\section{The helium monoanion}\label{sec:05-helium-monoanion}

We prove that a helium nucleus does not bind three electrons
(Theorem~\ref{thm:helium-monoanion}). The lower bound must equal the
neutral helium energy $E=\AtomE{2}{2}$ exactly, so a numerical
approximation of $E$ does not suffice.

The proof splits the configuration space into two regions. In the interior
region $K_7$, all three electrons lie within distance $7$ of the nucleus.
In the exterior region $\Omega_7$, at least one electron lies farther away.
On $\Omega_7$, a localization argument bounds the Coulomb form below by $E$
plus a positive boundary term (Lemma~\ref{lem:helium-monoanion-exterior}).
On $K_7$, finite matrix computations bound the form below, up to the same
boundary term, in the two symmetry sectors of lowest energy
(Lemma~\ref{lem:he3-interior} in
Appendix~\ref{sec:10-helium-interior-certificate}). The boundary terms
cancel when the two bounds are added. The other symmetry sectors are
treated separately (Lemma~\ref{lem:helium-monoanion-complement}). All of
these steps use bounds on the ground state of neutral helium
(Lemma~\ref{lem:helium-monoanion-neutral}).

We use two standard results. The first is Temple's inequality for an
isolated ground level \cite[Theorem~4.13]{teschl2009}.

\begin{theorem}[Temple's inequality]\label{thm:temple}
Let $A$ be self-adjoint and bounded below, let $E=\inf\sigma(A)<b$, and
suppose $\sigma(A)\setminus\{E\}\subset[b,\infty)$.
For $v\in D(A)$ with $\|v\|=1$ put $\mu=\langle v,Av\rangle$.
If $\mu<b$, then
\[
 \mu-\frac{\|(A-\mu)v\|^2}{b-\mu}\leq E\leq\mu.
\]
\end{theorem}

The second is the local regularity result needed at the threshold, namely
the analyticity away from collisions stated in \cite[p.~3,
equations~(1.5)--(1.7) and the following
paragraph]{fournais2009}. The dilation $y_i=2x_i$, followed by division
of the equation by two, converts our Hamiltonian into that reference's
normalization.

\begin{theorem}[Off-collision analyticity]\label{thm:analyticity}
Let $N\geq1$ be an integer, $\Omega\subset(\mathbb R^3)^N$ be open,
$Z>0$, and $\lambda\in\mathbb R$.
Define the collision set
\[
 \Sigma=\bigcup_i\{x_i=0\}\;\cup\!\bigcup_{i<j}\{x_i=x_j\}.
\]
Every distributional solution on $\Omega$ of
\[
 \left(-\frac12\sum_i\Delta_i-Z\sum_i|x_i|^{-1}
             +\sum_{i<j}|x_i-x_j|^{-1}-\lambda\right)v=0
\]
is real analytic on $\Omega\setminus\Sigma$.
\end{theorem}

Three sets of data used below are supplied in the accompanying code/data
release of Set-up~\ref{setup:execution}. They are the coefficients of a
trial function for neutral helium, a table of rational bounds on its radial
tails, and a table of rational numbers used in the exterior estimate. The
interior matrix estimates are proved in
Appendix~\ref{sec:10-helium-interior-certificate}.

\begin{lemma}[Neutral spectral and spatial bounds]\label{lem:helium-monoanion-neutral}
Assume Set-ups~\ref{setup:atoms} and~\ref{setup:execution} for the
degree-fourteen neutral-helium calculation.
Let $E=\AtomE{2}{2}$ and let $(P_j)_{j=8}^{128}$ be the fixed rational
neutral tail table in the accompanying code/data release. Define
\[
 p_0(t)=
 \begin{cases}
  1,&0\leq t<1,\\
  P_{\lfloor8t\rfloor},&1\leq t<16,\\
  P_{128},&t\geq16,
 \end{cases}
 \qquad
 (M_1,M_2,M_3,M_4)
 =\frac{(1094533,2003681,5206311,17861539)}{10^6}.
\]
Then $E$ is a simple eigenvalue with a normalized rotationally invariant
spin-singlet eigenfunction $\Phi$, and
\[
 -2.903724674<E<-2.903724372,\qquad
 \AtomH{2}{2}\big|_{\Phi^\perp}\geq
       \left(E+\frac{67039}{100000}\right)I.
\]
For $i=1,2$, $k=1,2,3,4$, and $t\geq0$,
\[
 \|r_i^k\Phi\|\leq M_k,\qquad
 \|\mathbf1_{\{r_i\geq t\}}\Phi\|^2\leq p_0(t).
\]
The function $p_0$ is nonnegative, bounded by one, and nonincreasing.
\end{lemma}

