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Many-body Schroedinger operators, strong magnetic fields
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\begin{document}
\markboth{\scriptsize{BSY 19/11/99}}{\scriptsize{BSY 19/11/99}}
\title{\bf{Atoms in strong magnetic fields:\\The high field limit \\at
fixed nuclear charge}}
\author{\vspace{5pt} Bernhard Baumgartner$^{1}$, Jan Philip Solovej$^{2}$,
and Jakob
Yngvason$^{1}$\\
\vspace{-4pt}\small{$1.$ Institut f\"ur Theoretische Physik, Universit\"at
Wien}\\
\small{Boltzmanngasse 5, A-1090 Vienna, Austria}\\
\vspace{-4pt}\small{$2.$ Department of Mathematics, University of
Copenhagen}\\
\small{Universitetsparken 5, DK-2100 Copenhagen \O, Denmark}}
\date{\small\today}
\maketitle
\begin{abstract}
Let $E(B,Z,N)$ denote the ground state energy of an atom with $N$
electrons and nuclear charge $Z$ in a homogeneous magnetic field $B$.
We study the asymptotics of $E(B,Z,N)$ as $B\to \infty$ with $N$ and
$Z$ fixed but arbitrary. It is shown that the leading term has the
form $(\ln B)^2 e(Z,N)$, where $e(Z,N)$ is the ground state energy of
a system of $N$ {\em bosons} with delta interactions in {\em one}
dimension. This extends and refines previously known results for
$N=1$ on the one hand, and $N,Z\to\infty$ with $B/Z^3\to\infty$ on the
other hand.
\end{abstract}
\section{Introduction}
The effects of extremely strong magnetic fields (order of $10^9$ Gauss and
higher) on atoms and molecules are of considerable astrophysical as well as
mathematical interest and are far from being completely understood in spite of
many theoretical studies since the early seventies. We refer to \cite{LSY94}
and
\cite{RWHG} for a general discussion of this subject and extensive lists of
references.
An atom (ion) with $N$ electrons and nuclear
charge $Z$ in a homogeneous magnetic field ${\bf B}=(0,0,B)$ is (in
appropriate
units) usually
modeled by the nonrelativistic many-body Hamiltonian
\begin{equation}H_{B,Z,N}=\sum_{i=1}^N \left(H^{(i)}_{\bf A}-{Z\over \vert
x_{i}\vert}\right)+\sum_{i1$. The relevance of the $\delta$-function model for the ground
state of hydrogen in strong magnetic fields was noted already in \cite{Sp76}.
We also verify that the mean field limit of
$e(Z,N)$ agrees with (\ref{hs}):
\begin{thm}[Mean field limit]\label{2}If $Z,N\to\infty$ with $\lambda=N/Z$
fixed, then
\beq
e(Z,N)/Z^3\to \left\{ \begin{array}{l@{\quad}l}
-\mfr1/4\lambda+\mfr1/8\lambda^2
-\mfr1/{48}\lambda^3 & {\rm if}\quad \lambda< 2\\ \\-\mfr1/6 & {\rm if}\quad
\lambda\geq 2
\end{array}
\right.
\eeq
\end{thm}
Taken together, Theorems \ref{1} and \ref{2} lead to the same high $B$, high
$Z$ limit as Theorem 1.4
in
\cite{LSY94}, where $Z\to\infty$ and $B/Z^3\to\infty$ simultaneously
(the \lq\lq hyper-strong\rq\rq\ limit.)
We now describe briefly the strategy for the proof of these results and
introduce some notation that will be used throughout. The first step in the
proof of Theorem \ref{1} is a reduction to
the subspace ${\cal H}_{N}^0\subset
{\cal H}_{N}$ generated by wave functions in the
lowest Landau band. Let $\Pi^0_N$ denote the projector on ${\cal H}_{N}^0$.
(Its integral kernel is given by Eqs. (\ref{intkern})--(\ref{intkern2}) in
Section 5. Note that $\Pi^0_N$ and ${\cal H}_{N}^0$ depend on $B$.)
Let $E^{\rm conf}(B,Z,N)$ denote the ground state energy of
$\Pi^0_N H_{B,Z,N}\Pi^0_N$.
It is clear that
\beq E(B,Z,N)\leq E^{\rm conf}(B,Z,N),\eeq
and by Theorem 1.2 in \cite{LSY94},
\beq E^{\rm conf}(B,Z,N)\leq E(B,Z,N) (1-\delta(B,Z,N))\label{conf}\eeq
where $\delta(B,Z,N)\to 0$ for $B\to\infty$ with $Z,N$ fixed. Hence it
suffices to prove (\ref{limit}) with $E(B,Z,N)$ replaced by $E^{\rm
conf}(B,Z,N)$.
