\section{Electron counts}\label{sec:08-counts}

This section completes the proof of Theorem \ref{thm:main}. We first treat every
electron count above the dianions (Proposition~\ref{prop:high-counts}).
Lieb's bound covers all of these cases except lithium with six electrons,
which needs a separate argument. We then combine the results of
Sections~\ref{sec:04-helium-dianion}--\ref{sec:07-lithium-dianion}
(Theorem~\ref{thm:main}).

We use Lieb's bound~\cite{lieb1984} in the following form, which includes
a possible eigenvalue at the removal threshold. See
\cite[Section~2.1, equations~(3)--(6)]{nam2012}.

\begin{theorem}[Lieb's ionization bound]\label{thm:lieb}
Let $N\geq2$ be an integer and $Z>0$.
If $\AtomE{N}{Z}$ is an eigenvalue of $\AtomH{N}{Z}$, then $N<2Z+1$.
\end{theorem}

\begin{proposition}\label{prop:high-counts}
Let $N\geq2$ be an integer and $Z>0$. Suppose either $Z=2$ and $N\geq5$,
or $Z=3$ and $N\geq6$.
Then $\AtomE{N}{Z}=\AtomE{N-1}{Z}$ and this energy is not an $L^2$
eigenvalue of $\AtomH{N}{Z}$.
\end{proposition}

\begin{proof}
Lieb's bound treats all cases except $(N,Z)=(6,3)$. For that case we
combine three ingredients. The first is a pointwise inequality for the
weighted electron repulsion. The second is Lieb's argument with a quadratic
weight, as in the proof of Proposition~\ref{prop:quadratic-removal}, which
bounds the mean radius of an eigenfunction. The third is an upper bound on
the repulsion energy from the virial theorem and the Pauli principle.

By Theorem~\ref{thm:hvz}, $\AtomE{N}{Z}\leq\AtomE{N-1}{Z}$, and a
strict inequality makes $\AtomE{N}{Z}$ a discrete eigenvalue.
Consequently, absence of a ground eigenvalue implies both conclusions.
Theorem~\ref{thm:lieb} gives this absence for $Z=2$, $N\geq5$, and for
$Z=3$, $N\geq7$.

\emph{The six-particle geometric inequality.}
For $X=(x_1,\ldots,x_6)\in(\mathbb R^3)^6$, put
\[
 R(X)=\sum_{i=1}^6 r_i,\qquad
 v(X)=\sum_{i<j}\frac{r_i^2+r_j^2}{d_{ij}}.
\]
We claim that, almost everywhere,
\begin{equation}\label{eq:counts-geometry}
 v(X)\geq\frac{2657}{800}R(X).
\end{equation}
Work away from the collision and zero-radius sets. For each pair $i<j$,
\begin{equation}\label{eq:counts-pair}
 \frac{r_i^2+r_j^2}{d_{ij}}
 \geq \frac{1223}{1600}(r_i+r_j)
 +\frac{153}{100}\frac{x_i\cdot x_j}{r_i+r_j}
 +\frac{189}{200}
   \frac{3(x_i\cdot x_j)^2-r_i^2r_j^2}{(r_i+r_j)^3}.
\end{equation}
To verify this, set $s=r_i+r_j$, $\theta=d_{ij}/s$,
$z=1-\theta^2$ and $t=4r_ir_j/s^2$.
The triangle inequalities give $0<\theta\leq1$ and $z\leq t\leq1$.
The difference between the two sides of \eqref{eq:counts-pair}, divided
by $s$, is
\[
 F(t)=\frac{1-t/2}{\theta}-\frac{1223}{1600}
 -\frac{153}{100}\left(\frac z2-\frac t4\right)
 -\frac{189}{200}\left(\frac{3z^2}{4}-\frac{3zt}{4}+\frac{t^2}{8}\right).
\]
This is concave in $t$, and its endpoint values satisfy
\begin{align*}
 800\theta F(1)
 &= (1-\theta)(3\theta-2)^2(63\theta^2+147\theta+100),\\
 1600\theta F(z)
 &=800\left(\theta-\frac{239}{400}\right)^2+\frac{23837}{600}
   +801\left(\theta-\frac23\right)^2\left(\theta+\frac43\right)
   +189\theta^3(1-\theta^2).
\end{align*}
Both are nonnegative, proving \eqref{eq:counts-pair}.

