Information:The TeX file contains 5529 lines and 215064 characters. BODY %PLAINTEX \overfullrule=0pt \magnification=1100 \baselineskip=4ex \raggedbottom \font\eightpoint=cmr8 \font\fivepoint=cmr5 \headline={\hfill{\fivepoint EHLJPSJY---30/Aug/92 }} \def\fracone{\hbox{${1 \over 2}$}} \def\p{{\bf p}} \def\A{{\bf A}} \def\B{{\bf B}} \def\R{{\bf R}} \def\H{{\cal H}} \def\C{{\bf C}} \def\G{{\cal G}} \def\W{{\cal W}} \def\b{{\rm b}} \def\HS{{\rm HS}} \def\SS{{\rm SS}} \def\STF{{\rm STF}} \def\MTF{{\rm MTF}} \def\TF{{\rm TF}} \def\DM{{\rm DM}} \def\SDM{{\rm SDM}} \def\sc{{\cal C}} \def\E{{\cal E}} \def\const{{\rm const}} \def\lin{{\rm lin}} \def\loc{{\rm loc}} \def\spec{{\rm spec}} \def\supp{{\rm supp}} \def\mfr#1/#2{\hbox{${{#1} \over {#2}}$}} \def\uprho{\raise1pt\hbox{$\rho$}} \def\upchi{\raise1pt\hbox{$\chi$}} \def\dlambda{\lower1pt\hbox{$\lambda$}} \catcode`@=11 \def\eqalignii#1{\,\vcenter{\openup1\jot \m@th \ialign{\strut\hfil$\displaystyle{##}$& $\displaystyle{{}##}$\hfil& $\displaystyle{{}##}$\hfil\crcr#1\crcr}}\,} \catcode`@=12 \def\Q{{\rm Q}} \def\Tr{{\rm Tr}} \def\SS{{\rm SS}} \def\SSS{{\rm SSS}} \def\lanbox{{$\, \vrule height 0.25cm width 0.25cm depth 0.01cm \,$}} \font\tenmb=cmmib10 \def\vsigma{\hbox{\tenmb\char27}} \def\ge{\mathrel{\rlap{\lower.6ex\hbox{$>$}}\raise.35ex\hbox{$\s im$}}} \def\le{\mathrel{\rlap{\lower.6ex\hbox{$<$}}\raise.35ex\hbox{$\s im$}}} \def\go{\mathrel{\rlap{\lower.6ex\hbox{$\sim$}}\raise.35ex\hbox{$>$}}} \def\lo{\mathrel{\rlap{\lower.6ex\hbox{$\sim$}}\raise.35ex\hbox{$<$}}} \def\boxit#1{\thinspace\hbox{\vrule\vtop{\vbox{\hrule\kern1pt \hbox{\vphantom{\tt/}\thinspace{\tt#1}\thinspace}}\kern1pt\hrule}\vrule} \thinspace} %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \centerline{\bf ASYMPTOTICS OF HEAVY ATOMS IN HIGH MAGNETIC FIELDS:} \centerline{\bf I. LOWEST LANDAU BAND REGIONS} \vskip 1truecm \centerline{Elliott H. Lieb$^{(1),(2),}$\footnote{*}{\eightpoint Work partially supported by U.S. National Science Foundation grant PHY90-19433 A01}, Jan Philip Solovej$^{(2),}$\footnote{**}{\eightpoint Work partially supported by U.S. National Science Foundation grant DMS 92-03829} and Jakob Yngvason$^{(3),}$\footnote{***}{\eightpoint Work partially supported by the Hereus-Stiftung and the Research Fund of the University of Iceland. }} \bigskip \noindent{\it $^{(1)}$Department of Physics, Jadwin Hall, Princeton University, P.O. Box 708, Princeton, NJ, 08544\hfil\break $^{(2)}$Department of Mathematics, Fine Hall, Princeton University, Princeton NJ, 08544\hfil\break $^{(3)}$Science Institute, University of Iceland, Dunhaga 3, IS--107 Reykjavik, Iceland} \bigskip \bigskip {\narrower{\it Abstract:\/} The ground state energy of an atom of nuclear charge $Ze$ in a magnetic field $B$ is evaluated exactly to leading order as $Z\to\infty$. In this and a companion work [28] we show that there are 5 regions as $Z\to\infty$: $B\ll Z^{4/3}$, $B\sim Z^{4/3}$, $Z^{4/3}\ll B\ll Z^3$, $B\sim Z^3$, $B\gg Z^3$. Regions 1,2,3,4 (and conceivably 5) are relevant for neutron stars. Different regions have different physics and different asymptotic theories. Regions 1,2,3 are described by a simple density functional theory of the semiclassical Thomas-Fermi form. Here we concentrate mainly on regions 4,5 which cannot be so described, although 3,4,5 have the common feature (as shown here) that essentially all electrons are in the lowest Landau band. Region 5 does have, however, a simple non-classical density functional theory (which can be solved exactly). Region 4 does not, but, surprisingly it can be described by a novel density {\it matrix } functional theory!\bigskip} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%% {\narrower\baselineskip=3.5ex {\bf Table of Contents} \item{I.} Introduction \dotfill 2 \item{II.} The Super-strong Magnetic Field Density Functional \dotfill 12 \item{III.} The Hyper-strong Magnetic Field Density Functional \dotfill 23 \item{IV.} The Density Matrix Functional \dotfill 30 \item{V.} Upper Bound to the Quantum Energy \dotfill 45 \item{VI.} Proof of Concentration in the Lowest Landau Band \dotfill 53 \item{VII.} Reduction to a One-body Problem \dotfill 57 \item{VIII.} Limit of Quantum Mechanics \dotfill 62 \item{IX.} Ionization and Binding of Atoms \dotfill 68 \item{} Bibliography \dotfill 72 \bigskip } \noindent {\bf I. INTRODUCTION} The observed characteristics of pulsars agree well with the theory that they are in fact rapidly rotating neutron stars endowed with a surface magnetic field of the order $10^{12}-10^{13}$ G or even stronger [1]. This field results, presumably, from the trapping of magnetic field lines during the gravitational collapse from which the neutron star emerged, and it is huge by earthly standards. The strongest stationary fields yet produced in the laboratory are of the order $10^6$ G, whereas fields as large as $10^7$ G can, so far, only be sustained for a few milliseconds [2,3]. More importantly, the magnetic field of a neutron star is large compared with the natural atomic unit $B^*=m_e^2e^3c/\hbar^{3}=2.35\times 10^9$ G, which is the field for which the lowest cyclotron (or Landau) energy of an electron, $\mfr1/2\hbar Be/m_ec$, is equal to the Rydberg energy, $\mfr1/2m_ee^4/\hbar^2$, of a hydrogen atom. In fields of such strength the magnetic forces are at least as important as the Coulomb forces and have a decisive effect on the structure of matter. Since $B$ is $10^3$--$10^4$ times as large as the natural unit $B^*$, an asymptotic analysis in $B$ is certainly called for. The interior of a neutron star is presumably composed of highly dense nuclear matter with an exotic mixture of elementary particles at the core. On the other hand the outermost surface layer, through which all information about the interior has to pass, is believed to consist mainly of iron atoms. Iron has a nuclear charge number $Z$ and an electron number $N$ equal to $26$ --- which is large enough to justify an asymptotic treatment. The properties of heavy atoms in high magnetic fields are thus of considerable interest in astrophysics and have been studied by a number of authors using a variety of approximation methods [4-26] since the discovery of pulsars in the late sixties. Among the pioneering papers on this subject those of Kadomtsev [4], Ruderman [6] and Mueller, Rau and Spruch [7] should particularly be mentioned. The review article of Ruderman [12] summarizes the heuristic picture that emerged from these early investigations. Although the above mentioned works provide valuable insights, there remained considerable uncertainty about the status of the approximations used, and it seemed desirable to supplement them with precise statements about the qualitatively different regimes of parameters and the exact leading approximation in each regime. In [27] we announced the results of such an analysis of the ground state properties of the atomic Hamiltonian with a homogeneous magnetic field of strength $B$ in the asymptotic limit when $Z$, $N$, and $B$ tend to infinity. The present paper contains the details of this work for the case that the field $B$ is strong enough to confine the electrons to the lowest Landau band in leading order. The remaining cases, with partial overlap with the one treated here, are the subject of a separate publication [28]. {F}rom the mathematical point of view the present paper is the more interesting part of our work. It contains a novel construct --- the single-particle density matrix theory of Sect.~IV --- that is half way between the full quantum theory and its semiclassical approximation. It reduces the problem to the computation of a fixed, finite number of functions instead of (as in Hartree-Fock theory) a number that tends to infinity as the size of the atom tends to infinity. It resolves a problem that at first seemed to us to be beyond the reach of exact asymptotic analysis. The $Z\to\infty$ limit of the ground state of ordinary matter without a magnetic field was studied rigorously in [29-31] and shown to be described exactly by Thomas-Fermi (TF) theory. Since TF theory is already fairly accurate for iron with $Z=26$, the leading contribution for $Z\to\infty$ to the energy in a magnetic field can be expected to be relevant also for the surface properties of a neutron star. Before we turn to a discussion of this asymptotic, heavy atom, regime we should mention that there also exists considerable physics literature on hydrogen and other light atoms in high magnetic fields, cf., e.g., [32-37]. The mathematically rigorous paper [38] also deals mainly with hydrogen but contains several interesting theorems for atoms of arbitrary size. For the convenience of the reader we now give a brief summary of our main results including the cases treated in [28]. For a more detailed survey we refer to [27] and [39]. Our starting point is the Pauli Hamiltonian for an $N$-electron atom with nuclear charge $Ze$ in a magnetic field with vector potential $\A$ such that curl $\A={\bf B}$: $$H_N=\sum_{i=1}^N\left\{ (\p^i+\A(x^i))^2+{\vsigma}^i\cdot{\bf B} -Z\vert x^i\vert^{-1}\right\}+\sum_{1\leq i 0$. As before $\beta = B/Z^{4/3}$. Our conclusion is that there is some constant $\delta = \delta (\lambda^{2/3}\beta, \Lambda)$ such that $\delta (\lambda^{2/3}\beta, \Lambda) \rightarrow 0$ as $\lambda^{2/3}\beta \rightarrow \infty$ and $$ (1 - \delta) E^{\rm Q}(N,Z,B) \geq E^{\rm Q}_{\rm conf} (N,Z,B). \eqno(1.16) $$ Since $E^{\rm Q}_{\rm conf} \geq E^{\rm Q}$ (by definition), we have in particular that $E^{\rm Q}/E^Q_{\rm conf}\to1$ as $ \lambda^{2/3}\beta \rightarrow \infty$.} Theorem~1.2 amounts to the assertion that the ground state energy can be calculated (to leading order in $\beta^{-1}$) by confining all the electrons to the lowest Landau band. It is not asserted that $\Pi^N_0 \Psi = \Psi$ for the true ground state of $H_N$. One immediate consequence should be noted right away: In the lowest Landau band the spin points in a fixed direction and hence {\it we can forget about the spin} when discussing regions 3, 4 and 5. Henceforth, the arena of the game will therefore simply be $\bigwedge\limits^N_{} L^2({\bf R}^3)$ rather than $\bigwedge\limits^N_{} L^2({\bf R}^3;{\bf C}^2)$. Let us now consider the ground state wave function $\Psi$ of an atom with large $N$ and $Z$ in a field such that $B/Z^{4/3}$ is also large but $B/Z^3$ not necessarily small. According to Theorem 1.2 we may, to leading order, assume that all electrons are in the lowest Landau band. In this band the kinetic energy associated with the motion along the field (including the spin contribution) is zero, which means that % $$\Pi_0 H_\A\Pi_0=-{\partial^2\over \partial x_3^2}\Pi_0\eqno(1.17)$$ % Anticipating the fact that all exchange and correlation terms can be neglected for $Z$ large enough (which, however, is by no means easy to prove when $B\gg Z^5$), we can thus reasonably expect that $E^{\rm Q}=(\Psi, H_N\Psi)$ is correctly given to leading order by the expression % $$ \int -{\partial^2\over \partial x_3^2} K_\Psi({x_{\perp}},x_3;{x_{\perp}},x_3^\prime) \vert_{x^\prime_3=x_3} d x -Z\int \uprho_\Psi(x)\vert x\vert^{-1}d x +D(\uprho_\Psi,\uprho_\Psi),\eqno(1.18)$$ where $$K_\Psi(x;x^\prime)=N\int \Psi(x,x^2,\dots,x^N) \overline{\Psi(x^\prime,x^2,\dots,x^N)} dx^2\dots dx^N,\eqno(1.19)$$ $$\uprho_\Psi(x)=K_\Psi(x;x).\eqno(1.20)$$ The important point to note is that since only differentiation along the $x_3$ direction enters in the kinetic energy (1.17), it is {\it not} the full one-body density matrix $K_\Psi(x;x^\prime)$ that is relevant but only its ``semi-diagonal'' part $$\Gamma_{x_\perp}^\Psi(x_3,x^\prime_3)= K_\Psi({x_{\perp}},x_3;{x_{\perp}},x_3^\prime)$$ with respect to the variables perpendicular to the field. Furthermore, the kernel $\Pi_0(x,x^\prime)$ has the property that $\Pi_0(x_\perp,x_3;x_\perp,x^\prime_3)=(B/2\pi)\delta(x_3- x^\prime_3)$ from which it follows easily that for $\Psi$ in the lowest Landau band and each value of $x_\perp$ one has (see Lemma~4.1) $0\leq \Gamma^\Psi_{x_{\perp}}(x_3,x_3^\prime)\leq (B/2\pi)I$ as operators on $L^2(\R)$. The physics behind this inequality is the Pauli principle and the fact that the density of states per unit area perpendicular to the field is $B/(2\pi)$ in the lowest Landau band. The above considerations suggest a functional that does not depend on the density alone, but rather on functions $ \Gamma:x_{\perp}\mapsto\Gamma_{x_{\perp}}(x_3,y_3)$ from $\R^2$ to density {\it matrices} on $L^2(\R)$, satisfying the condition $$0\leq\Gamma_{x_{\perp}}\leq(B/2\pi)I\quad\hbox{as an operator on\ } L^2({\bf R}).\eqno(1.21)$$ The functional is defined in analogy with (1.18) as $${\cal E}^{\rm DM}[\Gamma]:=\int_{{\bf R}^2} {\rm Tr}_{L^2({\bf R})} [-\partial_3^2\Gamma_{x_{\perp}}]d x_{\perp} -Z\int|x|^{-1}\uprho_{\Gamma}(x)dx +D(\uprho_\Gamma,\uprho_\Gamma),\eqno(1.22) $$ where $\uprho_{\Gamma}(x)=\Gamma_{x_{\perp}}(x_3,x_3)$ and $\partial_3=\partial/\partial x_3$. Since the functional only contains derivatives with resect to the $x_3$ direction it might appear to be unbounded below; it turns out, however, that the \lq\lq hard core\rq\rq\ condition (1.21) prevents collapse. We define $$ E^{\rm DM}(N,Z,B):=\inf\{{\cal E}^{\DM}[\Gamma]:\int \uprho_\Gamma\leq N,\ \Gamma\ \hbox{satisfies}\ (1.21)\}.\eqno(1.23) $$ The DM theory has two nontrivial parameters, $\lambda=N/Z$ and $\eta:=B/(2\pi Z^3)$; the energy obeys the {\bf scaling relation} $$\openup1\jot\tabskip0pt \halign to\displaywidth{#\hfill\tabskip0pt plus1fil &\tabskip0pt$\hfill#\ ${}&$#\hfill$\tabskip0pt plus1fil &\llap#\tabskip0pt\cr Reg. 4&E^{\rm DM}(N,Z,B)=&Z^3 E^{\rm DM}(N/Z,1,B/Z^3).&(1.24)\cr}$$ The main limit theorem for strong fields, to be proved in Sect.~VIII, can now be stated as follows {\bf 1.3. THEOREM (Energy asymptotics for regions 3,4,5).} {\it Let $N/Z$ be fixed and suppose $B/Z^{4/3} \rightarrow\infty$ as $Z \rightarrow \infty$. Then $$E^{\rm Q}(N,Z,B)/E^{\rm DM}(N,Z,B) \rightarrow 1 \qquad \hbox{as} \ \ Z \rightarrow \infty.$$} Notice that there is an overlap in the parameter regions covered by Theorems 1.1 and 1.3. In fact, both theorems apply to region 3 described above. The functional $\E^{\rm DM}$ has a unique minimizer for a given $N$, and the associated variational problem amounts to finding, for each value of $x_\perp$, the eigenfunctions of a {\it one-dimensional\/} Schr\"odinger operator with a potential that has to be determined in a self consistent way from the charge density $\uprho_\Gamma$. The important point is that the number of eigenfunctions that have to be considered at each point $x_\perp$ is {\it finite} and depends only on the parameters $\lambda$ and $\eta$, but is {\it independent of} $N$. This makes the problem vastly more simple than the corresponding minimization problem for the full one-body density matrix, which is totally unmanageable as $N\to\infty$. In fact, for large values of $B/Z^3$ the minimization problem for $\E^{\rm DM}$ can easily be attacked on the computer since only the first few eigenfunctions have to be taken into account. Conversely, if $B/Z^3\to 0$, the number of eigenfunctions tends to infinity and we recover the semiclassical theory described by $\E^{\rm STF}$. In Theorem~4.6 we show that there is a critical value for $B/Z^3$, above which at most {\it one} eigenfunction enters for each fixed $x_\perp$. In this case $\E^{\rm DM}$ collapses to a density functional, since the kinetic term $\int{\rm Tr}_{L^2({\bf R})} (-\partial_3^2\Gamma_{x_{\perp}})d x_{\perp}$ can be replaced by $\int(-\partial_3\sqrt{\uprho_\Gamma(x)})^2 d x$. We shall refer to this special form of the density matrix functional as the {\bf super-strong functional} and denote it by $\E^{\rm SS}$. More precisely, this is the functional of the density $\uprho$ given by $$\E^{\SS} [\uprho] := \int \left( {\partial \over \partial x_3} \sqrt{\uprho (x)} \right)^2 dx - Z\int \vert x \vert^{-1} \uprho (x) dx + D(\uprho,\uprho). \eqno(1.25)$$ As a historical remark we note that essentially the same density functional was written down in 1971 by Kadomtsev and Kudryavtsev [8] who estimated its ground state energy by a simple variational ansatz. In Sect.~III we show that as $B/Z^3\to\infty$ the energy $\E^{\SS}$ converges, after an appropriate rescaling, to the energy of a functional of a {\it one-dimensional} density, called the {\bf hyper-strong functional}. It is defined as $$\E^{\HS} [\uprho] := \int_{\R} \left( {d \over dx} \sqrt{\uprho (x)} \right)^2 dx - \uprho (0) + \mfr1/2 \int_{\R} \uprho (x)^2 dx, \eqno(1.26)$$ where $\uprho$ is now a density depending on the one-dimensional variable $x\in\R$. In Theorem 3.1 we show that the minimizer of $\E^{\HS}$ can be evaluated {\it exactly}. Combining the solution with Theorem 1.3 we thus obtain the following result for the quantum mechanical ground state energy in region~5. (See (3.7) and (8.10).) {\bf 1.4. THEOREM (Energy asymptotics for region 5).} {\it If $Z\to\infty$ and $B/Z^3\to\infty$ with $N/Z=\lambda$ fixed, then $$E^{\rm Q}(N,Z,B)/(Z^3[\ln(B/Z^3)]^2)\rightarrow \cases{-{1\over 4}{\lambda} +{1\over 8}{\lambda}^2 -{1\over 48}{\lambda }^3, & for $\lambda\leq2$\cr&\cr -{1\over6},& for $\lambda\geq2$}.$$} This is one of the few cases in atomic physics where the quantum mechanical ground state can be evaluated in closed form. One of the remarkable corollaries of Theorem 1.4, which will be proved in Sect.~IX, is the following --- which we call a theorem because of its physical importance. In a normal $B = 0$ atom [42-44], the maximum number of electrons, $N^{\rm Q}_c$, that can be bound to a nucleus, satisfies $N^{\rm Q}_c/Z \rightarrow 1$ as $Z \rightarrow \infty$. The situation is the same in regions 1, 2, 3 %zzz where a TF type theory continues to be valid and atoms are spherical; {\it cf.\/} Theorems~3.18 and 3.23 in [31]. %zzz The asymptotic neutrality result in [42] can be extended to regions 1, 2, 3 by using the methods in [44]. The situation in regions 4 and 5 is very different (although we must admit we prove this assertion only for region 5): {\it the ionization can be proportional to $Z$ itself.\/} In these two regions the atom also happens to be non- spherical. Of course this ionization result applies only to a single atom and cannot be expected to hold for bulk matter because the Coulomb repulsion of very many ionized atoms would eventually be overwhelming. {\bf 1.5. THEOREM (Ionization of atoms).} {\it The maximal number, $N^{\rm Q}_c$, of electrons that can be bound to an atom of nuclear charge $Z$, defined by $$N^{\rm Q}_c = \max \{ N: E^{\rm Q} (N,Z,B) < E^{\rm Q} (N-1,Z,B)\}, \eqno(1.27)$$ satisfies $$\liminf N^{\rm Q}_c/Z \geq 2 \quad \hbox{as} \quad Z \rightarrow \infty \quad \hbox{and} \quad B/Z^3 \rightarrow \infty. \eqno(1.28)$$} We {\it conjecture\/} that the $Z \rightarrow \infty$ limiting value of $N^{\rm Q}_c/Z$ (and $N^{\DM}_c/Z$ in the density matrix theory) is an increasing function of $B/Z^3$. In any event, $N^{\rm Q}_c/Z = 2$ for $B \gg Z^3$ and this is probably the largest value of that ratio. Theorem 1.5 should not be confused with a theorem of Lieb [45] that $N^{\rm Q}_c < 2 Z+1$, which is not relevant here because that theorem does not include the $\vsigma \cdot \B$ term of the Pauli Hamiltonian. This point, which we confess confused us for some time, is discussed in Sect.~IX. Another property that holds for regions 4 and 5, but not 1, 2, 3, is that {\it atomic binding energies are on the scale of the atomic energy itself.} In a TF type theory atoms do not bind together [29-31] but in region 5 they do bind, and presumably also in the DM theory in region 4. This translates into the following (also proved in Sect.~IX). {\bf 1.6. THEOREM (Strong atomic binding).} {\it Let $E^{\rm Q}_\b (Z_1,Z_2,B)$ denote the binding energy (defined in Sect.~IX) of a neutral molecule consisting of two nuclei of charges $Z_1$ and $Z_2$. It is always true that $\lim E^{\rm Q}_\b (Z_1, Z_2, B)/[E^{\rm Q} (Z_1, Z_1, B) + E^{\rm Q} (Z_2, Z_2, B)] > 0$ as $Z_{1,2} \rightarrow \infty$ and $B/Z_{1,2}^3 \rightarrow \infty$, provided $Z_1/Z_2$ is bounded away from 0 and $\infty$. In particular, if $Z_1 = Z_2 = Z$ then $$E^{\rm Q}_\b (Z, Z, B)/2 \vert E^{\rm Q} (Z,Z,B) \vert \rightarrow 3, \hskip 10pt as\ Z\rightarrow\infty. $$} In view of the somewhat conflicting statements in the literature about atomic binding in a strong magnetic field (see e.g. [6, 9, 10, 12, 16, 20, 26]), we emphasize that this is a {\it theorem} about quantum mechanics in the limit considered. It does not necessarily contradict the Hartree-Fock calculations of [16] and [20] that seem to indicate that iron is weakly bound in fields of the order $10^{12}-10^{13}$ G. \medskip In this overview we have concentrated on the properties of the energy for brevity. In all cases, however, corresponding results for the electron density can also be proved. The precise statements can be found in Theorems~8.1 and 8.2 in Sect.