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Stabilitity of Matter, Large Coulomb systems, CLR estimates, 
Graf-Schenker inequality, Coherent states, Thomas-Fermi theory, 
Thermodynamic limit.
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\documentclass[12pt]{article}
\usepackage{a4,amsthm,amsfonts,amsmath
}   
\newcommand{\R}{{\mathord{\mathbb R}}}
\newcommand{\Z}{{\mathord{\mathbb Z}}}
\newcommand{\N}{{\mathord{\mathbb N}}}
\newcommand{\C}{{\mathord{\mathbb C}}}
\newtheorem{co}{Corollary}
\newtheorem{lm}{Lemma}
\newtheorem{q}{Theorem}

\newcommand{\tf}{{\rm TF}}
\newcommand{\lt}{{\rm LT}}
\newcommand{\cf}{{C^\beta_\tf}}
\newcommand{\cfi}{{C_\tf}}
\newcommand{\dq}{{{\cal D}_0}}
\newcommand{\da}{{{\cal D}_\alpha}}
\newcommand{\fe}{{{\cal E}_{\tf,{\cal R,Z}}}}
\newcommand{\fee}{{{E}_{\tf,{\cal R,Z}}}}
\newcommand{\feb}{{{\cal E}^\beta_{\tf,{\cal R,Z}}}}
\newcommand{\febe}{{{E}^\beta_{\tf,{\cal R,Z}}}}
\newcommand{\feba}{{{E}^\beta_{{\tf,{\cal R}_\alpha,{\cal
        Z}_\alpha}}}}
\newcommand{\febaa}{{{\cal E}^\beta_{{\tf,{\cal R}_\alpha,{\cal
        Z}_\alpha}}}}
\newcommand{\feta}{{{E}_{\tf,t^{-1}{\cal R}_\alpha,{\cal
        Z}_\alpha}}}
\newcommand{\fea}{{{E}_{\tf,{\cal R}_\alpha,{\cal
        Z}_\alpha}}}
\newcommand{\febao}{{{E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal
        Z}_\alpha}}}
\newcommand{\febh}{{{\cal E}^{\beta/2}_{\tf,{\cal R},{\cal Z}}}}
\newcommand{\febeh}{{{E}^{\beta/2}_{\tf,{\cal R},{\cal Z}}}}
\newcommand{\he}{\hen}
\newcommand{\hen}{{H_{{\cal R,Z},N}}}
\newcommand{\z}{{\overline{Z}}}


\begin{document}


\title{On the asymptotic exactness of Thomas-Fermi theory in the thermodynamic
limit\thanks{Work supported in parts by the EU TMR-grant FMRX-CT
  96-0001.
The authors are grateful to Aarhus University where this work was initiated.
}
} 

\author{Pedro Balodis Matesanz \& Jan Philip
  Solovej\thanks{J.P.~Solovej
is also supported by  MaPhySto -- Centre for Mathematical Physics and 
Stochastics, funded by a grant from The Danish National Research
Foundation and by a grant from the Danish Natural Science Research
Council}\\  \\
Department of Mathematics\\ University of Copenhagen\\
Universitetsparken 5\\
DK-2100 Copenhagen \O\\
E-mails: balpedro@math.ku.dk\quad\&\quad solovej@math.ku.dk
}

\date{September 1998}


\maketitle

\begin{abstract}
In this paper we obtain a new version of stability of matter, which in 
particular shows that Thomas-Fermi theory is asymptotically correct 
in the limit of large nuclear charges uniformly in the number of
nuclei.
As a consequence we give a new lower bound on the volume of matter
with an improved dependence on the nuclear charges.
\end{abstract}
\newpage

{\bf \large Contents:}
\begin{description}
\item[\ref{sec:int}.]  Introduction.\dotfill\ref{sec:int}
\item[\ref{sec:GS}.]  Graf \& Schenker electrostatic inequality.\dotfill\pageref{sec:GS}
\item[\ref{sec:loc}.]  Localizing the Hamiltonian in  tiles.\dotfill\pageref{sec:loc}
\item[\ref{sec:tf}.]  Some results about Thomas-Fermi Theory.\dotfill\pageref{sec:tf}
\item[\ref{sec:kin}.]  Estimating the Kinetic Energy.\dotfill\pageref{sec:kin}
\item[\ref{sec:many}.]  Estimating the energy of tiles with too many nuclei.\dotfill\pageref{sec:many}
\item[\ref{sec:main}.]  Proving the main result.\dotfill\pageref{sec:main}
\item[\ref{sec:density}.]  Estimating the density.\dotfill\pageref{sec:density}
\item[\ref{sec:app}.]  Appendix: The Lieb-Thirring inequality 
(Theorem~\ref{thm:LT}).\dotfill\pageref{sec:app}
\item[]  References\dotfill\pageref{sec:ref}
\end{description}

\section{Introduction.}\label{sec:int}


One of the most celebrated results in mathematical physics
is the Theorem on Stability of Matter. 
This result, which was originally proved by Dyson and 
Lenard~\cite{DL}, states that the binding energy per particle, for a 
a system of charged quantum particles, where either the positive or
negatively charged particles are fermions, is bounded 
independently of the number of particles. 

Another and maybe more intuitively understandable
formulation of this result is that the volume occupied by the
particles increases at least linearly in the number of particles. 

Since the original work of Dyson and Lenard there has been numerous 
improvements and generalizations. 
In particular the work of Lieb and Thirring~\cite{LT} (see also the
review \cite{L2}) established the
connection between stability of matter
and the semi-classical Thomas-Fermi theory and greatly improved the
numerical constants found by Dyson and Lenard. 
Among other proofs of stability of matter we can mention the work of 
Federbush~\cite{Fed} and the recent 
proof of Graf~\cite{G}, which we shall use in the present work. 

The result on stability of matter
has been generalized to include relativistic effects
\cite{C,FL,LY,LSS}, classical magnetic fields
\cite{Fef,LLS}, and quantized fields \cite{FFG}.

In the present work we are concerned with another type of
generalization of stability of matter. We are interested in the 
correct dependence in the physical parameters.
One of the remarkable features of macroscopic matter is 
that the mean atomic spacing is nearly independent of the 
type of atoms that the matter consists of. Or put differently
the mean atomic spacing
is nearly independent of the nuclear charges. 

In Thomas-Fermi theory distances scale as the $-1/3$ 
power of the nuclear charge, i.e., 
$Z^{-1/3}$. One would therefore naively expect that in macroscopic
matter the volume
per particle would behave as $Z^{-1}$. This is 
however in stark contrast to the near independence of $Z$ 
which is found experimentally. 

The Lieb-Thirring proof of stability of matter implies (see 
\cite{LT}) a lower bound on the volume per 
atom of the form $Z^{-1}$ for large $Z$. 
Our main goal here is to show that this is indeed not optimal. 
We prove that there exists $\delta'>0$ such that 
the volume per atom is bounded below by $Z^{-1+\delta'}$.
Based on the experimental evidence one would
hope to prove $\delta'=1$, we are, however, very far 
from such a result (see Theorem~\ref{thm:volume}).  

Our proof is based on first showing that Thomas-Fermi 
theory not only gives a bound on the energy, but that this bound 
is indeed asymptotically correct in the limit of large nuclear
charges. The bound of volume of matter is arrived at by a careful study
of Thomas-Fermi theory. 

The fact that Thomas-Fermi theory is asymptotically correct in the
limit of large nuclear charge is a classical 
result due to Lieb and Simon~\cite{LS}.

The new feature is that we establish this asymptotics uniformly
in the number of nuclei allowing us to use the Thomas-Fermi 
approximation independently of the number of nuclei. 

We consider matter formed by $M$ nuclei of charges $Z_j\geq 1$, 
$j=1,\ldots,M$  located at 
positions $R_j\in\R^3$. We denote  
${\cal R}=(R_1,\ldots,R_M)$ and ${\cal Z}=(Z_1,\ldots,Z_M)$.
We consider these nuclei as static and consider the 
non-relativistic Hamiltonian of $N$
electrons moving in the electric potential of the nuclei.
The Hamiltonian is
\begin{equation}
\label{eq:ham}
\hen=
\sum^N_{i=1}\left\{-\Delta_i-\sum^M_{j=1}\frac{Z_j}{|x_i-R_j|}\right\}+
\sum_{1\le i<j\le N}\frac1{|x_i-x_j|}+\sum_{1\le i<j\le M}
\frac{Z_iZ_j}{|R_i-R_j|}
\end{equation}


Our improvement on stability of matter is as follows.

\begin{q}[Stability of Matter]\label{thm:stability}
Let $\z=\frac1M \sum^M_{i=1}Z_i$.  Assume there are $0<a\le A<\infty$ such that the charges satisfy
$a\overline{Z}\le Z_j\le A\overline{Z}$ for all j and that $\z\geq1$. 
Then
there is $\delta>0$ and $C>0$ a finite constant only depending on $a, A$
such that
\begin{equation}
\inf_{{\cal R}:\,\#{{\cal R}}=M}\inf_{\Psi\in {\cal H},\,
\|\Psi\|=1}\left\langle\Psi,\he
\Psi\right\rangle_{\cal H}\ge -\cfi\sum^M_{j=1}Z_j^{7/3}
-CM\overline{Z}^{7/3-\delta}
\end{equation} 
Here the Hilbert space where the Hamiltonian acts is
${\cal H}=\bigwedge\limits^N L^2(\R^3;\C^2)$ (Fermion space).
\end{q}
In Theorem~\ref{thm:stability}, the constant $\cfi$ 
is the corresponding Thomas-Fermi constant (see Sect.~\ref{sec:tf} 
below for the definition)
in the case $Z=1$, $M=1$. It is worth noting that the kind of estimate
given in this theorem is essentially optimal. In fact, by taking the 
nuclei very far apart and using the Theorem of Lieb and Simon on the
exactness
of Thomas-Fermi theory for one nucleus we obtain
\begin{equation}
\inf_{{\cal R}:\,\#{{\cal R}}=M}\inf_{\Psi\in {\cal H},\,
\|\Psi\|=1}\left\langle\Psi,\he\Psi\right\rangle_{\cal H}
\le -\cfi\sum^M_{j=1}Z_j^{7/3}+CM\overline{Z}^{7/3-\delta}
\end{equation} 

As an application of Theorem~\ref{thm:stability}  we shall 
derive our results on the total volume and
density of matter. 
This is the content of the next two theorems. The first is
really a strengthening of Theorem~\ref{thm:stability}, 
which requires an estimate
on the ``pressure'' (the extra energy we have for confining
particles in a bounded set). The estimate is a generalization to the
case of many nuclei
of a result by Brezis and Lieb~\cite{BL} on
the long-range potentials in Thomas-Fermi theory. The last
theorem follows straightforwardly from that, and shows that for nuclear
configurations giving us an energy close to the minimal possible one,
almost all nuclei are very far from each other (on the natural scale
of Thomas-Fermi theory, i.e., $\overline{Z}^{-1/3}$) and therefore
that matter under such conditions should fill a volume in space
much larger than const$M\overline{Z}^{-1}$, which is the estimate
that the simpler model of matter given by Thomas-Fermi theory
predicts.