We note in passing that (\ref{conf}) also holds for bosons. In fact, it will
become evident in the sequel that Theorem \ref{1} is independent of the
statistics of the particles.
To study $E^{\rm conf}(B,Z,N)$ the next step is to introduce
a Hamiltonian for the motion parallel
to the magnetic field with the coordinates perpendicular to the
magnetic field
as parameters.
We write the variables $x_{i}\in\R^3$ as
$x_{i}=(x^\perp_{i},z_{i})$, where $ x^\perp_{i}\in\R^2$ and
$z_{i}\in \R$ are respectively the components perpendicular and
parallel to the field.
Moreover, we write $(x_{1},\dots,x_{N})=(\xoperp,\underline z)$
with $\xoperp=(x^\perp_1,\dots,x^\perp_N)\in \R^{2N}$ and
$\underline z=(z_{1},\dots,z_{N})\in\R^N$.
In the lowest Landau band the part of (\ref{ha}) associated with the motion
perpendicular to the field is exactly canceled by the spin contribution and
only the
part corresponding to the motion along the field remains. Hence
\beq \Pi^0_N H_{B,Z,N}\Pi^0_N=\Pi^0_N H_{Z,N}\Pi^0_N\eeq
with
\beq H_{Z,N}=\sum_{i=1}^N \left(-\partial^2/\partial z_{i}^2-{Z\over
\vert
x_{i}\vert}\right)+\sum_{i0$) defined as
\beq V_{B,r}(z)={L(B)}^{-1}(B^{-1}L(B)^2r^2+z^2)^{-1/2}.\label{vbr}
\eeq
Let $E_{Z,N}(\xoperp)$ and $e^B_{Z,N}(\yoperp)$ denote the ground
state energies of $H_{Z,N}(\xoperp)$ and $h^B_{Z,N}(\yoperp)$ respectively. In
order to avoid discussions about the domains of the Hamiltonians, which in
fact
depend on whether some of the parameters $x^\perp_i$ (resp.\ $y^\perp_i$)
coincide,
we define the ground state energies in terms of quadratic forms in the same
way
as (\ref{ezn}):
\beqa E_{Z,N}(\xoperp)&=&\inf \{\langle\Psi,H_{Z,N}(\xoperp)\Psi\rangle\,:\,
\Psi\in
C^{\infty}_{0}(\R^{N}),\, \Vert\Psi\Vert_{2}=1\},\\
e^B_{Z,N}(\yoperp)&=&\inf \{\langle\Psi,h^B_{Z,N}(\yoperp)\Psi\rangle\,:\,
\Psi\in
C^{\infty}_{0}(\R^{N}),\, \Vert\Psi\Vert_{2}=1\}.
\eeqa
These energies are connected by the scaling relation
\beq
E_{Z,N}(B^{-1/2}\yoperp)/L(B)^{2}=e^B_{Z,N}(\yoperp).\label{scal}\eeq
In the next section we show that with the choice $L(B)\sim \ln B$ the
potential
$V_{B,r}(z)$ converges for each $r>0$ in the sense of distributions to the
delta
function as $B\to 0$. This is the heuristic basis of Theorem \ref{1}. Since
the
convergence is not uniform in $r$, however, more is needed for a rigorous
proof.
In particular, one needs estimates on the $r$-dependence of the convergence
$V_{B,r}(z)\to \delta(z)$. These estimates, stated in Lemmas \ref{lm:delta1}
and \ref{cl:delta2} in the next section, can be regarded as variants
of Propositions 3.3 and 3.4 in \cite{LSY94} and the Appendix in \cite{JY},
adapted to the problem at hand. They are included here for completeness.
The upper bound on the energy, given in Section 3, is a straight-forward
variational calculation. The lower bound is more subtle.
An important ingredient needed is the superharmonicity of the energy
$E_{Z,N}(\xoperp)$ in the variables $x^\perp_i$.
This result, established in Theorem \ref{thm:super}, generalizes a
corresponding
result
(Proposition 2.3) in \cite{LSY94}.
Superharmonicity implies that the lowest value of $E_{Z,N}(B^{-1/2}\yoperp)$
for $|y^\perp_i|\geq \varepsilon$ with $\varepsilon>0$ is obtained at the
boundary of the variable range, i.e., when either $|y^\perp_i|= \varepsilon$
or
$|y^\perp_i|\to\infty$.
Variables tending to infinity can be ignored, since $V_{B,r}(z)\to 0$ for
$r\to\infty$, so by this result one may in (\ref{exp}) restrict the attention
to
wave functions localized where $|x^\perp_i|\leq \hbox{(\rm const.)}B^{-1/2}$.
On
the other hand, the requirement that only wave functions in the lowest
Landau band are taken into account in (\ref{exp}) plays the role of a 'hard
core
condition' that prevents collapse, since such wave functions cannot be
concentrated on shorter scales than $O(B^{-1/2})$. This statement is made
precise in Lemma \ref{5.3}.