~VIII. We conclude the introduction with a few remarks about the organization of the paper. It was found convenient to discuss first the properties of all the asymptotic functionals and then the quantum mechanical limit theorems. For the former we start with the functional $\E^{\SS}$, although it is in reality a special case of $\E^{\DM}$. The main reason is that we think it is easier to assimilate the latter after some acquaintance with the former. Also, it turns out that certain results about $\E^{\SS}$ are needed in our discussion of the density matrix functional. We proceed by considering the hyper-strong functional $\E^{\HS}$, and finally the density matrix functional $\E^{\DM}$. We turn to the quantum mechanical limit theorems in Sect.~V by deriving a variational upper bound for the quantum mechanical energy. In the next section we prove the result on confinement in the lowest Landau band (Theorem 1.2). The most difficult problem, treated in Sect.~VII, is to estimate the negative exchange energy from below, because the Lieb-Oxford inequality [46], although universally valid, is not strong enough if $B$ is of order higher than $Z^5$. Our exchange bound is given in Theorem~7.1. After the problem has been reduced in this way to the study of a one-body Hamiltonian the completion of the proof of Theorem~1.3 turns out to be fairly simple. We thank I.~Fushiki, E.~H.~Gudmundsson, M.~Loss, C.~J.~Pethick, M.~Ruderman and L.~Spruch for valuable comments. JY is grateful for hospitality at Princeton University, Institute for Theoretical Physics, G\"ottingen, and I.~H.~E.~S., Bures-sur-Yvette. %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf II. THE SUPER-STRONG MAGNETIC FIELD DENSITY FUNCTIONAL} In this section we introduce a density functional $\E^{\SS} [\uprho]$ that is very different from the Thomas-Fermi functional discussed in [28] and in the Introduction. It involves derivatives of the density $\uprho (x)$, and in some respects resembles a density functional appropriate for bosons rather than fermions. Nevertheless, as we shall show in Sect.~VIII, this functional correctly gives the limiting energy for an atom in a magnetic field $B \gg Z^3$. Our functional $\E^{\SS}$ involves a parameter $\eta=B/(2\pi Z^3)$. The functional makes sense for all $\eta > 0$, and we shall first undertake the usual task of proving existence and uniqueness of a minimizing density $\uprho$. However, as the motivation for $\E^{\SS}$ comes from quantum mechanics with $\eta$ large, the $\E^{\SS}$ theory is physically significant mainly in that case, where the theory is asymptotically exact if $Z\to\infty$ as well. In Sect.~III, we show how to extract the $\eta \rightarrow \infty$ limit of the super-strong theory. Oddly enough, this limiting theory (which we call the {\bf hyper-strong theory}) can be solved {\it exactly}. One point worth noting is that the errors, in this hyper-strong theory, turn out to be {\it non}-negligible in the sense that they are typically approximately of the order $1/\vert \ln \eta \vert$. Thus, while the hyper-strong limit is the correct mathematical limit for $B/Z^3 \rightarrow \infty$, it is worthwhile studying the somewhat more complicated functional $\E^{\SS}$ without taking the limit $\eta \rightarrow \infty$. This claim is supported by the fact, which we show in Sects.~IV and VIII, that the super-strong theory is also correct (as $Z\to\infty$) for large but finite values of the parameter $\eta$, without taking the limit $\eta\to\infty$. Another reason for studying the super-strong theory, even in the limit as $\eta\to\infty$, is that it contains more information than the hyper-strong theory; in particular it tells us about the cross-sectional density in the atom, whereas the hyper-strong theory deals only with the cross-sectional integral of the density. The super-strong energy functional is defined by (1.25). Note that the first term (the kinetic energy) involves only the derivative in the $x_3$ direction. The last two terms are familiar. The first term is the kinetic energy in the $x_3$ direction --- the kinetic energy in the $x_\bot = (x_1, x_2)$ direction being absent on account of the assertion in Theorem~1.2 that all the electrons can be assumed to be in the lowest Landau band, and hence have zero kinetic energy in the $x_\bot$ direction. As usual, we have the condition on the particle number $$\int \uprho = N, \eqno(2.1)$$ but now (1.25) has to be supplemented by another, strange looking condition: $$\int_{\R} \uprho (x_\perp,x_3) dx_3 \leq {B \over 2 \pi} \ {\rm for \ each \ } x_\bot = (x_1, x_2). \eqno(2.2)$$ This condition comes from the fact, to be explicated in Theorem~4.6, that for large $B/Z^3$ at most one eigenstate (the ground state, of course) in the $x_3$ direction is occupied by an electron. That fact, together with (1.21), leads to (2.2). The domain of definition of $\E^{\SS}$ is the set $\sc^{\SS}$ of nonnegative measurable functions $\uprho$ satisfying the conditions \item{(a)} $\int \uprho (x) dx < \infty$ \item{(b)} $\int \limits_\R \uprho (x_\bot, x_3) dx_3 \leq C$ for some $C < \infty$ independent of $x_\bot$ \hfill (2.3) \item{(c)} The distributional derivative $\partial \sqrt{\uprho}/\partial x_3$ is a function in $L^2 (\R^3)$. \medskip\noindent The functional $\E^{\SS}$ is easily shown to be strictly convex on $\sc^{\SS}$. Note, however, that $\E^{\SS}$ is {\it not} convex as a functional of $\psi = \sqrt{\uprho}$. The proof of the convexity is the same as in [31], Theorem 7.1. As a preliminary to our theorems here, we note some properties of functions in $\sc^{\SS}$. (i) For almost every $x_\bot$ the function $\psi : = \sqrt{\uprho}$ is a H\"older continuous function of $x_3$ with exponent 1/2. This is so because $\psi (x_\bot, \cdot)$ is in $H^1 (\R)$ for almost every $x_\bot \in \R^2$. To be precise $$\eqalignno{\vert \psi (x_\bot, a) - \psi (x_\bot, b) \vert &= \left\vert \int^b_a (\partial \psi/\partial x_3) dx_3 \right\vert \leq \left\vert\int^b_a dx_3 \right\vert^{1/2} \left\vert \int^b_a (\partial \psi /\partial x_3)^2 dx_3 \right\vert^{1/2} \cr &\leq \vert a-b \vert^{1/2} \left\vert \int (\partial \psi /\partial x_3)^2 dx_3 \right\vert^{1/2} .\qquad&(2.4)\cr}$$ (ii) $\uprho \in L^3 (\R^3)$ and, for some universal constant $K$, $$\int \limits_{\R^3} \uprho^3 \leq K C^{2} \int \limits_{\R^3} (\partial \psi /\partial x_3)^2 ,\eqno(2.5)$$ where $C$ is the constant in (2.3). Inequality (2.5) follows easily from the Sobolev inequality in $\R$, $K\int (\partial f /\partial x)^2 \geq \left( \int f^6 \right) \left( \int f^2 \right)^{-2}$. After scaling, we see from (2.2) that the $C$ of interest in our problem can be taken to be unity. The following lemma is useful for proving the existence of minimizers for $\E^{\SS}$. {\bf 2.1. LEMMA (Energy of one-dimensional Coulomb problem).} {\it For $Z > 0$ and $a > 0$, let $\widehat h_{Z,a}$ be the operator on $L^2 (\R)$ given by $$\widehat h_{Z,a} = - {d^2 \over dx^2} - {Z \over \sqrt{x^2 + a^2}}.$$ Then the eigenvalues, $-\widehat \mu_n$, satisfy $$\eqalignno{\widehat \mu_1 &\leq Z^2 \{ 1 + [\sinh^{-1} (1/Za)]^2 \} \cr \widehat \mu_{2n} \ \hbox{\it and}\enskip \widehat \mu_{2n+1} &\leq Z^2/4n^2, \quad n=1, \dots .\cr}$$} \indent {\it Proof:} By scaling, we see that $\widehat h_{Z,a} = Z^2$ times an operator that depends only on $Za$. Therefore we can assume $Z = 1$. With $T:= \int_\R (d \psi /d x)^2$ and $\psi \in H^1 (\R)$, (which we can assume to be real), $$\eqalignno{(\psi, \widehat h_{1,a} \psi) &= T - \int_\R \psi (x)^2 (x^2 + a^2)^{-1/2} dx \cr &=T - \int \limits_{\vert x \vert \leq 1} {d \over dx} \left[ \int^x_0 (y^2 + a^2)^{1/2} dy \right] \psi^2 (x) dx - \int \limits_{\vert x \vert \geq 1} \psi (x)^2 (x^2 + a^2)^{-1/2} dx \cr &\geq T + 2 \int \limits_{\vert x \vert \leq 1} \left[ \int^x_0 (y^2 + a^2)^{-1/2} dy \right] \psi (x) {d \psi (x) \over dx} dx \cr &\quad - \int^1_0 (y^2 + a^2)^{-1/2} dy [\psi (1)^2 + \psi (-1)^2] - \int_\R \psi^2 \cr &\geq T - 2\sinh^{-1} (a^{-1}) \int_\R \left\vert \psi (x) {d \psi (x) \over dx} \right\vert dx - \int_\R \psi^2 \cr &\geq T - 2 \sinh^{-1} (a^{-1}) I^{1/2} T^{1/2} - I }$$ with $I = \int \psi^2$. The fourth line was obtained from the third by noting that $\psi (1)^2 \leq 2 \int^\infty_1 \psi (x) [d\psi (x)/dx] dx$. The bound on $\widehat \mu_1$ is obtained by minimizing the last expression with respect to $T$. To estimate $\widehat \mu_{2n}$ we first replace $-Z (\vert x \vert^2 + a^2)^{-1/2}$ with the lower bound $-Z \vert x \vert^{-1}$. Thus, if $\psi_{2n}$ denotes the $2n$-th normalized eigenfunction, $$-\mu_{2n} \geq ( \psi_{2n} ,\ ( p^2 - Z \vert x \vert^{-1}) \psi_{2n} ).$$ Since the potential $x \mapsto Z (\vert x \vert^2 + a^2)^{-1/2}$ is symmetric about $x = 0$, the eigenfunction $\psi_{2n}$ must have a node at $x = 0$. The eigenvalues of the operator on the right side of this inequality, when restricted to functions with a node at the origin, are precisely the same as the eigenvalues of the hydrogen atom restricted to radial functions. Thus we obtain $\widehat \mu_{2n} \leq Z^2/4n^2$. Since $\widehat \mu_{2n+1} \leq \widehat \mu_{2n}$, the lemma follows. \lanbox {\bf 2.2. THEOREM (Unique minimizer for $\E^{\SS}$).} {\it For each $N,Z$ and $B > 0$ there is a unique (and hence axially symmetric) $\uprho^{\SS} \in \sc^{\SS}$ satisfying the conditions (2.2) and $$\int \uprho^{\SS} \leq N, \eqno(2.6)$$ and such that $$\E^{\SS} [\uprho^{\SS}] = E^{\SS} (N,Z,B): = \inf \big\{ \E^{\SS} [\uprho] \,:\ \uprho \in \sc^{\SS}, \int \uprho \leq N, \ \ \uprho \ {\rm satisfies} \ (2.2) \big\}. \eqno(2.7)$$ The energy $E^{\SS}(N,Z,B)$ is a monotonically nonincreasing, convex function of $N$, and in fact $$E^{\SS} (N,Z,B) = \inf \{ \E^{\SS} [\uprho] \,: \ \uprho \in \sc^{\SS}, \int \uprho = N, \ \uprho \ {\rm satisfies \ } (2.2) \}.$$ Moreover, if $\uprho^{(n)}$ is any minimizing sequence for (2.7) (i.e. $\E^{\SS} [\uprho^{(n)}] \rightarrow E^{\SS} (N,Z,B)$ and $\int \uprho^{(n)} (x) dx \leq N, \int \uprho^{(n)} (x) dx_3 \leq B/2\pi)$, then each of the three terms in the energy (1.25) converges to the corresponding term for $\uprho^{\SS}$.} {\it Remark:} $\uprho^{\SS}$ depends on $N$, $Z$ and $B$, but our notation suppresses this for simplicity. {\it Proof:} We begin by finding a lower bound for $\E^{\SS} [\uprho]$ that behaves like $NZ^2\left[\ln(B/Z^2N)\right]^2$ for $B/(Z^2N)$ large. To do so we ignore the third (repulsion) term in (1.25). For each $\uprho \in \sc^{\SS}$ satisfying (2.1) and (2.2) we define $$\lambda (\uprho; x_\bot) = \int_\R \uprho (x_\bot, x_3) dx_3 \leq B/2\pi.$$ By Lemma 2.1 we have that $$\E^{\SS} [\uprho] \geq - \int_{\R^2} \lambda (\uprho; x_\bot) Z^2 \{ 1 + [\sinh^{-1} (1/(Z \vert x_\bot \vert)]^2 \} dx_\bot.$$ We know that $\int_{\R^2} \lambda (\uprho; x_\bot) dx_\bot = N$. Since the factor in braces $\{ \ \}$ is a decreasing function of $\vert x_\bot \vert$, a lower bound for $\E^{\SS} [\uprho]$ is obtained as follows: We have $$\E^{\SS} [\uprho] \geq - {Z^2B \over 2\pi} \int \limits_{\vert x_\bot \vert < \sqrt{2N/B}} \left\{ 1 + \left[\sinh^{-1} (1/Z \vert x_\bot \vert) \right]^2 \right\} dx_\bot .\eqno(2.8)$$ The function $\sinh^{-1}(t)$ behaves like $t$ for small $t$ and like $\ln t$ for large $t$. From this we conclude that the expression (2.8) is bounded below\footnote{$^\dagger$}{\eightpoint Throughout we shall use (const.) to denote any positive universal constant.} by $-\hbox{(const.)}NZ^2[1+[\ln (B/Z^2N)]^2]$. Now fix $N$ and $B$ and let $\uprho^{(1)}, \uprho^{(2)}, \dots$ be a minimizing sequence in $\sc^{\SS}$, i.e. $\E^{\SS} [\uprho^{(n)}] \rightarrow E^{\SS} (N,Z,B)$ as $n \rightarrow \infty$ and $\int \uprho^{(n)} \leq N$ for each $n$. The kinetic energy $T[\uprho^{(n)}]:= \int (\partial \sqrt{\uprho^{(n)}} /\partial x_3)^2$ is bounded. The reason is that we can write $\E^{\SS} [\uprho] := \mfr1/2 T[\uprho] + \widetilde \E^{\SS} [\uprho]$. By the same argument as above, $\widetilde \E^{\SS} [\uprho]$ is bounded below, and hence $\mfr1/2 T[\uprho^{(n)}]$ must be bounded above. By inequality (2.5) the $L^3 (\R^3)$ norm of $\uprho^{(n)}$ is uniformly bounded in $n$. By the Banach-Alaoglu Theorem there is a subsequence (which we continue to denote by $\uprho^{(n)}$) that converges weakly in $L^3 (\R^3)$ to some function $\uprho^\infty \in L^3 (\R^3)$. (Of course $\uprho^\infty \geq 0$ since weak limits of nonnegative functions are nonnegative; and $\int\uprho^\infty\leq N$ since otherwise, for some ball $K$ we would have $N<\int_K\uprho^\infty=\lim_{n\to\infty}\int_K\uprho^{(n)}\leq N$.) By Mazur's theorem we can take convex combinations of the $\uprho^{(n)}$'s so that the new sequence (which we denote again by $\uprho^{(n)}$) converges {\it strongly} to $\uprho^\infty$ in $L^3 (\R^3)$. Moreover, we can require that the new $\uprho^{(n)}$ be a convex combination only of the old subsequence starting with $n$, i.e., the subsequence $\uprho^{(n)},\uprho^{(n+1)},\uprho^{(n+2)},\ldots$. Since $\uprho^{(n)}$ is a bounded sequence in $L^1 (\R^3)$, we can also demand that $\uprho^{(n)}$ converges strongly to $\uprho^\infty$ in $L^p (\R^3)$ for some fixed $3 \geq p > 1$. We choose $p = 6/5$. Since $\uprho \mapsto \E^{\SS} [\uprho]$ is convex, this new, strongly convergent sequence is also minimizing. By the standard proof of the Riesz-Fischer theorem that $L^p$ is complete we can pass to a further subsequence and demand that the following holds: \medskip \item{(i)} $\lim \limits_{n \rightarrow \infty} \uprho^{(n)} (x) = \uprho^\infty (x)$ for almost every $x$. \item{(ii)} There is a function $G \in L^3 (\R^3) \cap L^{6/5} (\R^3)$ such that $\uprho^{(n)} (x) \leq G(x)$ for all $x$ and $n$. \smallskip \noindent By (i) and Fatou's Lemma, $\uprho^\infty$ satisfies condition (2.2). Let us investigate the convergence properties of the three terms $T [\uprho] - A[\uprho] + R[\uprho]$ that define $\E^{\SS} [\uprho]$ according to (1.25). Clearly, $$\lim \limits_{n \rightarrow \infty} A[\uprho^{(n)}] = A[\uprho^\infty]$$ thanks to the strong convergence of $\uprho^{(n)}$ to $\uprho^\infty$ in $L^3 (\R^3)$. (Actually, weak convergence would suffice here.) Next $$\lim \limits_{n \rightarrow \infty} R[\uprho^{(n)}] = R[\uprho^\infty]$$ by dominated convergence (since $\uprho^{(n)} \leq G$ and $G \in L^{6/5} (\R^3)$; but $(x,y) \mapsto$\hfill\break $G(x) G(y) V \vert x-y \vert^{-1}$ is in $L^1 (\R^3 \times \R^3)$ by Young's inequality). The difficult term is $T[\uprho^{(n)}]$. We claim that $\uprho^\infty \in \sc^{\SS}$ and that $$\liminf \limits_{n \rightarrow \infty} T[\uprho^{(n)}] \geq T[\uprho^\infty]. \eqno(2.9)$$ To prove (2.9) we use the fact that $$\left\{ \int \limits_{\R^3} \vert \partial \phi /\partial x_3 \vert^2 \right\}^{1/2} = \sup \left\{ \int \limits_{\R^3} \phi (\partial f/\partial x_3) \,: \ f \in C^\infty_0 (\R^3) \ \hbox{and} \ \int \limits_{\R^3} f^2 = 1 \right\}$$ for all functions $\phi \in L^2 (\R^3)$. Implicit in this equation is the assertion that the finiteness of the right side implies that the distributional derivative of $\phi$ is an $L^2 (\R^3)$ function. Now, for each fixed $f \in C^\infty_0 (\R^3)$ with $\int f^2 = 1$ we have $$\liminf \limits_{n \rightarrow \infty} \left\{ \int (\partial \sqrt{\uprho^{(n)}} /\partial x_3)^2 \right\}^{1/2} \geq \liminf \limits_{n \rightarrow \infty} \left\vert \int \sqrt{\uprho^{(n)}} \partial f/\partial x_3 \right\vert = \left\vert \int \sqrt{\uprho^\infty} \partial f/\partial x_3 \right\vert,$$ where the last equality is a consequence of dominated convergence. This implies (2.9). We have shown that $E^{\SS} (N,Z,B) = {\mathop{\lim}_{n \rightarrow \infty}} \E^{\SS} [\uprho^{(n)}] \geq \E^{\SS} [\uprho^\infty]$, from which it is evident that $\uprho^\infty$ is a minimizer. The uniqueness property follows trivially from the strict convexity of $\uprho \mapsto \E^{\SS} [\uprho]$. In fact this convexity also implies the convexity of $E^{\SS} (N,Z,B)$ in $N$. The monotonicity of $E$ is a trivial consequence of its definition (2.7). The fact that $E^{\SS}$ agrees with the energy defined using the strong condition $\int \uprho = N$ follows easily from the fact that extra charge can always be ``added at infinity'' as in ref. [30]. It remains to prove that when $\uprho^{(n)}$ is any minimizing sequence then the three terms in (1.25) converge to those for $\uprho^{\SS}$. First, by the Banach-Alaoglu theorem, $\uprho^{(n)} \rightharpoonup \widehat{\uprho}$ weakly in $L^3 (\R^3)$ for some subsequence and some $\widehat{\uprho}$. By the proof we just gave, $\widehat{\uprho}$ is minimizing. But the minimizer is unique and hence $\widehat{\uprho} = \uprho^{\SS}$. We also remark that the uniqueness implies that $\uprho^{(n)} \rightharpoonup \uprho^{\SS}$ without the necessity of passing to a subsequence. By the weak convergence of $\uprho^{(n)}$ to $\uprho^{\SS}$, $\int Z \vert x \vert^{-1} \uprho^{(n)} (x) dx$ converges to $\int Z \vert x \vert^{-1} \uprho^{\SS} (x) dx$ and hence the middle term in (1.25) behaves properly. We conclude that $$\lim \limits_{n \rightarrow \infty} T[\uprho^{(n)}] + R[\uprho^{(n)}] = T [\uprho^{\SS}] + R[\uprho^{\SS}]$$ and therefore it suffices to prove that $\liminf \limits_{n \rightarrow \infty} T[\uprho^{(n)}] \geq T[\uprho^{\SS}]$ and $\liminf \limits_{n \rightarrow \infty} R[\uprho^{(n)}] \geq R[\uprho^{\SS}]$. By Mazur's theorem there are convex combinations (call them $\widetilde {\uprho}^{(n)}$) of the $\uprho^{(n)}$'s so that $\widetilde{\uprho}^{(n)}$'s converge strongly in $L^3 (\R^3)$ to $\uprho^{\SS}$ and satisfy (i) and (ii) above. As we saw, this implies that $\lim {\mathop{\inf}_{n \rightarrow \infty}} T[\widetilde{\uprho}^{(n)}] \geq T[\uprho^{\SS}]$. However, $\uprho \mapsto T[\uprho]$ is convex and this convexity implies that $\lim {\mathop{\inf}_{n \rightarrow \infty}} T[\uprho^{(n)}] \geq \lim {\mathop{\inf}_{n \rightarrow \infty}} T[\widetilde{\uprho}^{(n)}]$, which is the desired conclusion. The same strategy proves that $\lim {\mathop{\inf}_{n \rightarrow \infty}} R[\uprho^{(n)}] \geq R[\uprho^{\SS}]$. \lanbox The $\E^{\SS}$ functional has a natural scaling which is given by defining a scaled density by $\widetilde{\uprho}(x)=Z^{-4}\uprho(Z^{-1}x)$. Then $\E^{\SS}[\uprho]=Z^3\E^{\SS}[\,\widetilde{\uprho}\,]$. Moreover, if $\uprho$ satisfies (2.1) and (2.2) then $\widetilde{\uprho}$ satisfies $\int_\R\widetilde{\uprho}dx_3\leq B/(2\pi Z^3)$ and $\int \widetilde{\uprho}\leq N/Z$. We can then conclude that the energy $E^{\SS}$ satisfies the {\bf scaling relation} $E^{\SS} (N,Z,B) = Z^3 E^{\SS} (N/Z, 1, B/Z^3)$ (compare (1.24)). There are thus two non-trivial parameters in the theory $\lambda = N/Z$ and $\eta = B/(2\pi Z^3)$. Having proved the existence of a unique minimizer, we wish next to establish some of its properties. To begin with we can note, that there is a number $N^{\SS}_c$ (which {\it a-priori} might be $+\infty$) such that the energy $E^{\SS}$ is a strictly decreasing function of $N$ for $N < N^{\SS}_c$ and is constant for $N\geq N^{\SS}_c$. Whenever $N \leq N^{\SS}_c$ the minimizer satisfies $\int \uprho^{\SS} = N$ and not $\int \uprho^{\SS} < N$. When $N > N^{\SS}_c$ the minimizer satisfies $\int \uprho^{\SS} = N^{\SS}_c$. Given $\uprho^{\SS}$ (for $N \leq N^{\SS}_c$) we can form the function $$\eqalignno{\phi^{\SS}_{x_\bot} (x_3) &= Z \vert x \vert^{-1} - \vert x \vert^{-1} * \uprho^{\SS} \cr &=Z(|x_\bot|^2+x_3^2)^{-1/2}- \int\uprho^{\SS}(y)(|x_\bot-y_\bot|^2+|x_3-y_3|^2)^{-1/2}dy. \qquad&(2.10)\cr}$$ As long as $x \not= 0$, $\phi^{\SS}_{x_\bot}$ is a continuous function of $x = (x_\bot, x_3)$. This follows easily from the facts that $\uprho \in L^3 (\R^3)$ and $|x|^{-1} \in L^{3/2}_{\loc} (\R^3)$. For each $x_\bot \not= 0$ there is a well defined Schr\"odinger equation associated with the energy functional $$\W_{x_\bot} (\psi) := \int_\R (\partial \psi /\partial x_3)^2 - \int_\R \phi^{\SS}_{x_\bot} (x_3) \psi (x_3)^2 dx_3 .