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\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 PHY9019433 A01},
Jan Philip Solovej$^{(2),}$\footnote{**}{\eightpoint
Work partially supported by U.S. National Science Foundation grant
DMS 9203829}
and Jakob Yngvason$^{(3),}$\footnote{***}{\eightpoint Work
partially supported
by the HereusStiftung 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, IS107
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 ThomasFermi
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 nonclassical 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}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%
{\narrower\baselineskip=3.5ex
{\bf Table of Contents}
\item{I.} Introduction \dotfill 2
\item{II.} The Superstrong Magnetic Field Density Functional \dotfill 12
\item{III.} The Hyperstrong 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 Onebody 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 [426] 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 singleparticle 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 HartreeFock 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 [2931] and shown to be described
exactly by
ThomasFermi (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.,
[3237].
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 onebody density
matrix $K_\Psi(x;x^\prime)$ that is relevant
but only its ``semidiagonal'' 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
$${\cal E}^{\rm DM}[\Gamma]:=\int_{{\bf R}^2} {\rm Tr}_{L^2({\bf
R})}
[\partial_3^2\Gamma_{x_{\perp}}]d x_{\perp}
Z\intx^{1}\uprho_{\Gamma}(x)dx
+D(\uprho_\Gamma,\uprho_\Gamma),\eqno(1.22)
$$
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 onedimensional\/} 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 onebody 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 superstrong
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 onedimensional} density, called the
{\bf hyperstrong 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 onedimensional
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
[4244], 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
$$N^{\rm Q}_c = \max \{ N: E^{\rm Q} (N,Z,B) < E^{\rm Q} (N1,Z,B)\},
\eqno(1.27)$$
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 [2931]
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
HartreeFock 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 hyperstrong 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 LiebOxford 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 onebody 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., BuressurYvette.
%%%%%%%%%%%%%%%%%%%%%%%%
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\bigskip
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf II. THE SUPERSTRONG MAGNETIC FIELD DENSITY FUNCTIONAL}
In this section we introduce a density functional $\E^{\SS} [\uprho]$
that is very different from the ThomasFermi 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 superstrong
theory.
Oddly enough, this limiting theory (which we call the {\bf
hyperstrong theory}) can be solved {\it exactly}. One point worth
noting
is that the errors, in this hyperstrong 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 hyperstrong 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 superstrong 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 superstrong
theory,
even in the limit as $\eta\to\infty$, is that it contains more
information
than the hyperstrong theory; in particular it tells us about the
crosssectional density in the atom, whereas the hyperstrong
theory
deals only with the crosssectional integral of the density.
The superstrong 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
$$\eqalignno{\vert \psi (x_\bot, a)  \psi (x_\bot, b) \vert &=
\left\vert
\int^b_a (\partial \psi/\partial x_3) dx_3 \right\vert \leq
\left\vert\int^b_a dx_3 \right\vert^{1/2} \left\vert \int^b_a
(\partial
\psi /\partial x_3)^2 dx_3 \right\vert^{1/2} \cr
&\leq \vert ab \vert^{1/2} \left\vert \int (\partial \psi /\partial
x_3)^2
dx_3 \right\vert^{1/2} .\qquad&(2.4)\cr}$$
(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 onedimensional 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 BanachAlaoglu 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 RieszFischer 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
$G(x) G(y) V \vert xy \vert^{1}$ 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 BanachAlaoglu 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 nontrivial 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 apriori} 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
$$\eqalignno{\phi^{\SS}_{x_\bot} (x_3) &= Z \vert x \vert^{1} 
\vert x
\vert^{1} * \uprho^{\SS} \cr
&=Z(x_\bot^2+x_3^2)^{1/2}
\int\uprho^{\SS}(y)(x_\boty_\bot^2+x_3y_3^2)^{1/2}dy.