Write $I_3$ for the $3\times3$ identity matrix and
$\|\cdot\|_{\mathrm{HS}}$ for the Hilbert--Schmidt norm. Set
$\mathsf Q_i=\sqrt3\,x_ix_i^{\mathsf T}-r_i^2 I_3/\sqrt3$. Then
$\operatorname{tr}(\mathsf Q_i\mathsf Q_j)
=3(x_i\cdot x_j)^2-r_i^2r_j^2$ and
$\|\mathsf Q_i\|_{\mathrm{HS}}^2=2r_i^4$.
Expanding the squares in the following Laplace integrals gives
\begin{align*}
 0&\leq\int_0^\infty\left|\sum_{i=1}^6e^{-ur_i}x_i\right|^2\,du
   =\frac R2+2\sum_{i<j}\frac{x_i\cdot x_j}{r_i+r_j},\\
 0&\leq\int_0^\infty\frac{u^2}{2}
       \left\|\sum_{i=1}^6e^{-ur_i}\mathsf Q_i\right\|_{\mathrm{HS}}^2\,du
   =\frac R4+2\sum_{i<j}
       \frac{3(x_i\cdot x_j)^2-r_i^2r_j^2}{(r_i+r_j)^3}.
\end{align*}
Summing \eqref{eq:counts-pair} therefore gives
\[
 v\geq\left(5\frac{1223}{1600}
             -\frac{153}{400}-\frac{189}{1600}\right)R
   =\frac{2657}{800}R,
\]
as asserted.

\emph{A radius bound valid at threshold.}
Suppose that a normalized $\Psi\in\FermionSpace{6}$ satisfies
$\AtomH{6}{3}\Psi=E\Psi$, where $E=\AtomE{6}{3}$.
It belongs to the Coulomb form domain. Write
$\rho=|X|=(\sum_i r_i^2)^{1/2}$, and choose a smooth nonincreasing
$\chi:[0,\infty)\to[0,1]$ equal to one on $[0,1]$ and zero on
$[2,\infty)$. For $L>0$, set $\chi_L(X)=\chi(\rho/L)$,
$f_L=\chi_L\Psi$ and $K=\|\chi'\|_\infty$.
These cutoffs preserve antisymmetry.

For each label $i$, $r_i f_L$ is antisymmetric in the other five labels.
Since $E\leq\AtomE{5}{3}$, the quadratic form of the five-electron
Hamiltonian in these labels minus $E$ is nonnegative. The omitted
coordinate and its spin are integrated as parameters. Test the weak
eigenvalue equation
with $\rho^2\chi_L^2\Psi$ and split off these six nonnegative forms.
Using $\Delta\rho^2=36$ gives
\begin{align*}
 &\frac12\sum_{i=1}^6\int r_i^2|\nabla_i f_L|^2
       -9\|f_L\|^2+\int(v-3R)|f_L|^2\\
 &\qquad\leq\frac12\int
       \left(\rho^2|\nabla\chi_L|^2
              +2\chi_L X\cdot\nabla\chi_L\right)|\Psi|^2
 \leq2K^2\int_{\rho\geq L}|\Psi|^2.
\end{align*}
The last inequality uses $X\cdot\nabla\chi_L\leq0$ and
$\rho^2|\nabla\chi_L|^2\leq4K^2$ on the cutoff annulus.
All multipliers in the weak equation are bounded with bounded derivatives
and compact support, so the identity uses no moment assumption on $\Psi$.
Integration by parts, first for smooth functions and then by $H^1$
approximation on compact support, gives for each label $i$ and such a
function $f$
\[
 \int r_i^2|\nabla_i f|^2
 \geq\|x_i\cdot\nabla_i f\|^2
 =\left\|x_i\cdot\nabla_i f+\frac32f\right\|^2
   +\frac94\|f\|^2.
\]
Together with \eqref{eq:counts-geometry}, this yields
\[
 \frac{257}{800}\int R|f_L|^2
 \leq\frac94\|f_L\|^2+2K^2\int_{\rho\geq L}|\Psi|^2.
\]
Fatou's lemma and $f_L\to\Psi$ in $L^2$ now give
\begin{equation}\label{eq:counts-radius}
 0<S:=\int R|\Psi|^2\leq\frac{1800}{257}.
\end{equation}

\emph{Incompatible repulsion bounds.}
Define the finite expectations
\[
 T=\frac12\sum_{i=1}^6\int|\nabla_i\Psi|^2,\qquad
 U=\int\sum_{i=1}^6\frac{|\Psi|^2}{r_i},\qquad
 W=\int\sum_{i<j}\frac{|\Psi|^2}{d_{ij}}.
\]
Their finiteness follows from the Coulomb form domain and Hardy's
inequality. The fifteen pairs satisfy
\[
 \sum_{i<j}\frac1{d_{ij}}
 \geq\sum_{i<j}\frac1{r_i+r_j}
 \geq\frac{15^2}{5R}=\frac{45}{R}.
\]
Cauchy--Schwarz against the probability density $|\Psi|^2$, followed by
\eqref{eq:counts-radius}, gives
\begin{equation}\label{eq:counts-repulsion-lower}
 W\geq45\int R^{-1}|\Psi|^2\geq\frac{45}{S}
 \geq\frac{257}{40}.
\end{equation}