\eqno(2.11)$$ The ground state energy is $$-\mu^{\SS}_1 (x_\bot) := \inf \{ \W_{x_\bot} (\psi) \,: \int \vert \psi \vert^2 = 1 \} \eqno(2.12)$$ and we note that $\mu^{\SS}_1 (x_\bot) \geq 0$ (since $\phi^{\SS}_{x_\bot} (x_3) \rightarrow 0$ as $x = (x_\bot, x_3) \rightarrow \infty$). We also note that whenever $\mu_1^{\SS} (x_\bot) > 0$ there is a unique, nonnegative minimizing $\psi$ (denoted by $\psi_{x_\bot}$) for (2.12). It satisfies the Schr\"odinger equation $$-\psi^{\prime\prime} - \phi^{\SS}_{x_\bot} \psi = - \mu^{\SS}_1 (x_\bot) \psi. \eqno(2.13)$$ Since $\phi^{\SS}_{x_\bot} (x_3)$ is axially symmetric, ie., depends only on $\vert x_\bot \vert$ it is evident that $\mu^{\SS}_1 (x_\bot)$ and $\psi_{x_\bot} (x)$ are axially symmetric in $x_\bot$. The following property of $\mu^{\SS}_1 (x_\bot)$ will play an important role in our discussion of $\uprho^{\SS}$. {\bf 2.3. PROPOSITION (Subharmonicity of $\mu^{\SS}_1 (x_\bot)$).} {\it For each $N \geq 0$ the function $x_\bot \mapsto \mu^{\SS}_1 (x_\bot)$ is a subharmonic function on $\R^2 \setminus \{ 0\}$ and is a decreasing convex function of $\vert x_\bot \vert$. It is strictly decreasing on the set $\{ x_\bot,: \mu^{\SS} (x_\bot) > 0 \}$. This allows us to define $$r^{\SS} = \sup \{ r \,: \ \mu^{\SS}_1 (x_\bot) > 0 \ {\rm for} \ \vert x_\bot \vert < r \}. \eqno(2.14)$$ When $\vert x_\bot \vert > r^{\SS}$ there are no $L^2 (\R)$ solutions to (2.13).} {\it Proof:} Subharmonicity is a property that only has to be checked locally. If $\mu^{\SS} (x^0_\bot) = 0$ then $\mu^{\SS}_1$ obviously satisfies the mean value inequality at $x^0_\bot$. Thus, we can henceforth assume $\mu^{\SS}_1 (x^0_\bot) > 0$, in which case there is surely a minimizing $\psi$ for the $x^0_\bot$ problem (2.13). We denote it here simply by $\psi^0$. Consider the following function, for $x = (x_\bot, x_3) \in \R^3$. It is the energy of a one-dimensional wire, with one-dimensional charge density $\psi^0 (x)^2$, as a function of its position in the field $\phi^{\SS}$. Namely, $$F(x) = \int \limits_\R \phi^{\SS}_{x_\bot} (x_3 + y) \psi^0 (y)^2 dy. \eqno(2.15)$$ We observe that $F$ is subharmonic on $\R^3 \setminus L$, where $L = \{ (x_\bot, x),: x_\bot = 0 \}$ is the vertical line through the origin. This follows straightforwardly from (2.10). For $x_\bot$ close to our base point $x^0_\bot$, we shall use the function $\psi^0 (\cdot + x_3)$, with $x_3$ arbitrary, as a variational function in (2.11) for the energy $-\mu^{\SS}_1 (x_\bot)$. Thus $$G(x_\bot):= \mu^{\SS}_1 (x_\bot^0) - \mu^{\SS}_1 (x_\bot) \leq \W_{x_\bot} (\psi^0 (\cdot + x_3)) - \W_{x_\bot^0} (\psi^0) = F(x^0_\bot, 0) - F(x_\bot, x_3) .\eqno(2.16)$$ This inequality holds for every $x_3 \in \R$. The right side of (2.16) is a superharmonic function of $x = (x_\bot, x_3)$ on $\R^3 \setminus L$ and we set $$H(x_\bot) := \inf \limits_{x_3} [F(x_\bot^0, 0) - F(x_\bot, x_3)].$$ It is not hard to prove by verifying the mean value inequality, that the infimum over $x_3$ of a continuous superharmonic function on $\R^3$ is superharmonic on $\R^2$. We conclude that $G$ in (2.16) is bounded above by the superharmonic function $H$ and that $G(x^0_\bot) = H(x^0_\bot) = 0$. Hence $G$ satisfies the mean-value inequality around $x^0_\bot$ appropriate for a superharmonic function (since $H$ does). It is immediate that $\mu^{\SS}_1 (x_\bot)$ is a strictly monotone decreasing function of $\vert x_\bot \vert$ on the set where $\mu^{\SS}_1 (x_\bot) > 0$. This follows from the fact that $\mu^{\SS}_1$ is axially symmetric, $\mu^{\SS}_1 \geq 0$ and $\mu^{\SS}_1$ vanishes at infinity. On any open set of the form $\Omega_a := \{ x_\bot\,: \vert x_\bot \vert > a \}$, $\mu^{\SS}_1$ cannot attain its supremum (by the strong maximum principle) unless $\mu^{\SS}_1$ is a constant function. Since $\Delta \mu^{\SS}_1 \geq 0$ we have (with $r = \vert x_\bot \vert$) $${d^2 \over dr^2} \mu^{\SS}_1 (r) + {1 \over r} {d \over dr} \mu^{\SS}_1 (r) \geq 0$$ and hence $\mu^{\SS}_1$ is a convex function of $\vert x_\bot \vert$. It remains to show that there is no $L^2 (\R)$ solution to (2.13) with $\mu^{\SS}_1 (x_\bot) = 0$ and with $\vert x_\bot \vert > r^{\SS}$. If there is a solution to (2.13) for some point $x^0_\bot$ satisfying these conditions we can form $F(x)$ as in (2.15). This $F$ is subharmonic and $F(x_\bot, x_3) - F(x^0_\bot, 0)$ achieves its maximum (namely zero) at an interior point $(x^0_\bot, 0)$ of the open set $A = \{ (x_\bot, x_3),: \vert x_\bot \vert > r^{\SS} \}$. Then $F$ is constant in $A$. But that means $F(x^0_\bot, y) = F(x^0_\bot, 0)$ for all $y$, i.e. we can take $\psi^0$, translate it vertically by $y$, and still have a ground state solution to (2.13). This would violate uniqueness of the ground state -- among other things. \lanbox Let us now consider (for fixed $B$ and $N$) the following {\bf linearized problem} defined by the functional $$\E^{\SS}_{\lin} [\uprho] := T[\uprho] - \int \uprho (x) \phi^{\SS}_{x_\bot} (x_3) dx \eqno(2.17)$$ with $$T[\uprho] = \int (\partial \sqrt{\uprho} /\partial x_3)^2. \eqno(2.18)$$ We seek the energy and a minimizer for $$e (N):= \inf \{ \E^{\SS}_{\lin} [\uprho] \,: \ \uprho \in \sc^{\SS}, \ \uprho \ {\rm satisfies \ (2.2) \ and \ } \int \uprho \leq N \}. \eqno(2.19)$$ Note that the same $N$ appears twice; once in (2.17) and once in $\int \uprho \leq N$. Problem (2.19) has a simple solution. Define a radius $R^{\SS}$ using (2.14), by $$R^{\SS} = \min (\sqrt{2N/B} ,\ r^{\SS}). \eqno(2.20)$$ The required solution is $$\widehat{\uprho} (x_\bot, x_3) = {B \over 2 \pi} \cases{\psi_{x_\bot} (x_3)^2, &for $\vert x_\bot \vert \leq R^{\SS}$ \cr 0, &otherwise \cr},$$ where $\psi_{x_\bot}$ is the normalized solution to (2.13). We then have $$e (N) = - \int \limits_{\vert x_\bot \vert \leq R^{\SS}} \mu^{\SS}_1 (x_\bot) dx_\bot.\eqno(2.21)$$ If $R^{\SS} = \sqrt{2N/B} \leq r^{\SS}$ our solution is clearly unique and has $\int \widehat{\uprho} = N$. We note here that the positive level sets of $\mu^{\SS}_1$ all have zero measure since $\mu^{\SS}_1$ is strictly decreasing. When $R^{\SS} = r^{\SS} < \sqrt{2N/B}$ our minimizer has $\int \widehat{\uprho} < N$. It is also unique since there are no $L^2 (\R)$ solutions to (2.13) with $\vert x_\bot \vert > r^{\SS}$. The next theorem implies that the case $r^{\SS} < \sqrt{2N/B}$ does not occur when $N \leq N^{\SS}_c$. {\bf 2.4. THEOREM (Equivalence of the linear and nonlinear problems).} {\it Let $\uprho^{\SS}$ be the unique minimizer for the nonlinear problem as in Theorem 2.2. Then $\uprho^{\SS}$ is the unique minimizer $\widehat{\uprho}$ for the linear problem defined by (2.19).} {\it Remarks:} (1). The theorem tells us that for $N \leq N^{\SS}_c$, the minimizer $\uprho^{\SS}$ satisfies the Euler-Lagrange equation $$-{\partial^2 \over \partial x^2_3} \sqrt{\uprho^{\SS}} - \phi^{\SS}_{x_\bot} \sqrt{\uprho^{\SS}} = - \mu^{\SS}_1 (x_\bot) \sqrt{\uprho^{\SS}} .\eqno(2.22)$$ Here $-\mu^{\SS}_1 (x_\bot)$ is the ground state energy of the Schr\"odinger operator with potential $\phi^{\SS}_{x_\bot}$. (2). For $N \leq N^{\SS}_c$ we have $\int \uprho^{\SS} = N$ and thus the alternative $R^{\SS} < \sqrt{2N/B}$ is not possible (for otherwise $\int \uprho^{\SS} < N$ as discussed above). Uniqueness of the solution of the linearized problem thus precludes the possibility that $r^{\SS} < \sqrt{2N/B}$. We conclude that the support of $\uprho^{\SS}$ (as far as the $x_\bot$ direction is concerned) is {\it exactly} the cylinder $\vert x_\bot \vert \leq \sqrt{2N/B}$. For each $x_\bot$ in this cylinder, the bound (2.2) is saturated. It is remarkable that this holds for all $N \leq N^{\SS}_c$ and $B$. (For $N\geq N^{\SS}_c$ the radius is, of course, $\sqrt{2N^{\SS}_c/B}$.) The atom has a sharp surface; there is no ``exponential decay'' in the $x_\bot$ direction as might have been expected. {\it Proof:} For $\delta$ positive and small, define $\uprho^\delta = (1 - \delta) \uprho^{\SS} + \delta \widehat{\uprho}$, where $\widehat{\uprho}$ is the minimizer for the linear problem. We wish to compute the energy $\E^{\SS} [\uprho^\delta]$ to leading order in $\delta$. The two potential energy terms are trivial. The attraction is linear in $\delta$ while the repulsion is quadratic. The kinetic energy $T[\uprho^\delta]$ is more problematic. Note, by convexity of $\uprho \mapsto T[\uprho]$, that $T[\uprho^\delta] \leq (1 - \delta) T[\uprho^{\SS}] + \delta T[\widehat{\uprho}]$. Hence we have $$\E^{\SS} [\uprho^\delta] \leq E^{\SS} (N,Z,B) + \delta \Bigl(\E^{\SS}_{\lin} [\widehat{\uprho}] - \E^{\SS}_{\lin} [\uprho^{\SS}]\Bigr) + O (\delta^2).$$ If $\uprho^{\SS}$ is not a minimizer for $\E^{\SS}_{\lin}$ we would be able to lower $E^{\SS}$ by using $\uprho^\delta$ in place of $\uprho^{\SS}$, for some small $\delta$. This is a contradiction. \lanbox {\bf 2.5. COROLLARY.} {\it The derivative of the function $E^{\SS} (N,Z,B)$ with respect to $N$ is equal to the value of $-\mu^{\SS}_1 (x_\perp)$ at the edge of the atom, i.e., if $\vert x_\bot \vert = R^{\SS}$ then} $${\partial E^{\SS} (N,Z,B) \over \partial N} = - \mu^{\SS}_1 (x_\bot)=:- \mu^{\SS}_1 (R^{\SS}).$$ \indent {\it Proof:} Choose two numbers $N_\pm$ satisfying $0 \lambda_c$. {\bf 3.1. THEOREM (Solution of the hyper-strong problem).} {\it The hyper-strong problem (3.5) has a minimizer if and only if $\lambda \leq 2$. It is unique and given by $\uprho^{\HS} = \psi^2$ with $$\eqalignii{\psi (x) &= {\sqrt 2 (2 - \lambda) \over 4 \sinh [\mfr1/4 (2 - \lambda) \vert x \vert + c]} \quad &\hbox{for} \ \lambda < 2 \cr \psi (x) &= \sqrt 2 (2 + \vert x \vert)^{-1} \quad &\hbox{for} \ \lambda = 2, \cr}\eqno(3.6)$$ with $\tanh c = (2 - \lambda)/2$. The energy is} $$E^{\HS} (\lambda) = \E^{\HS} [\psi^2] = - \mfr1/4 \lambda + \mfr1/8 \lambda^2 - \hbox{${1 \over 48}$} \lambda^3 \eqno(3.7)$$ \indent {\it Proof:} The existence of a minimizing $\uprho$ with $\int \uprho \leq \lambda$ is particularly simple. Since $\uprho (x) \leq 2 \left\{ \int \uprho \right\}^{1/2} \{ T[\uprho] \}^{1/2} \leq 2 \sqrt \lambda T[\uprho]^{1/2}$, we have that $\E^{\HS} [\uprho] \geq T - 2 \sqrt \lambda \sqrt T$ which is bounded below. This also shows that $T[\uprho^{(n)}]$ is bounded for any minimizing sequence. Also, $\int (\uprho^{(n)})^2$ remains bounded since $H^1 (\R) \subset L^4 (\R)$. The lower semicontinuity of $\E^{\HS} [\uprho]$ is trivial and we omit the proof. The uniqueness of the minimizer follows from the strict convexity of $\uprho \mapsto \int \uprho^2$. The Euler-Lagrange equation is easy to derive (by replacing $\psi = \sqrt {\uprho}$ by $\psi + \alpha f$ with $f \in C^\infty_0$ and $\alpha$ small). We omit the details. The distributional equation is $$- \psi^{\prime\prime} (x) - \psi (0) \delta (x) + \psi (x)^3 = - \mu_\lambda \psi (x) \eqno(3.8)$$ with $\delta (x)$ being Dirac's delta function at the origin. We want a solution to (3.8) with $\psi (x) \geq 0$ and $\int \psi^2 = \lambda$. Eq.\ (3.8) is equivalent to $$-\psi^{\prime\prime} (x) + \psi (x)^3 = - \mu_\lambda\psi (x) \eqno(3.9)$$ with the boundary condition $\psi^\prime (0) /\psi (0) = - 1/2$. By standard ordinary differential equation methods, we can note that the solution to (3.9) is unique once the three numbers $\psi^\prime (0)$, $\psi (0)$ and $\mu_\lambda$ are specified. Thus, if we find a three parameter family of solutions, it must contain the correct one. We also note that a solution, $\psi$, to (3.8) which is in $H^1 (\R)$ must be a minimizer for $\E^{\HS} [\uprho]$. This is so because $\psi$ is then a critical point for $\E^{\HS}$, and a critical point of a convex function must be a minimizer. Equation (3.9) is integrable in the form $(\psi')^2=\mfr1/2\psi^4+ \mu_\lambda\psi^2$. The solution has the form $$\psi (x) = {a \over \sinh (b \vert x \vert + c)} $$ with $a,b,c > 0$. Then (3.9) and the delta-function condition at the origin are satisfied if $$a^2 = 2b^2, \ \ \mu_\lambda = b^2 \ \ {\rm and} \ \ 2b = \tanh c,$$ the latter being the delta function condition at the origin. Thus $b$ can range over the interval $\left( 0, \mfr1/2 \right)$. We find that $$\lambda = {2a \over b} \int^\infty_0 [\sinh (x + c)]^{-2} dx = 2 (1 - \tanh c)$$ This is the solution for $\lambda < 2$. The energy is given by $$E^{\HS} (\lambda) = - \mu_\lambda \lambda - \mfr1/2 \int^\infty_{- \infty} \psi^4 (x) dx. \eqno(3.10)$$ This yields (3.7) since $\int^\infty_c [\sinh x]^{-4} dx = [2 - 3 \coth c + (\coth c)^3]/3$. When $\lambda = \lambda_c = 2$ there is also a solution, obtained by taking the limit $b \rightarrow \infty$. This is given in (3.6). We have obtained a minimizer for $0 \leq \lambda \leq 2$. Because (3.9) is integrable, we have found all $H^1 (\R)$ solutions to (3.8) and hence there can be no solution for $\lambda > 2$. There is also an indirect proof of this fact. Were there a solution for $\lambda > 2$, it would mean that $\E^{\HS} [\uprho]$ has a minimizing $\uprho$ with $\lambda > 2$. But $E^{\HS} (\lambda)$ is seen to have a zero derivative at $\lambda = 2$ and we know that $E^{\HS} (\lambda)$ is convex and nonincreasing. Therefore, any minimizer with $\int \uprho > 2$ would have to obey $\E^{\HS} [\uprho] = E^{\HS} (2)$. But such a minimizer cannot exist because $\uprho \mapsto \E^{\HS} [\uprho]$ is {\it strictly} convex. \lanbox {\bf 3.2. COROLLARY (Virial inequalities).} {\it In analogy with Lemma 2.6, define the three components of the energy to be $$T^{\HS} = \int \limits_\R \left( {\partial \over \partial x} \sqrt{\uprho^{\HS} (x)} \right)^2 dx, \ \ A^{\HS} = \uprho^{\HS} (0), \ \ R^{\HS} = \mfr1/2 \int \limits_\R \uprho^{\HS} (x)^2 dx,$$ where $\uprho^{\HS}$ is the minimizer. Then $$\eqalignno{T^{\HS} &= \vert E^{\HS} \vert = \mfr1/4 \lambda^2 - \mfr1/8 \lambda^2 + \hbox{${1 \over 48}$} \lambda^2 \cr A^{\HS} &= \mfr1/2 \lambda - \mfr1/8 \lambda^2 \leq 3 \vert E^{\HS} \vert \ \ \hbox{and} \ \ R^{\HS} = \mfr1/8 \lambda^2 - \hbox{${1 \over 24}$} \lambda^3 \leq \vert E^{\HS} \vert. \cr}$$ In particular, $R^{\HS} = T^{\HS} = \vert E^{\HS} \vert$ and $A^{\HS} = 3 \vert E^{\HS} \vert$ when $\lambda = \lambda_c = 2$.} We turn to the relationship between $E^{\SS}$ and $E^{\HS}$ as $\eta \rightarrow \infty$. First, we have to introduce the appropriate scaling for the functions $\uprho\in\sc^{\SS}$. {\bf Scaling}. If $\uprho\in \sc^{\SS}$, $x = (x_\bot, x_3)$, we define $\uprho_\eta$ by $$\uprho (x) = Z^4 \eta \ln (\eta) \uprho_\eta (Z\eta^{1/2} x_\bot, Z \ln (\eta) x_3). \eqno(3.11)$$ We then have $$\E^{\SS} [\uprho] : = Z^3 [\ln \eta]^2 \E^{\SS}_\eta [\, \uprho_\eta \, ] \eqno(3.12)$$ where the function $\E^{\SS}_\eta$ is defined as follows: $$\E^{\SS}_\eta [\,\uprho_\eta\,] : = \int \left( {\partial \sqrt{\uprho_\eta} \over \partial x_3} \right)^2 dx - \int \uprho_\eta (x) V_\eta (x) dx + \mfr1/2 \int\int \uprho_\eta (x) V_\eta (x-y) \uprho_\eta (y) dx dy \eqno(3.13)$$ with $$V_\eta (x) : = {1 \over \vert \ln \eta\vert} (x^2_3 + \eta^{- 1}|\ln\eta|^2 x^2_\bot)^{-1/2},\eqno(3.14)$$ The conditions (2.1) and (2.2) become $$\int \limits_{\R^3} \uprho_\eta (x) dx = \lambda = N/Z \eqno(3.15)$$ and $$\int \limits_\R \uprho_\eta (x_\bot, x_3) dx_3 \leq 1. \eqno(3.16)$$ Corresponding to $\E^{\SS}_\eta$ there is an energy $$E^{\SS}_\eta (\lambda): = \inf \{ \E^{\SS}_\eta (\uprho_\eta) \,: \ \uprho_\eta \in \sc^{\SS}, \int \uprho_\eta \leq \lambda, \ \uprho_\eta \ {\rm satisfies} \ (3.16) \}.$$ This energy is related to $E^{\SS}$ by the scaling given in (3.12), i.e., $$E^{\SS} (N,Z,B) = Z^3 [\ln \eta]^2 E^{\SS}_\eta (\lambda).$$ We shall denote the minimizing density by $\uprho^{\SS}_\eta$. It is obtained from the unscaled minimizer $\uprho^{\SS}$ by the scaling in (3.11). We have, by Corollary 2.5, that $$-\mu^{\SS}_1 (R^{\SS}) = {\partial E^{\SS} \over \partial N} = Z^2 [\ln \eta]^2 {\partial E^{\SS}_\eta (\lambda) \over \partial \lambda}.\eqno(3.17)$$ We shall now prove that $E^{\SS}_\eta (\lambda) \rightarrow E^{\HS} (\lambda)$ as $\eta \rightarrow \infty$. First we need the following estimate. {\bf 3.3. PROPOSITION.} {\it For any choice of $\lambda,\ T$ and $R > 0$ there is a constant $C(\lambda, T, R)$ such that if $\eta > 3$ $$\left\vert \overline{\uprho} (0) - \int_{\R^3} V_\eta \uprho \right\vert \leq C(\lambda, T, R) \left( {1 + \vert \ln \vert \ln \eta \vert \vert \over \vert \ln \eta \vert } \right) \eqno(3.18)$$ when $\uprho \in \sc^{\SS}, \int \uprho \leq \lambda, \int \uprho (x_\bot, x_3) dx_3 \leq 1$, $T[\uprho] \leq T$ and $\uprho (x_\bot, x_3) = 0$ for $\vert x_\bot \vert > R$. We can take $C(\lambda, T, R) = \lambda + 8 \sqrt{2} \lambda^{1/4} T^{3/4} + (\const.) TR (1 + \vert \ln R \vert).$} {\it Remark:} If $\uprho$ satisfies the first four of the conditions above then the function $\uprho_y (x) = \uprho (x + y)$ also does so for all $y \in \R^3$. However, $\int V_\eta \uprho_y \rightarrow 0$ as $\vert y \vert \rightarrow \infty$ while $\overline{\uprho}_y (0)$ is independent of $y$ provided $y_3 = 0$. Therefore $C(\lambda, T, R)$ cannot be independent of $R$. {\it Proof:} We write the difference on the left side of (3.18) as $A_1 + A_2 + A_3$ with $$\eqalignno{A_1 &= - \int \limits_{\vert x_3 \vert \geq 1} V_\eta (x) \uprho (x) dx, \qquad A_2 = \int \limits_{\vert x_3 \vert \leq 1} V_\eta (x) [\uprho (x_\bot, 0) - \uprho (x)] dx, \cr A_3 &= \int\limits_{\R^2} \Biggl\{ 1 - \int \limits_{\vert x_3 \vert \leq 1} V_\eta (x_\bot, x_3) dx_3 \Biggr\} \uprho (x_\bot, 0) dx_\bot. \qquad&(3.19)\cr}$$ Since $\vert V_\eta (x) \vert \leq 1/\vert \ln \eta \vert$ for $\vert x_3 \vert \geq 1$, we have $\vert A_1 \vert \leq \lambda /\vert \ln \eta \vert$. To estimate $A_2$ we first note the following inequality for $\psi (x) : = \sqrt{\uprho (x)}$. $$\psi (x_\bot, x_3) \leq \left\{ \int^{x_3}_{-\infty} (\partial \psi^2 /\partial x_3) dx_3 \right\}^{1/2} \leq \sqrt 2 \lambda (x_\bot)^{1/4} T(x_\bot)^{1/4} \eqno(3.20)$$ where $T(x_\bot) = \int_\R (\partial \psi /\partial x_3)^2 dx_3$ and $\lambda (x_\bot) = \int_\R \psi^2 dx_3$. Combining this with (2.4) we obtain $$\vert \uprho (x_\bot, x_3) - \uprho (x_\bot, 0) \vert \leq 2 \sqrt 2 \sqrt{x_3} \lambda (x_\bot)^{1/4} T (x_\bot)^{3/4} \eqno(3.21)$$ and hence $$\eqalignno{\vert A_2 \vert &\leq 2 \sqrt 2 \int \limits_{\vert x_3 \vert \leq 1} V_\eta (x) \sqrt{x_3} \lambda (x_\bot)^{1/4} T(x_\bot)^{3/4} dx \cr &\leq {2 \sqrt 2 \over \vert \ln \eta \vert} \left( \int \limits_{\,\,\vert x_3 \vert \leq 1} {dx_3 \over \sqrt{x_3}} \right) \int \lambda (x_\bot)^{1/4} T(x_\bot)^{3/4} dx_\bot \leq {8 \sqrt 2 \over \vert \ln \eta \vert} \lambda^{1/4} T^{3/4}. \cr}$$ Finally, we carry out the $x_3$- integration in (3.19) and obtain, using (3.20) and $\lambda (x_\bot) \leq 1$: $$\eqalignno{\vert A_3 \vert &= \left\vert \int \biggl[ {2 \over \vert \ln \eta \vert} \sinh^{-1} \Bigl(\eta^{1/2} /(\vert \ln \eta \vert \vert x_\bot \vert )\Bigr) - 1 \biggr] \uprho (x_\bot , 0) dx_\bot \right\vert \cr &\leq 2 T \Biggl\{\int \limits_{\vert x_\bot \vert \leq R} \biggl[ {2 \over \vert \ln \eta \vert} \sinh^{-1}\Bigl( \eta^{1/2} /(\vert \ln \eta \vert \vert x_\bot \vert)\Bigr) - 1 \biggr]^2 dx_\bot \Biggr\}^{1/2} \cr &\leq ({\rm const.}) {T \over \vert \ln \eta \vert} \left\{ \int_{\vert x_\bot \vert \leq R} \Bigl[(\ln \vert x_\bot \vert)^2 + (\ln \vert \ln \eta \vert)^2 \Bigr] dx_\bot \right\}^{1/2}. \cr}$$ The last inequality follows from $$\sinh^{-1} (1/\varepsilon \vert x_\bot \vert) = \ln (1/\varepsilon) + \ln (1/\vert x_\bot \vert) + \ln (1 + \sqrt{1 + \varepsilon^2 \vert x_\bot \vert^2}).$$ Moreover, $\ln (\eta^{-1/2} \vert \ln \eta \vert) = - \mfr1/2\vert \ln \eta \vert + O (\ln \vert \ln \eta \vert),$ so that $$\eqalignno{&\left[ {2 \over |\ln\eta|} \sinh^{-1}\Bigl(\eta^{1/2}/(\vert \ln \eta \vert \vert x_\bot \vert)\Bigr) - 1 \right]^2\cr &= {4 \over \vert \ln \eta \vert^2} \biggl[\ln (1/\vert x_\bot \vert) + O (\ln \vert \ln \eta \vert) + \ln (1 + \sqrt{1 + \eta^{-1} \vert \ln \eta \vert^2 \vert x_\bot \vert^2)}\biggr]^2 \cr &\leq ({\rm const.}) {1 \over \vert \ln \eta \vert^2} \Bigl[(\ln \vert x_\bot \vert)^2 + (\ln \vert \ln \eta \vert )^2\Bigr]. \qquad \qquad \hbox{\lanbox} \cr}$$ As a corollary of Proposition 3.3 we obtain {\bf 3.4. PROPOSITION:} {\it For $C(\lambda, T, R)$ and $\uprho$ as in Proposition 3.2 one has} $$\left\vert \int \limits_{\R^3} \int \limits_{\R^3} \uprho (x) V_\eta (x-y) \uprho (y) dx dy - \int \limits_\R \overline{\uprho} (x)^2 dx \right\vert \leq \lambda C (\lambda, T, 2R) \left( {1 + \vert \ln \vert \ln \eta \vert \vert \over \vert \ln \eta \vert} \right). $$ \indent {\it Proof:} Write $y = x - \xi$ and apply (3.18) to the $\xi$- integration for fixed $x$. \lanbox {\bf 3.5. THEOREM ($E^{\HS}$ is the limit of $E^{\SS}_\eta$).