\begin{q}[Pressure in macroscopic matter]\label{thm:pressure}
Let ${\cal R,Z}$ be a configuration of nuclei and charges fulfilling 
the conditions of Theorem 1 and such that 
$$\inf{\rm spec}\,\he\le 
 -C_{\rm TF}\sum^M_{j=1}Z_j^{7/3}+C_0M\overline{Z}^{7/3-\delta}$$
for a (fixed) constant $C_0$. Then there exists $C,c>0$ 
depending only on $a,A$, and $C_0$ such that 
\begin{equation}
  \inf{\rm spec}\,\he\ge
-C_{\rm TF}\sum^M_{j=1}Z_j^{7/3}-CM\overline{Z}^{7/3-\delta}
+c\overline{Z}^{7/3}\sum^M_{j=1}\Gamma\left(
\overline{Z}^{1/3}\delta(R_j)\right)
\end{equation}
where $\Gamma(t)=\min
\left\{t^{-1},t^{-7}\right\},\,t>0$ and $\delta(R_j)=
\min_{i\ne j}|R_j-R_i|$ is the distance from $R_j$ to 
the nearest other nuclei.
\end{q}

\begin{q}[The volume of matter]\label{thm:volume}
With the conditions in Theorem~\ref{thm:pressure}, 
we have that ${\cal R}$ should satisfy,
for $0<\delta_1<\delta/7$ and $k$ finite
\begin{equation}
  \#\left\{j:\delta(R_j)\le \overline{Z}^{-1/3+\delta_1}
\right\}\le
kM\overline{Z}^{-(\delta-7\delta_1)}=o(M),\ \overline{Z}\mapsto\infty
\end{equation}
Another interpretation of this result is as follows: Let us call
$${\rm volume}=\sum^M_{j=1}\left|B\left(R_j,\frac12\delta(R_j)\right)\right|
;\ B({\rm x},r)=\left\{{\rm y}\in\R^3:|{\rm x}-{\rm y}|\le r\right\}$$
(Above, and hereafter, $|G|$
stands for Lebesgue measure of $G$).
Then we have the estimate
\begin{equation}
  \label{eq:th3'}
  {\rm volume}\ge k'M\overline{Z}^{-1+3\delta_1}
\end{equation}
where $k'>0$ is a universal constant.
\end{q}


We shall prove these theorems using a tiling of space into simplices
and an electrostatic inequality developed by Graf~\cite{G} and
Graf-Schenker~\cite{GS}
which enables us to localize the electrostatic interactions 
into the tiles and
to ignore the  electrostatic
interactions between these tiles up to some 
error which can be controlled.
This reduces our problem to bounded regions, and for these we shall
use Thomas-Fermi theory in order to estimate 
the maximal number of nuclei which is allowed in order to have 
non-positive 
energy. We ignore the tiles with positive energy, in the  tiles 
which may have non-positive energy, we
show how to reduce the asymptotic estimates to that of the whole space.


\section{Graf \& Schenker (GS) electrostatic inequality.}
\label{sec:GS}
The setting we shall use for the GS inequality is as follows: 
Let $Q=[0,1]^3$ be the unit cube in $\R^3$. The cube $Q$ can be written 
as a union of 24 congruent tetrahedra. (To see this first note 
that there are 6 pyramids with top
at the center of the cube and base equal to one of the 
faces of the cube. Each pyramid consists of 4 congruent tetrahedra.)
Denote by $\dq$ the open interior of one of these tetrahedra.
We then have a ``tiling'' $T=\{\da\}_{\alpha\in\N}$, i.e,
a collection of disjoint tetrahedra all congruent to $\dq$ such that
$\bigcup_{\alpha\in \N}\overline\da=\R^3$.
We shall also need to consider the tiling 
$T_l=\{l\da\}_{\alpha\in\N}$ of scale $l>0$, 
(which will be chosen later to be 
$\overline{Z}^{-1/3+\delta}$).

Given a rotation $R\in SO(3)$ and a $y\in Q$
we denote by $R(T_l+ly)$ the tiling 
$\{R(l\da+ly)\}$.

 

We are now ready to state the result of Graf and Schenker.
Given points $x_1,\ldots,x_K\in\R^3$. We then consider the function
$$
\delta_{R,y}(x,x')=\left\{
  \begin{array}{ll}
1 &{\textrm{if}\ x,x'\ \textrm{belong to the same tetrahedron of }
R(T_l+ly)}\\
0 &\textrm{otherwise}
  \end{array}\right.
$$
for $R\in SO(3)$ and $y\in Q$.


\begin{q}[GS inequality]\label{thm:graf}
There is a $C>0$ such that for all $K\in\N$, all $(x_1,\ldots,x_K)\in
\R^{3K}$
, all $(z_1,\ldots,z_K)\in\C^K$, and any $l>0$ we have 
\begin{equation}
\label{eq:elect}
  \sum_{
    \begin{array}{c}
1\le i,j\le K\\ i\ne j
    \end{array}}\frac{z_i\overline{z_j}}{|x_i-x_j|}\ge\left\langle
\sum_{
    \begin{array}{c}
1\le i,j\le K\\ i\ne j
\end{array}}\frac{z_i\overline{z_j}}{|x_i-x_j|}\delta_{R,y}(x_i,x_j)\right\rangle
-\frac Cl\sum^K_{i=1}|z_i|^2
\end{equation}
For any function $f$ on $SO(3)\times Q$ 
we have here defined its {\it average} over translations and rotations
\begin{equation}
\label{eq:ave}
  \langle f\rangle:=\int_{SO(3)\times Q}f(R,y)d\mu(R)dy
\end{equation}
where $d\mu(R)$ stands for Haar measure on $SO(3)$.
\end{q}
We refer to \cite{G} and \cite{GS} for a proof of this inequality.

\section{Localizing the Hamiltonian into the tiles.}\label{sec:loc}
 

The inequality of Graf and Schenker allows us to localize
the potential energy into tiles. 
We shall also localize  the kinetic energy. Since we are asking for
estimates from {\it below} it is natural to do this by 
Neumann-bracketing. 


\begin{lm}[Localization estimate]\label{thm:decoupling}
Corresponding to a tiling $R(T_l+ly)$ with $R\in SO(3)$, $y\in Q$
we define
the Neumann Laplacians $-\Delta_\alpha$ for the tile 
$l\da(R,y):=R(l\da+ly)$, $\alpha\in\N$.
and in terms of this the Hamiltonians
\begin{eqnarray}\label{eq:bra}
H_{\alpha,N'}&:=
&\sum^{N'}_{i=1}\left\{-\Delta^i_{\alpha}-\sum^{M}_{j=1}
\frac{Z_j\chi_{\alpha}(R_j)}{|x_i-R_j|}\right\}+
\sum_{1\le i<j\le N'}\frac{1}{|x_i-x_j|}\\
&&+\sum_{1\le i<j\le M}
\frac{Z_iZ_j\chi_{\alpha}(R_i)\chi_{\alpha}(R_j)}{|R_i-R_j|}\nonumber
\end{eqnarray}
acting on $\bigwedge\limits^{N'}L^2(l\da(R,y);\C^2)$. 
Here $\chi_\alpha$ is the
characteristic function of $l\da(R,y)$. 
Then we have the following decoupling inequality:
\begin{equation}
\label{eq:dec}
\inf{\rm spec}\hen\ge
\inf_{N_\alpha:\sum_\alpha N_\alpha=N}\left\{
\sum_\alpha\Bigl\langle 
\inf{\rm spec}H_{{\alpha,N_\alpha}}\Bigr\rangle\right\}
-\frac Cl(N+\sum^M_{j=1}Z_j^2)
\end{equation}
\end{lm}
\begin{proof}

Given $\tilde{\alpha}=(\alpha_1,\ldots,\alpha_N)\in\N^N$. 
We consider the subset $A_{\tilde{\alpha}}(R,y)$ of $\R^{3N}$
given by 
$$
  A_{\tilde{\alpha}}(R,y)=\{(x_1,\ldots,x_N)\in\R^{3N}\ :\ 
  x_j\in l{\cal D}_{\alpha_j}(R,y),\
        j=1,\ldots,N\}
$$
We let $N_\alpha$, $\alpha\in\N$ denote the number of $j$ such that 
$\alpha_j=\alpha$. The sets $A_{\tilde{\alpha}}(R,y)$ corresponding 
to different $\tilde{\alpha}$ are disjoint and we 
can write $\R^{3N}$ as the union
$$
   \R^{3N}=\bigcup\limits_{\tilde{\alpha}}\overline{A_{\tilde{\alpha}}(R,y)}
$$
We denote by $\Psi_{\tilde{\alpha}}=\Psi\chi_{{}_{A_{\tilde{\alpha}}}}$.
We first investigate the kinetic energy
$$
        (\Psi, \sum_{j=1}^N-\Delta_j\Psi)=
        \sum_{\tilde{\alpha}}
        \sum_{j=1}^N\int_{A_{\tilde{\alpha}}(R,y)}|\nabla_j\Psi|^2
        =\sum_{\tilde{\alpha}}
        \sum_{\alpha=1}^\infty\left\langle
        \sum_{j:\alpha_j=\alpha}\left(\Psi_{\tilde{\alpha}},-\Delta_\alpha^j
        \Psi_{\tilde{\alpha}}\right)\right\rangle
$$
The left side is independent of $R$ and $y$ and we may therefore
average over rotated and translated tiles. 
{F}rom the Graf-Schenker inequality we have
\begin{eqnarray*}
  \lefteqn{\Bigl(\Psi, \Bigl[-\sum^M_{j=1}\sum_{i=1}^N\frac{Z_j}{|x_i-R_j|}
+\sum_{1\le i<j\le N}\frac1{|x_i-x_j|}+\sum_{1\le i<j\le M}
\frac{Z_iZ_j}{|R_i-R_j|}\Bigr]\Psi\Bigr)}\\
&\geq&\biggl\langle
\biggl(\Psi, 
\Bigl[-\sum^M_{j=1}\sum_{i=1}^N\frac{Z_j\delta_{R,y}(x_i,R_j)}{|x_i-R_j|}+
\sum_{1\le i<j\le N}\frac{\delta_{R,y}(x_i,x_j)}{|x_i-x_j|}
\\&&{}+\sum_{1\le i<j\le M}
\frac{Z_iZ_j\delta_{R,y}(R_i,R_j)}{|R_i-R_j|}\Bigr]\Psi\biggr)\biggr\rangle
-\frac Cl(N+\sum^M_{j=1}Z_j^2)\\
&=&\sum_{\tilde{\alpha}}\sum_{\alpha=1}^\infty\biggl\langle
\biggl(\Psi_{\tilde\alpha}, 
\Bigl[-\sum^M_{j=1}\sum_{i=1}^N\frac{Z_j\chi_\alpha(x_i)\chi_\alpha(R_j)}{|x_i-R_j|}+
\sum_{1\le i<j\le N}\frac{\chi_\alpha(x_i)\chi_\alpha(x_j)}{|x_i-x_j|}
\\&&{}+\sum_{1\le i<j\le M}
\frac{Z_iZ_j\chi_\alpha(R_i)\chi_\alpha(R_j)}{|R_i-R_j|}\Bigr]\Psi_{\tilde\alpha}\biggr)\biggr\rangle-\frac
Cl(N+\sum^M_{j=1}Z_j^2)
\end{eqnarray*}
Note that, although $\Psi_{\tilde\alpha}$ is not antisymmetric
in all variables, it is
antisymmetric in the variables belonging to the same tile. 
Hence
$$
  (\Psi,\hen\Psi)\geq\sum_{\tilde{\alpha}}\sum_{\alpha=1}^\infty\biggl\langle
  \inf{\rm spec}H_{{\alpha,N_\alpha}}\biggl\rangle\|\Psi_{\tilde\alpha}\|^2
-\frac
Cl(N+\sum^M_{j=1}Z_j^2)
$$
Noting that $\sum_{\tilde{\alpha}}\|\Psi_{\tilde\alpha}\|^2=1$ 
we conclude (\ref{eq:dec}).
\end{proof}


\section{Some results about Thomas-Fermi Theory}\label{sec:tf}

In this section we shall prove a result purely about Thomas-Fermi theory. 
It is closely related
to the results (and ideas) of Brezis and Lieb in \cite{BL} about the  
asymptotic behavior of many-body potentials in Thomas-Fermi theory.
In this section we shall assume that we deal with neutral systems.
The effect of screening in neutral systems is that long-range
interactions are much smaller than for non-neutral systems.