An elementary two-electron trial function bounds $T$ from below.
For $\alpha>0$ and $x\in\mathbb R^3$, set
$\phi_\alpha(x)=(\alpha^3/\pi)^{1/2}e^{-\alpha|x|}$.
The normalized determinant formed by pairing $\phi_\alpha$ with the
two orthonormal spin states has energy at $Z=3$ equal to
$\alpha^2-6\alpha+5\alpha/8$. Indeed, the angular average of
$|x-y|^{-1}$ at fixed radii $\xi,\eta>0$ is
$1/\max\{\xi,\eta\}$, and hence its
repulsion integral is
\[
 32\alpha^6\int_0^\infty \xi e^{-2\alpha\xi}
       \left(\int_0^\xi \eta^2e^{-2\alpha\eta}\,d\eta\right)\,d\xi
 =\frac{5\alpha}{8}.
\]
Choosing $\alpha=43/16$ and using Theorem~\ref{thm:hvz} successively gives
\[
 E\leq\AtomE{2}{3}\leq-\frac{1849}{256}<-\frac{25}{4}.
\]
The normalized dilations $\Psi_a(X)=a^9\Psi(aX)$, $a>0$, have energy
$a^2T+a(-3U+W)$, minimized at $a=1$. Thus
$2T-3U+W=0$, $E=-T$, and $T>25/4$.

By antisymmetry, each normalized spin orbital has occupation at most one.
Hence the one-particle density operator of $\Psi$ has trace six and lies
between zero and the identity. For each $\zeta>0$,
Theorem~\ref{thm:hydrogen} therefore gives
\[
 T-\zeta U\geq
       2\left(-\frac{\zeta^2}{2}\right)
       +4\left(-\frac{\zeta^2}{8}\right)
       =-\frac32\zeta^2.
\]
Optimizing in $\zeta$ yields $U\leq\sqrt{6T}$.
Put $\tau=\sqrt T>5/2$. Since $3\sqrt{3/2}<15/4$,
\[
 W=3U-2T\leq2\tau\left(3\sqrt{\frac32}-\tau\right)
   <2\tau\left(\frac{15}{4}-\tau\right)<\frac{25}{4};
\]
the last quadratic is decreasing for $\tau\geq5/2$ and equals $25/4$
at $\tau=5/2$. This contradicts \eqref{eq:counts-repulsion-lower}.
Thus the six-electron ground energy is not an eigenvalue, and HVZ gives
the remaining equality $\AtomE{6}{3}=\AtomE{5}{3}$.

\emph{Proof of Theorem \ref{thm:main}:}
Theorems~\ref{thm:helium-monoanion} and~\ref{thm:helium-dianion} give
$\AtomE{3}{2}=\AtomE{4}{2}=\AtomE{2}{2}$ and exclude eigenfunctions
at or below this energy for three and four electrons.
Proposition~\ref{prop:high-counts} gives
$\AtomE{N}{2}=\AtomE{N-1}{2}$ for every $N\geq5$ and excludes an
eigenfunction at that energy. Induction proves
$\AtomE{N}{2}=\AtomE{2}{2}$ for all $N\geq3$.
Theorem~\ref{thm:lithium-dianion} similarly starts the lithium induction
at $\AtomE{5}{3}=\AtomE{4}{3}$, including threshold exclusion, and
Proposition~\ref{prop:high-counts} supplies every $N\geq6$.
Hence $\AtomE{N}{3}=\AtomE{4}{3}$ for all $N\geq5$, with no
eigenfunction at that energy. Since no eigenvalue can lie below the
spectral infimum, these statements give both asserted kernel exclusions.

\emph{Strict binding and the counts.}
Theorem~\ref{thm:hydrogen} and Lemma~\ref{lem:helium-neutral-cap} give
\[
 \AtomE{2}{2}<-\frac{29}{10}<-2=\AtomE{1}{2}<\AtomE{0}{2}=0.
\]
The preceding equalities exclude every $N\geq3$ from the helium
strict-binding set. That set is exactly $\{1,2\}$, so
$\AtomCount{2}=2$.
Propositions~\ref{prop:lithium-lower} and~\ref{prop:lithium-upper} give
\[
 \AtomE{3}{3}-\AtomE{4}{3}
 >(-7.493955380)-(-7.497020240)=0.003064860.
\]
Thus four belongs to the lithium strict-binding set, whereas the energy
equalities exclude every $N\geq5$. Its supremum is therefore four,
irrespective of which smaller electron counts belong to that set.
\end{proof}