} {\it For all $\lambda \geq 0$ $$\lim_{\eta\to\infty}Z^{-3}[\ln(B/Z^3)]^{-2}E^{\SS} (N,Z,B) = \lim \limits_{\eta \rightarrow \infty} E_\eta^{\SS} (\lambda) = E^{\HS} (\lambda).$$ Furthermore, if $\uprho^{\SS}_\eta$ and $\uprho^{\HS}$ denote the super-strong and hyper-strong minimizers respectively, we have for all $\lambda \geq 0$ $$\overline{\uprho}^{\SS}_\eta \rightarrow \uprho^{\HS} \ \ \hbox{in} \ L^1_{\loc} (\R).$$ If $\lambda \leq 2$ we get $$\sqrt{\overline{\uprho}^{\SS}_\eta} \rightarrow \sqrt{\uprho^{\HS}} \ \ \hbox{in} \ H^1 (\R).$$ } \indent {\it Proof:} Let $\upchi_\lambda : \R^2 \rightarrow \R^+$ be given by $\upchi_\lambda (x_\bot) = 1/\lambda$ for $\vert x_\bot \vert \leq \sqrt{\lambda /\pi}$ (which is the rescaled version of the upper bound $\sqrt{2N/B}$ on the cylindrical radius of the atom) and $\upchi_\lambda (x_\bot) = 0$ otherwise. For each function $\widehat{\uprho} \in \sc^{\HS}$ with $\int_\R \widehat{\uprho} \leq \lambda$ we define $\uprho \in \sc^{\SS}$ by $\uprho (x_\bot, x_3) = \upchi_\lambda (x_\bot) \widehat{\uprho} (x_3)$. In terms of (3.1), $\overline{\uprho} = \widehat{\uprho}$. Also, $\int \limits_{\R^3} \uprho \leq \lambda$ and $\int \limits_\R \uprho dx_3 \leq 1$. Furthermore, $T[\uprho] = \int \limits_\R (d \sqrt{\widehat{\uprho}} /dx)^2$. Using Propositions 3.3 and 3.4 we have $\E^{\SS}_\eta [\uprho] \leq \E^{\HS} [\widehat{\uprho}] +$ $(1 + \lambda) C(\lambda, T, 2R) (1 + \vert \ln \vert \ln \eta \vert \vert)/\vert \ln \eta \vert$. Therefore, $\limsup\nolimits_{\eta \rightarrow \infty} E^{\SS}_\eta (\lambda) \leq E^{\HS} (\lambda)$. To derive the inequality $\liminf\nolimits_{\eta \rightarrow \infty} E_\eta (\lambda) \geq E^{\HS} (\lambda)$ we can take any $\uprho \in \sc^{\SS}$ and define $\overline{\uprho} \in \sc^{\HS}$ by (3.1). Evidently $\overline{\uprho}$ satisfies all the requisite conditions for the $\E^{\HS}$ minimization problem. The rest follows as in the preceding paragraph, but with the additional information provided by (3.2) that $T[\uprho] \geq T [\overline{\uprho}]$. To prove the convergence of the densities we first notice that the convergence of the energies and Propositions 3.3 and 3.4 imply that $\E^{\HS} [\overline{\uprho}^{\SS}_\eta] \rightarrow E^{\HS} (\lambda)$ as $\eta \rightarrow \infty$. This fact, together with the uniqueness of the minimizer for $\E^{\HS}$, implies that our theorem will be established if we prove that for any {\it minimizing} sequence $\uprho_n$, with $\int \uprho_n = \lambda$, there is a subsequence $\uprho_{n_k}$ such that $\sqrt{\uprho_{n_k}} \rightarrow \sqrt{\uprho^{\HS}}$ in $L^2_{\loc} (\R)$ and in $H^1 (\R)$ if $\lambda \leq 2$ as $k \rightarrow \infty$. Since $T[\uprho_n] = \int (d \sqrt{\uprho_n}/dx)^2$ is bounded (see the proof of Theorem 3.1), the sequence $\sqrt{\uprho_n}$ is bounded in $H^1 (\R)$. Thus, by passing to a subsequence (that we again denote $\uprho_n$) we can assume, by the Banach-Alaoglu theorem, that $\sqrt{\uprho_n}$ converges weakly in $H^1(\R)$ to some function $\sqrt{\uprho_\eta}$ which, by lower semicontinuity, minimizes $\E_{\HS}$, i.e., $\uprho_\eta = \uprho^{\HS}$. Weak convergence in $H^1 (\R)$ implies that we can find a subsequence that converges strongly in $L^2_{\loc} (\R)$. To show strong convergence in $H^1 (\R)$ we only have to conclude that the $H^1 (\R)$ norm of $\sqrt{\uprho_n}$ converges to the $H^1 (\R)$ norm of $\sqrt{\uprho^{\HS}}$. Since $\int \uprho_n = \lambda$ and $\int \uprho^{\HS} = \lambda \leq 2,$ we have that $\sqrt{\uprho_n}$ converges strongly to $\sqrt{\uprho^{\HS}}$ in $L^2 (\R)$. It is therefore enough to prove that $T[\uprho_n]$ converges to $T[\uprho]$. Because of the strong $L^2 (\R)$ convergence we can pass to a subsequence such that $\uprho_n (x)$ converges to $\uprho^{\HS} (x)$ for almost every $x \in \R$. But since $T[\uprho_n]$ is bounded, this convergence must, in fact, occur for {\it every} $x \in \R$. Thus $\uprho_n (0) \rightarrow \uprho^{\HS} (0)$. Also, lower semicontinuity in the $H^1 (\R)$ topology (which we also used in the beginning of Theorem 3.1) means that $$\liminf \limits_{n \rightarrow \infty} \int (d \sqrt{\uprho_n}/dx)^2 dx \geq \int (d \sqrt{\uprho^{\HS}}/dx)^2 dx \eqno(3.22)$$ and $$\liminf \limits_{n \rightarrow \infty} \int \uprho_n (x)^2 dx \geq \int \uprho^{\HS} (x)^2 dx. \eqno(3.23)$$ (Note that (3.23) is also a consequence of Fatou's Lemma.) Since $\E^{\HS} [\uprho_n]$ converges to $E^{\HS} (\lambda)$, we see that both (3.22) and (3.23) must be equalities. But (3.22) (with equality) is precisely what was needed to complete the proof. \lanbox Since $\lambda\mapsto E^{\HS}(\lambda)$ is strictly decreasing for $\lambda\leq 2$ we immediately conclude from Theorem~3.5 that the following version of Theorem 1.5 holds in the SS theory. {\bf 3.6. COROLLARY.} {\it As $B/Z^3\to\infty$ we get for the critical electron number $N^{\SS}_c$ the lower bound $ \liminf_{B/Z^3\to\infty} N^{\SS}_c/Z\geq2$.} %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf IV. THE DENSITY MATRIX FUNCTIONAL} In this section we present a third functional which is more complicated than the functionals $\E^{\SS}$ and $\E^{\HS}$. It is not a functional of a density $\uprho$, but depends on a density matrix of a new type that we shall now describe. Each density matrix we shall consider is a {\it one-dimensional}, single-particle density matrix that is defined for each value of $x_\bot \in \R^2$. It will be denoted by $\Gamma_{x_\bot}$ or, if we need to include the one-dimensional coordinates explicitly, as $\Gamma_{x_\bot} (x_3, x^\prime_3)$. Our new functional is defined on the convex set $\G^{\DM}_B$ of $L^2 (\R)$-operator valued functions $\Gamma: \R^2 \ni x_\bot \mapsto \Gamma_{x_\bot}$ that satisfy the conditions $$\eqalignno{\hbox{(a)} \ \ &x_\bot \mapsto (f, \Gamma_{x_\bot} f)\ \hbox{is measurable for all} \ f \in L^2 (\R) \cr \hbox{(b)} \ \ &\Gamma_{x_\bot} \ \hbox{is a positive semidefinite trace class operator on} \ L^2 (\R) \ \hbox{for almost all} \ x_\bot \in \R^2 \cr \hbox{(c)} \ \ &0 \leq \Gamma_{x_\bot} \leq (B/2\pi)I \ \hbox{for almost all} \ x_\bot \in \R^2&\qquad(4.1) \cr \hbox{(d)} \ \ &\int\limits_{\R^2} \Tr_{L^2 (\R)} [(1 - \partial^2/\partial x^2_3) \Gamma_{x_\bot} ] dx_\bot < \infty.}$$ Conditon (d) requires some explanation. The expression $T (x_\bot):= \Tr_{L^2 (\R)} [-\partial^2_3 \Gamma_{x_\bot}]$ is {\it defined} to be $\sum\nolimits_n \alpha_n \int_\R \vert df_n (x_3)/dx_3 \vert^2 dx_3$ where $0 \leq \alpha_n \leq B/2\pi$ and $f_n$ are the eigenvalues and normalized eigenvectors of $\Gamma_{x_\bot}$. Our notation suppresses the dependence of $\alpha_n$ and $f_n$ on $x_\bot$ and we neither assume nor prove here that these quantities are $x_\bot$-measurable. It is important, however, and easy to prove, that $T(x_\bot)$ is a measurable function; condition (d) then requires it to be in $L^1 (\R^2)$. First of all, it is easy to check (by Fubini's Theorem for sums) that $$T(x_\bot) = \sum\nolimits_k \left( {d \phi_k \over dx_3}, \Gamma_{x_\bot} {d \phi_k \over dx_3} \right) \eqno(4.2)$$ where $\phi_1, \phi_2, \dots$ are $C^\infty_0 (\R)$ functions that form an orthonormal basis for $L^2 (\R)$. {F}rom (4.2) and condition (a), $x_\bot \mapsto T(x_\bot)$ is evidently measurable. Since $\Gamma_{x_\bot}$ is trace class we can define the map $\G_B^{\DM}\ni\Gamma\mapsto\uprho_{\Gamma}\in L^1(\R^3)$ by $$\uprho_{\Gamma}(x): = \Gamma_{x_\bot} (x_3,x_3) = \sum \limits_n \alpha_n \vert f_n (x_3)\vert^2.\eqno(4.3)$$ {\it Remark:} To show that $\uprho_{\Gamma}$ is a measurable function on $\R^3$ it is enough to show that $$\uprho_{\Gamma}(x_\bot,x_3)= \lim_{k\to\infty}(\zeta_k^{(x_3)},\Gamma_{x_\bot}\zeta_k^{(x_3)} ), \eqno(4.4)$$ where $\zeta_k^{(x_3)}={k\over 2}\upchi_{[x_3-k^{-1},x_3+k^{- 1}]}$ ($\upchi_A$ denotes the characteristic function of the set $A$). The reason (4.4) suffices is that $(\uprho^{(x_3)}_k, \Gamma_{x_\bot} \uprho^{(x_3)}_k) = \sum\nolimits_{m,n} (\uprho^{(x_3)}_k, \phi_m) (\phi_m, \Gamma_{x_\bot} \phi_n) (\phi_n, \uprho^{(x_3)}_k)$, where $\phi_1, \phi_2, \dots$ are an ($x_\bot$ independent) orthonormal basis as before. The first and last factors are measurable functions of $x_3$, while the middle factor is a measurable function of $x_\bot$. The product is then a measurable function on $\R^3$. Finally, we note that the pointwise limit (i.e., $k \rightarrow \infty$) of measurable functions is measurable. Now we turn to the proof of (4.4). Using the $x_\bot$ dependent eigenfunctions $f_n$, we can write $$(\zeta_k^{(x_3)},\Gamma_{x_\bot}\zeta_k^{(x_3)})=\sum_n\alpha_n |(f_n,\zeta_k^{(x_3)})|^2.$$ Introducing the maximal function $M_{f_n}$ of $f_n$ we have (by definition) that $|(f_n,\zeta_k^{(x_3)})|\leq M_{f_n}(x_3)$ and the maximal inequality $ \int M_{f_n}(x_3)^2\leq\|f_n\|^2=1$. Since $(f_n,\zeta_k^{(x_3)})\to f_n(x_3)$ for almost every $x_3$ as $k\to\infty$ by Lebesgue's Theorem, we can use dominated convergence to conclude that $$\lim_{k\to\infty}\sum_n\alpha_n|(f_n,\zeta_k^{(x_3)})|^2= \sum_n\alpha_n|f_n(x_3)|^2=\uprho_{\Gamma}(x).$$ The density $\uprho_\Gamma$ is a non-negative function with the properties $$\eqalign{ \hbox{(a)}\ \ &\int\uprho_{\Gamma}(x)dx=\int_{\R^2}\Tr_{L^2(\R)}\left[\Gamma _{x_\bot} \right]dx_\bot \cr \hbox{(b)}\ \ &\int_\R\left({\partial \sqrt{\uprho_{\Gamma}(x)}\over\partial x_3} \right)^2dx_3\leq\Tr_{L^2(\R)}\left[-\mfr\partial^2/{\partial x^2_3}\Gamma_{x_\bot}\right] \ \hbox{for almost all}\ x_\bot\in \R^2.}\eqno(4.5)$$ Property (b) is easily seen by again writing $\Gamma_{x_\bot}=\sum_n\alpha_nf_n\otimes\overline{f}_n$. Then by a standard argument in the theory of distributions, and using the fact that the $f_n$'s are in $H^1 (\R)$, we have that for almost every $x_\bot$ $$\left|{\partial \sqrt{\uprho_{\Gamma}(x)}\over\partial x_3} \right|= (\sqrt{\uprho_{\Gamma}})^{-1} \left|{\partial\uprho_\Gamma\over\partial x_3}\right| \leq (\sqrt{\uprho_{\Gamma}})^{-1}\sum_n\alpha_n|f_n| \left|{\partial f_n\over\partial x_3}\right|,$$ with the understanding that the right side is zero where $\uprho_\Gamma = 0$. Since $\uprho_{\Gamma}=\sum_n\alpha_n|f_n|^2$, property (b) in (4.5) follows by estimating the sum above by the Cauchy-Schwarz inequality. We point out that the set $\sc^{\SS}$ defined in the previous section can be thought of as a subset of $\bigcup_{B>0}\G^{\DM}_B$. In fact, if $\uprho \in \sc^{\SS}$ we can define $\Gamma^{\uprho}_{x_\bot}$ by the integral kernel $\Gamma^{\uprho}_{x_\bot}(x_3,x'_3)=\sqrt{\uprho(x_\bot,x_3)\uprho(x_\bot, x_3')}$. The conditions (2.3) (with $C=B/2\pi$) on $\uprho$ translate into the conditions (4.1) on $\Gamma^{\uprho}$. Notice that in general $\uprho_\Gamma$ does not satisfy condition (b) of (2.3) with $C=B/2\pi$. This property would follow if we knew that $\Gamma_{x_\bot}$ is of rank one for almost all $x_{\bot}$, i.e., we can make the following identification $$\sc^{\SS}= \bigcup\limits_{B>0} \{\Gamma\in\G^{\DM}_{B}\,:\, \Gamma_{x_\bot}\ \hbox{has rank at most one for almost all}\ x_\bot\}.$$ To motivate the definition $\G^{\DM}_B$ we consider an antisymmetric many-body wave function $\Psi$ in the lowest Landau band in the sense that $\Pi^N_0 \Psi = \Psi$, where $\Pi^N_0$ is the projection of the physical Hilbert space $\bigwedge \limits^N_1 L^2 (\R^3)$ onto the subspace where all the electrons are in the lowest Landau band. For such a function $\Psi$ we define $$\Gamma^{\Psi}_{x_\bot} (x_3, x^\prime_3) = N \int \Psi (x_\bot, x_3; x^{2} , \dots , x^{N}) \overline{\Psi (x_\bot, x_3^\prime; x^{2}, \dots, x^{N})} dx^{2} \dots dx^{N}. \eqno(4.6)$$ {\bf 4.1 LEMMA.} {\it If the normalized many-body wave function $\Psi$ satisfies $\Pi^N_0 \Psi = \Psi$ and $(\Psi, \sum\nolimits_i H^{i}_\A \Psi) < \infty$ then $\Gamma^{\Psi}$ defined in (4.6) is in $\G^{\DM}_B$.} {\it Proof:} Consider the kernel of the usual one-particle density matrix $K_\Psi$ for $\Psi$ given in (1.19). Then $K_\Psi$ is an operator in $L^2 (\R^3)$ satisfying $0 \leq K_\Psi \leq I$ and $\Pi_0 K_\Psi \Pi_0 = K_\Psi$ and we have $\Gamma^{\Psi}_{x_\bot} (x_3, x^\prime_3) = K_\Psi(x_\bot, x_3; x_\bot, x^\prime_3)$. Conditions (b) and (d) of (4.1) are now clearly satisfied by $\Gamma^{\Psi}$. In order to prove (c) we first notice that for all $x_\bot \in \R^2$ we can find a sequence $(u_n)$ in $L^2 (\R^2)$ converging in the sense of distributions to the Dirac delta function $\delta_{x_\bot}$ at $x_\bot \in \R^2$ such that for all $f \in L^2 (\R)$ the sequence $\Pi_0 (u_n \otimes f)$ converges in $L^2 (\R^3)$. This follows from the expression (1.14) for $\Pi_0$. Furthermore, $(u_n \otimes f, \Pi_0 (u_n \otimes f)) \rightarrow (B/2\pi) \Vert f \Vert^2_{L^2 (\R)}$ as $n \rightarrow \infty$. Then condition (c) indeed follows from $$(u_n \otimes f, K_\Psi (u_n \otimes f)) = (u_n \otimes f, \Pi_0 K_\Psi \Pi_0 (u_n \otimes f)) \leq (u_n \otimes f, \Pi_0 (u_n \otimes f))$$ for all $x_\bot \in \R^2$ in the limit as $n \rightarrow \infty$. \lanbox We now define our density matrix functional on the set $\G_B^{\DM}$ by (1.22) Because of the convexity of the kernel $\vert x-y \vert^{-1}$, this is a convex functional on the (convex) set $\G^{\DM}_B$. Since two different density matrices might have the same density the functional is not strictly convex. The energy for this density matrix functional is, as usual, defined to be $$E^{\DM}(N,Z,B)=\inf\{\E^{\DM}[\Gamma]\, :\, \Gamma\in\G_B^{\DM},\ \int\uprho_{\Gamma}\leq N\}. \eqno(4.7)$$ Notice that the dependence of the energy on $B$ is only through the set $\G_B^{\DM}$. That the energy is bounded below, i.e., $E^{\DM}(N,Z,B) > -\infty$, follows from the following lemma by ignoring the positive repulsive term, i.e., the last term in (1.22). {\bf 4.2 LEMMA.} {\it If $\Gamma\in\G_B^{\DM}$ we have for almost every $x_\bot$ $$\int_{\bf R}\uprho_{\Gamma}(x)^3dx_3\leq \left({3\over\pi^2}\right)B^2 \Tr_{L^2(\R)}\left[-{\partial^2\over\partial x_3^2}\Gamma_{x_\bot}\right]. \eqno(4.8)$$ If, furthermore, $\int\uprho_\Gamma\leq N$ and $\hat{h}_{Z,x_\bot}$ is the one-dimensional operator on $L^2(\R)$ given in Lemma 2.1, i.e., $$\hat{h}_{Z,x_\bot}=-{\partial^2\over\partial x_3^2}-{Z\over|x|}= -{\partial^2\over\partial x_3^2}-{Z\over\sqrt{|x_\bot|^2+x_3^2}}, \eqno(4.9)$$ we have the bounds $$\int_{\R^2}\Tr_{L^2(\R)}\left[\hat{h}_{Z,x_\bot}\Gamma_{x_\bot }\right]dx_\bot \geq -\mfr5/6(\mfr\pi/2)^{2/5}Z^{6/5}N^{3/5}B^{2/5}\eqno(4.10)$$ and $$\int_{\R^2}\Tr_{L^2(\R)}\left[\hat{h}_{Z,x_\bot}\Gamma_{x_\bot }\right]dx_\bot \geq -2NZ^2\left[\ln(2^{-1/2}Z^{-1}N^{- 1/2}B^{1/2}+\mfr1/2)\right]^2 -NZ^2.\eqno(4.11)$$ } \indent {\it Proof:} The bound (4.8) is a special case of the one-dimensional Lieb-Thirring inequality [40]. To prove (4.8) we notice that since $\Gamma_{x_\bot}\leq{B\over2\pi}I$ we get for all $\alpha>0$ that $$\Tr_{L^2(\R)}\left[-{\partial^2\over\partial x_3^2}\Gamma_{x_\bot}\right] -\alpha\int_{\R}\uprho_{\Gamma}(x)^3dx_3$$ is bounded below by the sum of the negative eigenvalues of the one-dimensional operator $-{\partial^2\over\partial x_3^2} -\alpha\uprho_{\Gamma}(x)^2$ multiplied by $B/2\pi$. The Lieb-Thirring inequality implies that the sum of the negative eigenvalues is in turn bounded below by $-{4\over 3}\int_{\R}\left(\alpha\uprho_{\Gamma}(x)^2\right)^{3/2}dx_3$. Thus by optimizing over $\alpha$ $$\eqalign{\Tr_{L^2(\R)}\left[-{\partial^2\over\partial x_3^2}\Gamma_{x_\bot}\right] &\geq\left(\alpha-{2\over3\pi}B\alpha^{3/2}\right)\int_{\R} \uprho_{\Gamma}(x)^3dx_3\cr &\geq {\pi^2\over3}B^{- 2}\int_{\R}\uprho_{\Gamma}(x)^3dx_3.\cr}$$ The second bound (4.10) also follows from the Lieb-Thirring inequality. In fact, for all $R>0$ we get by using the Lieb-Thirring inequality for $|x_\bot|R$ follows from $$\int_{|x_\bot|>R}\Tr_{L^2(\R)}\left[\hat{h}_{Z,x_\bot}\Gamma_{x _\bot}\right] dx_\bot \geq -{Z\over R}\int\uprho_{\Gamma}(x)dx\geq-{NZ\over R}.$$ Optimizing (4.12) over $R$ gives (4.10). Although the estimate (4.10) is always correct it is only interesting when $B\lo NZ^2$. When $B\go NZ^2$ the last estimate (4.11) is better than (4.10). To prove (4.12) we appeal to Lemma 2.1 on the one- dimensional Coulomb problem. Let the first two eigenvalues of $\hat{h}_{Z,x_\bot}$ be denoted by $-\hat{\mu}_1(|x_\bot|)$, $-\hat{\mu}_2(|x_\bot|)$. Then since $\hat{\mu}_1$ is a monotone decreasing function of $|x_\bot|$, we see that the optimal (in the sense of minimizing the left side of (4.11)) density matrix $\Gamma$ is obtained by concentrating as many particles as possible in the lowest eigenstate of $\hat{h}_{Z,x_\bot}$ in a disk $|x_\bot|0$, we get from (1.22) and (4.18) that $$\eqalignno{ \E^{\DM}[(1-\delta)\Gamma^{\DM}&+\delta\Gamma_0]= (1-\delta) \E^{\DM}_{\rm lin}[\Gamma^{\DM}]+\delta\E^{\DM}_{\rm lin}[\Gamma_0] - D(\uprho^{\DM}, \uprho^{\DM}) \cr &+ \delta^2 D(\uprho_{\Gamma_0} - \uprho^{\DM}, \uprho_{\Gamma_0} - \uprho^{\DM}). \qquad&(4.22)}$$ Hence if $\int\uprho_{\Gamma_0}\leq N$ and $\E^{\DM}_{\rm lin}[\Gamma^{\DM}]>\E^{\DM}_{\rm lin}[\Gamma_0]$ we can choose $\delta$ small enough to conclude that $$\E^{\DM}[(1-\delta)\Gamma^{\DM}+\delta\Gamma_0]< \E^{\DM}_{\rm lin}[\Gamma^{\DM}] - D(\uprho^{\DM}, \uprho^{\DM}) =\E^{\DM}[\Gamma^{\DM}].$$ This contradicts the fact that $\Gamma^{\DM}$ is a minimizer for $\E^{\DM}$. We thus conclude that $\Gamma^{\DM}$ is a minimizer for $\E^{\DM}_{\rm lin}$. We have proved: {\bf 4.4 THEOREM.} {\it A minimizer $\Gamma^{\DM}\in\G^{\DM}_B$ for $\E^{\DM}$ under the constraint $\int\uprho^{\DM}\leq N$ is also a minimizer for the linearized functional $\E^{\DM}_{\rm lin} $ (with the same constraint). } We are now in a position to complete the proof of Theorem~4.3. In fact, we can use the characterization of $\Gamma^{\DM}$ as a minimizer for the linear functional to show that it is unique. It is essential here that we have already argued that the density $\uprho^{\DM}$ is unique. {\it Proof of the uniqueness part of Theorem~4.3:} It is easy to describe the minimizer for the linear problem in terms of eigenvalues and eigenvectors for the one-dimensional operator $h^{\DM}_{x_\bot}$. Let $-\mu_1(x_\bot),-\mu_2(x_\bot),\ldots\leq0$ denote the eigenvalues of $h^{\DM}_{x_\bot}$ and let $e_{x_\bot}^{(1)}, e_{x_\bot}^{(2)},\ldots\in L^2(\R)$ be the corresponding normalized eigenvectors. Then the minimizer $\Gamma^{\DM}$ is given by filling levels, i.e., corresponding to $N$ there is a $-\mu\leq0$, {\bf the chemical potential}, such that if we denote $i_0(x_\bot)=\max\{i\, :\, \mu_i(x_\bot)>\mu\}$ we have $$\Gamma^{\DM}_{x_\bot}={B\over2\pi}\sum_{i\leq i_0}e^{(i)}_{x_\bot}\otimes \overline{e^{(i)}_{x_\bot}}+\lambda(x_\bot)e^{(i_0(x_\bot)+1)}_{x_ \bot}\otimes \overline{e^{(i_0(x_\bot)+1)}_{x_\bot}},\eqno(4.23)$$ where $0\leq\lambda(x_\bot)\leq{B\over2\pi}$ gives the filling of the last level. It is only non-zero on the set where the eigenvalue of the last level is equal to the chemical potential, i.e., $\mu_{i_0(x_\bot)+1}(x_\bot)=\mu$ . The uniqueness of $\Gamma^{\DM}$ follows if we can prove that the function $\lambda$ is unique. This is true because $\lambda$ is given from the unique density by the relation $$\uprho^{\DM}(x)={B\over2\pi}\sum_{i\leq i_0}|e^{(i)}_{x_\bot}(x_3)|^2 +\lambda(x_\bot)|e^{(i_0(x_\bot)+1)}_{x_\bot}|^2.\eqno(4.24)$$ \lanbox The condition $\int \Tr_{L^2 (\R)} [\Gamma^{\DM}_{x_\bot}] dx_\bot \leq N$ implies that the rank of $\Gamma^{\DM}_{x_\bot}$ is finite for almost every $x_\bot$. Furthermore, there are some cases in which it is easy to establish that the rank has a finite bound independent of $x_\bot$. Since the potential $\phi^{\DM}_{x_\bot}(x_3)$ tends to zero as $|x_3|$ tends to infinity the eigenvalues $\mu_1, \mu_2, \dots$ can accumulate only at zero and we see from (4.23) that the finiteness of the rank holds uniformly in $x_\bot$ if $\mu > 0$. Another case for which we can make the assertion is $N>Z$. Here, there is necessarily some radius $R$ such that the potential $\phi_{x_\bot}^{\DM}(x_3)$ is bounded above by $-C/|x|$ for some $C>0$ and for $|x|>R$. Then, by any one of a variety of arguments, the number of {\it negative} eigenvalues of $h_{x_\bot}^{\DM}$ is bounded by a constant depending only on $R$ and $C$. In exactly the same way as in Proposition 2.3 we have that $\mu_1(x_\bot)$ is a decreasing convex function of $|x_\bot|$. We can conclude from this that $\Gamma_{x_\bot}$ must vanish outside the ball $\{x_\bot\,:\, |x_\bot|0} \G^{\DM}_B$ consisting of density matrices of rank at most one. Furthermore, it is clear from the definitions (1.25) and (1.22) of the functionals $\E^{\SS}$ and $\E^{\DM}$ that $\Gamma\in\sc^{\SS}$ implies $$\E^{\DM}[\Gamma]=\E^{\SS}[\uprho_{\Gamma}].