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Mathematics
\vskip 2em
{\BBigrm An Upper Bound for the Number of} \par 
{\BBigrm Electrons in a Large Ion}
\vskip2em
Asymptotic neutrality/excess charge/quantum mechanics
\vskip2em

{\Bigrm C. L. Fefferman and L. A. Seco}\vskip1em
Department of Mathematics, Princeton University, Princeton NJ 08544.
\vskip1em
\vfil
\eject






{\bf Abstract}
Let $E(Z,N)$ be the ground-state energy of $N$ quantized electrons and
a single nucleus of charge $Z$. For fixed $Z$, $E(Z,N)$ is independent
of $N$ for $N\ge N_{\rm critical}(Z)$. Physically, this means that at
most $N_{\rm critical}$ electrons can bind to the nucleus. We prove
that $N_{\rm critical}\le Z+CZ^a$ with $a=0.84$.

\vfil
\eject


Consider the Hamiltonian for a nucleus of charge $Z$ and $N$ quantized
electrons,

$$H_{Z,N} = \sum_{i=1}^N \biggl [(-\lapl _{x_i}) - {Z \over {|x_i|}} \biggr
] + {1 \over 2} \sum_{i \ne j} {1 \over {|x_i - x_j|}} 
 = -\lapl + V_{\rm Coulomb}.$$

The ground state energy is then
$$ E(Z) = \inf_N E(Z,N) =\inf_{N}\,\inf\Biggl\{
       \Bigl < H_{Z,N}\,\psi\, , \,\psi \Bigr > \quad|\quad
\psi\in \wedge_{i=1}^N (L^2(\reals{3})\otimes {\bf C}^q),\quad
{\lnorm{\psi}_2 = 1}\Biggr\}. $$

For each $Z$, call $N(Z)$ the smallest number for which $E(Z)=E(Z,N)$.

It is an interesting problem to obtain sharp estimates for $N(Z)$. The
sharpest known result appears in (1). In particular, $N(Z)/Z \to 1$ as
$Z\to\infty$, although there were no estimates for the rate of
convergence.
Our main result is the following:

\smallskip
{\bf\Bigrm Theorem:}

$$ N(Z) = Z + O(Z^\alpha)  \qquad{\rm for\ some\ } \alpha<1$$

For the proof we will be interested only in the case $Z\le N\le 2Z$.
Recall that $E(Z)/Z^{\frac 7,3}\to 1$ as $Z\to\infty$.
\smallskip
{\Bigrm\bf Definitions:} 
{\parindent=20pt
\item{1.}
Given a ball of center $0$ and radius $R$,
call $N_{\sss R}=N_{\sss R}(x_1,\ldots,x_N)$ 
the number of $x_i$ that belong to it.

\item{2.}We say that $\est(\bar\epsilon,\epsilon,R)$
holds if for a nucleus of charge $Z$ at the origin and $N$ quantized
electrons {\sl confined to the ball} $B(0,R)$ we have
$$ \Bigl< H_{Z,N}\,\psi, \psi\Bigr> \ge
E_0(Z) + {\bar\epsilon Z \over R}\Bigl(N - (1+\epsilon)Z\Bigr)  $$
where 
$$E_0(Z) = \inf_{0\le N\le (1+\epsilon_\sharp)Z}\,E(Z,N)  $$
for $\epsilon_\sharp$ to be picked later.\vskip0em
By {\sl N quantized electrons confined to the ball $B(0,R)$} we mean 
that the support of
$\psi$ is included in the set $\{{\bf x}| N_R({\bf x}) = N\}$.


\item{3.}Fix once and for all an 
even approximation to the identity, $\phi$, supported in
$B(0,Z^{\frac -2,3})$. 
Given points $x_1,\ldots,x_N$ set $\rho(x)=\sum_i\phi(x-x_i)$.