The lower bound is obtained in Section 5 by combining Theorem \ref{thm:super},
Lemma \ref{5.3} and the convergence of the potentials $V_{B,r}$ discussed in
Section 2. It is noteworthy that this lower bound
holds also for bosonic statistics while the upper bound holds for fermionic
statistics, so that altogether the convergence of $E(Z,N,B)/(\ln B)^2$ to
$e(Z,N)$ is independent of the statistics.
In section 6 we discuss the delta-function model (\ref{dh}) and in
particular prove Theorem \ref{2}. In the course of the proof we compare
(\ref{dh}) with another model, whose ground state energy can be explicitly
calculated. This model provides an upper bound
for the ground state energy of (\ref{dh}) and has the same mean field limit.
The Hamiltonian for this model is
\begin{equation}\label{symm}
\widetilde{h}_{Z,N}=\sum_{i=1}^{N}\left( p_{i}^{2}-\delta
(z_{i})\right) +\frac{1}{2Z}\sum_{i0$ is small enough,
where $c_{d+1}$ is the volume of the unit ball in $\R^{d+1}$.
For $x\in U$ it follows from the
lower semicontinuity of $f$ that
we have either $g(x)=b$ or there exists $t\in\R$ such that
$g(x)=f(x,t)$. In the first case we obviously have
\begin{eqnarray}\label{eq:g}
c_{d+1}r^{d+1}g(x)
\geq 2\int_{|x-y|\leq r}g(y)\sqrt{r^2-(x-y)^2}dy
\end{eqnarray}
since $g(y)\leq b$ for all $y$.
If $g(x)**0$
and in fact this limit holds in the topology of
$C^\infty_0(U)$.
Thus if $\phi\geq0$
we have
$$
\int_U g(x)\Delta\phi(x)dx
=C^{-1}\lim_{r\to0}r^{-(d+3)}
\int_{|x-y|\leq r} g(x)
(\phi(y)-\phi(x))\sqrt{r^2-(x-y)^2}dydx\leq0
$$
by the inequality (\ref{eq:g}). Hence $\Delta g\leq 0$.
\end{proof}
\begin{thm}[Superharmonicity of the energy]\label{thm:super}
On the set ${\cal A}$ defined in (\ref{eq:aset}) the function $\xoperp\mapsto
E_{Z,N}(\xoperp)$ is
superharmonic in each of the variables
$x^\perp_i$, $i=1,\ldots,N$
independently.
\end{thm}
\begin{proof}
We follow closely the proof of
Prop.~2.3 in \cite{LSY94}, which stated the superharmonicity of
the ground state energy of a one-body operator which
can be considered as a mean field approximation of $H_{Z,N}(\xoperp)$.
It is clearly enough to prove that $E_{Z,N}(\xoperp)$ is
superharmonic in $x^\perp_1$ (on the region $x^\perp_1\ne0$)
for $x^\perp_2,\ldots,
x^\perp_N$ fixed.
We shall prove this by showing that $x^\perp_1\mapsto
E_{Z,N}(\xoperp)$
satisfies the mean value inequality around any given point
$x^\perp_{1,0}$. Let $\xoperp_0=(x^\perp_{1,0},x^\perp_2,\ldots,
x^\perp_N)$.
Choose a sequence of $L^2$ normalized functions
$\Psi_n\in C^\infty_0(\R^N)$ such that
$\langle\Psi_n,H_{Z,N}(\xoperp_0)\Psi_n\rangle\to E_{Z,N}(\xoperp_0)$ as
$n\to\infty$.
For $w\in \R$ denote by $\Psi^{(w)}_n$ the function
$$
\Psi^{(w)}_n(z_1,\ldots,z_N)=\Psi_n(z_1-w,z_2,\ldots,z_N).
$$
We clearly have
$$
\inf_{w\in\R}
\langle\Psi^{(w)}_n,H_{Z,N}(\xoperp_0)\Psi^{(w)}_n\rangle\to
E_{Z,N}(\xoperp_0)
\quad\text{as }n\to\infty.
$$
If $x^\perp_1$ is close to $x^\perp_{1,0}$ we shall use
$\Psi^{(w)}_{n}$ as a trial function for $H(\xoperp)$.
We then obtain
$$
E_{Z,N}(\xoperp)\leq \liminf_n\inf_{w\in\R}
\langle\Psi^{(w)}_n,H_{Z,N}(\xoperp)\Psi^{(w)}_n\rangle.
$$
Hence
\begin{equation}\label{eq:wnbound}
E_{Z,N}(\xoperp)-E_{Z,N}(\xoperp_0)\leq \liminf_n\left[
\inf_{w\in\R}\langle\Psi^{(w)}_n,H_{Z,N}(\xoperp)\Psi^{(w)}_n\rangle
-\inf_{v\in\R}\langle\Psi^{(v)}_n,H_{Z,N}(\xoperp_0)\Psi^{(v)}_n\rangle\right]
.