\eqno(4.27)$$ In the proof of the following theorem we shall show that if the parameter $\eta=B/(2\pi Z^3)$ is large enough then the minimizer $\Gamma^{\DM}$ of $\E^{\DM}$ is, indeed, in the subset $\sc^{\SS}$. It then follows from (4.27) that $E^{\DM}$ and $E^{\SS}$ are identical. Heuristically, the reason that the minimizer $\Gamma$ belongs to $\sc^{\SS}$ is that if $B/Z^3$ is large we can without violating condition (b) of (4.1), allow all the particles to be in the lowest eigenstate of the one-dimensional operator $h^{\DM}_{x_\bot}$. {\bf 4.6. THEOREM. (Equality of $E^{\DM}$ and $E^{\SS}$ for large $B$).} {\it There exists a universal constant $\eta_c > 0$ such that if $B/Z^3 \geq \eta_c$ then $$E^{\DM} (N,Z,B) = E^{\SS} (N,Z,B). \eqno(4.28)$$ Furthermore, the minimizer $\Gamma^{\DM}$ has at most one eigenstate for each $x_\bot$ and $$\Gamma^{\DM}_{x_\bot} (x_3, x^\prime_3) = \sqrt{\uprho^{\SS} (x_\bot, x_3) \uprho^{\SS} (x_\bot, x^\prime_3)}. \eqno(4.29)$$} \indent {\it Proof:} It is clear that $E^{\DM} \leq E^{\SS}$ since we can use the density matrix on the right side of (4.29) as a trial density for $\E^{\DM}$ and use (4.27). To prove the lower bound we write $$\E^{\DM} [\Gamma] = \int_{\R^2} \Tr_{L^2 (\R)} [h^{\SS}_{x_\bot} \Gamma_{x_\bot}] dx_\bot - D(\uprho^{\SS} , \uprho^{\SS}) +D(\uprho^{\SS} - \uprho_{\Gamma} ,\uprho^{\SS}- \uprho_\Gamma)\quad, \eqno(4.30)$$ where $$h^{\SS}_{x_\bot} = - {\partial^2 \over \partial x^2_3} - \phi^{\SS}_{x_\bot} (x_3), \eqno(4.31)$$ and $\phi^{\SS}_{x_\bot}$ is given in (2.10). Using that $\vert x-y \vert^{-1}$ is a positive definite kernel we get $$\E^{\DM} [\Gamma] \geq \int_{\R^2} \Tr_{L^2 (\R)} [h^{\SS}_{x_\bot} \Gamma_{x_\bot}] dx_\bot - D(\uprho^{\SS}, \uprho^{\SS}) \quad. \eqno(4.32)$$ If we can prove that the minimizer on the set $\G^{\DM}_B$ of the following linear functional $$\Gamma \mapsto \int_{\R^2} \Tr_{L^2 (\R)} [h^{\SS}_{x_\bot} \Gamma_{x_\bot}] dx_\bot \eqno(4.33)$$ (supplemented by the constraint $\int\uprho_{\Gamma}\leq N$) has rank at most one for all $x_\bot$, then we know from Theorem 2.4 that it must be given by $\sqrt{\uprho^{\SS} (x_\bot, x_3) \uprho^{\SS} (x_\bot, x^\prime_3)}$. Thus the right side of (4.32) would be bounded below by $E^{\SS}$ and the theorem follows. To prove that the minimizer of (4.33) has rank one when $B/Z^3$ is large enough we shall treat small values of $N$ and large values of $N$ differently. We first consider the case of small $N$. If $-\mu^{\SS}_1 (x_\bot), - \mu^{\SS}_2 (x_\bot)$ denote the first and second eigenvalue of $h^{\SS}_{x_\bot}$ we shall show that if $\vert y_\bot \vert, \vert x_\bot \vert \leq R^{\SS}$ (the cylindrical radius of the $\SS$ atom) then for $B/Z^3$ large enough $$\mu^{\SS}_1 (x_\bot) > \mu^{\SS}_2 (y_\bot). \eqno(4.34)$$ This will imply that the minimizer of (4.33) has rank one. If we neglect the positive term $\uprho^{\SS} * \vert x \vert^{-1}$ in $h^{\SS}_{x_\bot}$ we see from Lemma 2.1 on the energy of the one-dimensional Coulomb problem that $$\mu^{\SS}_2(y_\bot) \leq Z^2/4. \eqno(4.35)$$ However, from Proposition 2.3 we know that $\mu^{\SS}_1 (x_\bot) \geq \mu^{\SS}_1 (R^{\SS})$. We have the scaling relation (3.17), namely $Z^{-2} [\ln \eta]^{-2} \mu^{\SS}_1 (R^{\SS}) = - {dE_\eta \over d \lambda}$, and thus we have $${\mu^{\SS}_1 (x_\bot) \over \mu^{\SS}_2 (y_\bot)} \geq {\mu^{\SS}_1 (R^{\SS}) \over Z^2/4} = - 4 [\ln \eta]^2 {dE_\eta \over d \lambda}. \eqno(4.36)$$ However, since $E_\eta (\lambda) \rightarrow E^{\HS} (\lambda)$ as $\eta \rightarrow \infty$ (see Theorem 3.5) and $\lambda \mapsto E_\eta (\lambda)$ is a convex function we conclude that $${dE_\eta (\lambda) \over d \lambda} \rightarrow {dE^{\HS} (\lambda) \over d \lambda} \ \hbox{as} \ \eta \rightarrow \infty.$$ If $\lambda$ is strictly smaller than 2 the derivative ${dE^{\HS} (\lambda) \over d \lambda}$ is bounded away from zero. Indeed, from (3.7) ${dE^{\HS} \over d \lambda} = - {1 \over 16} \lambda^2 + {1 \over 4} \lambda - {1 \over 4}$. Thus (4.36) implies that if $N/Z$ is strictly less than 2 and $\eta$ is large enough then (4.34) holds. We now turn to the case of large $N$, i.e. $N/Z$ not strictly less than 2. We shall, in fact, consider $N/Z$ strictly greater than 1. In this case we shall show that the operator $h^{\SS}_{x_\bot}$ has at most one negative eigenvalue for all $x_\bot \not= 0$ provided $\eta$ is large enough. This will then again imply that the minimizer of (4.33) has rank one. As we have argued in the proof of Lemma 2.1 the second eigenvalue $-\mu^{\SS}_2 (x_\bot)$ of $h^{\SS}_{x_\bot}$ is identical to the lowest eigenvalue of the {\it three-dimensional} operator $$H_{x_\bot} = - \Delta + V_{x_\bot}, \eqno(4.37)$$ where $$V_{x_\bot} (y) = - Z (\vert x_\bot \vert^2 + \vert y \vert^2)^{- 1/2} + \int \uprho^{\SS} (x^\prime) (\vert x_\bot - x^\prime_\bot \vert^2 + \vert x^\prime_3 - \vert y \vert \vert^2)^{-1/2} dx^\prime, \eqno(4.38)$$ is a spherically symmetric potential. We shall show that if $N/Z$ is strictly greater than one and if $B/Z^3$ is large enough then the operator $H_{x_\bot}$ has no negative eigenvalues. To do this we first observe that if $\lambda = N/Z = 1 + \delta$ for some $\delta > 0$ we can find a constant $c_\delta > 0$ such that if $B/Z^3$ is large enough the inequality $$\int_\Omega \uprho^{\SS} > (1 + \delta/2)Z \eqno(4.39)$$ holds for the integral of $\uprho^{\SS}$ on the set $$\Omega = \{ (x_\bot, x_3)\, :\, \vert x_\bot \vert < R^{\SS}\hbox{ and } \vert x_3 \vert < c_\delta Z^{-1} [\ln \eta]^{-1} \}. \eqno(4.40)$$ The estimate (4.39) follows from the scaling relation (3.11): $$\uprho^{\SS} (x) = Z^4 \eta \ln(\eta) \uprho^{\SS}_\eta (Z \eta^{1/2} x_\bot, Z \ln (\eta) x_3) \eqno(4.41)$$ and the fact proved in Theorem 3.5 that $\overline{\uprho}^{\SS}_\eta \rightarrow \overline{\uprho}^{\HS}$ in $L^1_{\loc} (\R)$ as $\eta \rightarrow \infty$. {F}rom (4.39) we get the following lower bound on the potential $V_{x_\bot}$ $$V_{x_\bot} (y) \geq - {Z \over \vert y \vert} + {Z (1 + \delta /2) \over \sqrt{4(R^{\SS})^2 + (\vert y \vert + c_\delta Z^{-1} [\ln \eta]^{-1})^2}}.$$ Hence if $\vert y \vert > {3 \over \delta} c_\delta Z^{-1} [\ln \eta]^{-1}$ and $\eta$ is large enough (depending only on $\delta$) we get $V_{x_\bot} (y) \geq 0$. We thus have the following estimate for $B/Z^3$ large enough $$V_{x_\bot} (y) \geq \cases{-{Z \over \vert y \vert}, &$\vert y \vert < {3 \over \delta} c_\delta Z^{-1} [\ln \eta]^{-1}$ \cr 0 &otherwise \cr} . \eqno(4.42)$$ We now estimate the number of negative eigenvalues of $H_{x_\bot}$ using the Cwikel, Lieb, Rosenbljum bound [47-49] and (4.42). The number of negative eigenvalues of $H_{x_\bot}$ is at most $$0.1156 \int \vert V_{x_\bot} (y) \vert^{3/2}_- dy \leq 0.1156 \int \limits_{\vert y \vert < 3\delta^{-1} c_\delta Z^{- 1} [\ln \eta]^{-1}} Z^{3/2} \vert y \vert^{-3/2}dy \leq c_\delta {8 \pi \over \delta} (\ln \eta)^{-3/2}. $$ Thus for $\eta$ large enough we find that $H_{x_\bot}$ has no negative eigenvalues and the theorem follows. \lanbox {\it Remark:} We point out that in the above proof it is really $\ln (B/Z^3)$ that must be large. Since it is hard to estimate $B/Z^3$ from knowledge of its logarithm, it is very hard to give a reasonable estimate on the numerical value of $\eta_c$ and we have not attempted to do so. The reason for the logarithm in the above proof is that we we use the functional $\E^{\HS}$, which only approximates $\E^{\SS}$ when $\ln\eta$ is large (see Proposition~3.3). However, it is reasonable to believe that $\E^{\SS}$ {\it is valid even for smaller $\eta$.} We proceed to discuss the viral inequalities for the density matrix functional. These inequalities are identical to the corresponding inequalities for the super-strong functional given in Lemma 2.6. Since the proof is a simple generalization of the proof of Lemma 2.6 we shall leave it to the reader. {\bf 4.7. LEMMA (Virial inequalities).} {\it Define $T^{\DM}: = \int_{\R^2} \Tr_{L^2 (\R)} \left[ - {\partial^2 \over \partial x^2_3} \Gamma^{\DM}_{x_\bot}\right] dx_\bot$, $$A^{\DM}: = \int Z \vert x \vert^{-1} \uprho^{\DM} (x) dx, R^{\DM}: = \mfr1/2 \int\int \uprho^{\DM} (x) \vert x-y \vert^{-1} \uprho^{\DM} (y) dx dy,$$ where $\uprho^{\DM} = \uprho_{\Gamma^{\DM}}$. Then $E^{\DM} = T^{\DM} - A^{\DM} + R^{\DM}$ and the following inequalities hold: $$R^{\DM} \leq A^{\DM} - 2T^{\DM} \quad\hbox{and}\quad T^{\DM} \leq A^{\DM} - 2R^{\DM}. \eqno(4.43)$$ These inequalities imply $$A^{\DM} \leq 3 \vert E^{\DM} \vert,\quad T^{\DM} \leq \vert E^{\DM} \vert\quad\hbox{and}\quad R^{\DM} \leq \vert E^{\DM} \vert. \eqno(4.44)$$} {\bf 4.8. THEOREM.} {\it We have the following estimates on the energy $E^{\DM}$ $$E^{\DM} (N,Z,B) \geq - {5 \over 12} \left( {\pi \over 2} \right)^{2/5} Z^{6/5} N^{3/5} B^{2/5} \eqno(4.45)$$ and $$E^{\DM} (N,Z,B) \geq - NZ^2 \left[ \ln \bigl( 2^{-1/2} Z^{-1} N^{- 1/2} B^{1/2} + \mfr1/2 \bigr) \right]^2 - NZ^2. \eqno(4.46)$$} \indent {\it Proof:} According to (4.43) $R^{\DM} \geq - \mfr1/2 (T^{\DM} - A^{\DM})$. Hence $E^{\DM} \geq \mfr1/2 (T^{\DM} - A^{\DM})$. The estimates (4.45) and (4.46) then follow from (4.10) and (4.11). \lanbox Combining the estimates (4.45--46) we get with $\lambda = N/Z$ $$E^{\DM} (N,Z,B) \geq - C_\lambda \cases{Z^{9/5} B^{2/5} &if $B/Z^3 \leq 1$ \cr Z^3 (1 + [\ln (BZ^{-3})]^2) &if $B/Z^3 \geq 1$ \cr}. \eqno(4.47)$$ We know from (1.24) that $E^{\DM} (N,Z,B) = Z^3 E^{\DM} (N/Z, 1, B/Z^3)$ (notice that the bound (4.47) respects this scaling). Since $E^{\DM} (N/Z, 1, B/Z^3) = E^{\SS} (N/Z, 1, B/Z^3)$ for large $B/Z^3$ we get from Theorem 3.5 the asymptotic relation $$E^{\DM} (\lambda, 1, 2\pi\eta) \approx [\ln \eta]^2 E^{\HS} (\lambda) \eqno(4.48)$$ for large $\eta$. Or stated differently, $$E^{\DM} (N,Z,B) /Z^3 [\ln (B/Z^3)]^2 E^{\HS} (N/Z) \rightarrow 1, \eqno(4.49)$$ as $B/Z^3 \rightarrow \infty$. Recall that according to (3.7) the function $E^{\HS} (\lambda)$ is a third degree polynomial for $0 \leq \lambda \leq 2$. We conclude this section with an estimate on the density $\uprho^{\DM} = \uprho_{\Gamma^{\DM}}$. This estimate will be needed in Sect.~V to give an upper bound on the true quantum energy. {\bf 4.9. PROPOSITION (Bound on $\uprho^{\DM}$).} {\it The density $\uprho^{\DM}$ satisfies the bound $$\eqalignno{ \int_\R \biggl( {\partial \over \partial x_3} \sqrt{\uprho^{\DM}} (x) \biggr)^2 dx_3 &\leq \left( {2B \over \pi} \right) Z^2 \left\{ 1 + {\pi^2 \over 12} + [ \sinh^{-1} ({1\over2Z \vert x_\bot \vert})]^2 \right\} \cr &\leq (\const.) B Z^2 \{ 1 + [\ln(Z \vert x_\bot \vert)]^2 \}. \qquad&(4.50)\cr}$$} \indent {\it Proof:} To simplify the notation we write $\uprho$ and $\Gamma$ instead of $\uprho^{\DM}$ and $\Gamma^{\DM}$. We recall the estimate (4.5b) $$\int_\R \left( {\partial \over \partial x_3} \sqrt{\uprho (x)} \right)^2 dx_3 \leq \Tr_{L^2 (\R)} \left[ - {\partial^2 \over \partial x^2_3} \Gamma_{x_\bot} \right].$$ We are thus left with estimating $\Tr_{L^2 (\R)} \left[ - {\partial^2 \over \partial x^2_3} \Gamma_{x_\bot}\right]$. Since $\Gamma$ minimizes the linear functional $\E_L$ (see Theorem~4.4) we conclude that for all $x_\bot$ $$\Tr_{L^2 (\R)} \left[ \biggl( - {\partial^2 \over \partial x^2_3} - \phi_{x_\bot} \biggr) \Gamma_{x_\bot} \right] \leq 0$$ where $\phi_{x_\bot}$ was defined in (4.20). Thus $$\eqalignno{\mfr1/2 \Tr_{L^2 (\R)} \left[ - {\partial^2 \over \partial x^2_3} \Gamma_{x_\bot} \right] &\leq - \Tr_{L^2 (\R)} \left[ \biggl( - \mfr1/2 {\partial^2 \over \partial x^2_3} - \phi_{x_\bot} \biggr) \Gamma_{x_\bot} \right] \cr &\leq - \Tr_{L^2 (\R)} \left[ \biggl( - \mfr1/2 {\partial^2 \over \partial x^2_3} - Z (\vert x_\bot \vert^2 + x^2_3)^{-1/2} \biggr) \Gamma_{x_\bot} \right] \cr &\leq - {B \over 2 \pi} \left( \hbox{sum of negative eigenvalues of} \ - \mfr1/2 {\partial^2 \over \partial x^2_3} - Z (\vert x_\bot \vert^2 + x^2_3)^{-1/2} \right), \cr}$$ where we have used that $0 \leq \Gamma_{x_\bot} \leq {B \over 2 \pi} I$. We now again appeal to Lemma 2.1 on the one-dimensional Coulomb problem to estimate the sum of the negative eigenvalues of $\mfr1/2 \widehat h_{2Z, x_\bot} = - \mfr1/2 {d^2 \over dx^2_3} - Z (\vert x_\bot \vert^2 + x^2_3)^{-1/2}$. We obtain $$\eqalignno{\Tr_{L^1 (\R)} \left[ - {\partial^2 \over \partial x^2_3} \Gamma_{x_\bot} \right] &\leq {B \over 2 \pi} (2Z)^2 \left\{ 1 + \biggl[ \sinh^{-1} \biggl( {1 \over 2Z \vert x_\bot \vert} \biggr) \biggr]^2 + 2 \sum \limits^\infty_{n=1} {1 \over 4 n^2} \right\} \cr &= {2B \over \pi} Z^2 \left\{ 1 + {\pi^2 \over 12} + \biggl[ \sinh^{- 1} \biggl( {1 \over 2Z \vert x_\bot \vert} \biggr) \biggr]^2 \right\}. \cr}$$ We have here used the formula $\mathop{\sum}^\infty_{n=1} n^{-2} = \pi^2 / 6$. \lanbox %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf V. UPPER BOUND TO THE QUANTUM ENERGY} This section begins the study of the quantum mechanical many-body problem with the goal --- achieved in Sect.~VIII --- of showing that the functionals studied in the earlier sections do, indeed, give the correct asymptotics as $Z\to\infty$ and $B/Z^{4/3}\to\infty$. There are $N$ electrons with the atomic Hamiltonian $H_N$ given in (1.1). The first goal is a variational upper bound to the ground state energy (1.2). Although the bound is valid for all $B$ and $Z$, it is only in the asymptotic regime $Z \rightarrow \infty$ and $B/Z^{4/3} \rightarrow \infty$ that it agrees with the true energy to leading order. This bound is far from the best obtainable with our methods; little effort will be made here to optimize constants. We emphasize, however, that our bound is a little more than just a bound on $E^{\rm Q}(N,Z,B)$; it is, in fact, a bound on $E^{\rm Q}_{\rm conf}(N,Z,B)$. In other words, our variational (or comparison) functions have the property that all the electrons are in the lowest Landau band. Our final conclusion is stated in Theorem 5.1 in which our bound is related to the density matrix energy $E^{\DM}$ plus correction terms. To show that the corrections are of lower order than $E^{Q}$, we need a simple-minded, independent upper bound to $E^{\rm Q}$. This is provided in Theorem~5.3. This theorem also bounds $E^{\rm Q}$ in terms of the energy of the independent electron model (without repulsion). In Theorem 5.2, we show that there is a simple modification of the density matrix functional with the property that it gives an upper bound to the energy without the need for correction terms. This modified functional (the smeared density matrix functional $\E^{\rm SDM}$) has the defect, however, of not being easily related to a lower bound for $E^{\rm Q}$. Thus, $E^{\SDM}$ may be computationally useful, but only after an independent proof (via $E^{\DM}$) shows that $E^{\SDM}$ is asymptotically exact. We shall use the variational principle of Lieb [50] which asserts that the true quantum-mechanical ground state energy, $E^{\rm Q}$, satisfies $$E^\Q \leq \Tr \big[ ( H_\A - Z \vert x \vert^{-1})K \big] + \mfr1/2 \int\int k (x,x) k(y,y) \vert x-y \vert^{-1} dx dy \eqno(5.1)$$ Here, $H_\A = [\vsigma \cdot (\p + \A)]^2$ and $K = K (x, s; x^\prime s^\prime)$ is any density matrix (a nonnegative, self-adjoint, trace class operator on $L^2 (\R^3; \C^2)$) satisfying \itemitem{(i)} $K \leq$ I. \itemitem{(ii)} $\Tr K \leq N$. \hfill(5.2) \smallskip\noindent In other words, $K$ is a $2 \times 2$ matrix of operators on $L^2 (\R^3)$ and $k$, which appears in (5.1), denotes the kernel of the operator obtained from the trace of this $2 \times 2$ matrix, i.e., $k (x,x^\prime) = \sum \nolimits_s K(x, s; x^\prime, s)$. The ``diagonal'', $k(x,x)$ is well defined as an $L^1 (\R^3)$ function because $k$ is the kernel of a trace class operator. The symbol Tr denotes the trace on $L^2 (\R^3, \C^2)$, i.e., the matrix trace as well as the $L^2 (\R^3)$ trace. Note that in [50] there is another contribution to the integrand in (5.1). It is the ``exchange term'' and it is omitted here because it is negative. We shall choose $K$ to have all spins ``down'', i.e. $K(x, s; x^\prime, s^\prime) = \delta (s = -\mfr1/2) \delta (s^\prime = -\mfr1/2) k(x,x^\prime)$, and (5.2) becomes $$\eqalignno{({\rm i}^\prime) \ \ &0 \leq \int \int \overline{f(x)} f(x^\prime) k(x,x^\prime) dx dx^\prime \leq \int \vert f(x) \vert^2 dx \quad \hbox{for all} \ f \in L^2 (\R^3) \cr ({\rm ii}^\prime) \ \ &\int k(x,x) dx \leq N. \qquad&(5.3)\cr}$$ Our choice of the kernel $k$ is $$k(x,x^\prime) = {2 \pi \over B} \int \Pi^\bot_0 (x_\bot, y_\bot) \Gamma^{\DM}_{y_\bot} (x_3, x^\prime_3) \Pi^\bot_0 (y_\bot, x^\prime_\bot) dy_\bot. \eqno(5.4)$$ Here $\Pi^\bot_0$ is the kernel of the projection onto the lowest Landau level in $L^2 (\R^2; dx_\bot)$ (i.e. the kernel in (1.14) with the $\delta$-function and $P^\downarrow$ omitted). It is given by $$\Pi^\bot_0 (x_\bot, y_\bot) = {B \over 2 \pi} \exp \big\{ {i \over 2} (x_\bot \times y_\bot) \cdot \B - \mfr1/4 (x_\bot - y_\bot)^2 B \big\}. \eqno(5.5)$$ We have constructed $K$ such that $\Pi_0K=K$. If we look at the proof [50] of the variational principle, we see trivially that the right side of (5.1) is a bound to the energy in which the condition $\Pi_0^N\Psi=\Psi$ is added as a constraint, i.e., it is a bound on $E^{\rm Q}_{\rm conf}(N,Z,B)$. The function $\Gamma^{\DM}$ in (5.4) is the integral kernel for the minimizer of the density matrix energy functional (1.22) on the set $\G^{\DM}_B$ satisfying $\int_{\R^2} \Tr_{L^2 (\R)} [\Gamma^{\DM}_{x_\bot}] dx_\bot \leq N$. One easily checks that (i$^\prime$) and (ii$^\prime$) of (5.3) are satisfied. Indeed, since $0 \leq \Gamma^{\DM}_{y_\bot} \leq \left(B / 2 \pi\right) I$ we find $$\eqalignno{\int\int \overline{f(x)} k(x,x^\prime) f(x^\prime) dx dx^\prime &\leq \int \limits_{\R^2} \left( \int \limits_\R \biggl\vert \int \limits_{\R^2} \Pi^\bot_0 (y_\bot, x_\bot) f(x) dx_\bot \biggr\vert^2 dx_3 \right) dy_\bot \cr &\leq \int \vert f(x) \vert^2 dx. \cr}$$ In the last inequality we have used the fact that $\Pi^\bot_0$ is the kernel of a projection. Moreover, recalling the definition of the density $\uprho^{\DM} (x) := \Gamma^{\DM}_{x_\bot} (x_3, x_3)$, we have $$\eqalignno{k(x,x) &= \int \big\vert \Pi^\bot_0 (x_\bot, y_\bot) \big\vert^2 \uprho^{\DM} (y_\bot, x_3) dy_\bot \cr &= \int m(x_\bot - y_\bot) \uprho^{\DM} (y_\bot, x_3) dy_\bot =: m *_\bot \uprho^{\DM} (x), \qquad&(5.6)\cr}$$ with $m$ being the Gaussian $$m(x_\bot) = {B \over 2 \pi} \exp (-\mfr1/2 \vert x_\bot \vert^2 B), \eqno(5.7)$$ with normalization $\int m = 1$. Equation (5.3 (ii$^\prime$)) follows immediately from (5.6) and $\int \uprho^{\DM} \leq N$. Note the introduction of the {\it convolution operator\/} $*_\bot$ in (5.6), which is convolution over the $x_\bot$ variables {\it only}. Let us now consider the various terms in (5.1). Writing $H_\A = - \partial^2_3 + H^\bot_\A$ and using that $H^\bot_\A \Pi^\bot_0 = 0$ we obtain $$\Tr [H_\A k] = \int_{\R^2} \Tr_{L^2 (\R)} [-\partial^2_3 \Gamma^{\DM}_{x_\bot}] dx_\bot . \eqno(5.8)$$ In the term $$\Tr (-Z \vert x \vert^{-1} k) = -Z \int k(x,x) \vert x \vert^{-1} dx \eqno(5.9)$$ and in the last term of (5.1) we would like to replace $k(x,x)$ by $\uprho^{\DM} (x)$, since, with that replacement, the right side of (5.1) is precisely the energy $\E^{\DM} [\Gamma^{\DM}].$ In other words, $$\eqalignno{E^\Q &\leq \hbox{Right side of (5.1)} \ = E^{\DM} + a - b, \cr a &= Z \int \vert x \vert^{-1} [\uprho^{\DM} (x) - k(x,x)] dx, \qquad&(5.10)\cr b &= \int\int \vert x-y \vert^{-1} [\uprho^{\DM} (x) \uprho^{\DM} (y) - k(x,x) k(y,y)] dx dy. \cr}$$ We claim that $b \geq 0$, and hence that it can be omitted from further consideration. To see this, we can note (by a simple change of variables of integration) that $b = (\uprho^{\DM}, Q \uprho^{\DM})$ where $Q$ is the quadratic form whose kernel is $\vert x-x^\prime \vert^{-1} - (m*_\bot \vert x \vert^{-1} *_\bot m) (x-x^\prime)$, i.e., $$Q(x,x^\prime) = \vert x-x^\prime \vert^{-1} - \int \limits_{\R^2} \int \limits_{\R^2} m (x_\bot - y_\bot) m (x^\prime_\bot - y^\prime_\bot) \bigl[ (x_3 - x^\prime_3)^2 + (y_\bot - y^\prime_\bot)^2 \bigr]^{- 1/2} dy^\prime_\bot dy_\bot.$$ Since $Q$ is a function only of $x-x^\prime$, we can evaluate $b$ in terms of the three-dimensional Fourier transform $\widehat{\uprho^{\DM}}$ and the two-dimensional Fourier transform $\widehat m$. A simple calculation shows that up to a constant that depends on normalization convention, $$b = \int \vert \widehat{\uprho^{\DM}} (p) \vert^2 \vert p \vert^{- 2} [1 - \vert \widehat m (p_\bot) \vert^2] dp.