\item{4.}Take a smooth function $\chi_{\sss R}$ that is equal to $1$ if $|x|<R$
and $0$ if $|x|>2R$,
and define $N_{\chi,R}(x)=\sum_{i=1}^N\phi(x-x_i)\cdot\chi_{\sss
R}(x)$.
Obviously, $N_{\sss R/2} \le N_{\chi,R} \le N_{\sss 2R} $.

\item{5.}If $\tfd$ is the Thomas-Fermi density,
$N_{\chi,\sss R}^{\sss TF} = \int_{B(0,R)} \tfd(x)\,\chi(x)\,dx$.
Note that if $R=Z^{{\frac -1,3}+\gamma}$, then $Z\ge N_{\chi,\sss R}^{\sss
TF}\ge
Z-cZ^{1-3\gamma}$, $\gamma>0$.

}
\smallskip
{\bf Key Result: }

$$ \Bigl<H_{Z,N}\,\psi,\psi\Bigr>
\ge E_0(Z) + C{|\bigl< N_{\chi}\psi,\psi\bigr> - N_{\chi}^{\sss TF}|^2\over R}
- O(Z^{{\frac 7,3}-b})  $$
for any $\psi$; in particular we do not assume that there are any
number of electrons confined to any ball.

{\Bigrm\bf Proof of Key Result: } 

It follows from W. Hughes (2) and H. Siedentop-R. Weikard (3) that if we set
$$ K(x_1,\ldots,x_N) = \2int{(\rho-\tfd)(x)(\rho-\tfd)
(y)\over|x-y|}dx\,dy  $$
then $H_{Z,N}\ge E_0(Z) + K(x_1,\ldots,x_N) - O(Z^{{\frac 7,3}-b})$.

Since
$$K=\int_{\reals{3}}|\xi|^{-2}|\hat\rho(\xi)-\hat\tfd(\xi)|^2d\xi $$
and
$$N_{\chi}-N_{\chi}^{\sss TF} = \int(\rho-\tfd)\chi\,dx=
\int(\hat\rho-\hat\tfd)\hat\chi\,d\xi$$
Cauchy-Schwarz yields
$$
|N_{\chi}-N_{\chi}^{\sss TF}|^2 \le \int|\chi(\xi)|^2\cdot|\xi|^2\,d\xi
\int|(\hat\rho-\hat\tfd)|^2\cdot|\xi|^{-2}\,d\xi \le CR\cdot K$$
And Cauchy-Schwarz again gives the result.

\medskip
{\bf Corollary 1:}
For any $\beta_1$, $\beta_2$,
satisfying 

$$2\beta_1+\beta_2<b  \eqno({\rm ineq}\,1)$$

for some $\smallc$  depending
on $b-2\beta_1-\beta_2$ and sufficiently large $Z$,
$\est(\bar\epsilon_0,\epsilon_0,R_0)$ holds for
$\epsilon_0 \ge Z^{-\beta_1}$, $\bar\epsilon_0 \le {\scriptstyle c}
Z^{-\beta_1}$ and $R_0 \le {1 \over 2}Z^{\beta_2 - {1\over 3}}$.

{\bf Proof: }Use the key result for $R=2R_0$. Note that $N_{R/2}=N$. 
If $N \ge Z+Z^{1-\beta_1}$, since
$N_{\chi}^{\sss TF}\le Z$, we have
$$C{|N-Z|^2\over R}-O(Z^{{\frac 7,3}-b})\ge C'{|N-Z|^2\over R} $$
and therefore
$${\scapro H_{Z,N}\psi,\psi;}\ge E_0(Z)+C'{|N-Z|^2\over R}\ge E_0(Z)+{
{\scriptstyle c}Z^{1-\beta_1}\over R}(N-Z-Z^{1-\beta_1})$$

{\bf Corollary 2: }
If $R=Z^{{\frac -1,3}+\gamma_2}$ and
$\Bigl<N_{\sss 2R}\psi,\psi\Bigr> < Z-cZ^{1-\gamma_1}$, and
$$3\gamma_2>\gamma_1\qquad  7\gamma_2<b \eqno({\rm ineq}\,2).$$
Then
$${\scapro H_{Z,N}\psi,\psi;} \ge E_0(Z)+cZ^{{\frac 7,3}-7\gamma_2}$$
$\smallc$ becomes $0$ as $3\gamma_2-\gamma_1$ and $7\gamma_2-b$ go to
$0$.