\end{equation}
The potential appearing in $H_{Z,N}(\underline{x}^\perp)$,
i.e.,
$$
W_{Z,N,\xoperp}(z_1,\ldots,z_N)=-\sum_{i=1}^N{Z\over \sqrt{
z_i^2+(x^\perp_i)^2}}+\sum_{i0$ small enough.
\end{proof}
\section{Lower Bound}
The first lemma in this section concerns the ground state energy
$e^B_{Z,N}(\yoperp)$ of $h^B_{Z,N}(\yoperp)$ and does not use
superharmonicity.
\begin{lm}[Lower bound on $e^B_{Z,N}(\yoperp)$] \label{lm:eblower} Let ${\cal
K}$
be a compact subset of
the set
${\cal A}$
given in (\ref{eq:aset}). Then
\beq \liminf_{B\to\infty}\inf_{\yoperp\in{\cal K}}
e^B_{Z,N}(\yoperp)\geq e(Z,N).\eeq
\end{lm}
\begin{proof} To avoid problems at points $\yoperp$ with
$y_{i}^\perp-y_{j}^\perp=0$ for some $i,j$, we replace the repulsive
potential $V_{B,|y_{i}^\perp-y_{j}^\perp|}(z_{i}-z_{j})$ by the
smaller potential $V_{B,|y_{i}^\perp-y_{j}^\perp|+1}(z_{i}-z_{j})$.
We denote the
corresponding Hamiltonian by $\tilde h^B_{Z,N}(\yoperp)$ and its
ground
state energy by $\tilde e^B_{Z,N}(\yoperp)$. It is obvious that
$e^B_{Z,N}(\yoperp)\geq \tilde e^B_{Z,N}(\yoperp)$, so a lower bound
on
$\tilde
e^B_{Z,N}(\yoperp)$ gives a lower bound on $e^B_{Z,N}(\yoperp)$.
Let $\Psi$ be a normalized, symmetric wavefunction in $C^\infty_0({\R}^N)$.
Since $\langle \Psi,h_{Z,N}
\Psi\rangle\geq e(Z,N)$ we have to estimate the matrix
elements of the difference $\tilde h^B_{Z,N}(\yoperp)-h_{Z,N}$. Using
Lemmas \ref{lm:delta1} and \ref{cl:delta2}, together with the
H\"older inequality for the integration over $z_2,\dots,z_N$ and
$z_3,\dots,z_N$ respectively, we obtain
\beqa
\left |\langle \Psi, \tilde h^B_{Z,N}
\Psi\rangle-\langle \Psi,h_{Z,N}
\Psi\rangle\right|\leq L(B)^{-1}(ZN+N(N-1))\nonumber \\ \times
\left[r_{\rm min}^{-1}+8{} T_\Psi^{3/4} (2r_{\rm max}+1)^{1/2}
\right]\label{error}
\eeqa
where $r_{\rm min}$ and $r_{\rm max}$ are respectively the minimum and
the maximum value of $|y^\perp_i|$, $i=1,\dots,N$, with $\yoperp\in{\cal K}$,
and
\beq T_\Psi=N\int|\partial_{z}
\Psi(z,z_2,\dots,z_N)|^2dz dz_2\cdots dz_N\eeq
is the kinetic energy of $\Psi$. Now if $\Psi^{B}_{\yoperp,n}$,
$n=1,2,\dots$ is a minimizing sequence of
normalized wave functions for $\tilde h^B_{Z,N}(\yoperp)$, then we may assume
that the
corresponding kinetic energy is uniformly bounded in $n$,
$B$ and $\yoperp\in{\cal K}$.
In fact, we may assume that $\langle\Psi^{B}_{\yoperp,n},
\tilde h^B_{Z,N}(\yoperp)\Psi^{B}_{\yoperp,n}\rangle$
is a bounded sequence. If we use the bound
from Lemma 2.1 in [LSYa], we obtain
\beq
\langle\Psi^{B}_{\yoperp,n},
\tilde h^B_{Z,N}(\yoperp)\Psi^{B}_{\yoperp,n}\rangle
\geq \frac{T_n}{2}-\frac{1}{2}\left(\frac{2Z}{L(B)}\right)^2
\left(1+\left[\sinh^{-1}\{(2Z)^{-1}B^{1/2}\}\right]^2\right),
\eeq
where we have saved half of the
the kinetic energy $T_n$ of $\Psi^{B}_{\yoperp,n}$.
For large $B$, $L(B)^{-1}\left[\sinh^{-1}\{(2Z)^{-1}B^{1/2}\}\right]$
is bounded and hence we see that $T_n$ is bounded. The error term
(\ref{error})
with $\Psi=\Psi^{B}_{\yoperp,n}$ thus tends to zero as $B\to\infty$,
uniformly in $n$, and the lemma is established.