\qquad&(2.10)\cr}$$
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 onedimensional wire, with onedimensional 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 meanvalue 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
EulerLagrange
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 hyperstrong problem).} {\it
The
hyperstrong 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 EulerLagrange 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 deltafunction 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:
$$\E^{\SS}_\eta
[\,\uprho_\eta\,] :
= \int \left( {\partial \sqrt{\uprho_\eta} \over
\partial x_3} \right)^2 dx  \int \uprho_\eta (x) V_\eta (x) dx +
\mfr1/2
\int\int \uprho_\eta (x) V_\eta (xy) \uprho_\eta (y) dx dy
\eqno(3.13)$$
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}
$$\left\vert \int \limits_{\R^3} \int \limits_{\R^3} \uprho (x)
V_\eta
(xy) \uprho (y) dx dy  \int \limits_\R \overline{\uprho} (x)^2 dx
\right\vert \leq \lambda C (\lambda, T, 2R) \left( {1 + \vert \ln
\vert \ln
\eta \vert \vert \over
\vert \ln \eta \vert} \right). $$
\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 superstrong and hyperstrong
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
BanachAlaoglu 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
onedimensional}, singleparticle 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 onedimensional 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 $\zeta_k^{(x_3)}={k\over 2}\upchi_{[x_3k^{1},x_3+k^{ 1}]}$
($\upchi_A$ 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
$$\lim_{k\to\infty}\sum_n\alpha_n(f_n,\zeta_k^{(x_3)})^2=
\sum_n\alpha_nf_n(x_3)^2=\uprho_{\Gamma}(x).$$
The density $\uprho_\Gamma$ is a nonnegative 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
$x_\bot$
$$\left{\partial \sqrt{\uprho_{\Gamma}(x)}\over\partial x_3}
\right=
(\sqrt{\uprho_{\Gamma}})^{1}
\left{\partial\uprho_\Gamma\over\partial x_3}\right
\leq (\sqrt{\uprho_{\Gamma}})^{1}\sum_n\alpha_nf_n
\left{\partial f_n\over\partial x_3}\right,$$
with the understanding that the right side is zero where
$\uprho_\Gamma =
0$.
Since $\uprho_{\Gamma}=\sum_n\alpha_nf_n^2$, property (b) in
(4.5)
follows by estimating the sum above by the CauchySchwarz
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
manybody 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 manybody 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 oneparticle 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 $\vert xy \vert^{1}$, 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 onedimensional operator on $L^2(\R)$ given in Lemma 2.1, i.e.,
$$\hat{h}_{Z,x_\bot}={\partial^2\over\partial x_3^2}{Z\overx}=
{\partial^2\over\partial x_3^2}{Z\over\sqrt{x_\bot^2+x_3^2}},
\eqno(4.9)$$
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 onedimensional
LiebThirring 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
onedimensional operator ${\partial^2\over\partial x_3^2}
\alpha\uprho_{\Gamma}(x)^2$
multiplied by $B/2\pi$. The LiebThirring 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 LiebThirring
inequality.
In fact, for all $R>0$ we get by using the LiebThirring inequality
for
$x_\botR$ 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 $x_\bot0$, 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 onedimensional 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 nonzero 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
$\{x_\bot\,:\, x_\bot0}
\G^{\DM}_B$ 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 onedimensional 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 $\vert xy \vert^{1}$ 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
onedimensional 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 threedimensional} 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 [4749] 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 superstrong 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 $T^{\DM}: =