$$ (Note that the reality of $\uprho^{\DM}$ and $m$ implies that $\widehat{\uprho^{\DM}} (p) = \overline{\widehat{\uprho^{\DM}} (-p)}$ and $\widehat m (p_\bot) = \overline{\widehat m (-p_\bot)}.$) Now $m$ is a nonnegative function (a Gaussian, in fact) and $\int m = 1$; hence $\vert \widehat m (p) \vert \leq 1$ for all $p$, and we conclude that $b \geq 0$. Next, we turn our attention to bounding $a$ from above. We pick an $r > 0$ and write $a = a_1 + a_2$ with $$a_1 = Z \int \limits_{\vert x \vert \leq r} \vert x \vert^{-1} (\uprho^{\DM} (x) - k(x,x)) dx$$ and $$a_2 = Z \int \limits_{\vert x \vert \geq r} \vert x \vert^{-1} (\uprho^{\DM} (x) - k(x,x)) dx.$$ We use two methods to estimate $a_1$. The first uses the Lieb- Thirring inequality in one dimension [40] and it is accurate and useful when $B \ll Z^2 N^2$. If we recall that $0 \leq \Gamma^{\DM}_{x_\bot} \leq (B/2\pi) I$, this inequality can be stated as (see (4.8)) $T^{\DM} \geq (\pi^2/3) B^{- 2} \int (\uprho^{\DM})^3$. Therefore $$\eqalignno{a_1 &\leq Z \int \limits_{\vert x \vert \leq r} \vert x \vert^{-1} \uprho^{\DM} (x) dx \leq Z B^{2/3} \left( \int \uprho^3 B^{-2} dx \right)^{1/3} \left( \int \limits_{\vert x \vert \leq r} \vert x \vert^{-3/2} dx \right)^{2/3} \cr &\leq 4 \cdot 3^{-1/3} ZB^{2/3} T^{1/3} r \leq 4 \cdot 3^{-1/3} ZB^{2/3} r \vert E^{\DM} \vert^{1/3}, \qquad&(5.11)\cr}$$ where we have used the virial inequality $T^{\DM} \leq \vert E^{\DM} \vert$ from (4.44). The other estimate, that gives small errors for $B \gg Z^2 N^{1/6}$, uses the inequality $\uprho \leq 2 \int \limits_\R \sqrt{\uprho} \vert \partial \sqrt{\uprho} /\partial x_3 \vert dx_3$ combined with the bound (4.50) $$\int \limits_\R (\partial \sqrt{\uprho^{\DM}}/\partial x_3)^2 dx_3 \leq (\const.)BZ^2 \{ 1 + [\ln (Z \vert x_\bot \vert)]^2\}.$$ This gives $$\eqalignno{a_1 &\leq Z \int \limits_{\vert x \vert \leq r} \vert x \vert^{-1} \uprho^{\DM} (x) dx \cr &\leq Z \int \limits_{\vert x_\bot \vert \leq r} \left( \int \limits_{\vert x_3 \vert \leq r} \vert x \vert^{-1} dx_3 \right) \left( 2 \int \sqrt{\uprho^{\DM} (x)} \ \left\vert \partial \sqrt{\uprho^{\DM} (x)} / \partial x_3 \right\vert dx_3 \right) dx_\bot \cr &\leq 4Z \left( \int \uprho^{\DM} (x) dx \right)^{1/2} \left\{ \int \limits_{\vert x_\bot\vert \leq r} [ \sinh^{-1} (\vert x_\bot \vert r^{- 1})]^2 \int \limits_\R \left(\partial \sqrt{\uprho^{\DM} (x)} /\partial x_3 \right)^2 dx_3 dx_\bot \right\}^{1/2} \cr &\leq (\const.) B^{1/2} Z^2 N^{1/2} r \{ 1 + [\ln (Zr)]^2 \}^{1/2}. \qquad&(5.12)\cr}$$ To bound $a_2$ we write it as $$a_2 = \int \limits_{\vert x \vert \geq r} \uprho^{\DM} (x) \phi (x) d x$$ with $$\phi (x) = Z \int_{\R^2} m (y_\bot) \left[ \vert x \vert^{-1} - \mfr1/2 (\vert x \vert^2 + \vert y_\bot \vert^2 - 2 x_\bot \cdot y_\bot)^{- 1/2} - \mfr1/2 (\vert x \vert^2 + \vert y_\bot \vert^2 + 2 x_\bot \cdot y_\bot)^{-1/2} \right] dy_\bot,$$ where we have used $m(y_\bot) = m(-y_\bot)$. Since the function $t \mapsto t^{-1/2}$ is convex, and $(1 + s)^{-1/2} \geq 1 - \mfr1/2 s$, we have $$\eqalignno{\mfr1/2 &(\vert x \vert^2 + \vert y_\bot \vert^2 - 2x_\bot \cdot y_\bot)^{-1/2} + \mfr1/2 (\vert x \vert^2 + \vert y_\bot \vert^2 + 2 x_\bot \cdot y_\bot)^{-1/2} \cr &\geq (\vert x \vert^2 + \vert y_\bot \vert^2)^{-1/2} \geq \vert x \vert^{-1} (1 - \mfr1/2 \vert y_\bot \vert^2 \vert x \vert^{-2}). \cr}$$ Hence, $$\phi (x) \leq \mfr1/2 Z \vert x \vert^{-3} \int \limits_{\R^2} m (y_\bot) y^2_\bot dy_\bot = Z B^{-1} \vert x \vert^{-3},$$ and we obtain the desired bound for $a_2$ by using the virial inequality (4.44): $$\eqalignno{a_2 &\leq ZB^{-1} \int \limits_{\vert x \vert \geq r} \uprho^{\DM} (x) \vert x \vert^{-3} dx \cr &\leq B^{-1} \left( \int \uprho^{\DM} (x) Z \vert x \vert^{-1} dx \right) \sup \limits_{\vert x \vert \geq r} \vert x \vert^{-2} \cr &\leq r^{-2} B^{-1} \vert E^{\DM} \vert . \qquad&(5.13)\cr}$$ We now add (5.11) and (5.13) and optimize the bound with respect to $r$. This gives $$\eqalignno{a_1 + a_2 &\leq (2^{5/3} 3^{-1/9} + 2^{2/3} 3^{-2/9}) Z^{2/3} B^{1/9} \vert E^{\DM} \vert^{5/9} \cr &\leq (2^{5/3} 3^{-1/9} + 2^{2/3} 3^{-2/9}) (5/12)^{5/9} (\pi /2)^{2/9} Z^{4/3} N^{1/3} B^{1/3} \cr &= 2.76 Z^{4/3} N^{1/3} B^{1/3} \qquad&(5.14)\cr}$$ where we have used the estimate (4.45). This bound is of lower order than $\vert E^{\DM} \vert$ provided $B \ll Z^2 N^2$. We can also add (5.12) and (5.13) and choose $r = B^{-1/2} E^{1/3} Z^{-2/3} N^{-1/6}$ (this is not optimal). In this case we obtain, this time using (4.46): $$\eqalignno{a_1 + a_2 &\leq (\const.) Z^{4/3} N^{1/3} \vert E^{\DM} \vert^{1/3} \big\{ 1 + (\const.) [\ln (B \vert E^{\DM} \vert^{-2/3} Z^{-2/3} N^{1/3})]^2 \big\}^{1/2} \cr &\leq (\const.) Z^2 N^{2/3} \big\{ 1 + \vert \ln N \vert^2 + [\ln (B/NZ^2)]^2 \big\}^{5/6} \qquad&(5.15)\cr}$$ which we shall show is small compared to $\vert E^{\Q} \vert$ if $B \gg Z^2 N^{1/16}$ (see Theorem~5.3 and (5.24)). The conclusion we have reached is that $E^\Q \leq E^{\DM}$ plus corrections (which we shall show below to be of lower order by a power of $Z$ for large $Z$). It can be stated as follows. {\bf 5.1. THEOREM (Upper bound in terms of $E^{\DM}$).} {\it For all $N, B$ and $Z$ the true quantum ground state energy $E^\Q$ and the confined energy $E^{\rm Q}_{\rm conf}$ are related to the energy $E^{\DM}$ by $$E^\Q (N,Z,B)\leq E^{\rm Q}_{\rm conf}(N,Z,B) \leq E^{\DM} (N,Z,B) + R_{\rm U}(N,Z,B).$$ where $R_{\rm U} (N,Z,B)$ is the lesser of the following two quantities: $$2.76 Z^{4/3} N^{1/3} B^{1/3} \ \ \hbox{and} \ \ (\const.)Z^2 N^{2/3} \big\{ 1 + \vert \ln N \vert^2 + [\ln (B/NZ^2)]^2 \big\}^{5/6} . \eqno(5.16)$$ } In fact, our calculation here shows more than Theorem 5.1. Let us define a {\bf smeared density matrix functional} by means of (1.22), but with the Coulomb attraction (not the repulsion) changed from $-Z \vert x \vert^{-1}$ to the function $$-Z m *_\bot \vert x \vert^{-1} = -Z \int \limits_{\R^2} m (x_\bot - y_\bot) [x^2_3 + (x_\bot - y_\bot)^2]^{-1/2} dx_\bot.$$ We denote the corresponding energy, as in (4.7) by $E^{\SDM} (N, Z, B)$. {\bf 5.2. THEOREM (Upper bound in terms of $E^{\SDM}$).} {\it For all $N, B$ and $Z$, the three energies satisfy $$\eqalignno{E^\Q (N,Z,B) &\leq E^{\SDM} (N,Z,B) \leq E^{\DM} + R_{\rm U} (N,Z,B) \qquad&(5.17)\cr E^{\DM} (N,Z,B) &\leq E^{\SDM} (N,Z,B) \qquad&(5.18)\cr}$$ with $R_{\rm U}(N,Z,B)$ as in Theorem 5.1.} {\it Proof:} (5.17) is simply a transcription of our calculations above. Inequality (5.18) follows from the definitions of $E^{\SDM}$ and $E^{\DM}$ as infima. For each $y_\bot\in \R^2$ define $\Gamma^{y_\bot}$ by $\Gamma^{y_\bot}_{x_\bot} = \Gamma^{\SDM}_{x_\bot-y_\bot}$. (Here $\Gamma^{\SDM}$ is the minimizer for $E^{\SDM} (N,Z,B)$. If none exists then we have to use a minimizing sequence in an obvious way.) Then $E^{\DM} (N,Z,B) \leq \E^{\DM} [\Gamma^{y_\bot}]$. However, a simple calculation shows that $$\int \limits_{\R^2} m (y_\bot) \E^{\DM} [\Gamma^{y_\bot}] dy_\bot = \E^{\SDM} [\Gamma^{\SDM}]. \qquad\qquad \hbox{\lanbox}$$ The next theorem gives a simple bound for $E^{\rm Q}(N,Z,B)$ in terms of the independent particle model. {\bf 5.3 THEOREM (Simple upper bounds)}. {\it Let $E_{(1)}^{\rm Q}(N,Z,B)$ be the ground state energy for the Hamiltonian (1.1) without the last (repulsion) term, i.e., for $ H^{(1)}_N:=\sum_1^N\left[H_{\bf A}^i-Z/|x^i|\right]$. Then, if $n\leq N$, $$ E^{\rm Q}(N,Z,B)\leq E^{\rm Q}_{(1)}(n,Z-\mfr1/2 n,B).\eqno(5.19) $$ A bound for $E^{\rm Q}_{(1)}$ is given by $$ E^{\rm Q}_{(1)}(N,Z,B)\leq-(\const.)\cases{ n^{3/5}Z^{6/5}B^{2/5},&for $B\leq nZ^2$ \cr nZ^2,&for $ B\geq nZ^2$}, \eqno(5.20) $$ with $n=\min\{N,Z\}$.} {\it Proof:} Let $K_{(1)}$ denote the one-particle reduced density matrix for the ground state of $H^{(1)}_n$. [Note: $n$ and not $N$. Note also that if there are several ground states, then choose any one. If there is none, then our proof can easily be carried out with a minimizing sequence, $K^\varepsilon_{(1)}$ with $\varepsilon\to 0$, in place of $K_{(1)}$.]. This operator satisfies $0\leq K_{(1)}\leq I$ and $\Tr K_{(1)}=n$. Let $\uprho_{(1)}(x)=K_{(1)}(x,x)$ as usual. A very important fact is that $\Phi(x):=\int|x-y|^{-1}\uprho_{(1)}(y)dy$ has its maximum at $x=0$ (because otherwise we could lower the energy of our state by moving the nucleus from $x=0$ to another point $x\in\R^3$ at which $\Phi(x)<\Phi(0)$; this argument is used again later in the $N$-body context just before (6.10), where a lengthier discussion is given). Thus, we have $\Tr \left[\vert x \vert^{-1} K_{(1)}\right] = \Phi (0) \geq \int f (x) \Phi (x) dx$ for any $f: \R^3 \rightarrow \R^+$ with $\int f \leq 1$. We now have, by the variational principle (5.1) and with $f:= \mfr1/2 \uprho_{(1)}$, $$\eqalignno{E^{\rm Q} (N,Z,B) &\leq E^{\rm Q}f(n,Z,B) \leq \Tr \left[H_\A K_{(1)}\right] - Z \Phi (0) + \int f \Phi\cr &\leq \Tr \left[\left(H_\A - (Z - \mfr1/2 n) \vert x \vert^{-1}\right) K_{(1)}\right] = E^{\rm Q}_{(1)} (n, Z - \mfr1/2 n, B)\quad. &(5.21)}$$ The first inequality states simply that $E^{\rm Q} (N,Z,B)$ is monotone in $N$ (because we can always remove particles to ``infinity''). Inequality (5.21) proves (5.19). To bound $E^{\rm Q}_{(1)} (N,Z,B)$ we use the same variational principle (5.1) (without the repulsion term) and choose, as before, a variational density matrix $K = k \otimes$ (all spins down) with $k (x,y) = g(x)g(y)\Pi_0^\bot(x_\bot,y_\bot) h(x_3 - y_3)$, where $\Pi_0^\bot$ is in (5.5), $g:\R^3\mapsto\R^+$ and $h(x) = \int^\mu_{-\mu}\exp [2 \pi i p x] dp$ for some number $\mu > 0$. We demand that $g$ satisfies the condition: $\int g^2\leq \pi N/\mu B$. Since $\Pi_0^\bot(x_\bot,x_\bot)=B/2\pi$ and $h(0) = 2 \mu$, this condition states that $n:=\Tr K = (2 \mu) (B/2\pi)\int g^2\leq N$, which means that we are computing an upper bound for $E^{\rm Q}_{(1)} (n,Z,B)$. Since $E^{\rm Q}_{(1)} (N,Z,B)$ is a decreasing function of $N$ the fact that $n\leq N$ is acceptable. We have to check that $K\leq I$ as an operator. Since $\Pi_0^\bot\leq I$ on $L^2(\R^2)$ and $h(x_3 - y_3) \leq I$ as an operator on $L^2 (\R)$ (Plancherel) we will, indeed, have $K \leq I$ if $0 \leq g(x) \leq 1$ for all $x \in \R^3$. Now we can compute the right side of (5.1) $= \Tr\left[(H_\A - Z/\vert x \vert)K\right]$. Using $H_{\bf A}\Pi_0^\bot=0$, we find that $$ E^{\rm Q}_{(1)} (N,Z,B)\leq {B \over 2 \pi} \int^\mu_{-\mu} p^2 dp \int g^2 + {B \mu \over \pi}\int|\nabla g|^2 -Z{B \mu \over \pi}\int |x|^{-1}g(x)^2dx .\eqno(5.22)$$ Now we fix $R>0$ and choose $g$ to be a smooth, radial function with $g(x)= 1$ for $|x|2R$ and $g(x) \leq 1$ everywhere. The condition on $\int g^2$, is $\mu BR^3 = (\const.)n$. We then obtain $$ E_{(1)}^{\rm Q}(N,Z,B) \leq (\const.) \left[\mu^2 + R^{-2} -(\const.)Z/R\right] \leq (\const.) n [n^2 R^{-6} B^{-2} + R^{-2} - (\const.)Z/R]. $$ We now choose $n = \min \{ N,Z \}$ and $R = (\const.)\max\{Z^{-1/5} n^{2/5} B^{-2/5},Z^{-1}\}$. \lanbox {\it Remarks}: (i). By omitting the repulsion term in $H_N$, we have the trivial {\it lower bound} $E^{\rm Q} (N,Z,B) \geq E^{\rm Q}_{(1)} (N,Z,B)$. Since $E^{\rm Q}_{(1)} (N,Z,B)$ is trivially convex in $N$ (because it is the sum of the $N$ lowest eigenvalues), we have $E^{\rm Q}_{(1)} (N,Z,B) \geq (N/n) E^{\rm Q}_{(1)} (n,Z,B)$ with $n = \min \{ N,Z \}$. Thus $$\max \{ 1, N/Z \} E^{\rm Q}_{(1)} (n,Z,B) \leq E^{\rm Q} (N,Z,B) \leq E^{\rm Q}_{(1)} (n, Z - \mfr1/2 n, B). \eqno(5.23)$$ (ii). In the next section, eq. (6.6), we will encounter the operator $H_\A - \delta Z /\vert x \vert$ with $\delta > 1$ . However, we will need a {\it lower bound} only for $h (\delta): = \Pi_0 (H_\A - \delta Z /\vert x \vert) \Pi_0 = \Pi_0 (p^2_3 - \delta Z/\vert x \vert) \Pi_0$. This can easily be achieved as follows. By a scaling $x_3 \rightarrow \delta^{-1} x_3$ we find that $h(\delta)$ is unitarily equivalent to $\delta^2 \Pi_0 [p^2_3 - Z (x^2_3 + \delta^2 x^2_\bot)^{-1/2}] \Pi_0$ and this operator is greater than $\delta^2 h(1)$. %zzzBOTH (iii) and (iv) changed (iii). The upper bound in (5.20) for $B\leq nZ^2$ is, in fact, optimal ({\it cf.\/} the lower bound (4.45) to $E^{\DM}$). It has the correct asymptotic dependence --- as we prove in this paper. The bound for $B\geq nZ^2$ is not optimal. It misses a logarithmic factor. It is easy to get an upper bound on $E^{\DM}$ (and thus, by Theorem~5.1, on $E^\Q$ as well) with the correct logarithmic factor in the following way. Recall that $E^{\DM} (N,Z,B) = Z^3 E^{\DM} (\lambda, 1, B/Z^3)$ and that $E^{\DM} (\lambda, 1, 2\pi\eta)$ is a decreasing function of $\eta$ with the asymptotic form (4.48) or (4.49). This immediately implies that for $B/Z^3 \geq 1$ $$ E^{\DM} (N,Z,B) \leq - C_\lambda Z^3 (1 + [\ln (B/Z^3)]^2). \eqno(5.24)$$ (iv). A {\bf corollary} of Theorem~5.3 and (5.24) is that $R_{\rm U}$ is really of lower order than $E^{\rm Q}$ as $Z \rightarrow \infty$ with $N/Z$ held fixed; this follows from (5.16). It is important to note here that any upper bound to $E^{\rm Q}$ also gives a {\it lower} bound to $|E^{\rm Q}|$. Furthermore, once we have established that $E^{\DM}/E^{\rm Q}\to1$ as $Z\to\infty$ and $B/Z^{4/3}\to\infty$ ( see Theorem~1.3 and (8.6)) we will also know that $R_{\rm U}$ is of lower order than $E^{\DM}$. Theorem~5.3 and (5.24) will be needed again in (8.7) to show that $R_{\rm L}$ is also lower order. %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf VI. PROOF OF CONCENTRATION IN THE LOWEST LANDAU BAND} %yyy(WORDING!) It is often asserted that electrons in a large, constant magnetic field are confined to the lowest Landau band, but this is {\it not} true in the parameter regions 1 and 2, even if $B$ is large. It is also not true under the weaker assumption $Z^{-2}N^{2/3}B \ll 1$. We expect the confinement to occur, however, when $Z^{-2}N^{2/3}B \gg 1$ and that assertion is the content of Theorem~1.2, which we shall prove now. {\it Proof of Theorem~1.2 (Confinement to lowest Landau band): Part 1.} Let $\Pi^{i}_0$ be the projection onto the subspace of $\H = \bigotimes \limits^N L^2 (\R^3; \C^2)$ in which electron $i$ is in the lowest Landau band. Note $\bigotimes\limits^N$ and not $\bigwedge\limits^N$; at this point we are temporarily dropping symmetry constraints. The integral kernel for $\Pi^{i}_0$ is given in (1.14) (with $(x,y)$ replaced by $(x^i, y^i)$) and multiplied by the identity operator in the other variables. We let $\Pi^{i}_> := I - \Pi^{i}_0$ be the projection onto all the other bands. If $\alpha$ is any subset of the integers $\{ 1, 2, \dots , N \}$ we now define the projection $$\Pi^\alpha = \prod \limits_{i \in \alpha} \Pi^{i}_0 \prod \limits_{j \not\in \alpha} \Pi^{j}_>, \eqno(6.1)$$ i.e., $\Pi^\alpha$ is the projection in $\H$ onto the subspace in which particles $i \in \alpha$ are in the lowest Landau band while the remaining particles are in orthogonal bands. Clearly, ${\mathop{\sum}_\alpha} \Pi^\alpha = I$. Note that $\Pi^N_0 = \Pi^\alpha$ when $\alpha = \{ 1, 2, \dots , N \}$. We claim that for any $\varepsilon > 0$ $$H_N \geq \sum \limits_\alpha \Pi^\alpha H^\alpha \Pi^\alpha , \eqno(6.2)$$ where $H^\alpha$ is the modified Hamiltonian defined by $$\eqalignno{H^\alpha &= \sum \limits^N_{i=1} H^{i}_\A - (1 + \varepsilon) Z \sum \limits_{i \in \alpha} \vert x^i \vert^{-1} - (1 + \varepsilon^{-1}) Z \sum \limits_{i \not\in \alpha} \vert x^i \vert^{-1} + (1 - 3 \varepsilon) \sum \limits_{\scriptstyle i < j \atop \scriptstyle i,j \in \alpha} \vert x^i - x^j \vert^{-1} \cr &- (3 \varepsilon^{-1} -1) \sum \limits_{\scriptstyle i < j \atop \scriptstyle i,j \not\in \alpha} \vert x^i - x^j \vert^{-1} - (\mfr3/2 \varepsilon^{-1} + \mfr3/2 \varepsilon -1) \sum \limits_{i \in \alpha, j \not\in \alpha} \vert x^i - x^j \vert^{- 1}. \qquad&(6.3)\cr}$$ To prove this, consider the three kinds of terms in (1.1). For the first, ${\mathop{\sum}_i} H^{i}_\A$, the decomposition (6.2) is evidently an identity since each $H^{i}_\A$ commutes with each $\Pi^\alpha$. For the second term, we can write $$\vert x^i \vert^{-1} = (\Pi^{i}_0 + \Pi^{i}_>) \vert x^i \vert^{-1} (\Pi^{i}_0 + \Pi^{i}_>). \eqno(6.4)$$ The ``diagonal'' terms, i.e., $\Pi^{i}_0 \vert x^i \vert^{-1} \Pi^{i}_0$ and $\Pi^{i}_> \vert x^i \vert^{-1} \Pi^{i}_>$, yield, without further ado, the terms in (6.3) proportional to $Z$. The term in (6.3) proportional to $\varepsilon Z$ and $\varepsilon^{-1} Z$ come from the ``off-diagonal'' terms in (6.4) by using the Schwarz inequality as follows: $$- \Pi^{i}_> \vert x^i \vert^{-1} \Pi^{i}_0 -\Pi^{i}_0 \vert x^i \vert^{-1} \Pi^{i}_> \geq - {\varepsilon } \Pi^{i}_0 \vert x^i \vert^{-1} \Pi^{i}_0 - \varepsilon^{-1} \Pi^{i}_> \vert x^i \vert^{-1} \Pi^{i}_> . \eqno(6.5)$$ Thus we have proved (6.2) as far as the single particle potential, $-Z /\vert x \vert$, is concerned. The two-body terms $\vert x^i - x^j \vert^{-1}$ are only slightly more complicated. We can write $\vert x^i - x^j \vert^{-1}$ in a manner similar to (6.4) using $\Pi^{i}_0, \Pi^{i}_> \Pi^{j}_0$ and $\Pi^{j}_>$. Now there will be 16 terms instead of 4 as in (6.4). Again, we bound the ``off-diagonal'' terms by the Schwarz inequality; the result is (6.3). Our next goal is to bound $H^\alpha$ on functions $\Psi$ that are antisymmetric in the $\alpha$ variables (we denote the number of these by $0 \leq n_\alpha \leq N$) and also antisymmetric in the remaining $n_{\sim \alpha} = N - n_\alpha$ variables. Additionally, we require that the $n_{\sim \alpha}$ variables are in the second or higher Landau band, i.e., $\Pi^{i}_0 \Psi = 0$ for $i \not\in \alpha$. Conversely, $\Pi^{i}_0 \Psi = \Psi$ for $i \in \alpha$. {\it Part 2.} First we study the three terms in (6.3) that involve {\it only} the $\alpha$ variables. These are the first, second and fourth terms in (6.3) and (after relabelling the particles) they are equal to $$\widehat H^\alpha := (1 - 3 \varepsilon) H_{n_\alpha} + \varepsilon \sum \limits^{n_\alpha}_{i=1} \{ 3 H^{i}_\A - 4Z \vert x^i \vert^{-1} \}, \eqno(6.6)$$ where $H_{n_\alpha}$ is given by (1.1) with $N = n_\alpha$ there. The last sum in (6.6), when restricted to the lowest Landau band, can be bounded below by the true ground state energy $E^{\rm Q}(n_\alpha, Z, B)$ multiplied by a universal constant. In fact, from the remark (ii) following Theorem~5.3, eq. (5.19) (with $\delta=8/3$ there), we can estimate the sum (when restricted to the lowest Landau band) below by $3(8.2)^2 E^{\rm Q}_{(1)}(n_\alpha,Z/2,B)$. {F}rom Theorem~5.3 the energy $E^{\rm Q}_{(1)}(n_\alpha,Z/2,B)$ is bounded below by $E^{\rm Q}(n_\alpha, Z, B)$. Thus, we reach the following conclusion about $\widehat H^\alpha$: $$\Pi^\alpha \widehat H^\alpha \Pi^\alpha \geq \Pi^\alpha \{ (1 - 3 \varepsilon) E^{\rm Q}_{\rm conf} (n_\alpha, Z, B) + (\const.)\varepsilon E^{\rm Q}(n_\alpha, Z, B) \} \Pi^\alpha. \eqno(6.7)$$ Note the distinction between $E^{\rm Q}_{\rm conf}$ and $E^{\rm Q}$ in (6.7). {\it Part 3.