\end{proof}
\begin{lm}[Uniform bounds on $E_{Z,N}(\xoperp)$] \label{5.2}Let
$\varepsilon>0$.
Consider the set
\beq
{\cal C}^{B,\varepsilon}=\{\xoperp\ :\ \varepsilon B^{-1/2}\leq\
|x^\perp_i|,
\text{ for all }i=1,\ldots,N\}.\label{cbe}
\eeq
Then
\beq
\liminf_{B\to\infty}(\ln B)^{-2}\inf\{E_{Z,N}(\xoperp)\ : \
\xoperp\in {\cal C}^{B,\varepsilon}\}\geq e(Z,N).\label{unif}
\eeq
where $e(Z,N)$ as before denotes the 1-dimensional delta function atom
energy.
\end{lm}
\begin{proof}
Define the sets
$$
{\cal C}^{B,\varepsilon}_n=\{\xoperp\ :\ \varepsilon
B^{-1/2}\leq\
|x^\perp_i|\leq n,
\text{ for all }i=1,\ldots,N\}.
$$
Since ${\cal C}^{B,\varepsilon}_n$ is compact and $ E_{Z,N}$ is lower
semicontinuous (being superharmonic, in fact, superharmonic in each
variable) we may find
$\xoperp_n\in {\cal C}^{B,\varepsilon}_n$
such that
$$
E_{Z,N}(\xoperp_n)=\min\{E_{Z,N}(\xoperp)\ : \
\xoperp\in {\cal C}^{B,\varepsilon}_n\}.
$$
Clearly,
$$
\lim_{n\to\infty} E_{Z,N}(\xoperp_n)\to
\inf\{E_{Z,N}(\xoperp)\ : \
\xoperp\in {\cal C}^{B,\varepsilon}\}.
$$
By the superharmonicity of $E_{Z,N}(\xoperp)$ in each variable
$x^\perp_i$ we know that each coordinate
$x^\perp_{i,n}$ of the point $\xoperp_n$ satisfies either
$|x^\perp_{i,n}|=\varepsilon B^{-1/2}$ or
$|x^\perp_{i,n}|=n$. Moreover, since $E_{Z,N}(\xoperp)$ is invariant under
permutations of the
coordinates of $\xoperp$ we may assume that $|x^\perp_{1,n}|\leq
|x^\perp_{2,n}|\leq \dots\leq |x^\perp_{N,n}|$ for all $n$.
By possibly going to a subsequence
we may assume that there exists an integer $0\leq K\leq N$ such that
for $n$ large enough
$$
|x^\perp_{i,n}|=\left\{
\begin{array}{ll}\varepsilon B^{-1/2},&\text{ for }i=1,\ldots,K\\
n,&\text{ for }i>K\\
\end{array}\right. .
$$
Moreover, we may assume that $ x^\perp_{i,n}$ converges as
$n\to\infty$ for $i=1,\ldots,K$.
Since we may ignore the variables $x^\perp_{i,n}$, $i=K+1,\ldots,N$,
which tend to infinity we have
$$
\lim_{n\to\infty}E_{Z,N}(\xoperp_n)/E_{Z,K}(x^\perp_{1,n},\ldots,
x^\perp_{K,n})=1
$$
Since $E_{Z,K}(\xoperp)$ is lower
semicontinuous we conclude
that there exists a point $(x^\perp_{1,\infty},\ldots,
x^\perp_{K,\infty})\in\R^{2K}$ with
$|x^\perp_{i,\infty}|= \varepsilon B^{-1/2}$ for all $i=1,\ldots,K$
such that
$$
\inf\{E_{Z,N}(\xoperp)\ : \
\xoperp\in {\cal C}^{B,\varepsilon}\}=
E_{Z,K}(x^\perp_{1,\infty},\ldots,
x^\perp_{K,\infty}).
$$
By Lemma~\ref{lm:eblower} we have that
$$
\liminf_{B\to\infty}
\inf\{L(B)^{-2}E_{Z,K}(B^{-1/2}y^\perp_{1},\ldots,
B^{-1/2}y^\perp_K)\ : \ |y^\perp_{i}|= \varepsilon,
\text{ for all }i\}\geq e(Z,K).
$$
Since $K\leq N$ and hence $e(Z,K)\geq e(Z,N)$ we have proved
the lemma.