\int_{\R^2} \Tr_{L^2 (\R)} \left[  {\partial^2 \over
\partial x^2_3} \Gamma^{\DM}_{x_\bot}\right] dx_\bot$,
$$A^{\DM}: = \int Z \vert x \vert^{1} \uprho^{\DM} (x)
dx, R^{\DM}: = \mfr1/2 \int\int \uprho^{\DM} (x) \vert
xy \vert^{1} \uprho^{\DM} (y) dx dy,$$
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 (4.4546) 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 onedimensional 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
manybody 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 simpleminded,
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 quantummechanical ground state energy, $E^{\rm Q}$, satisfies
$$E^\Q \leq \Tr \big[ ( H_\A  Z \vert x \vert^{1})K \big] +
\mfr1/2
\int\int k (x,x) k(y,y) \vert xy \vert^{1} dx dy \eqno(5.1)$$
Here, $H_\A = [\vsigma \cdot (\p + \A)]^2$ and $K = K (x, s;
x^\prime
s^\prime)$ is any density matrix (a nonnegative, selfadjoint, 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,
$$\eqalignno{E^\Q &\leq \hbox{Right side of (5.1)} \ = E^{\DM} + a 
b, \cr
a &= Z \int \vert x \vert^{1} [\uprho^{\DM} (x)  k(x,x)] dx,
\qquad&(5.10)\cr
b &= \int\int \vert xy \vert^{1} [\uprho^{\DM} (x) \uprho^{\DM} (y)
 k(x,x) k(y,y)] dx dy. \cr}$$
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 $\vert xx^\prime \vert^{1}  (m*_\bot
\vert x
\vert^{1} *_\bot m) (xx^\prime)$, i.e.,
$$Q(x,x^\prime) = \vert xx^\prime \vert^{1}  \int \limits_{\R^2}
\int
\limits_{\R^2} m (x_\bot  y_\bot) m (x^\prime_\bot 
y^\prime_\bot)
\bigl[ (x_3  x^\prime_3)^2 + (y_\bot  y^\prime_\bot)^2 \bigr]^{
1/2}
dy^\prime_\bot dy_\bot.$$
Since $Q$ is a function only of $xx^\prime$, we can evaluate $b$
in
terms of the threedimensional Fourier transform
$\widehat{\uprho^{\DM}}$ and the
twodimensional 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}$$
with $R_{\rm U}(N,Z,B)$ as in Theorem 5.1.}
{\it Proof:} (5.17) is simply a transcription of our calculations
above.
Inequality (5.18) follows from the definitions of $E^{\SDM}$ and
$E^{\DM}$ as infima.
For each $y_\bot\in \R^2$ define $\Gamma^{y_\bot}$ by
$\Gamma^{y_\bot}_{x_\bot} = \Gamma^{\SDM}_{x_\boty_\bot}$.
(Here
$\Gamma^{\SDM}$ is the minimizer for $E^{\SDM} (N,Z,B)$. If none
exists
then we have to use a minimizing sequence in an obvious way.)
Then
$E^{\DM} (N,Z,B) \leq \E^{\DM} [\Gamma^{y_\bot}]$.
However, a simple calculation
shows that
$$\int \limits_{\R^2} m (y_\bot) \E^{\DM} [\Gamma^{y_\bot}]
dy_\bot =
\E^{\SDM} [\Gamma^{\SDM}]. \qquad\qquad \hbox{\lanbox}$$
The next theorem gives a simple bound for
$E^{\rm Q}(N,Z,B)$ in terms of the independent particle model.
{\bf 5.3 THEOREM (Simple upper bounds)}. {\it
Let $E_{(1)}^{\rm Q}(N,Z,B)$ be the ground state energy for the
Hamiltonian (1.1) without the last (repulsion) term, i.e.,
for $ H^{(1)}_N:=\sum_1^N\left[H_{\bf A}^iZ/x^i\right]$.
Then, if $n\leq N$,
$$
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)
$$
with $n=\min\{N,Z\}$.}
{\it Proof:} Let $K_{(1)}$ denote the oneparticle reduced
density matrix for the ground state of $H^{(1)}_n$.
[Note: $n$ and not $N$. Note also that if there are several
ground states, then choose any one. If
there is none, then our proof can easily be carried
out with a minimizing sequence, $K^\varepsilon_{(1)}$
with $\varepsilon\to 0$, in place of $K_{(1)}$.].
This operator satisfies $0\leq K_{(1)}\leq I$ and
$\Tr K_{(1)}=n$. Let $\uprho_{(1)}(x)=K_{(1)}(x,x)$ as
usual. A very important fact is that
$\Phi(x):=\intxy^{1}\uprho_{(1)}(y)dy$ has its
maximum at $x=0$ (because otherwise we could lower the
energy of our state by moving the nucleus from $x=0$ to another
point $x\in\R^3$ at which $\Phi(x)<\Phi(0)$; this argument
is used again later in the $N$body context just before (6.10),
where a lengthier discussion is given).
Thus, we have $\Tr \left[\vert x \vert^{1} K_{(1)}\right]
= \Phi (0) \geq \int f
(x) \Phi (x) dx$ for any $f: \R^3 \rightarrow \R^+$ with $\int f \leq
1$.