Suppose we have some configuration of $M$ nuclei of charges 
${\cal Z}=(Z_1,\ldots,Z_M)$ and positions
${\cal R}=(R_1,\ldots,R_M)$. The Thomas-Fermi model for this 
problem is defined by the following functional
on positive densities $\rho\in L^1(\R^3)\cap L^{5/3}(\R^3)$.
\begin{eqnarray}
\label{eq:ft}
\feb(\rho)&=&\beta\int_{\R^3}
\rho(x)^{5/3}dx-\int_{\R^3}V(x)\rho(x)dx\nonumber\\
&&+\frac12\iint_{\R^3\times\R^3}
\frac{\rho(x)\rho(y)}{|x-y|}dxdy+\sum^M_{l<s}\frac{Z_lZ_s}{|R_l-R_s|}
\end{eqnarray}
where the potential is $V(x)=
\sum^M_{i=1}\frac{Z_i}{|x-R_i|}$ and $\beta>0$ is a parameter.
In our units the physical value of this parameter is 
$\beta_{\rm phys}=\frac35(6\pi^2)^{2/3}$. We shall however also 
use Thomas-Fermi theory for other values of this parameter and we 
therefore allow it to be arbitrary for the moment.
The Thomas-Fermi energy is defined by 
\begin{equation}
\label{eq:energ}
  \febe:=\inf_{0\leq \rho\in L^1(\R^3)\cap L^{5/3}(\R^3)}
\feb(\rho)
\end{equation}

When $\beta=\beta_{\rm phys}$ we shall write $\febe$ 
simply as $\fee$. If $M=1$ and $Z=1$ 
we shall write $\cf$ instead of $|\febe|$ and if $\beta=\beta_{\rm phys}$
simply $\cfi$
(see also Theorems~\ref{thm:stability}--\ref{thm:volume}).

The energy satisfies the scaling properties
\begin{equation}
\label{eq:scaling}
\febe=\lambda{E}^{\lambda\beta}_{{\rm TF},\lambda{\cal R,Z}}
=\lambda^{7}{E}^{\beta}_{{\rm TF},\lambda{\cal R},\lambda^{-3}{\cal Z}}
\end{equation}
for any  $\lambda>0$.

It is known that there is a unique function $\rho$ which minimizes the
functional (\ref{eq:energ}). This function
fulfills the Thomas-Fermi equations
\begin{equation}
\label{eq:edp}
\begin{array}{rcl}
\frac53\beta\rho^{2/3}(x)&=&\phi(x)\\
\phi(x)&=& V(x)-\frac1{|y|}\ast\rho(x)  
\end{array}
\end{equation}
and is moreover the unique non-negative solution $\rho$
of (\ref{eq:edp}).
This function has the property that $\int\rho=\sum_{j=1}^MZ_j$, i.e., 
the density minimizing the functional (\ref{eq:ft}) corresponds to a neutral 
system.
For future use we shall denote by $\phi({\cal R,Z},x)$ the unique solution
$\phi$ of (\ref{eq:edp}), and $\phi^{\rm at}(R,Z,x)$ 
the one corresponding to a single ``atom''
of charge $Z$  located at $R\in\R^3$. 
We refer to the original paper \cite{LS} or the review
\cite{L1} for the proofs of these statements and for further results on 
Thomas-Fermi theory.

We
consider the following function:
\begin{equation}
\label{eq:tf1}
  f^\beta({\cal R,Z})=
\febe+\cf\sum^M_{J=1}Z_j^{7/3}
\end{equation}
Using the scaling properties of Thomas-Fermi theory we have
\begin{equation}
\label{eq:tf2}
  f^\beta({\cal R,Z})=\overline{Z}^{7/3}f^\beta\left(
\overline{Z}^{1/3}{\cal R},\overline{Z}^{-1}{\cal Z}\right)
\end{equation}

In \cite{BL} the asymptotics of $\lambda^7f^\beta(\lambda{\cal R},{\cal Z})$
for large $\lambda$ was studied in the case of a finite number of 
nuclei. Our goal here is to show an estimate on $f^\beta({\cal R,Z})$
similar to the asymptotics of \cite{BL} which holds uniformly in 
the number of nuclei.

\begin{lm}[Pressure in Thomas-Fermi Theory]\label{thm:pres}
Assume there is a constant $a>0$ such that 
$a\overline{Z}\le Z_j$ for all j. Then
there is a universal $c>0$ (in particular 
independent of $M$) such that for each ${\cal R}\in \R^{3M}$
we have   
\begin{equation}
\label{eq:pres}
f^\beta\left({\cal R,Z}\right)\ge c\beta^{-1}(a\overline{Z})^{7/3}
\sum^M_{j=1}\Gamma\left(
(a\overline{Z})^{1/3}\beta^{-1}\delta(R_j)\right)
\end{equation}
where $\Gamma(t)=\min\{t^{-1},t^{-7}\}$ and 
$\delta(R_j)=\min_{i\ne j}|R_i-R_j|$ is as before
the nearest neighbor distance.
Moreover,  $f^\beta$ is a non-decreasing function of all the $Z_i$ 
variables.
\end{lm}
\begin{proof} 
By the scaling property (\ref{eq:scaling}) it is enough to prove the
lemma for $Z_j\ge 1$ and $\beta=\beta_{\rm phys}$ and in this case we
omit the superscript. 
Then we have the following (a Feynman-Hellman type result)
\begin{equation}
\label{eq:fh1}
  \frac\partial{\partial Z_i}f\left({\cal R,Z}\right)=\lim_{x\mapsto R_i}
[\phi({\cal R,Z},x)-\phi^{\rm at}(R_i,Z_i,x)] 
\end{equation}
The right side of (\ref{eq:fh1}) is non-negative because of 
Teller's Lemma (see \cite{L1} Theorem~3.4) , which
implies that $f\left({\cal R,Z}\right)$ is non-decreasing in 
any of the $Z_i$
arguments. Therefore we have that
\begin{equation}
  \label{eq:fh2}
  f\left({\cal R,Z}\right)\ge f\left({\cal R},(1\ldots,1)\right)
\end{equation}
Let $H(\lambda)$ be defined by
\begin{equation}
  \label{eq:h1}
  H(\lambda)= f\left({\cal R},(\lambda ,\ldots,\lambda )\right)
\end{equation}
Applying Feynman-Hellman's formula again we obtain
\begin{equation}
  \label{eq:h2}
  \frac d{d\lambda}H(\lambda)=\sum^M_{i=1}\lim_{x\mapsto R_i}[\phi({\cal R},
(\lambda ,\ldots,\lambda ),x)-\phi^{\rm at}(R_i,\lambda ,x))]
\end{equation}
For any of the points $R_i$ we choose a {\it nearest neighbor}, which
shall be denoted $R'_i$, i.e.,
$|R_i-R'_i|=\delta(R_i)$. Then, by
Teller's Lemma once again, we have that
\begin{equation}
  \label{eq:h3}
  \frac d{d\lambda}H(\lambda)\ge \sum^M_{i=1}\lim_{x\mapsto R_i}
[\phi((R_i,R'_i),
(\lambda ,\lambda ),x)-\phi^{\rm at}(R_i,\lambda ,x))]
\end{equation}
Since $H(0)=0$, the estimate (\ref{eq:h3}) implies that
\begin{eqnarray}
  \label{eq:h4}
  H(1)&\ge&\int_0^1d\lambda \sum^M_{i=1}\lim_{x\mapsto R_i}
[\phi((R_i,R'_i),
(\lambda ,\lambda ),x)-\phi^{\rm at}(R_i,\lambda ,x))]\nonumber\\
&=&\frac12\int_0^1d\lambda \sum^M_{i=1}\left[\lim_{x\mapsto R_i}
[\phi((R_i,R'_i),
(\lambda ,\lambda ),x)-\phi^{\rm at}(R_i,\lambda ,x))]\right.\nonumber\\
&&\left.+\lim_{x\mapsto R'_i}
[\phi((R_i,R'_i),
(\lambda ,\lambda ),x)-\phi^{\rm at}(R'_i,\lambda ,x))]
\right]\\
&=&\frac12\sum^M_{i=1}f((R_i,R'_i),(1,1))\nonumber\\
&=&\frac12\sum^M_{i=1}f((0,R_i-R'_i),(1,1))\nonumber
\end{eqnarray}
In order to finish the proof, it is hence enough to show that there 
is a $c>0$
such that
\begin{equation}
  \label{eq:fa}
  c\Gamma(R)\le f((0,R),(1,1))
\end{equation}
If $|R|\ge 1$, (\ref{eq:fa}) follows from the analysis in \cite{BL}. 
On the
other hand, if $|R|\le1$ we will use the Thomas-Fermi equation directly to
analyze the behavior. 
We have by the Thomas-Fermi scaling
\begin{eqnarray*}
    \phi((0,R),(\lambda ,\lambda ),x) - \phi^{\rm at}(0,\lambda ,x)
    &=&\lambda^{4/3}\Big(
       \phi((0,\lambda^{1/3}R),(1,1),\lambda^{1/3}x)\\ && - 
       \phi^{\rm at}(0,1,\lambda^{1/3}x))\Big)
\end{eqnarray*}
By 
the Thomas-Fermi (\ref{eq:edp}) equation and the fact (see \cite{L1}
Corollary~3.6)
that $\phi((0,r),(1,1),x) \leq \phi^{\rm at}(0,1,x)
+\phi^{\rm at}(r,1,x)$ we have
\begin{eqnarray*}
 %\label{eq:psi}
 \lefteqn{\lim_{|x|\to0}\left[ \phi((0,r),(1,1),x) 
     - \phi^{\rm at}(0,1,x)\right]}&&\nonumber\\
 &=&\frac 1{|r|}+\int
\frac{(3\phi^{\rm at}(0,1,y))^{3/2} }{(5\beta_{\rm phys})^{3/2}|y|}dy-
\int\frac{(3\phi((0,r),(1,1),y))^{3/2} }{(5\beta_{\rm phys})^{3/2}|y|}dy
\nonumber\\
&\ge&\frac 1{|r|}+(1-\sqrt2)\int
\frac{(3\phi((0,r),(1,1),y))^{3/2}}{(5\beta_{\rm phys})^{3/2}|y|}dy-
\sqrt{2}\int\frac{(3\phi(y-r))^{3/2}}{(5\beta_{\rm phys})^{3/2}|y|}dy\\
&\ge&\frac 1{|r|}- C\nonumber
\end{eqnarray*}
for $r<1$.
The estimate (\ref{eq:fa}) then follows from the Feynman-Hellmann
identity (\ref{eq:fh1}).
\end{proof}

\section{Estimating the Kinetic Energy}\label{sec:kin}

Now, in order to compare the localized Hamiltonians with their localized
Thomas-Fermi counterparts, we need to establish some inequalities like those of
Lieb-Thirring. More specifically, we ask for estimates on
$\textrm{Tr}(-\Delta_0-V)_-^\alpha$ where $\Delta_0$ is the Neumann-Laplacian
on the tile $l\dq$ and $V$ is a real-valued function. We are interested
mainly in the cases $\alpha=0,1$. We point out that if we were dealing with
Dirichlet Laplacians, the problem can be automatically reduced to that of the
whole space just by extension. In the case at hand, this is not allowed, and
because of that we shall formulate and prove first a version of these
inequalities in the case of a cube. Finally we are dealing with tiles rather
than cubes, but the problem for the tiles can be reduced to that of the cubes
as explained in the appendix below. 
\begin{q}[Lieb-Thirring type estimate]\label{thm:LT}
Let $Q$ be the unit cube in $\R^3$ and $0\le\alpha$. 
If $V$ is a
real-valued function in $L^1_{\rm loc}(Q)$
for which $[V]_-\in L^{3/2+\alpha}(Q)$, 
then the following trace estimate holds:
\begin{equation}
\label{eq:th5}
\textrm{\rm Tr}\left(-\Delta_Q-V\right)_-^\alpha
\le C_\alpha\int_Q[V]_-^{3/2+\alpha}+
C_\alpha\left[\int_Q[V]_-\right]^\alpha
\end{equation}
Here $-\Delta_Q$ is the Neumann-Laplacian in $Q$. The
constant $C_\alpha$ depends only on $\alpha$.
The same result (possibly with a different constant $C_\alpha$) holds
if the cube $Q$ is replaced by the tile $\dq$.
\end{q}
We did not find a reference for this theorem and
we therefore include a proof
in the appendix. Using the theorem above we 
can provide an estimate for the kinetic energy of $N$
antisymmetric particles on a tile (of any scale $l>0$) in terms of the 
1-particle density  $\rho_\psi(x)$ which is defined as
\begin{equation}
\label{eq:1pd}
  \rho_\psi(x):=N\int_{(l\dq)^{N-1}}\|\psi(x,x_2,\ldots,x_N)\|^2_{\C^{2^N}}
dx_2,
\ldots,dx_N
\end{equation}
for
$\psi\in \bigwedge\limits^N(L^2(l\dq;\C^2))$ normalized.