\end{proof}
\begin{lm}[Wave functions in the lowest Landau band]\label{5.3}
If
$\Psi\in{\cal H}_N^0$ belongs to the lowest Landau band at field strength $B$,
then
$\int|\Psi(\xoperp,\underline{z})|^2
d\underline{z}$ is a bounded function of $\xoperp$ (possibly after a
modification on a null set) and for all $1\leq n\leq N$
\beq \label{lbound}
\sup_{x_1^\perp,\dots,x_n^\perp}\Big|\int\int|
\Psi(\xoperp,\underline{z})|^2
d\underline{z}dx_{n+
1}^\perp\cdots dx_{N}^\perp\Big |\leq \frac {B^n}{(2\pi)^n} \Vert
\Psi\Vert^2\eeq
\end{lm}
\begin{proof} The projector $\Pi^0_N$ on the lowest Landau band is the $N$-th
tensorial
power of the projector $\Pi^0$ that operates on $L^2({\mathbb R}^3;{\mathbb
C}^2)$ and is given by the integral kernel
\beq\label{intkern}\Pi^0(x,x')=\Pi_\perp^0(x^\perp,{x'}^\perp)\delta(z-z')
P^{\downarrow},
\eeq
where
\beq\label{intkern2}
\Pi_\perp^0(x^\perp,{x'}^\perp)=\frac B{2\pi}\exp\left\{\mfr{\rm
i}/2(x^\perp\times
{x'}^\perp)\cdot{\bf B}-\mfr 1/4(x^\perp-{x'}^\perp)^2B\right\}
\eeq
and $P^{\downarrow}$ is the the projector on vectors in ${\mathbb C}^2$ with
spin component $-1/2$. The kernel $\Pi_\perp^0(x^\perp,{x'}^\perp)$ is a
continuous function with
\beq\label{proj}\int
\Pi^0(x^\perp,u^\perp)\Pi^0(u^\perp,y^\perp)du^\perp=\Pi^0(x^\perp,y^\perp)
\eeq
and
\beq\label{bd}
\Pi^0(x^\perp,x^\perp)=\frac B{2\pi}
\eeq
for all $x^\perp$.
A wave function in the lowest Landau band has the representation
$\Psi=\Pi^0_N\Psi$. After writing $\Pi^0_N$ as an integral operator
(\ref{lbound}) follows from the Cauchy-Schwarz inequality, using (\ref{proj})
and (\ref{bd}).
\end{proof}
\begin{proposition}[Lower bound]
\beq\liminf_{B\to\infty}\frac{E(B,Z,N)}{(\ln B)^2}\geq e(Z,N).\eeq
\end{proposition}
\begin{proof} For fixed $B$ let $\Psi$ be a normalized wave function
in the
lowest Landau band. By (\ref{exp}) we have
\beq\langle\Psi,H_{Z,N}\Psi\rangle\geq\int E_{Z,N}(\xoperp)
\left(\int|\Psi(\xoperp,\underline
z)|^2d\underline z\right)d\xoperp.
\eeq
We split the integral over $\xoperp$ into an integral over ${\cal
C}^{B,\varepsilon}$ (defined in (\ref{cbe})) and its complement in
$\R^{2N}$. By Lemma
\ref{5.2}
we have only to
consider the latter. Using the estimate (\ref{eq:Ebounds}) the task is to
bound
terms
of the form
\beq\int_{|x^\perp_i|\leq \varepsilon B^{-1/2}}
(1+[\sinh^{-1}(Z|x_j^\perp|^{-1})]^2)\left(\int|\Psi(\xoperp,\underline
z)|^2d\underline z\right)d\xoperp
\eeq
from above. If $i=j$ we carry out the integration over all $x_k^\perp$
with $k\neq i$ and use Lemma \ref{5.3} for the remaining variable
$x^\perp_i$. For small $r$, $|\sinh^{-1} r^{-1}|\leq
{\rm(const.)}|\ln r|$
and the term can be estimated by
\beq {\rm(const.)}\int_{|x^\perp|\leq \varepsilon B^{-1/2}}(\ln
|x^\perp|)^2 B dx^\perp\leq {\rm(const.)}\varepsilon^2 (\ln B)^2.
\eeq
For $i\neq j$ we split the integration over $x^\perp_j$ into two
parts, namely $|x^\perp_j|\leq B^{-1/2}$ and $|x^\perp_j|\geq
B^{-1/2}$.
For the first part we obtain the following bound, after transforming variables
and using Lemma 5.3, this time for $n=2$,
\beq {\rm(const.)}\varepsilon^2\int_{|y^\perp_i|\leq 1,|y^\perp_j|\leq 1
}(\ln B^{-1/2}
|y_j^\perp|)^2 dy^\perp_i dy^\perp_j\leq {\rm(const.)}\varepsilon^2
(\ln B)^2.