We now have, by the variational principle (5.1) and with $f:=
\mfr1/2
\uprho_{(1)}$,
$$\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)$$
Now we fix $R>0$ and choose $g$ to be a smooth, radial function
with
$g(x)= 1$ for $x2R$ and $g(x) \leq 1$
everywhere.
The condition on $\int g^2$,
is $\mu BR^3 = (\const.)n$.
We then obtain
$$
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)$$
The ``diagonal'' terms, i.e., $\Pi^{i}_0 \vert x^i \vert^{1}
\Pi^{i}_0$ and $\Pi^{i}_> \vert x^i \vert^{1} \Pi^{i}_>$, yield,
without further ado, the terms in (6.3) proportional to $Z$. The
term in
(6.3) proportional to $\varepsilon Z$ and $\varepsilon^{1} Z$ come
from
the ``offdiagonal'' terms in (6.4) by using the Schwarz inequality
as
follows:
$$ \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)$$
Thus we
have proved (6.2) as far as the single particle potential, $Z /\vert
x
\vert$, is concerned.
The twobody terms $\vert x^i  x^j \vert^{1}$ are only slightly
more
complicated. We can write $\vert x^i  x^j \vert^{1}$ in a manner
similar
to (6.4) using $\Pi^{i}_0, \Pi^{i}_> \Pi^{j}_0$ and $\Pi^{j}_>$.
Now there will be 16 terms instead of 4 as in (6.4). Again, we
bound the
``offdiagonal'' terms by the Schwarz inequality; the result is (6.3).
Our next goal is to bound $H^\alpha$ on functions $\Psi$
that are antisymmetric in the $\alpha$ variables (we denote the
number of
these by $0 \leq n_\alpha \leq N$) and also antisymmetric in the
remaining
$n_{\sim \alpha} = N  n_\alpha$ variables. Additionally, we
require that the
$n_{\sim \alpha}$ variables are in the second or higher Landau
band, i.e.,
$\Pi^{i}_0 \Psi = 0$ for $i \not\in \alpha$. Conversely, $\Pi^{i}_0
\Psi =
\Psi$ for $i \in \alpha$.
{\it Part 2.} First we study the three terms in (6.3) that involve
{\it only} the
$\alpha$ variables. These are the first, second and fourth terms in
(6.3)
and (after relabelling the particles) they are equal to
$$\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)$$
Here $n: = n_{\sim \alpha} = N  n_\alpha$.
To obtain a lower bound to $\overline H^\alpha$ we start by using
the
decomposition in L\'evyLeblond's paper [51] for $n \geq 2$:
$$\eqalignno{\overline H^\alpha = nB/2 + {1 \over (n1)} \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(n1) \over 2 \varepsilon} \vert x^j  x^i \vert^{1}
\biggr\}. \qquad&(6.11)\cr}$$
To bound the operator in braces $\{\ \}$, we can set $x^i = 0$ (for
the
reason given in the preceding paragraph).
We also introduce a cutoff $R>0$ and
note that $x^{1}\geq vR^{1}$ where the potential
$v$ is given by $v(x)=x^{1}R^{1}$ for $x\leq R$ and $v=0$
if $x\geq R$.
Finally, we use the
LiebThirring inequality\footnote{$^\dagger$}
{\eightpoint \baselineskip=5ex
In [52] the LiebThirring inequality was originally
proved for $\DeltaV$. Because of
the diamagnetic inequality, the proof also works
when $\Delta={\bf p}^2$ is replaced by $({\bf p}+{\bf A})^2$,
as remarked by several authors.}
to bound the energy of $\sum \{ \mfr1/2 [\p+\A]^2
 V\}$ by $0.114 \int V^{5/2}$.
Here, the potential $V$ is
$$
V=((1+\varepsilon^{1})Z+3\varepsilon^{1}(n_\alpha
+(n1)/2))v\leq \varepsilon^{1}[2Z+3N]v.
$$
We find that the operator in $\{\ \}$ in (6.11) is bounded below by
$ 8\pi(0.114)R^{1/2}\varepsilon^{5/2}[2Z+3N]^{5/2}
\varepsilon^{1}[2Z+3N]n/R $.
We minimize this expression with respect to $R$. The result for
all $n = N  n_\alpha > 0$ is (after replacing $n/(n1)$ by 2)
$$\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)$$
The last inequality follows from the upper bound on $E^{\rm Q}(N, Z, B)$
given in Theorem~5.3.
