if $0<\delta<1/2$ then \begin{eqnarray*} \lefteqn{\Tr\left[\left(-\mfr{1}{2}\Delta-\phitfr \right)\gammar\right]}&&\\ &\geq&\mfr{3}{10}(3\pi^2)^{2/3}\int\left(\rhotfr\right)^{5/3} -\int\phitfr\rhotfr\\{}&& -\const\left(\delta r^{-7}+\delta^{-3/2}r^{-8}s\right) -\const s^{-2}r^{-3}\\ &=&\mfr{3}{10}(3\pi^2)^{2/3}\int\left(\rhotfr\right)^{5/3} -\int\phitfr\rhotfr -\const r^{-7+\frac{1}{3}}, \end{eqnarray*} where we have chosen $\delta=r^{-2/5}s^{2/5}$ and $s=r^{11/6}$, which agrees with $s\leq r$. If we insert this last estimate into (\ref{eq:meealower}) awe obtain $$ \ea\left[\gammar\right]\geq \etfr\left(\rhotfr\right) +\|\rhohfr-\rhotfr\|_{\rC}^2-\const r^{-7+\frac{1}{3}}. $$ If we compare this with (\ref{eq:meeaupper1}) we see that $$ \|\rhohfr-\rhotfr\|_{\rC}^2\leq\const r^{-7+\frac{1}{3}} +{\cal R}. $$ Finally, we use the Hardy-Littlewood-Sobolev inequality (\ref{eq:HLS}) and (\ref{eq:me5/3hf}) to conclude that \begin{eqnarray*} \lefteqn{\|\chiplus{r}\rhohf-\rhohfr\|_{\rC}}&&\\ &\leq& \const\|\chiplus{r}\rhohf-\rhohfr\|_{6/5} \leq \const\left(\int_{r<|y|<(1-\lambda)^{-1}r} \rhohf(y)^{6/5}dy\right)^{5/6}\\ &\leq& \const\left(\int_{r<|y|} \rhohf(y)^{5/3}dy\right)^{3/5} \left(\int_{r<|y|<(1-\lambda)^{-1}r} 1dy\right)^{7/30} \leq\const \lambda^{7/30} r^{-7/2}. \end{eqnarray*} We thus get from (\ref{eq:Restimate}) that \begin{eqnarray*} \lefteqn{\|\chiplus{r}\rhohf-\rhotfr\|_{\rC}\leq \|\chiplus{r}\rhohf-\rhohfr\|_{\rC}+\|\rhohfr-\rhotfr\|_{\rC}}&& \\ &\leq& \const r^{-\frac{7}{2}+\frac{1}{6}}+ \const r^{-7/2}(r\lambda^{-1}+\lambda+\lambda^{7/30}) \end{eqnarray*} which gives (\ref{eq:cnhfotf}) if we choose $\lambda=\min\{1/2, r^{5/7}\}$. \end{proof} We may now estimate ${\cal A}_3$ defined in (\ref{eq:A3defi}). \begin{lm}[Controlling ${\cal A}_3$]\label{lm:A3} Assume $N\geq Z$. Given constants $\varepsilon',{\sigma}>0$ there exists a constant $D>0$ depending only on $\varepsilon',{\sigma}$ such that for all $r$ with $\beta_0Z^{-1/3}\leq r\leq D$ for which (\ref{eq:phiasp}) holds for all $|x|\leq r$ we have for all $|x|\geq r$ that \begin{equation}\label{eq:A3} \left|{\cal A}_3(r,x)\right| \leq \ \const(|x|/r)^{1/12}r^{-4+\frac{1}{36}}. \end{equation} \end{lm} \begin{proof} We shall use the Coulomb norm estimate (\ref{eq:cn2}) with $f=\rhotfr-\chiplus{r}\rhohf$. We then immediately see from (\ref{eq:cn2}), Lemma~\ref{lm:cnhfotf}, and the fact, which follows from Lemma~\ref{lm:sommerfeldotf} and the TF equation (\ref{eq:tfeqgeneral}), that $\rhotfr(y)\leq \const|y|^{-6}$ , that $$ \left|{\cal A}_3(r,x)\right|\leq \const (\kappa |x|)^{1/5}r^{-21/5}+ \const\kappa^{-1}|x|^{-1/2}r^{-\frac{7}{2}+\frac{1}{6}}. $$ This gives (\ref{eq:A3}) if we choose $ \kappa= \const (r/|x|)^{7/12}r^{\frac{5}{36}}. $ \end{proof} \begin{proof}[Proof of Lemma~\ref{lm:me5/3}] For $|x|\geq \beta_0 Z^{-1/3}$ the estimate in (\ref{eq:me5/3}) follows from (\ref{eq:me5/3hf}). For $|x|\leq \beta_0 Z^{-1/3}$ we get from (\ref{eq:5/3}) that $$ \int_{|y|>|x|}\rhohf(y)^{5/3}\,dy\leq \int_{\R^3}\rhohf(y)^{5/3}\,dy \leq\const Z^{7/3}\leq \const |x|^{-7}. $$ \end{proof} \begin{proof}[End of proof of the iterative Lemma~\ref{lm:iterate}] Let $D>0$ depending on ${\sigma},\epsilon'$ be the smaller of the values $D$ occurring in Lemmas~\ref{lm:A1A2} and \ref{lm:A3}. We may without loss of generality assume that $D\leq1$. Given $\delta>0$. We consider $R_0R_0$. In order to prove (\ref{eq:itphi}) for $R_0<|x|0$ such that (\ref{eq:itphi}) is satisfied. \end{proof} \section{Proving the main results Theorems~\ref{thm:potentialestimate}, \ref{thm:radius}, \ref{thm:maxN}, and \ref{thm:ie}} \heading{Proof of main results}\label{sec:provingmain} The main result Theorem ~\ref{thm:maxN} on the maximal number of electrons $N$ is a simple consequence of Lemma~\ref{lm:meN} and Theorem~\ref{thm:me}. \begin{proof}[Proof of Theorem~\ref{thm:maxN}] \label{proof:maxN} We may of course assume that $N\geq Z$ and that $Z\geq 1$ (otherwise the result follows from Lieb's bound Theorem~\ref{thm:2z+1}). Then $\int\rhotf=Z$. We can then use Lemma~\ref{lm:meN} with $R$ chosen so small that $\CM\leq \CPhi R^{-4+\epsilon}$, because then (\ref{eq:phiasp}) holds with $\sigma=2\CPhi$ and $\epsilon'=\epsilon$. We conclude from Lemma~\ref{lm:meN} that $$ \int\rhohf\leq \int_{|x|R}\rhotf(x)-\rhohf(x)\,dx\right|&=& \left|\int_{|x|0$ such that, with the notation of Theorem~\ref{thm:me}, we have $\CM\leq \CPhi D^{-4+\epsilon}$. Then by Theorem~\ref{thm:me} we have that (\ref{eq:phiasp}) holds for all $|x|\leq D$ with $\sigma=2\CPhi$ and $\epsilon'=\epsilon$. We may now assume that $D\leq 1$ and that $D$ is smaller than the values for $D$ in Lemmas~\ref{lm:A1A2} and \ref{lm:cnhfotf} corresponding to the above choices of $\sigma$ and $\epsilon'$. Consider $$(\beta_0 Z^{-1/3})^{\frac{1}{1+\delta}}\leq |x||x|}\left(\rhotf(y)-\rhohf(y)\right)|x-y|^{-1} dy\right|. $$ The first term is controlled by the main estimate Theorem~\ref{thm:me}. If we use that according to Lemma~\ref{lm:me5/3} we have $\int_{|y|>|x|}\rhohf(y)^{5/3}dy\leq \const$ and that the same estimate holds for the TF density (see Lemma~\ref{lm:tfotfprelims}) we may estimate the second term above as follows. Using H\"older's inequality we have \begin{eqnarray*} \left|\,\,\int\limits_{|x|<|y|}\!\!\!\!\! \left(\rhotf(y)-\rhohf(y)\right)|x-y|^{-1}dy\right| &\leq& \const \left(\int_{|x-y|<1}\!\!\!\!\!|x-y|^{-5/2}dy\right)^{2/5}\\ &&{}+\int_{|x|<|y|}\left(\rhotf(y)+\rhohf(y)\right)dy \leq \const, \end{eqnarray*} where the last estimate follows from Lemma~\ref{lm:meN}. \end{proof} We end the paper by giving the proof of the bound on the ionization energy in Theorem~\ref{thm:ie} \begin{proof}[Proof of Theorem~\ref{thm:ie}] \label{proof:ie} Since the HF energy is a non-increasing function of $N$ we have that $0\leq E^\HF(Z-1,Z)-E^{\HF}(Z,Z)$. In order to prove an upper bound we shall construct a trial density matrix $\gamma$ for $\ehf$ with $\Tr[\gamma]\leq Z-1$. We then clearly have that $\ehf\left[\gamma\right]\geq E^\HF(Z-1,Z)$. Let $\theta_-$ be given in terms of appropriate $r,\lambda>0$ as in the beginning of the proof of Theorem~\ref{thm:outsideenergy}. We then choose as our trial matrix $$ \gammai=\theta_-\gammahf\theta_-, $$ where $\gammahf$ is the HF minimizer with $\Tr\left[\gammahf\right]=Z$. According to the definition of $\theta_-$ we have $$ \Tr\left[\gammai\right]\leq Z-\int_{|y|>r}\rhohf(y)dy. $$ We choose $\lambda=1/2$. Let $R>0$ be such that $\CM=\CPhi R^{-4+\varepsilon}$. We shall now choose $r$ satisfying $r\leq R$. Then according to Theorem~\ref{thm:me} we have that (\ref{eq:phiasp}) holds for $|x|\leq r$ with $\sigma=2\CPhi$ and $\epsilon'=\epsilon$. {F}rom Lemma~\ref{lm:meN} we therefore conclude that \begin{eqnarray*} \int_{|y|>r}\rhohf(y)dy&=&\int\rhohf-\int_{|y|r}\rhotf(y)\,dy\\ &\geq&\int_{|y|>r}\rhotf(y)\,dy-\const r^{-3+\epsilon} \end{eqnarray*} where we have used that $\int\rhohf=\int\rhotf=Z$. We may of course assume that $Z$ is larger than some fixed universal constant. For $Z$ less than a universal constant, the total energy $E^{\HF}(Z,Z)$ and hence the ionization energy $E^{\HF}(Z-1,Z)-E^{\HF}(Z,Z)$ are bounded by universal constants (see Theorem~\ref{thm:HFenergy}). We can therefore assume that $\beta_0Z^{-1/3}< R$ and we shall choose $\beta_0Z^{-1/3}r}\rhohf(y)dy&\geq& \const r^{-3} -\const r^{-3+\epsilon}. \end{eqnarray*} Thus we may choose $r$ to be a small enough universal number (assuming that $Z$ is large enough to allow $\beta_0Z^{-1/3}r}\rhohf(y)dy\geq1$ and hence that $ \Tr\left[\gammai\right]\leq Z-1$. {F}rom the estimate (\ref{eq:outsideenergy1}) in the proof of Theorem~\ref{thm:outsideenergy} we have $$ \ehf\left[\gammai\right]\leq\ehf\left[\gammahf\right]- \ea\left[\gammar\right]+{\cal R}=E^{\HF}(Z,Z)- \ea\left[\gammar\right]+{\cal R} $$ where as before $\gammar=\thetar\gammahf\thetar$, $\cal R$ is given in (\ref{eq:ioerror}), and the functional $\ea$ was defined in (\ref{eq:eadef}). It remains to prove that \begin{equation}\label{eq:gammarR} -\ea\left[\gammar\right]+{\cal R}\leq \const. \end{equation} As in (\ref{eq:Restimate}) we conclude that $ {\cal R}\leq \const r^{-5}\leq\const $, where we have used that $r$ is a universal constant. In order to estimate $\ea\left[\gammar\right]$ we note that, since $\Phihf{r}(y)$ is harmonic for $|y|>r$ and tends to 0 at infinity, we have that for all $|y|\geq r$ \begin{eqnarray*} \Phihf{r}(y)\leq |y|^{-1}r\sup_{|x|=r}\Phihf{r}(x) &\leq& |y|^{-1}r\sup_{|x|=r}|\Phitf{r}(x)| + |y|^{-1}\left(\CPhi r^{-3+\varepsilon}+r\CM\right) \\ &\leq& \const r^{-3} |y|^{-1}, \end{eqnarray*} where we have used the main estimate Theorem~\ref{thm:me} and the bound on $\Phitf{r}$ in Lemma~\ref{lm:Phitfrbound} with $\mutf=0$. If we use that $r$ is some universal constant we get $$ \Phihf{r}(y)\leq \const |y|^{-1}. $$ Using the Lieb-Thirring inequality (\ref{eq:LTdensity}) we see from the definition (\ref{eq:eadef}) of the auxiliary functional $\ea$ that \begin{eqnarray*} \ea\left[\gammar\right]&\geq& K_1\int\rhohfr(y)^{5/3}dy -\const \int\frac{\rhohfr(y)}{|y|}dy\\&&{} +\mfr{1}{2}\iint \rhohfr(x)|x-y|^{-1}\rhohfr(y)dx\,dy. \end{eqnarray*} Here again $\rhohfr=\theta_r^2\rhohf$ is the density corresponding to $\gammar$. It follows from standard atomic TF theory that $$ \inf_{\rho\geq 0}\left\{ K_1\int\rho(y)^{5/3}dy -\const \int\frac{\rho(y)}{|y|}dy +\mfr{1}{2}\iint \rho(x)|x-y|^{-1}\rho(y)dx\,dy\right\} $$ is some universal constant. Hence $ \ea\left[\gammar\right]\geq -\const $ and we have proved (\ref{eq:gammarR}). \end{proof} \begin{thebibliography}{99} \heading{References} \bibitem{Bach} V.~Bach, \newblock Error bound for the Hartree-Fock energy of atoms and molecules.\newblock {\em Commun. Math. 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