This function is an analogue of a charge distribution, and has the property
$\int_{l\dq}\rho_\psi=N$.

\begin{q}[Kinetic energy estimate for fermions on a tile]\label{thm:kin}
Let 
$\psi\in\bigwedge\limits^NL^2(l\dq;\C^2)$ normalized and define 
$\rho_\psi$ as before. Then, if
$$H_0=\sum^N_{i=1}(-\Delta^i_0)$$
we have
\begin{equation}
\label{eq:th6}
  \langle\psi,H_0\psi\rangle\ge
K_\lt\int_{l\dq}\rho_\psi(x)
^{5/3}dx-K_1|l\dq|^{-1}\int_{l\dq}\rho_\psi(x)^{2/3}dx
\end{equation}
where $K_\lt$ and $K_1$ are positive and finite absolute constants.
\end{q}
\noindent
\begin{proof} By a simple rescaling it is enough to 
consider the case $l=1$.
Let us define the potential $V(x)=-\rho_\psi(x)^{2/3}$ and the
{\it N}-particle Hamiltonian $H_N:=\sum^N_{i=1}(-\Delta_{0}^i+\lambda
V(x_i))$ where $\lambda>0$ is a parameter to be fixed later. 
This Hamiltonian
acts on $\bigwedge\limits^N(L^2(l\dq;\C^2))$, 
and it can be diagonalized by the eigenfunctions of
$H_V=-\Delta_{0}+V$ on $L^2(l\dq;\C^2)$. 
In fact, the lowest eigenvalue
of $H_N$ is just the sum of the first $N$ negative
eigenvalues of $H_V$ so we have from Theorem~\ref{thm:LT} that
$$\inf{\rm spec}H_N=-{\rm Tr}(H_V)_-\ge-C_{\rm LT}\lambda^{5/2}\int_{\dq}
\rho_\psi(x)^{5/3}dx-\frac{C_1\lambda}{|\dq|}\int_{\dq}
\rho_\psi(x)^{2/3}dx$$
{F}rom the variational principle
$$\langle\psi,H_0\psi\rangle-
\frac\lambda{|\dq|}\int_{\dq}\rho_\psi(x)^{5/3}dx=
\langle\psi,H_N\psi\rangle\ge\inf{\rm spec}H_N$$
and thus we get
\begin{equation}
\label{eq:k}
  \langle\psi,H_0\psi\rangle\ge(\lambda-C_{\rm LT}\lambda^{5/3})
\int_{\dq}\rho_\psi(x)^{5/3}dx-\frac{C_1\lambda}{|\dq|}
\int_{\dq}\rho_\psi(x)^{2/3}dx
\end{equation}
The theorem follows by choosing $\lambda$ appropriately.
\end{proof}
Theorem~\ref{thm:kin}  will be one of the main tools 
to provide a link between Quantum Theory and 
Thomas-Fermi theory. This is the subject of 
the next section.

\section{Estimating the energy of the tiles with too many nuclei.}
\label{sec:many}

The principal task we will pursue in this section is to prove that
the energy of the localized pieces of the Hamiltonian on  tiles is
positive if the tile contains a large enough number of nuclei,
depending only on the average charge $\overline{Z}$ of these nuclei 
and the scale $l$ of the tiles. We shall 
use the estimate for the
kinetic energy given by Theorem~\ref{thm:kin}, but we 
also need an estimate which allow
us to compare the electron-electron repulsion term
of the localized quantum Hamiltonians
with the corresponding term in the Thomas-Fermi functional (\ref{eq:ft}). 
Such an estimate is provided by the Lieb-Oxford inequality (see \cite{LO}).
\begin{lm}[Lieb-Oxford inequality] 
\label{thm:1pdm} Let $\psi\in L^2(\R^{3N};\C^{2^N})$ 
normalized and $\rho_\psi(x)$ the corresponding 1-particle  density
function. Then
\begin{equation}
\label{eq:oxf}
\left\langle\psi,
\sum^N_{i<j}\frac{1}{|x_i-x_j|}\psi
\right\rangle\ge
\frac12\iint\frac{\rho_\psi(x)\rho_\psi(y)}{|x-y|}dxdy
-1.68\int\rho_\psi(x)^{4/3}dx
\end{equation}
\end{lm}
We are now in a position to prove a lower bound on the 
localized Hamiltonians in terms of the {\it neutral}
Thomas-Fermi theory.
\begin{lm}[Lower bound in terms of TF theory]\label{lm:lbtf}
Let $H_{\alpha,N'}$ be the operator defined  in (\ref{eq:bra}) acting
acting on $\bigwedge\limits^{N'}L^2(l\da(R,y);\C^2)$.
We assume as before that $a\z\leq Z_j\leq A\z$ for all $j=1,\ldots,M$.
Then there exists a $c>0$ depending only on $a,A$ such that
\begin{equation}\label{eq:lbtf}
        \inf{ \rm spec}H_{\alpha,N'}\geq 
        {E}^{\beta_0}_{\tf,{\cal R}_\alpha,
        {\cal Z}_\alpha}-c N'(l^{-2}+1)
\end{equation}
where $\beta_0=K_{\rm LT}/2$ and ${\cal R}_\alpha$ and ${\cal Z}_\alpha$
are the coordinates and charges of the nuclei in the tile
$l\da(R,y)$.
\end{lm}
\begin{proof}
{}From the kinetic energy estimate (\ref{eq:th6})
and the Lieb-Oxford inequality (\ref{eq:oxf}) we get, 
for $\psi\in\bigwedge\limits^{N'}\left(L^2(l\da(R,y);\C^2)
\right)$ normalized that
  \begin{eqnarray}
\langle\psi,H_{\alpha,N'}\psi\rangle&\ge&K_{\rm LT}\int
\rho_\psi^{5/3}-\int V_\alpha(x)\rho_\psi(x)dx\nonumber\\
&&+\frac12\iint\frac{\rho_\psi(x)\rho_\psi(y)}
{|x-y|}dxdy+U_\alpha\\
&&-C\left(\int\rho_\psi^{4/3}+l^{-3}\int\rho_\psi^{2/3}\right)\nonumber
  \end{eqnarray}
where
  \begin{equation}\label{eq:uvalpha}
V_\alpha(x):=\sum^{M}_{j=1}\frac{Z_j}{|x-R_j|}\chi_\alpha(R_j)\quad
\hbox{and}\quad
U_\alpha:=\sum^{M}_{i<j}\frac{Z_iZ_j}{|R_i-R_j|}
\chi_\alpha(R_i)\chi_\alpha(R_j)
\end{equation}
Using H\"older's inequality and the fact that $\rho_\psi$ 
is supported in $l\da(R,y)$ we have
\begin{equation}\label{eq:holder}
        \int\rho_\psi^{4/3}\leq \left(\int\rho_\psi^{5/3}\right)^{1/2}
        (N')^{1/2}\quad\hbox{and}\quad 
        \int\rho_\psi^{2/3}\leq c(N')^{2/3}l\leq N' l
\end{equation}
Hence 
$$
        \frac{K_{\rm LT} }{2}\int\rho_\psi^{5/3}-
        C\left(\int\rho_\psi^{4/3}+l^{-3}\int\rho_\psi^{2/3}\right)
        \geq -cN'(l^{-2}+1)
$$
\end{proof}

We shall see now that if a tile contains too many nuclei then the
Thomas-Fermi energy ${E}^{\beta_0}_{\tf,{\cal R}_\alpha,
        {\cal Z}_\alpha}$ is positive. 
If on the other hand the tile contains so few nuclei 
that the Thomas-Fermi energy is negative 
we shall  in the next section prove
an estimate similar to (\ref{eq:lbtf}) but with the 
correct physical constant $\beta_{\rm phys}$ rather than $\beta_0$.

\begin{lm}\label{lm:tfpos}
Consider a tile $l\da(R,y)$ and let 
${\cal Z}_\alpha$ and ${\cal R}_\alpha$ 
denote the charges and positions of nuclei in this tile.
Assume as before that all the nuclear charges are bounded below
by $a\z$ and above by $A\z$. Let $M_\alpha$ denote the number 
of nuclei in the tile.
There is then a constant $c>0$ depending only on $a$, and $A$
such that for all $\beta>0$
$\feba\ge0$ if $M_\alpha\geq\max\{2,c\beta^{-3}l^3\z\}$.
\end{lm}
\begin{proof}
By the scaling property (\ref{eq:scaling}) of Thomas-Fermi theory
we need only consider the case $\beta=\beta_{\rm phys}$ and
$l=1$. {F}rom Lemma~\ref{thm:pres} we have
\begin{equation}
\label{eq:ge1}
  \fea\ge-\cfi M_\alpha (A\z)^{7/3}
 +c(a\z)^{7/3}\lambda^{-1}M'
\end{equation}
where 
$M'$ is the number of nuclei in the tile for which the distance to
the nearest other nuclei
in the tile is less than or equal to $\lambda(a\z)^{-1/3}$
for some $0<\lambda<1$.
We can place $M_\alpha-M'$ disjoint balls of radius $(a\z)^{-1/3}\lambda/2$ 
centered at all the remaining nuclei. Since $M_\alpha\geq2$ it is
clear that either $M_\alpha-M'=0$ or this radius is universally bounded. 
Thus we see that these balls cover a region of 
universally bounded volume.
We conclude that
$(M_\alpha-M')\leq ca\z\lambda^{-3}$.