In case $B \geq Z^3$ we retain the $B$ term in (6.12) and we see
(since $(N
 n_\alpha) \geq (N  n_\alpha)^{1/3}$) that (6.12) is actually
positive
(and hence can be neglected) if $Z$ and $\varepsilon$ are not too
small.
Recall that $Z \geq N/\Lambda \geq 1/\Lambda$, so $Z$ is never
small.
We can easily let
$\varepsilon \rightarrow 0$ as $\lambda^{2/3}\beta \rightarrow
\infty$.
\lanbox
%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%
\bigskip
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf VII. REDUCTION TO A ONEBODY PROBLEM}
Our goal here is to replace the twobody Coulomb repulsion $\sum
\nolimits_{i < j} \vert x^i  x^j \vert^{1}$ and attraction $Z \sum
\nolimits_i \vert x^i \vert^{1}$ by the following onebody
potential plus a
constant:
$$\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 cutoff perpendicular to the
field.
Indeed, we now introduce an additional cutoff parallel to the
field.
This additional cutoff 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 cutoff 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., $\int \int \overline f(x)
V_<
(xy) f(y) dx dy \geq
0$ for all functions $f$. As a consequence the following
holds for any $\uprho \in L^1 (\R^3)$ (see [53]).
$$\sum \limits_{i < j} V_< (x^i x^j) \geq \sum \limits_i \int
\uprho (y)
V_< (x^i  y) dy  \mfr1/2 N V_< (0)  \mfr1/2 \int \int \uprho (x)
\uprho
(y) V_< (xy) dx dy. \eqno(7.10)$$
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 (xx^i) V_< (xy) \delta (y  x^j) dx dy \cr
&= \mfr1/2 \int \int \left( \sum \limits_i \delta (xx^i)  \uprho
(x) \right)
V_< (xy) \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_< (xy) 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)$$
Inserting (7.14) and (7.15) into (7.13) gives (7.12). This finishes
the
proof of Lemma 7.2. \lanbox
{\it Continuation of the proof of Theorem 7.1.} We are now
prepared to
estimate the manybody Hamiltonian $H_N$. Using $\vert x
\vert^{1} \geq V(x)$
we get, for all $0 < \alpha < 1$,
$$\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)$$
with the $N1$ points $x^j, j \not= i$ fixed, in which case it is a
onebody operator. Since $H_\A \geq  (\partial /\partial x_3)^2$
we find
from Lemma 7.2 the following lower bound independent of $x^j, j
\not= i$
$$\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 (N1) (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
(N1)^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_< (xy) 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(N1)^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(xy)  V_< (xy)]
dy
\biggr\vert = \int \uprho^{\DM} (y) V_> (xy) 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_> (xy) 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(xy)  \vert xy \vert^{1}] dy
&=
\int \uprho^{\DM} (y) \exp (\vert xy \vert \delta^{1}) \vert xy
\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 xy \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}$$
We now make the choices $\delta = B^{2/3} Z^{1/3}, \ a = Z^{5/3}
[1 + \ln
(1 + BZ^{3})]^{1}$ and $\alpha = Z^{1/3}$. Recalling that $N =
\lambda Z$ we arrive at (7.6). This finishes the proof of Theorem
7.1.
\lanbox
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%
\bigskip
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
{\bf VIII. LIMIT OF QUANTUM MECHANICS}
In this section we shall complete the proof of the main result
that
the quantum energy $E^{\rm Q} (N,Z,B)$ can be approximated by the energy
$E^{\DM}
(N,Z,B)$. Unlike the semiclassical ThomasFermi approximation
studied in
our companion paper [28] the approximation of $E^{\rm Q}$ by $E^{\DM}$ is
quite
simple after having made the reduction to a onebody problem.
{\it Proof of Theorem~1.3
(Energy asymptotics for regions 3, 4, 5):\/}
Since Theorem 5.1 contains the appropriate upper bound on $E^Q$
we here
only have to address the question of a lower bound. Furthermore it
follows from Theorem~1.2 on the confinement to the lowest Landau
band
that we can replace $E^{\rm Q}$ by the ground state energy
$E^{\rm Q}f_{\rm
conf}$
of the confined Hamiltonian $\Pi^N_0 H_N \Pi^N_0$.