{\bf Proof: } Since $N_{\chi,R}^{\sss TF} \ge Z-cZ^{1-3\gamma_2}$
$${|N_R-N_{\chi,R}^{\sss TF}|^2\over R} \ge cZ^{{\frac 7,3}-7\gamma_2}.$$

\bigskip
\hskip10pt
Now, we study a system of $N$ quantized electrons in $B(0,R)$, and
$N'$ quantized electrons in $B(0,2R)-B(0,R/2)$, 
for $R>2Z^{{\frac -1,3}+\gamma}$,
with

$$\gamma>\gamma_2\qquad \gamma>\beta_2 \eqno(\rm{ineq}\,3)$$

Let's rewrite the Hamiltonian $H_{Z,N+N'}$ as
$$H_{Z,N+N'}= -\lapl_{x_i \ldots x_N}+V-\lapl_{\rm extra}+V_{\rm extra} $$
with

$$\eqalign{\lapl_{\rm extra} &= \lapl_{x'_1,\ldots,x'_{N'}}\cr
V &= -\sum_{i=1,\ldots,N}{Z\over |x_i|}+\repulsion\cr
V_{\rm extra}&= -\sum_{i=1,\ldots,N'}{Z\over |x'_i|} +
\sum_{{i=1,\ldots,N'}\atop{j=1,\ldots,N}}{1\over |x'_i-x_j|} +
{1\over 2}\sum_{{{i=1,\ldots,N'}\atop{j=1,\ldots,N'}}\atop{i\ne j}}{1\over |x'_i-x'_j|} \cr}$$


Corollary 2 now implies that

$${\scapro V_{\rm extra}\psi,\psi;}\ge {c\epsilon Z\over R}N' 
\eqno({\rm A})$$

or else $$\bigl<H_{Z,N+N'}\,\psi,\,\psi\bigr> \ge
E_0(Z)+cZ^{{\frac 7,3}-7\gamma_2}$$
since the fact that $\bigl<N_{R/Z^{\gamma-\gamma_2}}\psi,\psi\bigr>$
is at least $Z-Z^{1-\gamma_1}$ makes the 
interaction of the nucleus and of the electrons in 
$B(0,R/Z^{\gamma-\gamma_2})$ with a fixed electron $x'_j$ in
$B(0,2R)-B(0,R/2)$ contribute approximately with 
$(-Z+N_{R/Z^{\gamma-\gamma_2}})/ |x'_j| \ge -Z^{1-\gamma_1}/|x'_j|$ 
to the potential energy.
We omit the details and we simply point
out that you need enough electrons to almost cancel the effect of the
nucleus. Precisely,
$$ Z^{-\gamma_2}>\epsilon 
\eqno(\rm{ineq}\,4)$$

\bigskip
\indent We now use this estimate to go from
$\est(\bar\epsilon,\epsilon,R)$ to
$\est(\bar\epsilon,\epsilon',2R)$. To see this, simply take a
wave function $\psi$ living in $B(0,2R)$ and a partition of unity
$\theta_0$, $\theta_1$,
adapted to $B(0,R)$ and $B(0,2R)-B(0,R/2)$,
and set

$$\psi_{i_1,\ldots,i_N}(x_1,\ldots,x_N)
=\theta_{i_1}(x_1)\cdots\theta_{i_N}(x_N)\cdot\psi
(x_1,\ldots,x_N)$$