Hence
$$
        \fea\ge M_\alpha\z^{7/3}
\left( -\cfi A^{7/3}+ca^{7/3}\lambda^{-1}\right)
 -c(a\z)^{10/3}\lambda^{-4}
$$
and the lemma follows if we choose $\lambda$ small enough.
\end{proof}




\section{Proving the main result.}\label{sec:main}

We shall now use a semiclassical approximation to show that 
if the number of nuclei in a tile is small enough then 
one may use the correct physical constant $\beta_{\rm phys}$ 
instead of $\beta_0$ 
in the lower bound in (\ref{eq:lbtf}) .
The semiclassical approximation will be done using the method of
coherent states in a manner very similar to
the argument given in \cite{L1}. 

\begin{lm}[Lower bound in terms of physical TF theory]\label{lm:phtf}
Let $H_{\alpha,N'}$ be the operator defined  in (\ref{eq:bra}) acting
acting on $\bigwedge\limits^{N'}L^2(l\da(R,y);\C^2)$.
We assume as before that $a\z\leq Z_j\leq A\z$ for all $j=1,\ldots,M$.
Assume moreover that 
${E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}\leq0$,
where $\beta_0=K_{\rm LT}/2$ and ${\cal R}_\alpha$ and 
${\cal Z}_\alpha$
are the coordinates and charges of the nuclei in the tile
$l\da(R,y)$.
Then there exist constants $c,C>0$ 
depending only on $a,A$ such that if we choose $\delta=2/87$ 
and $l=\z^{-\frac{1}{3}+\delta}$
then
\begin{equation}\label{eq:phtf}
        \inf{ \rm spec}H_{\alpha,N'}\geq 
        {E}_{\tf,{\cal R}_\alpha,
        {\cal Z}_\alpha}-c N'\z^{\frac{4}{3}-\delta}- CM_{\alpha}\z^{\frac7{3}-\delta},
\end{equation}
where $M_\alpha$ is the number of nuclei in the tile.
\end{lm}
\begin{proof} Lemma~\ref{lm:tfpos} shows that the condition 
${E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}\leq0$
implies that $M_\alpha\leq\max\{2,Cl^3\z\}$.
We shall choose $l$ such that $l^3\z\geq 1$ (see the end of the proof)
we may therefore assume that  $M_\alpha\leq Cl^3\z$.

For $\psi\in \bigwedge\limits^{N'}\left(L^2(\da(R,y);\C^2)\right)$ normalized
we consider again the 1-particle density function
$\rho_\psi(x)$ defined in (\ref{eq:1pd}).
Using the Lieb-Oxford inequality 
(\ref{eq:oxf}) and the positivity of the Coulomb kernel, 
$$
D(f,g):=
\frac12\iint_{\R^3\times\R^3}\frac{f(x)\overline{g(y)}}{|x-y|}dxdy
$$
we find that for
any $0\leq\tilde{\rho}\in L^1(\R^{3})\cap
L^{5/3}(\R^{3})$ and $0<\epsilon<1$ we have
\begin{equation}
\label{eq:hae}
  \langle\psi,H_{\alpha,N'}\psi\rangle\ge\left\langle\psi,\sum^{N'}_{i=1}
h_i\psi\right\rangle-D(\tilde{\rho},\tilde{\rho})-
(1.68)\int\rho_\psi(x)^{4/3}
dx+\epsilon\langle\psi,H_0\psi\rangle+U_\alpha
\end{equation}
where we have introduced  the one-particle operator
$$
        h:=-(1-\epsilon)\Delta_\alpha-V_\alpha(x)+
|\cdot|^{-1}\ast\tilde{\rho}(x)
$$
and used $V_\alpha$ and $U_\alpha$ defined in (\ref{eq:uvalpha}).
In equation (\ref{eq:hae}) we have kept part of the kinetic energy 
$\langle \psi,H_0\psi\rangle=\sum^{N'}_{i=1}\langle\psi,-\Delta^i_\alpha
\psi\rangle$ in order to later use it to control errors. 



We choose $\tilde{\rho}$ to be the 
density that minimizes the Thomas-Fermi 
problem with parameter $\beta$ equal to $(1-\epsilon)\beta_{\rm phys}$,
i.e.,
$\febaa(\tilde{\rho})=\feba$.




It is not convenient to work with Neumann Laplacians, so 
our next goal is to replace them by Dirichlet Laplacians.
In order to do this we take a
partition of unity, with the following properties,
on $l\da(R,y)$
\begin{eqnarray}
0\le\Theta_t,\Xi_t&\le&1\\
\Theta_t^2+\Xi_t^2&=&1\\
\Theta_t(x)=1\textrm{ if dist}(x,\partial l\da)&\ge& t\\
  \Theta_t(x)=0\textrm{ if dist}(x,\partial l\da)&\le&\frac t2\\
|\nabla\Theta_t|_\infty+|\nabla\Xi_t|_\infty&\le&\frac Ct\\
\label{eq:ctd}
|\Delta\Theta_t|_\infty+|\Delta\Xi_t|_\infty&\le&\frac C{t^2}
\end{eqnarray}
For any
$\phi\in L^2(l\da(R,y);\C^2)$ we then have (omitting the parameters $R,y$)
  \begin{eqnarray}\label{eq:part}
\int_{l\da}|\nabla\phi|^2&=&\int_{l\da}\Theta_t^2 
|\nabla\phi|^2+\int_{l\da}\Xi_t^2 
|\nabla\phi|^2\nonumber\\
&=&\int_{l\da}|\nabla(\phi\Theta_t)|^2+
\int_{l\da}|\phi|^2\Theta_t\Delta\Theta_t
+\int_{l\da}|\nabla(\phi\Xi_t)|^2+
\int_{l\da}|\phi|^2\Xi_t\Delta\Xi_t\nonumber\\
&\ge&\int_{l\da}|\nabla(\phi\Theta_t)|^2
+\int_{l\da}|\nabla(\phi\Xi_t)|^2
-\frac C{t^2}\int_{l\da}|\phi|^2
  \end{eqnarray}




Let $\gamma$ be the one-particle density matrix 
on $L^2(l\da;\C^2)$
defined as the operator with integral kernel 
\begin{equation}
\label{eq:G0}
\gamma(x,y):=N'\int_{(l\da(R,y))^{N'-1}}\langle{\psi}(y,x_2,\ldots,
x_{N'}),\psi(x,x_2,\ldots,x_{N'})\rangle
dx_2\ldots dx_{N'}
\end{equation}
where $\langle\ ,\ \rangle$ 
here denotes the inner 
product (antilinear in the first variable) in $\C^{2^{N'}}$.
Because of the antisymmetry of $\psi$, the one-particle 
density matrix $\gamma$
satisfies the fundamental operator inequalities
\begin{equation}\label{eq:G}
  0\le \gamma\le{\bf 1};\ 0\le{\rm Tr}\,\gamma\le N'
\end{equation}
To any positive definite trace class operator $\gamma'$ on $L^2(\R^3;\C^2)$
we have a non-negative density
$\rho_{\gamma'}\in L^1(\R^3)$, defined by $\textrm{Tr}(\gamma' f)=
\int \rho_{\gamma'}(x) f(x)dx$ for any $f\in L^\infty(\R^3)$ identified 
as a multiplication operator on $L^2(\R^3;\C^2)$ (it is a simple exercise
in measure theory to see that this defines an $L^1$ function).
In the special case (\ref{eq:G0}) we have $\rho_\gamma (x)=\gamma(x,x)
=\rho_\psi(x)$.
Moreover (assuming that $\psi$ is in the operator domain of $H_0$)
we have that 
$\left\langle\psi,\sum^{N'}_{i=1}
h_i\psi\right\rangle={\rm Tr}(h\gamma)$
so in terms of $\gamma$ we can rewrite (\ref{eq:hae}) as
\begin{equation}
\label{eq:hae3}
  \langle\psi,H_{\alpha,N'}\psi\rangle\ge{\rm Tr}\left(
h\gamma\right)
-D(\tilde{\rho},\tilde{\rho})-
(1.68)\int_{\R^3}\rho_\psi(x)^{4/3}
dx+\epsilon\langle\psi,H_0\psi\rangle+U_\alpha
\end{equation}
By (\ref{eq:G}) $\gamma$ can be written as
\begin{equation}
\label{eq:G1}
\gamma=\sum_i\lambda_i\langle u_i,\cdot\rangle u_i;\ 0\le\lambda_i\le1
;\ \{u_i\} \textrm{ orthonormal set}
\end{equation}
where the functions $u_i$ belong to the Sobolev space $H^2(l\da(R,y);\C^2)$.
using this spectral representation together with (\ref{eq:part}) yields
\begin{equation}
  \label{eq:Tr}
  {\rm Tr}(h\gamma)\ge {\rm Tr}\left(h\Theta_t\gamma\Theta_t\right)+
{\rm Tr}\left(h\Xi_t\gamma\Xi_t\right)-\frac C{t^2}N'
\end{equation}
where in (\ref{eq:Tr}) both $\Theta_t$ and $\Xi_t$ are regarded as
multiplication operators. 

Let
$\gamma^{(1)}_t=\Theta_t\gamma\Theta_t$ and 
$\gamma^{(2)}_t=\Xi_t\gamma\Xi_t$.
Then $\gamma^{(1)}_t$ and $\gamma^{(2)}_t$ fulfill the bounds 
(\ref{eq:G}) if $\gamma$ does. We now notice that
\begin{equation}
  \label{eq:Tr2}
  {\rm Tr}\left(h\gamma^{(1)}_t\right)=
{\rm Tr}\left(h\Theta_t\gamma\Theta_t\right)
={\rm Tr}\left(\tilde{h}\Theta_t\gamma\Theta_t\right)
\end{equation}
where $\tilde{h}$ is defined as $h$ but with 
Dirichlet boundary conditions instead of Neumann boundary conditions.

Now (\ref{eq:hae3}), (\ref{eq:Tr}) and (\ref{eq:Tr2}) imply that 
\begin{eqnarray}
\label{eq:hae4}
  \langle\psi,H_{\alpha,N'}\psi\rangle&\ge&{\rm Tr}\left(
\tilde{h}\gamma^{(1)}_t\right)-D(\tilde{\rho},\tilde{\rho})+U_\alpha
-(1.68)\int_{\R^3}\rho_\psi(x)^{4/3}dx\nonumber\\
&&+\epsilon\langle\psi,H_0\psi\rangle+{\rm Tr}\left(h\gamma^{(2)}_t\right)
-\frac C{t^2}N'
\end{eqnarray}

Now we shall use coherent states given as follows: pick some function
$g\in H^1(\R^3)$ which is spherically symmetric and with
$\int_{\R^3}|g|^2=1$. Introduce a parameter $r>0$ and the 
family of functions
$g_r:= r^{-3/2}g(r^{-1}\cdot)$.
The coherent states we shall use are then given by
\begin{equation}
  f_{p,s;r}(x)=g_r(x-s){\rm e}^{ip\cdot x};\ p,s\in\R^3
\end{equation}
and let us introduce the projections
\begin{equation}
  \pi_{p,s;r}=\langle f_{p,s;r},\cdot\rangle f_{p,s;r}\otimes I;\ I=\left(
    \begin{array}{cc}
1&0\\0&1
    \end{array}\right)
\end{equation}
Then, for any $m\in L^2(\R^3;\C^2)$ a computation gives
  \begin{eqnarray}
\|m\|^2&=&(2\pi)^{-3}\int_{\R^3\times\R^3}\langle
m,\pi_{p,s;r}m\rangle dp\,ds\nonumber\\
\int_{\R^3}|\nabla m(x)|^2dx&=&(2\pi)^{-3}\int_{\R^3\times\R^3}|p|^2
\langle m,\pi_{p,s;r}m\rangle dp\,ds\nonumber\\
&&-\|m\|^2\int_{\R^3}|\nabla g_r(x)|^2dx\\
\int_{\R^3}|m(x)|^2\tilde{\phi}_r(x)dx&=&(2\pi)^{-3}\int_{\R^3\times\R^3}
\tilde{\phi}(s)\langle m,\pi_{p,s;r}m\rangle dp\,ds\nonumber
 \end{eqnarray}
where
$\tilde{\phi}_r(x)=|g_r|^2\ast\tilde{\phi}(x)$ and $\tilde\phi$ 
is any function  on $\R^3$. 
We shall now use this for the specific choice
$\tilde\phi(x)=V_\alpha(x)-|\cdot|^{-1}*\tilde{\rho}(x)$, i.e., the potential
appearing in the operators $h$ and $\tilde h$.
Introducing the operator 
$\tilde{h}_r=-(1-\epsilon)\Delta_i-\tilde{\phi}_r$
we  split the first term in (\ref{eq:hae4}) 
as follows
\begin{equation}\label{eq:split}
        {\rm Tr}\left(\tilde{h}\gamma^{(1)}_t\right)
        ={\rm Tr}\left(\tilde{h}_r\gamma^{(1)}_t\right)+a_r\left(\gamma^{(1)}_t\right)
\end{equation}
Here
$$
a_r(\gamma^{(1)}_t)=-\int_{\R^3}\left[V_\alpha(x)-|g_r|^2\ast V_\alpha(x)
-(\Psi(x)-|g_r|^2\ast\Psi(x))\right]
\rho_{\gamma^{(1)}_t}(x)dx
$$
where $\Psi=|\cdot|^{-1}*\tilde\rho$ and
$\rho_{\gamma^{(1)}_t}(x)$
is the density corresponding to the operator $\gamma^{(1)}_t$.