If $\Pi^N_0 \Psi = \Psi$ we know from Lemma~4.1 that
$\Gamma^{\Psi} \in
\G^{\DM}_B$. {F}rom Theorem 7.1 we also know that for such a
normalized
function $\Psi$
$$( \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)$$
In Theorem~5.3 we proved that
$E^{\rm Q}(N,Z,B) \leq  c_\lambda (Z^{9/5} B^{2/5} + 1) $
and, indeed this bound is of the correct order if $B \leq Z^3$. If we
compare this bound with the error
$R_{\rm L}$ in (8.4) we see that only when $B
%zzz
\gg Z^{13/6}$ can $R_{\rm L}$ be negligible. When the magnetic field is
in the
range $Z^{4/3} \ll B \lo Z^{13/6}$ we must use the standard
exchange
estimate (7.3) rather than (7.6).
If $\Psi$ is a normalized wave function with $\Pi^N_0 \Psi = \Psi$
we can
easily estimate the error in (7.3). In fact, from the LiebThirring
inequality (4.8)
$$\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)$$
which, when combined with (3.7), gives Theorem~1.4.
If we appeal to the strong ThomasFermi energy $E^{\STF}$ defined
in [7]
and [28]
we can also give the asymptotics for small values of $B/Z^3$.
\item{4)} If $N/Z = \lambda, B/Z^3 \rightarrow 0$ but $B/Z^{4/3}
\rightarrow \infty$ as $Z \rightarrow \infty$ then
$$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)$$
as $Z \rightarrow \infty$.
To prove (8.14) it is enough to consider $W\in C^{\infty}_0(\R^3)$.
To arrive at this conclusion we first notice that it
is clearly enough (see also Lemma III.4 in [30])
to consider test functions $W$ that are characteristic functions
of measurable sets. By outer regularity of the Lebesgue measure
we can restrict our attention to open sets. However, let $f_n\in
L^1_{\loc}(\R^3)$
be nonnegative functions satisfying
$\int f_n\widetilde W\to \int f\widetilde W$ for all
functions $\widetilde W\in C^{\infty}_0(\R^3)$ only.
If now $W$ is a characteristic function of an open set we can
choose functions
$W_\pm\in C^{\infty}_0(\R^3)$ such that $W_\leq W\leq W_+$ and
such that
$\int fW_\pm$ are as close to $\int fW$ as we please. Then for all
$\delta>0$ we
can choose $n$ so large that
$$
\int fW_\delta\leq \int f_nW_\leq \int f_nW \leq \int
f_nW_+
\leq \int fW_+ +\delta,
$$
and the convergence weakly in $L^1_{\loc}$ follows. It is important
that the functions $f_n$ are nonnegative for otherwise the
conclusion would
be wrong!
Weak convergence in $L^1(\R^3)$, which means that we consider
all $W\in L^\infty (\R^3)$ without restriction on the support,
follows (see again Lemma III.4 in [30])
if we have both weak convergence in $L^1_{\loc}$ and convergence
for the constant function $W=1$. This last condition simply means
that
$\lambda=N/Z$ (which is held fixed) converges to
$\int\uprho(x;\lambda,1,2\pi\eta)dx$.
This is of course true if $N\leq N^{\DM}_c$.
The standard method [30] of proving (8.14) is to introduce the
following perturbed Hamiltonian
$$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}Zx^{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}Zx^{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)}.
$$
The first term above was estimated in Proposition~4.9 the second
can be estimated
by the onedimensional LiebThirring inequality to be less than
$$
{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.$$
Thus (8.14) follows from (8.17) which is a consequence of (8.15).
\lanbox
Notice that the scaling in (8.13) is identical to the scaling
(3.11) introduced to study the hyperstrong limit.
In particular, if $W$ is independent of $x_\bot$, i.e.,
$W(x)=W(x_3)$,
on the support of $\uprho^{\DM}$
we have from Theorem~3.5 and Theorem~4.6
that the right side of (8.14) converges to
$\int_{\R}\uprho^{\HS}(x_3)W(x_3)dx_3$
as $\eta\to\infty$. This proves the following result.