} The remaining four terms in $H^\alpha$ can be bounded below by $$\widetilde H^\alpha = T^{\sim \alpha} - (1 + \varepsilon^{-1}) Z \sum \limits_{i \not\in \alpha} \vert x^i \vert^{-1} - 3 \varepsilon^{-1} \sum \limits_{i \not\in \alpha} \left\{ \sum\limits_{j \in \alpha} \vert x^i - x^j \vert^{-1} \right\} - 3 \varepsilon^{-1} \sum \limits_{\scriptstyle i < j \atop \scriptstyle i,j \not\in \alpha} \vert x^i - x^j \vert^{-1}, \eqno(6.8)$$ where $T^{\sim \alpha}$ is the kinetic energy operator (the sum of the $H^{i}_\A$'s) for the electrons {\it not in} $\alpha$. Let us note that for one particle $2B \Pi_> \leq \Pi_> H_\A \Pi_> = \Pi_> [(\p + \A)^2 + \vsigma \cdot \B] \Pi_>$. Thus, $\Pi_> (\p + \A)^2 \Pi_> \geq \Pi_> (2B - \vsigma \cdot \B) \Pi_>$. Hence, since $B \geq \pm\vsigma \cdot \B$ $$\Pi_> H_\A \Pi_> \geq \Pi_> [\mfr1/2 (\p + \A)^2 + \mfr1/2 (2B - \vsigma \cdot \B) + \vsigma \cdot \B] \Pi_> \geq \mfr1/2 \Pi_> [(\p + \A)^2 + B] \Pi_>. \eqno(6.9)$$ Because there are no kinetic energy operators in (6.8) for the $\alpha$ particles, the $x^j$'s for $j \in \alpha$ appearing in (6.8) can be taken to be {\it fixed points} in $\R^3$ whose value is adjusted to give the lowest possible energy for $\Pi^\alpha \widetilde H^\alpha \Pi^\alpha$. Given any admissible antisymmetric function $\Psi (x_{n_\alpha +1}, \dots , x_N)$, and corresponding density $\uprho_\Psi (x)$ (as given in (1.3)) with $x \in \R^3$, we can always translate $\Psi$ (i.e., $x \mapsto x + y$) so that the maximum of $\int \uprho_\Psi (x+z) \vert x \vert^{-1} dx$ occurs at $z =0$. Such a translation will minimize $- ( \Psi ,{\mathop{\sum}_i} \vert x^i \vert^{-1} \Psi )$. This, in turn, implies that the maximum of $\int \uprho_\Psi (x) \vert x - x^j \vert^{-1} dx$ occurs at $x^j = 0$. This observation, together with (6.9), means that inf spec $(\Pi^\alpha \widetilde H^\alpha \Pi^\alpha) \geq \inf \ {\rm spec} (\Pi^\alpha \overline H^\alpha \Pi^\alpha$), where $$\overline H^\alpha = nB/2 + {1 \over 2} \sum \limits^n_{i=1} [\p^i +\A (x^i)]^2 - [(1 + \varepsilon^{-1}) Z + 3 \varepsilon^{-1} n_\alpha] \sum \limits^n_{i=1} \vert x^i \vert^{-1} - 3 \varepsilon^{-1} \sum \limits_{1 \leq i < j \leq n} \vert x^i - x^j \vert^{-1}. \eqno(6.10)$$ Here $n: = n_{\sim \alpha} = N - n_\alpha$. To obtain a lower bound to $\overline H^\alpha$ we start by using the decomposition in L\'evy-Leblond's paper [51] for $n \geq 2$: $$\eqalignno{\overline H^\alpha = nB/2 + {1 \over (n-1)} \sum \limits^n_{i=1} \biggl\{ \sum \limits^n_{\scriptstyle j=1 \atop \scriptstyle j\not= i} &\mfr1/2 [\p^j + \A (x^j)]^2 - [(1 + \varepsilon^{-1}) Z + 3 \varepsilon^{-1} n_\alpha] \vert x^j \vert^{-1} \cr &- {3(n-1) \over 2 \varepsilon} \vert x^j - x^i \vert^{-1} \biggr\}. \qquad&(6.11)\cr}$$ To bound the operator in braces $\{\ \}$, we can set $x^i = 0$ (for the reason given in the preceding paragraph). We also introduce a cutoff $R>0$ and note that $-|x|^{-1}\geq -v-R^{-1}$ where the potential $v$ is given by $v(x)=|x|^{-1}-R^{-1}$ for $|x|\leq R$ and $v=0$ if $|x|\geq R$. Finally, we use the Lieb-Thirring inequality\footnote{$^\dagger$} {\eightpoint \baselineskip=5ex In [52] the Lieb-Thirring inequality was originally proved for $-\Delta-V$. Because of the diamagnetic inequality, the proof also works when $-\Delta={\bf p}^2$ is replaced by $({\bf p}+{\bf A})^2$, as remarked by several authors.} to bound the energy of $\sum \{ \mfr1/2 [\p+\A]^2 - V\}$ by $0.114 \int V^{5/2}$. Here, the potential $V$ is $$ V=((1+\varepsilon^{-1})Z+3\varepsilon^{-1}(n_\alpha +(n-1)/2))v\leq \varepsilon^{-1}[2Z+3N]v. $$ We find that the operator in $\{\ \}$ in (6.11) is bounded below by $ -8\pi(0.114)R^{1/2}\varepsilon^{-5/2}[2Z+3N]^{5/2} -\varepsilon^{-1}[2Z+3N]n/R $. We minimize this expression with respect to $R$. The result for all $n = N - n_\alpha > 0$ is (after replacing $n/(n-1)$ by 2) $$\overline H^\alpha \geq (N - n_\alpha) B/2 - 24 \varepsilon^{-2} (0.114 \pi/2)^{2/3} (N - n_\alpha)^{1/3} [2Z + 3N]^2. \eqno(6.12)$$ Our final lower bound on $\Pi^\alpha H^\alpha \Pi^\alpha$ is $K^\alpha \Pi^\alpha,$ where $K^\alpha$ is the number given by the sum of the right sides of (6.7) and (6.12) for $\widehat H^\alpha$ and $\overline H^\alpha$ respectively. {\it Part 4.} Our goal now is to show that the $\varepsilon$ in the previous inequalities can be chosen to depend (only) on $\lambda^{2/3}\beta = Z^{-2} N^{2/3} B$ and $\Lambda$ in such a way that $$K^\alpha \geq E^{\rm Q}_{\rm conf} (N,Z,B) + \delta (\lambda^{2/3}\beta, \Lambda) E^{\rm Q}(N, Z,B), \eqno(6.13)$$ for every $\alpha$, and where $\delta (\lambda^{2/3}\beta, \Lambda) \rightarrow 0$ as $\lambda^{2/3}\beta \rightarrow \infty$ for each fixed $\Lambda$. Equation (1.16) is an immediate consequence of this because $E^{\rm Q}(N, Z, B) = \inf \{ ( \Psi ,H_N \Psi )\, :\, ( \Psi , \Psi ) = 1 \}$, but $$( \Psi , H_N \Psi ) \geq \sum \limits_\alpha K^\alpha ( \Psi , \Pi^\alpha \Psi ) \quad \hbox{and} \quad ( \Psi , \Psi ) = \sum \limits_\alpha ( \Psi , \Pi^\alpha \Psi ) . \eqno(6.14)$$ To prove (6.13) we first note that $E^{\rm Q}_{\rm conf}$ and $E^{\rm Q}$ are both monotone nonincreasing in $N$ (because we can always move particles away to spatial infinity if we wish). Therefore we can replace $n_\alpha$ by $N$ in (6.7). The contribution of the right side of (6.7) is exactly of the correct form (provided we can let $\varepsilon \rightarrow 0$ as $\lambda^{2/3}\beta \rightarrow \infty$). In case $B \leq Z^3$ (and $\lambda^{2/3}\beta \geq 1$) the contribution of the right side of (6.12) can be usefully bounded below by omitting the $B$ term. The rest is bounded below (replacing $N - n_\alpha$ by $N$) by $$-(\const.) (1 + \mfr3/2 \Lambda)^2 \varepsilon^{-2} (\lambda^{2/3}\beta)^{-2/5} Z^{6/5} N^{3/5} B^{2/5} \geq (\const.) (1 + \mfr3/2 \Lambda)^2 \varepsilon^{-2} (\lambda^{2/3}\beta)^{-2/5} E^{\rm Q}(N, Z, B)$$ The last inequality follows from the upper bound on $E^{\rm Q}(N, Z, B)$ given in Theorem~5.3. In case $B \geq Z^3$ we retain the $B$ term in (6.12) and we see (since $(N - n_\alpha) \geq (N - n_\alpha)^{1/3}$) that (6.12) is actually positive (and hence can be neglected) if $Z$ and $\varepsilon$ are not too small. Recall that $Z \geq N/\Lambda \geq 1/\Lambda$, so $Z$ is never small. We can easily let $\varepsilon \rightarrow 0$ as $\lambda^{2/3}\beta \rightarrow \infty$. \lanbox %%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf VII. REDUCTION TO A ONE-BODY PROBLEM} Our goal here is to replace the two-body Coulomb repulsion $\sum \nolimits_{i < j} \vert x^i - x^j \vert^{-1}$ and attraction $-Z \sum \nolimits_i \vert x^i \vert^{-1}$ by the following one-body potential plus a constant: $$\Phi^{\DM} = - \sum \limits^N_{i=1} \phi^{\DM} (x^i) - D (\uprho^{\DM}, \uprho^{\DM}), \eqno(7.1)$$ where $\uprho^{\DM}$ is the density for the minimizer of $\E^{\DM}$. $$\phi^{\DM} (x):=\phi^{\DM}_{x_\bot}(x_3) = Z \vert x \vert^{-1} - \vert x \vert^{-1} * \uprho^{\DM} (x), \eqno(7.2)$$ and with a relative error that vanishes as $Z \rightarrow \infty$. Such a reduction is much more difficult in the $B \gg Z^3$ case than it is in the $B\ll Z^3$ case treated in our companion paper [28]. In the latter case, spherical symmetry is not broken, and it is sufficient to use a bound of the form (with $\Psi$ being the ground state of $H_N$, or an approximate ground state) $$(\Psi, \sum \nolimits_{i 0$ define the regularized Coulomb potential $$V (x) = \vert x \vert^{-1} [1- \exp (-\vert x \vert /\delta)]. \eqno(7.7)$$ Our notation suppresses the dependence of $V$ on $\delta$. Even though this regularization is spherically symmetric, the parameter $\delta^{-1}$ will essentially be the ultraviolet cut-off perpendicular to the field. Indeed, we now introduce an additional cut-off parallel to the field. This additional cut-off is most conveniently done on the Fourier transform of $V$, $$\widehat V (p) = \int \exp [-ip \cdot x] V(x) dx = 4 \pi (\vert p \vert^{-2} - (\vert p \vert^{2} + \delta^{-2})^{-1}). \eqno(7.8)$$ For $a > 0$ we define the cut-off potential $$\widehat V_< (p) = \cases{\widehat V (p) &,\ if $ \vert p_3 \vert \leq a^{-1}$ \cr 0 &,\ if $\vert p_3 \vert > a^{-1}$\cr}. \eqno(7.9)$$ Additionally, let $\widehat V_> (p) = \widehat V (p) - \widehat V_< (p)$. We shall prove that the main part of the repulsive energy comes from $\widehat V_< (p)$. Since $\widehat V_< (p) \geq 0$, its inverse Fourier transform $V_< (x)$ defines a positive semidefinite kernel, i.e., $\int \int \overline f(x) V_< (x-y) f(y) dx dy \geq 0$ for all functions $f$. As a consequence the following holds for any $\uprho \in L^1 (\R^3)$ (see [53]). $$\sum \limits_{i < j} V_< (x^i -x^j) \geq \sum \limits_i \int \uprho (y) V_< (x^i - y) dy - \mfr1/2 N V_< (0) - \mfr1/2 \int \int \uprho (x) \uprho (y) V_< (x-y) dx dy. \eqno(7.10)$$ This inequality can be seen by the following computation which is easily made rigorous by approximating the delta function by a sequence of smooth compactly supported functions. $$\eqalignno{\sum \nolimits_{i < j} V_< (x^i - x^j) &= \sum \limits_{i < j} \int \int \delta (x-x^i) V_< (x-y) \delta (y - x^j) dx dy \cr &= \mfr1/2 \int \int \left( \sum \limits_i \delta (x-x^i) - \uprho (x) \right) V_< (x-y) \left( \sum \limits_i \delta (y - x^i) - \uprho (y) \right) dxdy \cr &- {N \over 2} V(0) + \sum \limits_i \int V_< (x^i - y) \uprho (y) dy - \mfr1/2 \int \int \uprho (x) \uprho (y) V_< (x-y) dx dy. \cr}$$ Inequality (7.10) follows from the positivity of the kernel of $V_<$. In our application of (7.10) we shall take $\uprho = \uprho^{\DM}$. We must estimate $V_< (0)$ which, according to (7.10), is essentially the exchange energy per particle. {F}rom the formula for the inverse Fourier transform we get $$\eqalignno{V_< (0) &= (2 \pi)^{-3} \int \widehat V_< (p) dp = (2 \pi)^{-3} \int\limits_{\vert p_3 \vert \leq 1/a} \widehat V (p) dp \cr &= (2 \pi)^{-1} \int^{1/a}_{-1/a} \ln (1 + \vert p_3 \vert^{-2} \delta^{-2}) dp_3 \leq (2 \pi)^{-1} \int^{1/a}_{-1/a} \big[\ln (1 + a^2 \delta^{-2}) - 2 \ln (a \vert p_3 \vert)\big] dp_3 \cr &= \pi^{-1} a^{-1} (2 + \ln (1 + a^2 \delta^{-2})). \qquad&(7.11) \cr}$$ We turn now to $V_>$, the inverse Fourier transform of $\widehat V_>$. It follows from (7.7), (7.11) and the definition of $\widehat V_>$ that $V_> (0)$ is of order $\delta^{-1}$. We can therefore not use (7.10) with $V_>$ in place of $V_<$, since this would yield an estimate of the exchange energy per particle that is too large. Indeed, $\delta$ will be chosen smaller than the perpendicular radius, $\sim Z^{1/2} B^{- 1/2}$, of the atom. Thus, an exchange bound of the form $N\delta^{-1}$ will be larger than the total energy when $B$ is large enough, i.e., when $B \gg Z^5$. The moral of this is that if our error terms are to be useful the parameter $\delta$ must appear inside a logarithm. Hence the $V_>$ part of the kernel has to be treated differently from $V_<$. Our main estimate on $V_>$ is as follows. {\bf 7.2. LEMMA.} {\it If $\psi \in L^2 (\R)$ with $\int \vert \psi \vert^2 =1 $ then $$\sup \limits_{x_\bot} \left\vert \int \limits_\R V_> (x) \vert \psi (x_3) \vert^2 dx_3 \right\vert \leq (8 \pi^{-1} a)^{1/2} \ln (1 + a^2 \delta^{-2}) \left( \int \limits_\R \biggl( {d \psi (x_3) \over dx_3} \biggr)^2 dx_3 \right)^{3/4}. \eqno(7.12)$$} \indent {\it Proof of Lemma 7.2:} The supremum norm of the function $x_\bot \mapsto \int \limits_\R V_> (x) \vert \psi (x_3) \vert^2 dx_3$ can be estimated by the $L^1$ norm of its Fourier transform which is the function $p_\bot \mapsto (2 \pi)^{-1} \int \limits_\R \widehat V_> (p) \widehat{\vert \psi \vert^2} (p_3) dp_3$. Thus $$\eqalignno{\sup \limits_{x_\bot} \left\vert \int \limits_\R \widehat V_> (x) \vert \psi (x_3) \vert^2 dx_3 \right\vert &\leq (2 \pi)^{-3} \int \limits_{\R^2} \left\vert \int \limits_\R \widehat V_> (p) \widehat{\vert \psi \vert^2} (p_3) dp_3 \right\vert dp_\bot \cr &\leq (2 \pi)^{-3} \int \limits_{\R^2} \widehat V (p_\bot, a^{-1}) dp_\bot \int \limits_{\vert p_3 \vert \geq 1/a} \vert \widehat{\vert \psi \vert^2} (p_3) \vert dp_3, \qquad&(7.13)\cr}$$ where we have used that $\widehat V$ is positive and monotonically decreasing in $p_3$ for each fixed value of $p_\bot$. The first integral in (7.13) is easy to calculate: $$\int \limits_{\R^2} \widehat V (p_\bot, a^{-1}) dp_\bot = (2 \pi)^2 \ln (1 + a^2 \delta^{-2}). \eqno(7.14)$$ The second integral in (7.13) can be estimated by the Cauchy- Schwarz inequality. $$\eqalignno{\int_{\vert p_3 \vert \geq 1/a} &\left\vert \widehat{\vert \psi \vert^2} (p_3) \right\vert dp_3 \leq \left( \int_{\vert p_3 \vert \geq 1/a} \vert p_3 \vert^{-2} dp_3 \right)^{1/2} \left( \int \limits_\R \vert p_3 \vert^2 \left\vert \widehat{\vert \psi \vert^2} (p_3) \right\vert^2 dp_3 \right)^{1/2} \cr &= (4 \pi a)^{1/2} \left( \int \limits_\R \biggl( {d \over dx_3} \vert \psi \vert^2 \biggr)^2 dx_3 \right)^{1/2}. \cr}$$ Using $\sup \limits_x \vert \psi (x) \vert \leq \sqrt 2 \biggl( \int \bigl\vert {d \over dx_3} \psi \bigr\vert^2 dx_3 \biggr)^{1/4}$ we find $$\int_{\vert p_3 \vert \geq 1/a} \left\vert \widehat{\vert \psi \vert^2} (p_3) \right\vert dp_3 \leq (32 \pi a)^{1/2} \left( \int \biggl\vert {d \psi \over dx_3} \biggr\vert^2 dx_3 \right)^{3/4}. \eqno(7.15)$$ Inserting (7.14) and (7.15) into (7.13) gives (7.12). This finishes the proof of Lemma 7.2. \lanbox {\it Continuation of the proof of Theorem 7.1.} We are now prepared to estimate the many-body Hamiltonian $H_N$. Using $\vert x \vert^{-1} \geq V(x)$ we get, for all $0 < \alpha < 1$, $$\eqalignno{H_N &\geq \sum \limits^N_{i=1} (H^{i}_\A - Z \vert x^i \vert^{-1}) + \sum \limits_{1 \leq i < j \leq N} V(x^i - x^j) \cr &\geq \sum \limits^N_{i=1} ((1 - \alpha) H^{i}_\A - Z \vert x^i \vert^{-1}) + \sum \limits_{1 \leq i < j \leq N} V_< (x^i - x^j) \cr &+ \sum \limits^N_{i=1} \left[ \alpha H^{i}_\A + \mfr1/2 \sum \limits_{j \not=i} V_> (x^i - x^j) \right]. \qquad&(7.16)\cr}$$ We can estimate the operator $$\alpha H^{i}_\A + \mfr1/2 \sum \limits_{j \not= i} V_> (x^i - x^j)$$ with the $N-1$ points $x^j, j \not= i$ fixed, in which case it is a one-body operator. Since $H_\A \geq - (\partial /\partial x_3)^2$ we find from Lemma 7.2 the following lower bound independent of $x^j, j \not= i$ $$\eqalignno{\alpha H^{i}_\A &+ \mfr1/2 \sum \limits_{j \not= i} V_> (x^i - x^j) \geq \inf \limits_T \big\{ \alpha T - \mfr1/2 (N-1) (8 \pi^{- 1} a)^{1/2} \ln (1 + a^2 \delta^{-2}) T^{3/4} \big\} \cr &\geq - \mfr1/3 \big( \mfr3/8 \big)^4 \alpha^{-3} (8 \pi^{-1} a)^2 (N-1)^4 [\ln (1 + a^2 \delta^{-2})]^4. \qquad&(7.17)\cr}$$ Combining (7.10) (with $\uprho = \uprho^{\DM}$), (7.11), (7.16) and (7.17) we find for all $0 < \alpha < 1$ and all $a$ and $\delta > 0$ $$\eqalignno{H_N \geq &\sum \limits^N_{i=1} \biggl( (1 - \alpha) H^{i}_{\A} - Z \vert x^i \vert^{-1} + \int \uprho^{\DM} (y) V_< (x^i - y) dy \biggr) \cr &- \mfr1/2 \int \int \uprho^{\DM} (x) \uprho^{\DM} (y) V_< (x-y) dx dy - \mfr1/2 N \pi^{-1} a^{-1} (2 + \ln (1 + a^2 \delta^{-2})) \cr &- \mfr1/3 \big( \mfr3/8 \big)^4 \alpha^{-3} N(N-1)^4 (8 \pi^{-1} a)^2 [\ln (1 + a^2 \delta^{-2})]^4. \qquad&(7.18)\cr}$$ To arrive at the final estimate (7.6) we should like to replace $V_<$ in (7.18) by $\vert x \vert^{-1}$. We shall first replace $V_<$ by $V$ and then $V$ by $\vert x \vert^{-1}$. The error in the first replacement can be estimated with the aid of Lemma 7.2. Indeed, $$\eqalignno{\biggl\vert \int &\uprho^{\DM} (y) [V(x-y) - V_< (x-y)] dy \biggr\vert = \int \uprho^{\DM} (y) V_> (x-y) dy \cr &\leq (8 \pi^{-1} a)^{1/2} \ln (1 + a^2 \delta^{-2}) \int \limits_{\R^2} \left( \int \limits_\R (\partial \sqrt{\uprho^{\DM} (x)} /\partial x_3)^2 dx_3 \right)^{3/4} \left( \int \limits_\R \uprho^{\DM} (x) dx_3 \right)^{1/4} dx_\bot. \cr}$$ The integral $\left( \int \limits_\R \uprho^{\DM} \right)^{1/4}$ appears because of the normalization. Now, using H\"older's inequality, the estimate (4.5) and the virial inequality (4.44), $$\eqalignno{\int \uprho^{\DM} (y) V_> (x-y) dy &\leq (8 \pi^{-1} a)^{1/2} \ln (1 + a^2 \delta^{-2}) N^{1/4} \left( \int (\partial \sqrt{\uprho^{\DM}} /\partial x_3)^2 dx \right)^{3/4} \cr &\leq (8 \pi^{-1} a)^{1/2} \ln (1 + a^2 \delta^{-2}) N^{1/4} \vert E^{\DM} \vert^{3/4} .\qquad&(7.19)\cr}$$ Finally we control the replacement of $V$ by $\vert x \vert^{-1}$ this time using (4.8) and (4.44) $$\eqalignno{\int \uprho^{\DM} (y) [v(x-y) - \vert x-y \vert^{-1}] dy &= \int \uprho^{\DM} (y) \exp (-\vert x-y \vert \delta^{-1}) \vert x-y \vert^{-1} dy \cr &\leq \left( \int \uprho^{\DM} (y)^3 dy \right)^{1/3} \left( \int \exp \big( -\mfr3/2 \vert y \vert \delta^{-1} \big) \vert y \vert^{-3/2} dy \right)^{2/3} \cr &\leq (\const.) B^{2/3} \vert E^{\DM} \vert^{1/3} \delta. \qquad&(7.20)\cr}$$ Using (7.19) and (7.20) together with the estimate (4.46) on $E^{\DM}$ we get the following lower bound instead on (7.18). $$\eqalignno{H_N &\geq \sum \limits^N_{i=1} ((1 - \alpha) H^{i}_\A - \phi^{\DM} (x^i)) - \mfr1/2 \int \int \uprho^{\DM} (x) \uprho^{\DM} (y) \vert x-y \vert^{-1} dx dy \cr &- (\const.) \biggl[ Na^{-1} (2 + \ln (1 + a^2 \delta^{-2})) + \alpha^{- 3} N^5 a^2 (\ln (1 + a^2 \delta^{-2}))^4 \cr &+N^{5/4} a^{1/2} Z^{9/4} \ln (1 + a^2 \delta^{-2}) (1 +[\ln (BZ^{-3})]^2)^{3/4} \cr &+ \delta B^{2/3} NZ (1+ [\ln (BZ^{-3}) ]^2)^{1/3} \biggr]. \qquad&(7.21)\cr}$$ We now make the choices $\delta = B^{-2/3} Z^{1/3}, \ a = Z^{-5/3} [1 + \ln (1 + BZ^{-3})]^{-1}$ and $\alpha = Z^{-1/3}$. Recalling that $N = \lambda Z$ we arrive at (7.6). This finishes the proof of Theorem 7.1. \lanbox %%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf VIII. LIMIT OF QUANTUM MECHANICS} In this section we shall complete the proof of the main result that the quantum energy $E^{\rm Q} (N,Z,B)$ can be approximated by the energy $E^{\DM} (N,Z,B)$. Unlike the semi-classical Thomas-Fermi approximation studied in our companion paper [28] the approximation of $E^{\rm Q}$ by $E^{\DM}$ is quite simple after having made the reduction to a one-body problem. {\it Proof of Theorem~1.3 (Energy asymptotics for regions 3, 4, 5):\/} Since Theorem 5.1 contains the appropriate upper bound on $E^Q$ we here only have to address the question of a lower bound. Furthermore it follows from Theorem~1.2 on the confinement to the lowest Landau band that we can replace $E^{\rm Q}$ by the ground state energy $E^{\rm Q}f_{\rm conf}$ of the confined Hamiltonian $\Pi^N_0 H_N \Pi^N_0$. If $\Pi^N_0 \Psi = \Psi$ we know from Lemma~4.1 that $\Gamma^{\Psi} \in \G^{\DM}_B$. {F}rom Theorem 7.1 we also know that for such a normalized function $\Psi$ $$( \Psi, H_N \Psi ) \geq (1 - Z^{-1/3}) \int \limits_{\R^2} \Tr_{L^2 (\R)} \left[ - {\partial^2 \over \partial x^2_3} \Gamma^{\Psi}_{x_\bot} \right] dx_\bot - \int \phi^{\DM} (x) \uprho_\Psi (x) dx - D (\uprho^{\DM}, \uprho^{\DM}) - R_{\rm L} \eqno(8.1)$$ where $R_{\rm L} = (\const.) (1 + \lambda^5) (1 + Z^{8/3}) (1 + [\ln (B/Z^3)]^2)$. It is clear that if $( \Psi, H_N \Psi) \leq 0$ (which we can of course assume) then $\int\limits_{\R^2} \Tr_{L^2 (\R)} [\hat{h}_{Z, x_\bot} \Gamma^{\Psi}_{x_\bot}]$ $dx_\bot \leq 0$ where $\widehat h_{Z, x_\bot}$ was defined in (4.9). Hence $$\eqalignno{\mfr1/2 \int \limits_{\R^2} \Tr_{L^2 (\R)} \left[ - {\partial \over \partial x^2_3} \Gamma^{\Psi}_{x_\bot} \right] dx_\bot &= \int \limits_{\R^2} \Tr_{L^2 (\R)} \big[(\widehat h_{Z, x_\bot} - \mfr1/2 \widehat h_{2Z_1 x_\bot}) \Gamma^{\Psi}_{x_\bot} \big] dx_\bot \cr &\leq - \mfr1/2 \int \limits_{\R^2} \Tr_{L^2 (\R)} [\widehat h_{2Z, x_\bot} \Gamma^{\Psi}_{x_\bot}] dx_\bot \cr &\leq c_\lambda Z^3 (1 + [\ln (B/Z^3)]^2), \qquad&(8.2) \cr}$$ where the last inequality follows from (4.11). Thus from (8.1) we get the estimate $$\eqalignno{( \Psi , H_N \Psi) &\geq \int \limits_{\R^2} \Tr_{L^2 (\R)} \left[ \biggl( - {\partial^2 \over \partial x^2_3} - \phi^{\DM} \biggr) \Gamma^{\Psi}_{x_\bot} \right] dx_\bot - D (\uprho^{\DM}, \uprho^{\DM}) - (\const.)R_{\rm L} \cr &= \E_{\rm L}^{\DM} [\Gamma^{\Psi}] - D (\uprho^{\DM}, \uprho^{\DM}) - (\const.)R_{\rm L} \cr &\geq E^{\DM} - (\const.)R_{\rm L}. \qquad&(8.3)\cr}$$ The last inequality in (8.3) is a consequence of Theorem~4.4 since it follows from there that $$\E^{\DM}_{\rm L}[\Gamma^{\Psi}] \geq \E^{\DM}_{\rm L} [\Gamma^{\DM}] = \E^{\DM} [\Gamma^{\DM}] + D (\uprho^{\DM}, \uprho^{\DM}).$$ Since (8.3) holds for all normalized $\Psi$ satisfying $\Pi^N_0 \Psi = \Psi$ we can take the infimum. We obtain $$E^{\rm Q}_{\rm conf} (N,Z,B) \geq E^{\DM} (N,Z,B) - (\const.)R_{\rm L}. \eqno(8.4)$$ In Theorem~5.3 we proved that $E^{\rm Q}(N,Z,B) \leq - c_\lambda (Z^{9/5} B^{2/5} + 1) $ and, indeed this bound is of the correct order if $B \leq Z^3$. If we compare this bound with the error $R_{\rm L}$ in (8.4) we see that only when $B %zzz \gg Z^{13/6}$ can $R_{\rm L}$ be negligible. When the magnetic field is in the range $Z^{4/3} \ll B \lo Z^{13/6}$ we must use the standard exchange estimate (7.3) rather than (7.6). If $\Psi$ is a normalized wave function with $\Pi^N_0 \Psi = \Psi$ we can easily estimate the error in (7.3). In fact, from the Lieb-Thirring inequality (4.8) $$\hskip -.5truecm\int \uprho^{4/3}_\Psi (x) dx \leq \left( \int \uprho^3_\Psi (x) dx \right)^{1/6} \left( \int \uprho_\Psi (x) dx \right)^{5/6} \leq \left( {3 \over \pi^2} B^2 \int \Tr_{L^2(\R)} \biggl[ - {\partial^2 \over \partial x^2_3} \Gamma^{\Psi}_{x_\bot} \biggr] dx_\bot \right)^{1/6} N^{5/6}.$$ If we now proceed as in (8.2) but using the estimate (4.10) instead of (4.11) we get if $( \Psi , H_N \Psi ) \leq 0$ that $$\int \uprho^{4/3}_\Psi (x) dx \leq c_\lambda Z^{17/15} B^{2/5}. \eqno(8.5)$$ As explained in the beginning of Sect.~VII we combine (7.3) with the inequality $D(\uprho_\Psi, \uprho_\Psi) \geq 2D (\uprho_\Psi, \uprho^{\DM}) - D(\uprho^{\DM}, \uprho^{\DM})$ and arrive at $$\eqalignno{( \Psi , H_N \Psi ) &\geq \int \limits_{\R^2} \Tr_{L^2 (\R)} \left[ \biggl( - {\partial^2 \over \partial x^2_3} - \phi^{\DM} \biggr) \Gamma^{\Psi}_{x_\bot} \right] dx_\bot \cr &- D (\uprho^{\DM}, \uprho^{\DM}) - c_\lambda Z^{17/15} B^{2/5} \cr &\geq E^{\DM} - c_\lambda Z^{17/15} B^{2/5}.