We may then write 
\begin{eqnarray}
\label{eq:mo2}
  {\rm Tr}\left(\tilde{h}_r\gamma^{(1)}_t\right)&=&(2\pi)^{-3}\int_{\R^3\times\R^3}\left[(1-\epsilon)|p|^2
-\tilde{\phi}(s)\right]M(p,s)dp\,ds\nonumber\\
&&-{\rm Tr}\,\gamma^{(1)}_t\int_{\R^3}|\nabla g_r(x)|^2dx
\end{eqnarray}
where $M(p,s)={\rm Tr}\left(\gamma^{(1)}_t\pi_{p,s;r}\right)$. 
We have here considered $\gamma^{(1)}_t$ as an 
operator on all of $L^2(\R^3;\C^2)$.
The function $M(p,s)$ satisfies
  \begin{equation}
\label{eq:mo1}
0\le M(p,s)\le2
  \end{equation}

If we minimize (\ref{eq:mo2}) over the functions with the property 
(\ref{eq:mo1}) 
we find  that the
minimum is given by the ``bath-tub'' principle, which tell us that the
minimizer actually is the function
$M(p,s)=2\chi_{\{(p,s):\tilde{\phi}(s)-(1-\epsilon)|p|^2\ge0\}}$.
A straightforward calculation then shows that 
\begin{equation}
\label{eq:eg}
  {\rm Tr}\left(\tilde{h}_r\gamma^{(1)}_t\right)-D(\tilde{\rho},
\tilde{\rho})+U_\alpha\ge\feba
-C N'r^{-2}
\end{equation}


Returning to the estimate of the error term $a_r$ 
we note that $\Psi-|g_r|^2\ast\Psi\ge0$ since (being the
convolution of $|\cdot|^{-1}$ against some integrable function) $\psi$
is {\it superharmonic} and $|g_r|^2$ is spherically symmetric and
of integral one. Therefore 
\begin{eqnarray}
  a_r(\gamma^{(1)}_t)&\ge&-\int_{\R^3}\left(V_\alpha-|g_r|^2\ast V_\alpha\right)
\rho_{\gamma^{(1)}_t}\nonumber\\
&\ge&-\left\|V_\alpha-|g_r|^2\ast V_\alpha\right\|_{5/2}
\left\|\rho_{\gamma^{(1)}_t}\right\|_{5/3}
\end{eqnarray}
by H\"{o}lder's inequality.  
We will use the term $\epsilon\langle\psi,H_0\psi\rangle$ in (\ref{eq:hae4}) 
to control the errors $a_r(\gamma^{(1)}_t)$ and $-C\int\rho^{4/3}_\psi$ 
through the use of the Lieb-Thirring estimate Theorem~\ref{thm:kin}. 
If we write 
\begin{equation}
\label{eq:br}
b_r:=\epsilon\langle\psi,H_0\psi\rangle-(1.68)\int
\rho_\psi(x)^{4/3}dx-a_r(\gamma^{(1)}_t)
\end{equation} 
we find from Theorem~\ref{thm:kin} and H\"older's inequality 
(see also (\ref{eq:holder}) )that
\begin{equation}\label{eq:brestimate}
b_r\ge 
-C\epsilon^{-3/2}
\left\|V_\alpha-|g_r|^2\ast V_\alpha\right\|_{5/2}^{5/2}-C(\epsilon^{-1}+l^{-2})N'
  \end{equation}
where we have used the obvious fact that $\rho_{\gamma^{(1)}_t}\leq\rho_\psi$.
{}From Minkowski's inequality we have that
\begin{eqnarray*}
\left\|V_\alpha-|g_r|^2\ast V_\alpha\right\|_{5/2}&\le&
\sum^{M_{\alpha}}_{j=1}Z_j\left
\||\cdot-R_j|^{-1}-|g_r|^2\ast|\cdot-R_j|^{-1}\right\|_{5/2}\\
&\le&A\overline{Z}\sum^{M_{\alpha}}_{j=1}\left\||\cdot
-R_j|^{-1}-|g_r|^2\ast|\cdot-R_j|^{-1}\right\|_{5/2}
\end{eqnarray*}
but a computation gives
\begin{equation}
  \left\||\cdot-R_j|^{-1}-
|g_r|^2\ast|\cdot-R_j|^{-1}\right\|_{5/2}=Cr^{1/5}
\end{equation}
so we get
\begin{equation}\label{eq:conv}
\left\|V_\alpha-|g_r|^2\ast V_\alpha\right\|_{5/2}^{5/2}\le 
Cr^{1/2}(\overline{Z}M_\alpha)^{5/2}\le CM_{\alpha}r^{1/2}l^{9/2}
\overline{Z}^4
\end{equation}

It remains to estimate the term 
${\rm Tr}\,\left(h\gamma^{(2)}_t\right)$ in (\ref{eq:hae4}). First we note that
\begin{equation}
  \label{eq:hea4}
  {\rm Tr}\,\left(h\gamma^{(2)}_t\right)={\rm
    Tr}\,\left(h_t\gamma^{(2)}_t\right)
  \quad\hbox{with}\quad
  h_t=-(1-\epsilon)\Delta_\alpha-V_\alpha(x)\chi_{\{x:{\rm
      dist}(x,\partial (l\da))\le t\}}
\end{equation}
Now we will apply the Lieb-Thirring inequality  
from Theorem~\ref{thm:LT}
\begin{eqnarray}
  \label{eq:hea5}
{\rm Tr}\,\left(h_t\gamma^{(2)}_t\right)&\ge&-{\rm Tr}\,\left(h_t
\gamma^{(2)}_t\right)_-
\ge-{\rm Tr}\,(h_t)_-\nonumber\\
&\ge&-C_\lt\int_{l \da} V_\alpha(x)^{5/2}\chi_{\{x:{\rm dist}
(x,\partial(l\da))\le t\}}dx\\
&&-C_1l^{-3}\int_{l \da} V_\alpha(x)
\chi_{\{x:{\rm dist}(x,\partial(l\da))\le t\}}dx\nonumber
\end{eqnarray}
where in (\ref{eq:hea5}) we have used that $0\le\gamma^{(2)}_t\le {\bf1}$.
Again using Minkowski's inequality we conclude that
\begin{eqnarray}
  \label{eq:V1}
  \int_{l\da} V_\alpha(x)^{5/2}\chi_{\{x:{\rm dist}
(x,\partial(l\da))\le t\}}dx&\le& C\left(M_\alpha\z\right)^{5/2}t^{1/2}
\leq CM_\alpha t^{1/2}l^{9/2}\z^4\\
\label{eq:V2}
l^{-3}\int_{l\da} V_\alpha(x)
\chi_{\{x:{\rm dist}(x,\partial(l\da))\le t\}}dx&\le&CM_\alpha l^{-2}t\z
\end{eqnarray}

Finally we must deal with the fact that it is
$\beta=(1-\epsilon)\beta_{\rm phys}$ and not $\beta_{\rm phys}$
that appears in (\ref{eq:eg}).
We first note that it follows from the assumption 
that ${E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}\leq0$ 
that $\fea\leq CM_\alpha \z^{7/3}$. To see this note first that 
if we use the minimizer $\rho_{\beta_0}$ for the functional
${\cal E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}$   
as a trial density in the functional 
${\cal E}^{\beta_0/2}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}$ we
conclude that 
$$
  0\geq{E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}
  \geq \frac{\beta_0}{2}\int\rho_{\beta_0}^{5/3}
  +{E}^{\beta_0/2}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}
  \geq \frac{\beta_0}{2}\int\rho_{\beta_0}^{5/3}- C M_\alpha \z^{7/3}
$$
where the last estimate follows from the no-binding Theorem 
of Thomas-Fermi theory, i.e., the positivity of the 
function $f^{\beta_0/2}$ (see Lemma~\ref{thm:pres}).
Now using $\rho_{\beta_0}$ as a trial density in the physical
Thomas-Fermi functional ${\cal E}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}$ gives 
\begin{eqnarray*}
  \fea&\leq&{\cal E}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}
(\rho_{\beta_0})=(\beta_{\rm phys}-\beta_0)\int 
  \rho_{\beta_0}^{5/3}+{E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal
      Z}_\alpha}\\
  &\leq&(\beta_{\rm phys}-\beta_0)\int 
  \rho_{\beta_0}^{5/3} \leq 
  2(\beta_{\rm phys}-\beta_0)\beta_0^{-1}C M_\alpha \z^{7/3}
\end{eqnarray*}

We are now in a position to estimate  
$\feba$
in terms of the physical Thomas-Fermi energy $\fea$.
Indeed using the scaling (\ref{eq:scaling}) we see that 
\begin{eqnarray}
  \label{eq:err4}
\feba&=&\left(1-\epsilon\right)^{6}{E}_{\tf,{\cal R}
_\alpha,(1-\epsilon)^{-3}{\cal Z}_\alpha}\nonumber\\
&=&\left(1-\epsilon\right)^{6}\left\{f({\cal R}
_\alpha,(1-\epsilon)^{-3}{\cal Z}_\alpha)
-\cfi\left(1-\epsilon\right)^{-7}\sum^{M}_{j=1}Z_j^{7/3}
\chi_\alpha(R_j)
\right\}\nonumber\\
&\ge&\left(1-\epsilon\right)^{6}\left\{f({\cal R}
_\alpha,{\cal Z}_\alpha)
-\cfi\left(1-\epsilon\right)^{-7}\sum^{M}_{j=1}Z_j^{7/3}
\chi_\alpha(R_j)
\right\}\nonumber\\
&=&\left(1-\epsilon\right)^{6}\left\{\fea
-\cfi\left[\left(1-\epsilon
\right)^{-7}-1\right]\sum^{M}_{j=1}Z_j^{7/3}\chi_\alpha(R_j)
\right\}\nonumber\\
&\ge&\fea-CM_\alpha\overline{Z}^{7/3}[2-(1-\epsilon)^{6}
-(1-\epsilon
)^{-7}]
\end{eqnarray}
where in the third line of (\ref{eq:err4}) we have used 
that $f$ is a nondecreasing function of the nuclear charges.
together with the observation that $\left(1-\epsilon\right)
^{-3}>1$ and in
the bottom line the fact that $\fea\le CM_\alpha\z^{7/3} $
and $\left(1-\epsilon\right)^6<1$. 