\eeq
For the integral over $|x^\perp_j|\geq B^{-1/2}$ we
estimate $|\sinh^{-1}( Z|x^\perp_j|^{-1})|^2$ by its maximum
value,
$\leq {\rm(const.)}(\ln B)^2$ and obtain
for this part of the integral the upper bound
\beqa
\int_{|x^\perp_i|<\varepsilon B^{-1/2}}
(1+{\rm(const.)}(\ln B)^2)\left(\int|\Psi(\xoperp,\underline
z)|^2d\underline z\right)d\xoperp\nonumber\\
\leq \int_{|x^\perp_i|<\varepsilon B^{-1/2}}
(1+{\rm(const.)}(\ln B)^2)Bd x^\perp_i\leq
\varepsilon^2(1+c(\ln B)^2),
\eeqa
where we have used Lemma~\ref{5.3} again.
We see that (55) is bounded
above by ${\rm(const.)}(\varepsilon\ln B)^2$, for $B$ large enough.
Since $\varepsilon>0$ is arbitrary this completes the
proof.
\end{proof}
\section{The one-dimensional delta-function model}
We now want to study the the delta-function Hamiltonian
(\ref{dh}), in particular its mean field limit,
$N\to \infty $, $
Z\to \infty $, with $ \lambda =N/Z$ fixed.
For this it is convenient to make a scale transformation $z \rightarrow
z/Z$, which implies a unitary equivalence
\begin{equation}
h_{Z,N} \cong Z^2 \widehat{h}_{Z,N}
\end{equation}
with
\begin{equation}
\widehat{h}_{Z,N} = \sum_{i=1}^N \left( p_i^2 - \delta(z_i)\right) +
\frac{1%
}{Z} \sum_{i f^{\prime}(z)/f(z)$. This inequality can be
integrated to give $g(z)/g(0) > f(z)/f(0)$, a contradiction
to the assumption of the square-integrability of g(z).
Therefore we know that the Hamiltonian (\ref{htrb}) has
no negative eigenvalue. And so the operator inequality holds.
\end{proof}
The $\{W_b(z)\}$ and hence $\{ Z w_{Z,a,b}(z)\}$ are $\delta$-sequences in
the limit $b \rightarrow \infty$. All these functions are positive
definite,
and finite at the origin:
\begin{equation}
w_{Z,a,b}(0) < \frac{b}{2 Z^2\: a}.
\end{equation}
With this tool we can now deduce the lower bound for the many body
Hamiltonian:
\begin{proposition}
[Lower bound in the mean field limit] If $N,Z\to\infty$ with $\lambda=N/Z$
fixed, then
\beq\liminf e(Z,N)/Z^3\geq E^{\rm HS}(\lambda).\eeq
\end{proposition}
\begin{proof}
We use the operator inequality (\ref{opiq}) with $w_Z(z) := w_{Z,a,b}(z)$,
($a$ and $b$ will finally be chosen as appropriate powers of $N$)
to bound $\widehat{h}_{Z,N}$ from below. For each $\delta(z_i - z_j)$ we
use it twice; one time with $a p_i^2$, and a second time with
$a p_j^2$. Then we add these inequalities and divide by two:
\begin{eqnarray}
\widehat{h}_{Z,N} &=& \sum_{i=1}^N \left[ \left( 1 - a
\frac{N-1}{2}\right)p_i^2
- \delta(z_i) \right] + \sum_{i 0$ fixed the sequence $
Z w_Z(z)$ is a $\delta$-sequence, and $w_Z(0) \sim \lambda N^{2\varepsilon}/Z
\rightarrow 0$. If $\sigma (z)$ is smooth with
$|\sigma^{\prime}(z)|\leq \gamma $, then
$|N(\sigma * w_Z)(z) - \sigma (z)| \leq 2\gamma \lambda ^{2}
N^{-\varepsilon}$.
The one particle Hamiltonians $h(Z,N,\sigma)$, with
smooth $\sigma(z)$, converge as quadratic forms pointwise (i.e., for each test
function) to
\begin{equation}
h_{\lambda\sigma} = p^2 - \delta(z) + \lambda \sigma(z) .
\end{equation}
Moreover
\begin{equation}
\label{nearend}
h(Z,N,\sigma) \geq h_{\lambda\sigma} - (N^{-\varepsilon}/2)p^2 -%
2\delta \lambda ^{2} N^{-\varepsilon} -%
(\lambda ^{2}/2) N^{2\varepsilon -1}.
\end{equation}
Since the ground state energies of operators of the type
$\alpha p^2 + V$ are concave functions of $\alpha$ and hence continuous in
$\alpha$, the ground state energies of the right side
of (\ref{nearend}) converge in the limit $N \rightarrow \infty$.
The ground state energy of $h_{\lambda \sigma}$ is a concave
functional $e[\lambda\sigma]$, and the lower bound (\ref{lbd}),
when divided by the number of electrons $N$, gives
\begin{equation}\label{I}
\liminf_{N,Z \rightarrow \infty \atop N/Z = \lambda} \frac{1}{N}
\widehat{e}(Z,N) \geq e[\lambda\sigma] - \frac{\lambda }{2} \int
\sigma^2(z) dz =: \mathcal{I}_\lambda[\sigma].