\cr}$$ We can thus conclude that (8.4) holds with $R_{\rm L}$ being the lesser of the two quantities $$c_\lambda (Z^{17/15} B^{2/5} + 1) \quad \hbox{and} \quad c_\lambda (1 + Z^{8/3}) (1 + [\ln (B/Z^3)]^2). \eqno(8.6)$$ To complete the proof of Theorem~1.3 we have to show that both $R_{\rm U}$ from (5.16) and $R_{\rm L}$ from (8.6) are negligible compared to $E^{\rm Q}$. This immediately follows from Theorem~5.3 and (5.24). \lanbox We can now summarize the relations between the true quantum energy $E^{\rm Q}$ and our three approximating energies $E^{\DM}, E^{\SS}$ and $E^{\HS}$ as follows \item{1)} If $N/Z = \lambda$ and $B/(2\pi Z^3) = \eta$ are held fixed as $Z \rightarrow \infty$ then (the symbol $\approx$ here means that the ratio of the left side to the right side converges to one) $$E^{\rm Q} (N,Z,B) \approx Z^3 E^{\DM} (\lambda, 1, 2\pi\eta). \eqno(8.7)$$ \item{2)} If $N/Z = \lambda$ and $B/(2\pi Z^3) = \eta \geq \eta_c$ as $Z \rightarrow \infty$ then, from Theorem 4.6, $$E^{\rm Q} (N,Z,B) \approx Z^3 E^{\SS} (\lambda, 1, 2\pi\eta). \eqno(8.8)$$ \item{3)} If $N/Z = \lambda$ and $B/Z^3 \rightarrow \infty$ as $Z \rightarrow \infty$ then, from Theorem 3.5, $$E^{\rm Q} (N,Z,B) \approx Z^3 [\ln (B/(2\pi Z^3))]^2 E^{\HS} (\lambda), \eqno(8.9)$$ which, when combined with (3.7), gives Theorem~1.4. If we appeal to the strong Thomas-Fermi energy $E^{\STF}$ defined in [7] and [28] we can also give the asymptotics for small values of $B/Z^3$. \item{4)} If $N/Z = \lambda, B/Z^3 \rightarrow 0$ but $B/Z^{4/3} \rightarrow \infty$ as $Z \rightarrow \infty$ then $$E^{\rm Q} (N,Z,B) \approx Z^3 (B/Z^3)^{2/5} E^{\STF} (\lambda,1,1), \eqno(8.10)$$ which is just another way of writing (1.9). Because of Theorem~1.3 this last formula also holds for $E^{\DM}$: $$E^{\DM} (\lambda, 1, 2\pi\eta) \approx (2\pi\eta)^{2/5} E^{\STF} (\lambda,1,1), \eqno(8.11)$$ as $\eta \rightarrow 0$. A simple modification of the proof of Theorem~1.3 can be used in a standard way to prove that the quantum density in the large $Z$ limit can be approximated by the density of the minimizer for the density matrix functional. {\bf 8.1. THEOREM.} {\it Let $\uprho^{\DM} (x) = \uprho^{\DM} (x;N,Z,B)$ denote the density for the minimizer $\Gamma^{\DM}$ described in Theorem~4.3. We then have the following limit for the quantum density $\uprho^{\rm Q}$ as $Z \rightarrow \infty$ with $N/Z = \lambda$ and $B/(2\pi Z^3) = \eta$ held fixed $$Z^{-4} \uprho^{\rm Q} (Z^{-1} x) \rightharpoonup \uprho^{\DM} (x;\lambda, 1, 2\pi\eta), \eqno(8.12)$$ weakly in $L^1_{\loc} (\R^3)$. If $N\leq N^{\DM}_c$ the convergence is weakly in $L^1(\R^3)$. While the convergence in (8.12) is not uniform in $\eta$, neither for large nor small $\eta$, we can find a limit that is uniform for {\it large} $\eta$ by introducing $$ x_\eta:=\left(\eta^{-1/2} x_\bot,[\ln(\eta)]^{-1}x_3\right). $$ The weak $L^1_{\loc}(\R^3)$ limit that is uniform in $\eta$ for large $\eta$ is $$Z^{-4}\eta^{-1}[\ln(\eta)]^{-1} \uprho^{\rm Q} (Z^{-1}x_\eta) \rightharpoonup \eta^{-1}[\ln(\eta)]^{-1}\uprho^{\DM} (x_\eta;\lambda, 1, 2\pi\eta). \eqno(8.13)$$ } \indent {\it Proof:} The precise meaning of the limit in weak $L^1_{\loc}$ is that for each $W \in L^\infty (\R^3)$ with compact support $$Z^{-4}\eta^{-1}[\ln(\eta)]^{-1}\int \uprho^{\rm Q} (Z^{-1} x_\eta) W(x) dx \rightarrow \eta^{-1}[\ln(\eta)]^{-1}\int\uprho^{\DM} (x_\eta ; \lambda, 1, 2\pi\eta) W(x) dx. \eqno(8.14)$$ as $Z \rightarrow \infty$. To prove (8.14) it is enough to consider $W\in C^{\infty}_0(\R^3)$. To arrive at this conclusion we first notice that it is clearly enough (see also Lemma III.4 in [30]) to consider test functions $W$ that are characteristic functions of measurable sets. By outer regularity of the Lebesgue measure we can restrict our attention to open sets. However, let $f_n\in L^1_{\loc}(\R^3)$ be non-negative functions satisfying $\int f_n\widetilde W\to \int f\widetilde W$ for all functions $\widetilde W\in C^{\infty}_0(\R^3)$ only. If now $W$ is a characteristic function of an open set we can choose functions $W_\pm\in C^{\infty}_0(\R^3)$ such that $W_-\leq W\leq W_+$ and such that $\int fW_\pm$ are as close to $\int fW$ as we please. Then for all $\delta>0$ we can choose $n$ so large that $$ \int fW_--\delta\leq \int f_nW_-\leq \int f_nW \leq \int f_nW_+ \leq \int fW_+ +\delta, $$ and the convergence weakly in $L^1_{\loc}$ follows. It is important that the functions $f_n$ are non-negative for otherwise the conclusion would be wrong! Weak convergence in $L^1(\R^3)$, which means that we consider all $W\in L^\infty (\R^3)$ without restriction on the support, follows (see again Lemma III.4 in [30]) if we have both weak convergence in $L^1_{\loc}$ and convergence for the constant function $W=1$. This last condition simply means that $\lambda=N/Z$ (which is held fixed) converges to $\int\uprho(x;\lambda,1,2\pi\eta)dx$. This is of course true if $N\leq N^{\DM}_c$. The standard method [30] of proving (8.14) is to introduce the following perturbed Hamiltonian $$H_N (\alpha) = H_N + \alpha \sum \limits_{i=1} W_{Z,\eta}(x_i), $$ where the perturbing potential has the following scaling $$ W_{Z,\eta}(x):=Z^2(\ln\eta)^2 W \left(Z\eta^{1/2}x_\bot, Z(\ln\eta)x_3\right), $$ and the corresponding perturbed density matrix functional is $$\E^{\DM}_\alpha [\Gamma] = \E^{\DM} [\Gamma] + \alpha \int W_{Z,\eta} (x) \uprho_\Gamma (x) dx.$$ In terms of these we define energy functions $$E^{\rm Q}_\alpha (N,Z,B) = \inf \ \spec \ H_N (\alpha) $$ and $$E^{\DM}_\alpha (N,Z,B) = \inf \{ \E^{\DM}_\alpha [\Gamma] \big\vert \Gamma \in \G^{\DM}_B, \int_{\R^2} \Tr_{L^2 (\R)} [\Gamma_{x_\bot}] dx_\bot \leq N \}. $$ Since we are assuming that $W\in C^{\infty}_0(\R^3)$ the proof of Theorem~4.3 is applicable also in the case of the perturbed functional $\E^{\DM}_\alpha$ and shows the existence of a unique minimizer. It is clear that $E^{\DM}_\alpha$ satisfies the same scaling property as $E^{\DM}$, i.e., $E^{\DM}_\alpha (N,Z,B) = Z^3 E^{\DM}_\alpha (\lambda, 1, 2\pi\eta)$. It is also easy to see that for large values of $\eta$ the energy $E^{\DM}_\alpha (N,Z,B)$ can be estimated above and below by expressions of the form $(\const.)Z^3(\ln\eta)^2$, where the constants depend on $W$. We would like now to show that the proof of Theorem~1.3 extends to include the statement that the energies $E^{\rm Q}_\alpha$ and $E^{\DM}_\alpha$ satisfy $${E^{\rm Q}f_\alpha (N,Z,B) \over E^{\DM}_\alpha (N,Z,B)} \rightarrow 1 \eqno(8.15)$$ as $Z \rightarrow \infty$ uniformly in $\eta$ for large $\eta$. To do this we only need to argue that the unique minimizer for $\E^{\DM}_\alpha$ satisfies bounds similar to (4.50) and to the virial inequalities (4.44). These bounds were used in Sects.~V and VII and were essential in the proof of Theorem~1.3. \medskip \noindent1. {\it The estimate} (4.50): As in the proof of (4.50) we need to estimate $$ \eqalign{ -{B\over2\pi}&\left(\hbox{sum of negative eigenvalues of }- \mfr1/2{\partial^2 \over\partial x_3^2}-Z|x|^{-1}+\alpha W_{Z,\eta}(x)\right)\leq\cr -{B\over2\pi}&\left(\hbox{sum of negative eigenvalues of } -\mfr1/4{\partial^2\over\partial x_3^2}-Z|x|^{-1}\right)\cr -{B\over2\pi}&\left(\hbox{sum of negative eigenvalues of } -\mfr1/4{\partial^2\over\partial x_3^2}+\alpha W_{Z,\eta}(x)\right)}. $$ The first term above was estimated in Proposition~4.9 the second can be estimated by the one-dimensional Lieb-Thirring inequality to be less than $$ {B\over2\pi}\alpha^{3/2}\int |W_{Z,\eta}(x)|^{3/2} \leq Z^2(\ln\eta)^2\int_{\R} |W(Z\eta^{1/2}x_\bot,x_3)|^{3/2}dx_3. $$ Since $W$ has compact support the right side is bounded by the expression on the right side of (4.50) with a constant that now depends on $W$. \smallskip\noindent2. {\it The virial inequalities} (4.44): For the functional $\E^{\DM}_\alpha$ we can repeat the scaling argument that lead to the virial inequalities. Compared to the original virial inequalities we now get an extra term which depends on $W_{Z,\eta}$. By scaling, this term is of the form $Z^3(\ln\eta)^2$ multiplied by a constant which depends only on $W$. Hence we arrive at virial inequalities with constants depending on $W$. Having argued that (8.15) holds we can use the scaling property of $E^{\DM}_\alpha$ and rewrite (8.15) as the following limit which is uniform in $\eta$: $$Z^{-3} (\ln\eta)^{-2}E^{\rm Q}_\alpha (N,Z,B) \rightarrow (\ln\eta)^{-2} E^{\DM}_\alpha (\lambda, 1, 2\pi\eta). \eqno(8.16)$$ Both energies $E^{\rm Q}_\alpha$ and $E^{\DM}_\alpha$ are defined as infima of functions that are linear in $\alpha$ and are therefore themselves concave in $\alpha$. It then follows from (8.16) that $$Z^{-3}(\ln\eta)^{-2}{\partial \over \partial \alpha} E^{\rm Q}_\alpha (N,Z,B) \rightarrow (\ln\eta)^{-2}{\partial \over \partial \alpha} E^{\DM}_\alpha (\lambda, 1, 2\pi\eta), \eqno(8.17)$$ as $Z\to\infty$ unformly in $\eta$. We shall use (8.17) for $\alpha = 0$. {F}rom the concavity of $E^{\rm Q}_\alpha$ and $E^{\DM}_\alpha$ as functions of $\alpha$ we easily see that $$ \eqalign{Z^{-3}(\ln\eta)^{-2}{\partial \over \partial \alpha} E^{\rm Q}_\alpha (N,Z,B)\big\vert_{\alpha = 0} &= Z^{-3}(\ln\eta)^{-2} \int W_{Z,\eta}(x) \uprho^{\rm Q} (x) dx\cr &= Z^{-4}\eta^{-1}[\ln(\eta)]^{-1}\int \uprho^{\rm Q} (Z^{-1}x_\eta) W(x) dx.}$$ and $$(\ln\eta)^{-2}{\partial \over \partial \alpha} E^{\DM}_\alpha (\lambda, 1, 2\pi\eta) \big\vert_{\alpha = 0} = \eta^{-1}[\ln(\eta)]^{-1}\int \uprho^{\DM} (x_\eta; \lambda, 1, 2\pi\eta) W(x) dx.$$ Thus (8.14) follows from (8.17) which is a consequence of (8.15). \lanbox Notice that the scaling in (8.13) is identical to the scaling (3.11) introduced to study the hyper-strong limit. In particular, if $W$ is independent of $x_\bot$, i.e., $W(x)=W(x_3)$, on the support of $\uprho^{\DM}$ we have from Theorem~3.5 and Theorem~4.6 that the right side of (8.14) converges to $\int_{\R}\uprho^{\HS}(x_3)W(x_3)dx_3$ as $\eta\to\infty$. This proves the following result. {\bf 8.2. THEOREM.} {\it Let $\uprho^{\HS} (x) = \uprho^{\HS} (x; \lambda)$ be the minimizer for the hyper-strong functional given in Theorem~3.1 and let $N=\lambda Z$. Define the one-dimensional quantum density by $$\overline{\uprho}^{\rm Q} (x_3) = \int \limits_{|x_\bot|\leq \sqrt{2N/B}} \uprho^{\rm Q} (x_\bot, x_3) dx_\bot\quad.$$ (If $N\leq2Z$ we can, as a consequence of Corollary~3.6, let the integral extend over all of $\R^2$.) Then $$[Z^2\ln\eta)]^{-1} \overline{\uprho}^{\rm Q} ([Z \ln (\eta)]^{-1} x_3) \rightharpoonup \overline{\uprho}^{\HS} (x_3, \lambda),$$ weakly in $L^1_{\loc} (\R)$.} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bigskip %%%%%%%%%%%%%%%%%%%%%%%% \noindent {\bf IX. IONIZATION AND BINDING OF ATOMS} For reasons that no one really understands at present, enhanced binding of atoms is associated with enhanced ionization. By this we mean the following. In ordinary atoms (with $B = 0$) the maximal number of electrons, $N^\Q_c$, that can be bound to a nucleus of charge $Z$ satisfies $N^\Q_c /Z \rightarrow 1$ as $Z \rightarrow \infty$, as proved in [42]. Thus, the maximal ionization, $N^\Q_c - Z$, tends to zero on the scale of $Z$. At the same time, such an atom is well described to leading order by TF theory, which has the property, originally noted by Teller and proved in [30], that atoms never bind. More precisely, in TF theory the energy of a molecule containing two or more infinitely massive nuclei always has an energy that is not less than the sum of the constituent atomic energies. Thus for a real molecule, the binding energy must be of lower order than the atomic energies. We also note that TF theory gives spherically symmetric atoms. With a magnetic field the situation is unaltered in regions 1, 2 and 3 (see Sect.~I). In regions 4 and 5 things begin to change for the first time: \item{$\bullet$} (i) Atoms become non-spherical \item{$\bullet$} (ii) The maximal ionization is not zero on the scale of $Z$ \item{$\bullet$} (iii) Atoms bind together to form molecules with a binding energy $\sum E_{{\rm atom}} - E_{{\rm mol}}$ of the same order as the atomic energy $\sum E_{{\rm atom}}$ itself. As far as regions 4 and 5 go, we have proved assertion (i) in Theorem 4.5 for the density $\uprho^{\DM}$, and in Theorem 8.1 we proved that $\uprho^\Q$ is asymptotically the same as $\uprho^{\DM}$. Assertions (ii) and (iii) will not be proved here for all of region 4 but they are surely correct because they are resoundingly true in region 5, as we shall demonstrate now. We leave it as an {\it open problem} to verify (ii) and (iii) for the DM theory for all positive values of the parameter $\eta$. For large values of $\eta$, assertion (ii) was proved in the DM theory in Corollary 3.6 (recall Theorem 4.6) as a consequence of the study of the hyper-strong theory. We shall now use the hyper-strong theory to conclude assertions (ii) and (iii) for the quantum theory itself, as formulated in Theorems 1.5 and 1.6 in the Introduction. {\it Proof of Theorem 1.5. (Ionization of atoms):} If the theorem is false there is a number $\lambda < 2$ such that the following is true. There are two infinite sequences $Z_i$ and $B_i$ with $Z_i \rightarrow \infty$ and with $B_i /Z_i^3 \rightarrow \infty$ such that the ratio $E^\Q (N_i, Z_i, B_i) / E^\Q ([\lambda Z_i], Z_i, B_i) = 1$ for all integers $N_i > [\lambda Z_i]$. (Here $[\lambda Z]$ is the largest integer in $\lambda Z$.) Set $N_i = [2 Z_i]$. By Theorems 1.3, 3.5 and 4.6 the ratio converges, as $i \rightarrow \infty$, to $E^{\HS} (2) /E^{\HS} (\lambda) > 1$ by (3.7). This is a contradiction. \lanbox Before leaving the topic of ionization let us comment, as promised in Sect.~I, on the $2Z+1$ theorem [45]. That theorem is applicable to the case in which $H_\A$ is replaced by $(\p + \A)^2$, and states that the energy of an atom with $N$ electron is not smaller than the energy of an atom with $N-1$ electrons if $N \geq 2Z + 1$. This result can be translated into a statement about our Hamiltonian (1.1) in which the terms $\vsigma \cdot \B$ is included in $H_\A$. It says that if $N \geq 2Z+1$ then $E^\Q (N,Z,B) \geq E^\Q (N-1, Z,B) - B$. We cannot then conclude anything about $N^\Q_c$. In order to define the {\bf binding energy} that appears in Theorem 1.6 and in our discussion in this section and in Sect.~I, we proceed as follows. First we choose $K$ nuclei with charges $Z_1, Z_2, \dots , Z_K$ and locate them at $R_1, R_2, \dots , R_K$ in $\R^3$. (Theorem 1.6 concerns the special case $K = 2$, but here we shall consider the general case). The Hamiltonian $H_N$ in (1.1) is modified in two ways \item{(i)} $-Z \vert x \vert^{-1}$ is replaced by $-\sum \limits^K_{j=1} Z_j \vert x - R_j \vert^{-1}$. \item{(ii)} The nuclear Coulomb repulsion energy $\sum \limits_{1 \leq i < j \leq K} Z_i Z_j \vert R_i - R_j \vert^{-1}$ is added to $H_N$. \medskip\noindent The energy is defined by (1.2) as before, but this quantity is then minimized (more precisely, infimized) with respect to $R_1, \dots , R_K$. The final result is denoted by $E^\Q (N, Z_1, \dots , Z_K, B)$. The binding energy is defined to be $$E^\Q_\b (N, Z_1, \dots , Z_K, B) = \min \{ E^\Q (N^{\{a\}}, Z^{\{a\}}, B) + E^\Q (N^{\{b\}}, Z^{\{b\}}, B) \} - E^\Q (N, Z_1, \dots , Z_K, B), \eqno(9.1)$$ where the minimum is over all decompositions of $\{ 1,2, \dots , K\}$ into two subsets $\{a\}$ and $\{b\}$ and where $N^{\{a\}} + N^{\{b\}} = N$. A simpler and equally useful notion of binding energy is to restrict the three $N$'s appearing in (9.1) to be neutral, i.e., $N = \sum\nolimits^K_{j=1} Z, \ N^{\{a\}} = \sum \nolimits_{j \in \{a\}} Z, \ N^{\{b\}} = \sum \nolimits_{j \in \{b\}} Z$. (Incidentally, it is a fact about TF type theories [30] that if $N$ is neutral then the minimizing values of $N^{\{a\}}$ and $N^{\{b\}}$ are automatically neutral as well. This need not occur in the quantum theory or in the SS, DM or HS theories.) It is only for simplicity of exposition that Theorem 1.6 was restricted to neutral, diatomic molecules. The true state of affairs, from which Theorem 1.6 follows immediately, is that $E^\Q (N, Z_1, \dots , Z_K, B)$ can be calculated easily in the $B/Z^3 \rightarrow \infty$ limit from knowledge of $E^{\HS}(\lambda)$ --- as the following theorem shows. {\bf 9.1. THEOREM (Bound atoms are isocentric for large $B$).} {\it Fix $\lambda > 0$ and $K$. Consider a sequence of molecules with $K$ nuclei, electron number $N$ and magnetic field $B$ in which the parameters tend to infinity in the following way (with $Z: = \sum \nolimits^K_{j=1} Z_j): Z_i \rightarrow \infty, \ B/Z^3 \rightarrow \infty$ and $N/Z = \lambda$. Then $$\lim \limits_{Z \rightarrow \infty} \{ E^\Q (N, Z_1, \dots , Z_K, B) / E^\Q (N,Z,B) \} = 1.$$ Here, as usual, $E^\Q(N, Z, B)$ is the energy of an atom with nuclear charge $Z$.} {\it Proof:} We will derive upper and lower bounds of the necessary accuracy. {\it The lower bound\/} is trivial. For any choice of $R_1, \dots , R_K$ and electronic wave function $\Psi$, with spin included, we can define the electron density $\uprho_\Psi$ as in (1.3) and then define the potential $\Phi = \vert x \vert^{-1} * \uprho_\Psi$. Let $x_0 \in \R^3$ be such that $\Phi (x_0) = \sup_x \Phi (x)$. Clearly, the attractive electron-nuclear energy (see (i) above) is minimized if we now set $R_1 = R_2 = \dots = R_K = x_0$. At the same time, for a lower bound we can omit the positive nuclear repulsion (see (ii) above). The result is then $(\Psi, H^{(K)}_N \Psi) \geq (\Psi, H_N \Psi)$, where $H^{(K)}_N$ is our Hamiltonian for $K$ nuclei, and $H_N$ is the one nucleus Hamiltonian (1.1) with a nucleus of charge $Z$ located at $x_0$. By translation covariance, we can move $x_0$ to $0 \in \R^3$ without changing the energy. Thus, {\it for any\/} $Z_i$'s $$E^\Q (N,Z_1, \dots , Z_K, B) \geq E^\Q (N,Z,B).$$ {\it The upper bound\/} is a bit more complicated. We use Lieb's variational principle [50] (cf. Sect.~V) and take as variational density matrix the one given in (5.4) for an atom with electron number $N$ and nuclear charge $Z = \sum \nolimits^K_{j=1} Z_j$ (located at the origin, of course). Since we consider large $B/Z^3$ the density matrix in (5.4) is of the form given in (4.29) with density $\uprho^{\SS} (x)$. Next, we place our $K$ nuclei on the $x_\bot = 0$ axis, close to the origin and with a spacing $D = Z^{-1} (\ln \eta)^{-\varepsilon}$ with $\eta = B/(2\pi Z^3)$. Any fixed $1<\varepsilon <2 $ will do. The nuclear coordinates are thus $(0,0, \pm D), \ (0,0,\pm 2D)$, etc. The nuclear repulsion can be bounded by $\sum Z_i Z_j \vert R_i - R_j \vert^{-1} \leq \sum Z_i Z_j D^{-1} \leq \mfr1/2 Z^2 D^{-1} \leq Z^3 (\ln \eta)^\varepsilon$. By Theorem 1.4, therefore, the repulsive energy in our state is of lower order --- by a factor $(\ln \eta)^{\varepsilon -2}$. The remaining step is to show that the negative electron-nuclei energy (see (i) above) is, to leading order, the same as that for one nucleus of charge $Z$ placed at the origin $(0,0,0)$. We use Proposition 3.3. If $R$ is a point on the axis $x_\bot = 0$ we get, using the scalings (3.11) and (3.14), that $\phi(R):= \int Z\vert x - R \vert^{-1} \uprho^{\SS} (x) dx = Z^3 (\ln \eta)^2 \int V_\eta (x) \uprho^{\SS}_\eta (x + Z\ln \eta R) dx$. Thus from Proposition 3.3 with $\eta> 3$ $$Z^{-3} (\ln \eta)^{-2} \left\vert \int Z \vert x- R \vert^{-1} \uprho^{\SS} (x) - Z^3 (\ln \eta)^2 \overline{\uprho}^{\SS}_\eta (Z \ln \eta \vert R \vert) \right\vert \leq c_\lambda \left( {1 + \ln \vert \ln \eta \vert \over \ln \eta} \right),$$ Here we are using that $T[\uprho^{\SS}_\eta] \leq c^\prime_\lambda$, where $c_\lambda$ and $c^\prime_\lambda$ depend only on $\lambda$, and that the support of $\uprho^{\SS}_\eta$ is in the set $\{ x = (x_\bot, x_3): \vert x_\bot \vert \leq \sqrt{\lambda /\pi}\}$. Since $\int (d \sqrt{\overline{\uprho}^{\SS}_{\eta}}/dx_3)^2 dx_3 \leq T [\uprho^{\SS}_\eta] \leq c^\prime_\lambda$ we also have a bound on the Lipschitz constant of $\overline{\uprho}^{\SS}_{\eta}$, i.e., $$\vert \overline{\uprho}^{\SS}_\eta (Z \ln \eta \vert R \vert) - \overline{\uprho}^{\SS}_\eta (0) \vert < c^{\prime\prime}_\lambda Z \ln \eta \vert R \vert.$$ We choose $\vert R \vert$ to be less than $KD$, whence we obtain $$Z^{-3} (\ln \eta)^{-2} \vert \phi (R) - Z^3 (\ln \eta)^2 \overline{\uprho}^{\SS}_\eta (0) \vert < c^{\prime\prime}_\lambda \left( {1 - \ln \vert \ln \eta \vert \over \vert \ln \eta \vert} + (\ln \eta)^{1-\varepsilon} \right).$$ This permits us to compare $\phi(R_j)$ with $\phi (0)$ and we see that all are equal to leading order. 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