The lemma now follows if we combine all the estimates
(\ref{eq:hae4}), (\ref{eq:split}), (\ref{eq:eg}), (\ref{eq:br}),\-
(\ref{eq:brestimate}), (\ref{eq:conv}), (\ref{eq:hea5}), (\ref{eq:V1}),
(\ref{eq:V2}), and (\ref{eq:err4})
and choose $\epsilon=\z^{-\delta}$, $r=\z^{-57\delta/2}$, 
$l=\z^{-1/3+\delta}$, 
$t=\z^{-51\delta/2}$, where $\delta=2/87$.
\end{proof}


We are now ready to prove the main result Theorem~\ref{thm:stability}.
\begin{proof}[Proof of Theorem~\ref{thm:stability}]
We use a tiling as before with $l=\z^{-1/3+\delta}$.
For this tiling we apply (\ref{eq:dec}) 
together with the estimates (\ref{eq:lbtf})
and (\ref{eq:phtf}). If for a tile
${E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}\geq0$ we use 
(\ref{eq:lbtf}) and simply use the trivial estimate
$$
  {E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}\geq0
  \geq-C_\tf \sum_{j=1}^MZ_j^{7/3}\chi_\alpha(R_j)
$$ 
If ${E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal Z}_\alpha}\leq0$
we apply (\ref{eq:phtf})
together with the no-binding Theorem (the positivity of $f$ in 
Lemma~\ref{thm:pres}), i.e., 
$\fea\geq-C_\tf \sum_j^MZ_j^{7/3}\chi_\alpha(R_j)
$.
We then find from (\ref{eq:dec}) that 
$$
\inf{\rm spec}\hen\ge-C_\tf \sum_{j=1}^MZ_j^{7/3}-C\z^{4/3-\delta}N- 
CM\z^{\frac7{3}-\delta}-\frac Cl(N+\sum^M_{j=1}Z_j^2)
$$
It is a result of Lieb~\cite{L3} that 
if $N>N_0$ where $N_0$ is the greatest integer smaller than or equal to
$2\sum_{j=1}^MZ_j+M$ then 
$\inf{\rm spec}\hen\geq \inf{\rm spec} H_{{\cal R,Z},N_0}$. 
We may therefore without loss of generality assume that 
$N\leq c(\z+1)M$.
We hence conclude that 
$$
   \inf{\rm spec}\hen\ge-C_\tf \sum_{j=1}^MZ_j^{7/3}-CM\z^{7/3-\delta}.
$$
\end{proof}







\section{Estimating the density.}\label{sec:density}

So far, we were concerned with estimating from {\it below} the
quantum-mechanical energy uniformly 
with respect to the nuclear configurations. Now, it
turns out that this can be applied to give a new bound  on the
density of macroscopic objects. The observation is 
that in Theorem~\ref{thm:stability} we disregard an important
effect in Thomas-Fermi theory, namely, the positivity of the
pressure. This gives us a {\it positive} correction to the previous estimate.

Using
our estimate on the energy and the already mentioned result about the
positivity of the pressure, we shall now prove that
if we look at nuclear configurations which are close to those which give
the minimal possible energy, then the number of nuclei per unit
volume is bounded above by const$\overline{Z}^{1-\delta'}$. 
This 
is the content of Theorem~\ref{thm:pressure}.

We shall prove Theorem~\ref{thm:pressure} by applying 
Lemma~\ref{thm:pres}.  Lemma~\ref{thm:pres}
is,  however, purely a result
about Thomas-Fermi theory. 
We can therefore not use it directly 
since we need to know beforehand that given a particular nuclear
configuration the quantum energy can be well approximated from below
by the physical Thomas-Fermi model. 
We have proved this only for the tiles where the number of
nuclei and their configuration is such that the unphysical 
Thomas-Fermi energy is non-positive (see Lemma~\ref{lm:phtf}).
This is where 
the assumption that we look only at configuration 
with nearly minimizing energy 
plays its r\^{o}le. Since it allows us to show that
the number of ``bad'' tiles (with positive energy) should be very small. This
is the content of the next lemma.

\begin{lm}[Controlling the number of nuclei in bad tiles]\label{thm:few}
Consider again the tiling of scale 
$l=\overline{Z}^{-1/3+\delta}$ of $\R^3$ and
assume ${\cal R}\in\R^{3M}$ is a nuclear configuration of 
$M$ nuclei of charges $Z_j$, $j=1,\ldots,M$ for which the assumptions of 
Theorem~\ref{thm:pressure} hold. 
Let ${\cal A}=\{\alpha\ : \ {E}^{\beta_0}_{\tf,{\cal R}_\alpha,{\cal
    Z}_\alpha}\geq0
\}$, where ${\cal R}_\alpha$ and ${\cal Z}_\alpha$ are the positions
and charges of the nuclei in tile $l\da(R,y)$. Denote by $M_\alpha$ the
number of nuclei in the tile.
Then there exists $k>0$ depending only on $a,A$, and the constant
$C_0$ from  Theorem~\ref{thm:pressure} such that
\begin{equation}\label{eq:few}
  \left\langle \sum_{\alpha\in{\cal A}}M_\alpha\right\rangle\leq
kM\overline{Z}^{-\delta}
\end{equation}
\end{lm}
\begin{proof} As in the proof of Theorem~\ref{thm:stability}
we use (\ref{eq:dec}) 
together with the estimates (\ref{eq:lbtf})
and (\ref{eq:phtf}) and the result of \cite{L3} to conclude that
\begin{equation}
  \inf{\rm spec}\he\ge-\cfi
\left\langle\sum_{\alpha\not\in{\cal
      A}}\sum_{j=1}^MZ_j^{7/3}\chi_\alpha(R_j)
\right\rangle -
CM\overline{Z}^{7/3-\delta}
\end{equation}
Thus if the nuclei fulfill the assumption of
Theorem~\ref{thm:pressure} 
we immediately conclude (\ref{eq:few}).
\end{proof}

The following Lemma is of purely geometrical content, and. It
claims that if two
nuclei are nearest neighbors and are close enough, then the set of
$(R,y)\in SO(3)\times Q$ for which the two nuclei 
belong to the same tile of the tiling $\{l\da(R,y)\}$ has positive measure.
\begin{lm}\label{thm:prox}
Let $B({x}_0,r_0)\subset\dq$ be the maximal ball contained in 
$\dq$ and assume that ${x},{{x}'}\in\R^3$ 
satisfy $|{x}-{x}'|\le lr_0/2$. Then, if we call 
$$A_{x,x'}=\left\{({R},
{ y})\in SO(3)\times Q:{x},{x}'
{\rm\ belong\ to\ the\ same\ tile\ of\ }\{l\da(R,y)\}\right\}$$
we have the estimate
\begin{equation}
\label{eq:lm7}
\mu(A_{x,x'})\ge \frac{\pi}{6}r_0^3,\quad \mu=
{\rm\ measure\ on\ }SO(3)\times Q
\end{equation}
\end{lm}
\begin{proof} 
It is fairly easy to see that the claim is translationally invariant. 
We may therefore assume that ${x}={x}_0=0$.
Then $x,x'\in B(0,l r_0/2)$.
For $|y|\leq r_0/2$ and all $R\in SO(3)$ we have 
$$
 R^{-1}\left(B(0,r_0/2)\right)-y=B(0,r_0/2)-y\subset \dq
$$
Hence 
$
x,x'\in B(0,l r_0/2)\subset lR(\dq+y)=l\da(R,y)
$
and the result follows.
\end{proof}

We are now prepared to give the proof of Theorem~\ref{thm:pressure}.

\begin{proof}[Proof of Theorem~\ref{thm:pressure}]
We make the tiling
localization as before with a tiling of scale 
$l=\overline{Z}^{-1/3+\delta}$. Let $\cal A$ be defined as in 
Lemma~\ref{thm:few}.
We then find from (\ref{eq:dec}) 
together with(\ref{eq:lbtf})
and (\ref{eq:phtf}) and the result of \cite{L3} that
\begin{equation}\label{eq:bnob}
\inf{\rm spec}\hen\ge
\left\langle \sum_{\alpha\in{\cal A}}\febao\right\rangle+
\left\langle \sum_{\alpha\not\in{\cal A}}\fea  \right\rangle
-C M\z^{7/3-\delta}
\end{equation}

Denote the ratio $\beta_0/\beta_{\rm phys}=t\in(0,1]$ we then have by 
(\ref{eq:scaling}) and Lemma~\ref{thm:pres}
\begin{eqnarray}\label{eq:ebsmall}
 \febao
&=&t^{-1}\feta\nonumber\\
&\ge&-At^{-1}\cfi M_{\alpha}\overline{Z}^{7/3}+c
\overline{Z}^{7/3}
\sum_{j=1}^M\Gamma\left(t^{-1}\overline{Z}^{1/3}\delta_\alpha(R_j)\right)
\chi_\alpha(R_j)
\nonumber\\
&\ge&-At^{-1}\cfi M_{\alpha}\overline{Z}^{7/3}+c'\overline{Z}
^{7/3}\sum_{j=1}^M\Gamma\left(\overline{Z}^{1/3}\delta_\alpha(R_j)\right)
\chi_\alpha(R_j)
\end{eqnarray}
where $c'>0$ and  $\delta_\alpha(R_j)$ denotes the nearest neighbor
distance
among nuclei in the tile $l\da(R,y)$.
We have used the scaling properties of Thomas-Fermi theory
together with the fact $\Gamma(tx)\ge C(t)\Gamma(x)$ for
$0<t\le1$ and $C(t)>0$ (which is easily verified).

If we insert (\ref{eq:ebsmall}) into (\ref{eq:bnob}) 
and apply Lemma~\ref{thm:few} and Lemma~\ref{thm:pres}
we arrive at 
\begin{eqnarray}\label{eq:presss}
\inf{\rm spec}\hen&\ge& -\cfi\sum_{j=1}^MZ_j^{7/3}-C M\z^{7/3-\delta}
\nonumber\\
&&+c'\overline{Z}
^{7/3}
\sum_{j=1}^M\left\langle\sum_\alpha
\Gamma\left(\overline{Z}^{1/3}\delta_\alpha(R_j)\right)
\chi_\alpha(R_j)\right\rangle
\end{eqnarray}
If $\delta(R_j)\le lr_0/2$ 
we use Lemma~\ref{thm:prox} to conclude that
$$
   \left\langle\sum_\alpha\overline{Z}^{7/3}
   \Gamma\left(\overline{Z}^{1/3}\delta_\alpha(R_j)\right)
  \chi_\alpha(R_j)\right\rangle
  \geq c''\overline{Z}^{7/3}\Gamma\left(\overline{Z}^{1/3}\delta(R_j)\right)
$$

For a nucleus $R_j$
such that
$\delta(R_j)\ge lr_0/2$ we cannot argue as above, but in this case 
we have that
$\overline{Z}^{7/3}\Gamma(\overline{Z}^{1/3}\delta(R_j))\le 
\overline{Z}^{7/3-7\delta}\ll \overline{Z}^{7/3-\delta}$ and
this contribution can therefore be included 
at the expense of increasing the constant in front of the error term. 
This concludes the proof of Theorem~\ref{thm:pressure}.
\end{proof}
\vspace{4 mm}

\begin{proof}[Proof of Theorem~\ref{thm:volume}]
Follows immediately from Theorem~\ref{thm:pressure} noticing that
$$\sum^M_{j=1}
\Gamma\left(\overline{Z}^{1/3}\delta(R_j)\right)\ge \overline{Z}^{-7\delta_1}
\#\left\{R_j\in{\cal R}:\delta(R_j)\le
\overline{Z}^{-1/3+\delta_1}\right\}$$
\end{proof}

\appendix 
\section{Appendix: The Lieb-Thirring inequality (Th.~\ref{thm:LT}):} \label{sec:app}
\begin{proof}[Proof of Theorem~\ref{thm:LT}]
We shall look first at the case $\alpha=0$. 
In the case of the Laplacian on the whole space the corresponding
inequality, which holds without the last term, is the celebrated
CLR estimate, proved independently 
by Cwickel~\cite{Cw}, Lieb~\cite{L0}, and Rozenblum~\cite{R}. 
We are however not aware of a reference in the Neumann case. 
For completeness we therefore include a proof.
The method of Lieb uses the Feynman-Kac formula
and gives by far the best constant. In order not to introduce the 
Feynman-Kac formula on the cube we here appeal to Rozenblum's method.
 
The
appearance of the constant term is due to the fact that even 
if $V=0$, 0 is an
eigenvalue corresponding to a constant eigenfunction. 