For each $\psi_{i_1,\ldots,i_N}$ we have $N_1$ electrons living in
$B(0,R)$ and $N_2$ electrons living in $B(0,2R)-B(0,R/2)$, with
$N_1+N_2=N$;
now, assume $\est(\bar\epsilon,\epsilon,R)$ holds, apply Corollary 1
and 2 and estimate (A), and sum over all possible choices
of $i_1,\ldots,i_N$
to obtain

$$\eqalign{\Bigl<H_{Z,N}\psi,\psi\Bigr>&\ge
 E_0(Z) + {\bar\epsilon Z\over R}\Bigl(N_1-
(1+\epsilon) Z\Bigr) + {\scapro V_{\rm extra}\psi,\psi;}-{CN\over
R^2}\cr
&\qquad\ge 
 E_0(Z) + {\bar\epsilon Z\over R}\Bigl(N_1+N_2-
(1+\epsilon) Z\Bigr)  -{CN\over R^2}\cr
&\qquad= 
 E_0(Z) + {\bar\epsilon Z\over R}\Bigl(N-
(1+\epsilon) Z\Bigr)  -{CN\over R^2}\qquad(\bar\epsilon\le c\epsilon)\cr}$$
The terms $CN/R^2$ come from the Laplacian hitting $\theta_0$ and
$\theta_1$.
Again we omit the details.

Now observe that
$$\max\biggl(0,{\bar\epsilon Z\over R}\bigl(N-(1+\epsilon)Z
\bigr)-{CN\over R^2}
\biggr)
\ge {\bar\epsilon Z\over 2R}\bigl(N-(1+\epsilon')Z\bigr) $$

for $\epsilon'-\epsilon={4C\over \bar\epsilon RZ}$; we also need
that $R\ge {2C\over \bar\epsilon Z}$, which certainly holds in this
case.
This proves that $\est(\bar\epsilon,\epsilon',2R)$ holds.

\bigskip
\indent If
$$\beta_2>\gamma_2 \eqno({\rm ineq}\,5)$$
we can use corollary~1 and conclude that 
$\est(\bar\epsilon_0,\epsilon_n,2^n R_0)$ holds
for
$$\epsilon_{n+1}=\epsilon_n+{4C\over 2^n\bar\epsilon_0 R_0 Z}.$$

Note that
$ \epsilon_n \le C\epsilon_0=\epsilon_\sharp$, for $C$ independent of $n$.
Therefore, $\est(\bar\epsilon_0,C\epsilon_0,R)$ holds for all $R\ge R_0$
and so
$$ N(Z) - Z \le C\epsilon_0 \le C Z^{1-\beta_1}$$

The value for $\alpha=1-\beta_1$ 
depends on $b$ and on the parameters $\beta_1$, $\beta_2$, $\gamma_1$ and
$\gamma_2$ subject to the constraints (ineq$1$)---(ineq$5$). For each $b$,
the optimum value in the closure of this
set of parameters is attained for $\beta_1={\frac 3b,7}$.
The value for $\alpha$ is then $1-{\frac 3b,7}$.
>From Hughes and Siedentop-Weikard
it follows that we can take $b={\frac 21,{56}}$.
This allows us to take any $\alpha>{\frac 47,{56}}\approx 0.84$. 
Further improvement of $b$ will
lead to a better estimate for $\alpha$. What Hughes and Siedentop-Weikard
did
is something in fact much harder that just estimate $K$, and it is
clear that better estimates can be obtained; maybe we can take $b$
to be ${\frac 2,3}$; this would give the result for any 
$\alpha > {\frac 5,7}$.

\vskip1em
{\bf Acknowledgements.}The first author was partially supported by a National
Science Foundation Grant. The second author was supported by a Sloan
Foundation Graduate Dissertation Fellowship.
\vskip1em

\parindent13pt
\item{1.} Lieb, E. H,  Sigal, I., Simon, B. \& Thirring, W. (1988) 
``Approximate Neutrality of Large-Z Ions''
{\it Communications in Mathematical Physics} {\bf 116(4)} 635-644.\par
\item{2.} Hughes, W.  To appear in {\it Advances in Mathematics}.\par
\item{3.} Siedentop, H., Weikard, R. (1987) ``On the Leading Energy
Correction for the Statistical Model of the Atom: Interacting Case''
{\it Communications in Mathematical Physics} {\bf 112} 471-490\par

\end