\end{equation}
Inserting the mean field density $\rho $ for $\lambda \sigma$
(i.e., the minimizer of (64) which satisfies Eq.\ (3.8) of \cite{LSY94})
gives the mean field energy, divided by $\lambda $, as a
lower bound to the limit of the energy per electron.
\end{proof}
We remark that searching for the supremum of $\mathcal{I}_\lambda[\sigma]$
in (\ref{I})
also leads to the mean field equation of \cite{LSY94}: Assuming
$e(\lambda\sigma)=\langle\psi,h_{\lambda\sigma}\psi\rangle$ with a normalized
$\psi$ the variational condition on
$\sigma(z)$ for maximizing $\mathcal{I}_\lambda[\sigma]$ is
\begin{equation}
\sigma(z) = \psi^2(z).
\end{equation}
Inserting this into the Schr\"odinger equation
$h_{\lambda\sigma}\psi=\mu_\lambda\psi$ for $\psi$ gives
\begin{equation}
- \psi^{\prime\prime}(z) - \delta(z) \psi(0) + \lambda \psi^3(z) = -
\mu_\lambda \psi(z),
\end{equation}
i.e., Equation (3.8) in \cite{LSY94}.
Finally we remark that the energy per electron, $\widehat{e}(Z,N)/N$,
approaches the mean field limit monotonously. There is also a
subadditivity
property, which in the limit becomes concavity of $E^{\mathrm{HS}}(\lambda
)/\lambda $. These properties of the approach to a mean field hold in
some other cases too, as will be shown elsewhere \cite{B99}.
\section{Conclusions}
We have shown that the energy of an atom in a strong magnetic field $B$
approaches,
after division by $(\ln B)^2$, the energy of a many body Hamiltonian with
delta
interactions in one dimension as $B\to\infty$. This delta function model is
not
explicitly solvable, but an upper bound to the energy can be given in terms of
another model with the same mean field limit and
where we can explicitly calculate the ground state energy. In the latter model
an atom with nuclear charge $Z$ can bind up to $2Z$
electrons. Whether this represents the true state of affairs for the atomic
Hamiltonian in the $B\to\infty$ limit is an open problem.
\section*{Acknowledgements}
J.P.\ Solovej and J. Yngvason were supported in part by the EU
TMR-grant FMRX-CT 96-0001. J.P.S. was also supported in part by MaPhySto --
Centre for Mathematical Physics and Stochastics, funded by a grant from The
Danish National Research Foundation, and by a grant from the Danish Natural
Science Research Council.
\section*{Appendix}
We prove here that the ground state energy $\widetilde e(Z,N)$ of the
Hamiltonian (\ref{newh}) is independent of $N$ if $N\geq
2Z+1$.
\medskip
\noindent{\bf PROPOSITION (Maximal negative ionization for the comparison
model).} {\it If $N\geq 2Z+1$, then
\begin{equation}\widetilde e(Z,N)= \widetilde e(Z,N_{{\rm o}})
\end{equation}
where $N_{{\rm o}}$ is the largest integer strictly smaller than
$2Z+1$. Moreover, there is then no $L^2$-function with
$\widetilde e(Z,N)$ as an eigenvalue.}
\medskip
\noindent{\it Proof.} In the cone ${\cal M}$ defined by (\ref{cone}) we
consider the wave function
\begin{equation}
\check\psi(z_{1},\dots,z_{N})=\prod_{i=1}^{N_{{\rm o}}}
e^{-\kappa _{i}z_{i}}\prod_{j=N_{{\rm o}}+1}^{N}(1-\kappa_{j}z_{j}) ,
\end{equation}
with $\kappa_{n}$ defined by (\ref{kappa}). Since $\kappa_{j}\leq 0$
for $j\geq N_{\rm o}+1$, the function $\check\psi$ is strictly
positive. We extend $\check\psi$ symmetrically from
${\cal M}$ to all of $\mathbb R^N$ as a continuous function.
The jumps in the logarithmic derivatives of $\check\psi$ at the
boundary of ${\cal M}$ are not of the right size required
for an eigenfunction of $\widetilde h_{Z,N}$. But $\check\psi$ is an
eigenfunction of a slightly different operator:
\begin{equation}
\check h_{Z,N}\check\psi=\check e(Z,N)\check\psi
\end{equation}
with
\begin{equation}
\check e(Z,N)=-\sum_{i=1}^{N_{\rm o}}\kappa_{i}^2= \widetilde
e(Z,N_{\rm o})
\end{equation}
and
\begin{equation}
\check h_{Z,N}=\sum_{i=1}^{N}\left( p_{i}^{2}-\delta (z_{i})\right)
+%
\frac{1}{2Z}\sum_{i**