{\bf 8.2. THEOREM.} {\it Let $\uprho^{\HS} (x) = \uprho^{\HS} (x;
\lambda)$ be the minimizer for the hyperstrong functional given
in Theorem~3.1
and let $N=\lambda Z$.
Define the onedimensional quantum density by
$$\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 nonspherical
\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 hyperstrong theory. We shall now
use the hyperstrong 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 $N1$ 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 (N1, 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.$$
Here, as usual, $E^\Q(N, Z, B)$ is the energy of an atom with nuclear
charge $Z$.}
{\it Proof:} We will derive upper and lower bounds of the necessary
accuracy.
{\it The lower bound\/} is trivial. For any choice of $R_1, \dots ,
R_K$ and electronic wave function $\Psi$, with spin included, we can
define the electron density $\uprho_\Psi$ as in (1.3) and then define
the potential $\Phi = \vert x \vert^{1} * \uprho_\Psi$. Let $x_0 \in
\R^3$ be such that $\Phi (x_0) = \sup_x \Phi (x)$. Clearly, the
attractive electronnuclear energy (see (i) above) is minimized if we
now set $R_1 = R_2 = \dots = R_K = x_0$. At the same time, for a lower
bound we can omit the positive nuclear repulsion (see (ii) above). The
result is then $(\Psi, H^{(K)}_N \Psi) \geq (\Psi, H_N \Psi)$, where
$H^{(K)}_N$ is our Hamiltonian for $K$ nuclei, and $H_N$ is the one
nucleus Hamiltonian (1.1) with a nucleus of charge $Z$ located at
$x_0$. By translation covariance, we can move $x_0$ to $0 \in \R^3$
without changing the energy. Thus, {\it for any\/} $Z_i$'s
$$E^\Q (N,Z_1, \dots , Z_K, B) \geq E^\Q (N,Z,B).$$
{\it The upper bound\/} is a bit more complicated. We use Lieb's
variational principle [50] (cf. Sect.~V) and take as variational
density matrix the one given in (5.4) for an atom with electron number
$N$ and nuclear charge $Z = \sum \nolimits^K_{j=1} Z_j$ (located at the
origin, of course). Since we consider large $B/Z^3$ the density matrix
in (5.4) is of the form given in (4.29) with density $\uprho^{\SS}
(x)$. Next, we place our $K$ nuclei on the $x_\bot = 0$ axis, close to
the origin and with a spacing $D = Z^{1} (\ln \eta)^{\varepsilon}$
with $\eta = B/(2\pi Z^3)$. Any fixed $1<\varepsilon <2 $ will
do. The nuclear coordinates are thus $(0,0, \pm D), \ (0,0,\pm 2D)$,
etc. The nuclear repulsion can be bounded by $\sum Z_i Z_j \vert R_i 
R_j \vert^{1} \leq \sum Z_i Z_j D^{1} \leq \mfr1/2 Z^2 D^{1} \leq
Z^3 (\ln \eta)^\varepsilon$. By Theorem 1.4, therefore, the repulsive
energy in our state is of lower order  by a factor $(\ln
\eta)^{\varepsilon 2}$.
The remaining step is to show that the negative electronnuclei energy
(see (i) above) is, to leading order, the same as that for one nucleus
of charge $Z$ placed at the origin $(0,0,0)$. We use Proposition 3.3.
If $R$ is a point on the axis $x_\bot = 0$ we get, using the scalings
(3.11) and (3.14), that $\phi(R):= \int Z\vert x  R
\vert^{1} \uprho^{\SS} (x) dx = Z^3 (\ln \eta)^2 \int V_\eta (x)
\uprho^{\SS}_\eta (x + Z\ln \eta R) dx$. Thus from Proposition 3.3
with $\eta> 3$
$$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. Therefore, $\sum_j (Z_j/Z)\phi(R_j)$ equals
$\phi(0)$ to leading order. ~\lanbox
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%\def\ref#1{\par\vskip 12pt \noindent \hangafter=1 \hangindent 22.76pt[#1]}
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{\baselineskip=3ex\eightpoint\smallskip
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\def\it{\eightit}
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}
\end