On the orthogonal complement  of the constants the Neumann Laplacian
satisfies a Sobolev inequality similar to the Dirichlet Laplacian
(see e.g.~\cite{B}).
More precisely, there exists $C_S>0$ such that for all 
rectangular boxes $\tilde Q\subset \R^3$ with
eccentricity (the ratio between the longest and the shortest side) bounded
by 2 we have for all $\phi\in L^2(\tilde Q)$ with $\int_{\tilde
  Q}\phi=0$
that
$$\int_{\tilde{Q}}
|\nabla\phi|^2\ge C_S\left\{\int_{\tilde{Q}}|\phi|^6\right\}^{1/3}$$
Thus if 
\begin{equation}\label{eq:sobolevcondition}
  \int_{\tilde{Q}}[V]_-^{3/2}\le\left( 
    {C_S}\right)^{3/2}\equiv A
\end{equation}
we have for all $\phi$ with
$\int_{\tilde{Q}}\phi=0$
that 
\begin{eqnarray*}
\int_{\tilde{Q}}
|\nabla\phi|^2+\int_{\tilde{Q}}V|\phi|^2 &\ge&\int_{\tilde{Q}}
|\nabla\phi|^2-\int_{\tilde{Q}}[V]_-|\phi|^2\\
&\ge& C_S\left\{\int_{\tilde{Q}}|\phi|^6\right\}^{1/3}
-\left\{\int_{\tilde{Q}}[V]_-^{3/2}\right\}^{2/3}
\left\{\int_{\tilde{Q}}|\phi|^6\right\}^{1/3}\geq0
\end{eqnarray*}
Hence $-\Delta_{\tilde{Q}}+V\ge0$ in $\{\chi_{\tilde{Q}}\}^\perp$ if
$\int_{\tilde{Q}}[V]_-^{3/2}\le\left( 
{C_S}\right)^{3/2}\equiv A $. 

Now the idea, which goes back to Rozenblum \cite{R}, is
to try to cover the whole original unit cube by cubes of eccentricity bounded
by 2 (which from now on will be called {\it bricks}) such that in each of them
the condition (\ref{eq:sobolevcondition}) is satisfied. 
In order to do this we shall
employ the following special case of a general
covering lemma given by M. de Guzm\'{a}n~\cite{Gu}.
\begin{lm}[Covering Lemma]\label{thm:guzman} 
Let $J(G)$ be a real-valued function defined over the class 
of Borel subsets of a $d$-dimensional cube $Q$ which is lower-semiadditive 
(i.e $J(G_1\cup G_2)\ge
J(G_1)+J(G_2)$ if $G_1\cap G_2=\emptyset$) and continuous in measure (i.e if $G_t$ is a nested family of
sets continuous in measure, $t\rightarrow J(G_t)$ is continuous). Then, for
every integer $n\ge1$, there is a covering $\Xi$ of $Q$ by bricks $\tilde{Q}
\subset Q$ such that the number of bricks $\#\Xi\le n$, each point of the 
cube belongs to at most $d$ bricks and for any $\tilde{Q}\in\Xi$
\begin{equation}
  J(\tilde{Q})\le\frac{2^{d+1}}nJ(Q)
\end{equation}
\end{lm} 
We will use this Lemma~\ref{thm:guzman} for the function
\begin{equation}
  J(G)=\int_G[V]_-^{3/2}
\end{equation}
together with the following, which is a consequence of the min-max principle
for the eigenvalues of self-adjoint operators:
\begin{equation}
  n_-(-\Delta_Q+V)=\min\{\textrm{codim}\,{\cal V};\ {\cal V}\subset L^2(Q):
  -\Delta_Q+V>0 \textrm{ on }{\cal V}\}
\end{equation}
where $n_-(-\Delta_Q+V)$ is the number of non-positive 
eigenvalues of $-\Delta_Q+V$. Now there are two cases: if $\int_{{\tilde Q}}[V
]_-^{3/2}\le A$, then we have exactly one non-positive eigenvalue. If, on the
other hand, this is not satisfied, we choose $n$ such that 
$\frac{16}n\int_Q[3V]_-^{3/2}\le A$ and also being the smallest integer with
this property. Moreover, let $\tilde{Q}_1,\ldots,\tilde{Q_n}$
be the covering given by the Guzm\'{a}n Lemma
and let ${\cal V}$ be $\textrm{sp}\,\{\chi_{\tilde{Q_1}},\ldots,
\chi_{\tilde{Q_n}}\}^\perp$.  Since 
$1\leq\sum^n_{j=1}\chi_{\tilde{Q}_j}\le3$
we have, for $\phi\in{\cal V}$
\begin{eqnarray*}
\int_Q|\nabla\phi|^2+\int_QV|\phi|^2 &\ge& \int_Q|\nabla\phi|^2
\frac13\sum^n_{j=1}\chi_{\tilde{Q}_j}-\int[V]_-|\phi|^2\sum^n_{j=1}\chi_{\tilde{Q}_j}\\
&\geq&\frac13\sum^n_{j=1}
\left\{\int_{\tilde{Q}_j}|\nabla\phi|^2-\int_{\tilde{Q}_j}[3V]_-
 |\phi|^2\right\}\\
&\ge&0
\end{eqnarray*}
which follows since 
$$
  \int_{\tilde{Q}}[3V]_-^{3/2}=3^{3/2}J(\tilde{Q})
  \leq \frac{16}{n}3^{3/2}J(Q)\leq A
$$
by our choice of $n$ and this proves the $\alpha=0$ case. Hence 
\begin{equation}
n_-(-\Delta_Q+V)\le\mbox{codim }{\cal V}=n\leq\frac{3^{3/2}16}A\int_Q[V]_-^{3/2}+1 
\end{equation}
In order to deal with other $\alpha$ values, 
we need a bound on the
bottom of the spectrum. 
This is provided by the following: write $\phi\in H^1(Q)$
as $\phi=\phi_1+\phi_2$ where $\phi_1=(\int_Q\phi)\chi_Q$. We have
$|\phi|^2\le2|\phi_1|^2+2|\phi_2|^2$ and then, assuming also $\phi$ normalized
we get since $|\phi_1|^2\leq \|\phi\|^2\chi_Q$ (recall that $Q$ is a
unit cube)
  \begin{eqnarray}
\label{eq:lt}
\int_Q|\nabla\phi|^2+\int_QV|\phi|^2
&\ge& C_S\left\{\int_Q|\phi_2|^6\right\}^{1/3}\nonumber\\
&&-2\left\{\int_Q[V]_-^\frac3{2-2s}\right\}^\frac{2-2s}3\left\{\int_Q|\phi_2|^6
\right\}^\frac{1-s}3-2\int_Q[V]_-\nonumber\\
&\ge &\inf_{t>0}\left\{C_St-2\left\{\int_Q[V]_-^\frac3{2-2s}\right\}^\frac
  {2-2s}3t^{1-s}\right\} -2\int_Q[V]_-\nonumber\\
&\ge& -C(s)\left\{\int_Q[V]_-^\frac3{2-2s}\right\}^\frac{2(1-s)}{3s} 
-2\int_Q[V]_-
  \end{eqnarray}
where we have picked some $0<s<1$ in (\ref{eq:lt}). 
Now we can end the proof noticing that
\begin{equation}
\label{eq:lt1}
  \textrm{Tr}\,(-\Delta_Q+V)_-^\alpha=
\alpha\int^\infty_0n_{-\lambda}(-\Delta_Q+V) 
\lambda^{\alpha-1}d\lambda
\end{equation}
(here $n_{-\lambda}(-\Delta_Q+V)$ is the number of eigenvalues less
than
or equal to 
$\lambda$ ) and moreover
\begin{equation}
\label{eq:lt2}
  n_{-\lambda}(-\Delta_Q+V)\le n_-(-\Delta_Q-[V+\lambda]_-)
\end{equation}
which is an easy consequence of the minimax principle. Now 
(\ref{eq:lt1}), (\ref{eq:lt2})
together imply
\begin{equation}
\label{eq:lt3}
  \textrm{Tr}\,(-\Delta_Q+V)_-^\alpha\le\alpha\int_0^{|\inf
{\rm spec}(-\Delta_Q+V)|}
n_-(-\Delta_Q-[V+\lambda]_-)\lambda^{\alpha-1}d\lambda
\end{equation}
Now we choose $s=\frac{2\alpha}{3+2\alpha}$ and from 
(\ref{eq:lt}), (\ref{eq:lt1}), (\ref{eq:lt2}) we get
  \begin{eqnarray}
 \textrm{Tr}\,(-\Delta_Q+V)_-^\alpha&\le&\alpha
\int_0^{|\inf{\rm spec}(-\Delta_Q+V)|}\lambda^{\alpha-1}d\lambda\left\{
C_0\int_Q[V+\lambda]_-^{3/2}dx+1\right\}\nonumber\\
&\le&C_0\alpha\int_Qdx\int_0^{[V]_-}[V+\lambda]_-^{3/2}\lambda^{\alpha-1}
d\lambda+|\inf{\rm spec}(-\Delta_Q+V)|^\alpha\nonumber\\
&=&C_0\alpha\int_Qdx[V]_-^{\frac32+\alpha}
\int_0^1(1-\lambda)^{3/2}\lambda^{\alpha-1}d\lambda
+|\inf{\rm spec}(-\Delta_Q+V)|^\alpha\nonumber\\
&\le&C_\alpha\int_Q[V]_-^{3/2+\alpha}+2^\alpha\left\{\int_Q[V]_-\right\}^\alpha
  \end{eqnarray}
where in the last inequality we have used the elementary inequalities
$$(x+y)^\alpha\le x^\alpha+y^\alpha;\ x,y\ge0,\ 0<\alpha\le1$$
or
$$
        (x+y)^\alpha\leq 2^{\alpha-1}(x^\alpha+y^\alpha);
        \ x,y\ge0,\ 1<\alpha
$$
and this completes the proof for the unit cube. 

In order to prove the Theorem in the case of the tetrahedra 
described in the beginning of Sect.~\ref{sec:GS} we shall show that 
the Neumann eigenfunctions in a tetrahedra can be extended to Neumann 
eigenfunctions in the unit cube.
Note that we can get all 24 tetrahedra making up the unit cube 
by repeated reflections (through faces)
of one of the tetrahedra, say, $\dq$.
Moreover, it can be easily seen that it always takes an 
even number of reflections to return to a given tetrahedron.
Since an even number of reflections leaving $\dq$ invariant is 
the identity we see that any function on $\dq$ can  
be extended consistently to the whole
unit cube by reflections. 
 
If $\phi$ is an eigenfunction of $-\Delta_\dq-V$,
then $\phi\in H^2(\dq)$ (the Sobolev space of order 2) with 
$\left.\partial_N\phi\right|_{\partial\dq}=0$.
If we therefore define $\tilde \phi$ as the extension by reflection 
of $\phi$ to the whole unit cube $Q$ then $\phi\in H^2(Q)$
with $\left.\partial_N\phi\right|_{\partial Q}=0$.
Moreover, if $\tilde V$ is the reflected extension of $V$, 
then $\tilde\phi$ is a Neumann eigenfunction of
$-\Delta_Q-\tilde V$ with the same eigenvalue.
Thus 
\begin{eqnarray*}
  \textrm{Tr }(-\Delta_{\dq}-V)_-^\alpha&\leq& 
  \textrm{Tr }(-\Delta_{Q}-\tilde V)_-^\alpha\\
   &\leq& C_\alpha\int_Q [\tilde V]_-^{3/2+\alpha} +C_\alpha\left(\int_Q
     [\tilde V]_-\right)^{\alpha} \\
   &=&24  C_\alpha\int_{\dq}[V]_-^{3/2+\alpha} +C_\alpha\left(24 \int_{\dq}
     [V]_-\right)^{\alpha} 
\end{eqnarray*}
\end{proof}


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