\emph{Idea of the proof.}
We evaluate the energy and the residual of an explicit trial function in
exact arithmetic. Temple's inequality then encloses $E$, and a comparison
with hydrogenic operators gives the spectral gap. The trial function is
close to $\Phi$ in norm, which transfers its moment and tail bounds to
$\Phi$. The full proof is in
Appendix~\ref{sec:13-deferred-proofs-neutral}.

\begin{lemma}[Complementary sectors]\label{lem:helium-monoanion-complement}
Assume Set-ups~\ref{setup:atoms} and~\ref{setup:execution} for the
neutral calculation and the two rational pair-matrix checks.
Put $E=\AtomE{2}{2}$ and $\rho^2=r_1^2+r_2^2+r_3^2$.
Then the following inequalities hold on the fermionic $H^1$ form domain.
\begin{enumerate}
\item Every vector $u$ in a total-spin-$3/2$ sector, or in a
total-spin-$1/2$ sector of orbital angular momentum $L$ and parity
different from $(-1)^L$, satisfies
\[
 \mathfrak q_{3,2}[u]\geq-\frac{49}{18}\|u\|^2>E\|u\|^2
 \quad\text{if }u\ne0.
\]
\item Every vector $u$ in the closed sum of sectors of total orbital
angular momentum $L\geq2$ satisfies
\[
 \mathfrak q_{3,2}[u]-E\|u\|^2
       \geq\frac1{80}\int\rho^{-2}|u|^2.
\]
\end{enumerate}
\end{lemma}

\emph{Idea of the proof.}
The sectors in the first assertion contain no configuration with a doubly
occupied $1s$ shell. A hydrogenic comparison, corrected by the electron
repulsion, bounds all other configurations. For the second assertion, we
bound the two-electron operator of each pair from below, localize in the
ratios $r_i/\rho$, and use Hardy inequalities and the angular kinetic
energy. The full proof is in
Appendix~\ref{sec:13-deferred-proofs-complement}.

\begin{lemma}[Exterior trace at the neutral threshold]\label{lem:helium-monoanion-exterior}
Assume Set-ups~\ref{setup:atoms} and~\ref{setup:execution} for the fixed
neutral and exterior calculations. Let $E=\AtomE{2}{2}$,
$B_7=\{x\in\mathbb R^3:|x|<7\}$, $K_7=(B_7)^3$, and
$\Omega_7=\{(x_1,x_2,x_3):\max_i r_i>7\}$.
For antisymmetric $u\in H^1(\Omega_7;\mathbb C^8)$ let $q_>[u]$
be the Coulomb volume form of Set-up~\ref{setup:atoms}, restricted to
$\Omega_7$, with no boundary term. Then
\[
 q_>[u]-E\|u\|_{L^2(\Omega_7)}^2
 \geq\frac1{16}\int_{\partial K_7}|\operatorname{Tr}u|^2\,dS .
\]
The measure $dS$ is Euclidean surface measure on the three exposed
faces, and the trace is unrestricted. The inequality holds in every
spin, orbital-angular-momentum, and parity sector.
\end{lemma}

\emph{Idea of the proof.}
We localize according to which electrons have radii comparable to the
largest radius. When exactly one electron has this property, the spectral
gap of neutral helium from Lemma~\ref{lem:helium-monoanion-neutral}
controls its coupling to the other two electrons. Completing a square in
its radial variable then absorbs the localization error and leaves the
boundary term. In the other cases the form exceeds $E$ by a positive
amount. The full proof is in
Appendix~\ref{sec:13-deferred-proofs-exterior}.

\begin{theorem}\label{thm:helium-monoanion}
Under Set-ups~\ref{setup:atoms} and~\ref{setup:execution},
\[
 \AtomE{3}{2}=\AtomE{2}{2},\qquad \AtomH{3}{2}\geq-2.903724674\,I.
\]
For every real $\lambda\leq\AtomE{2}{2}$,
$\ker(\AtomH{3}{2}-\lambda I)=\{0\}$.
\end{theorem}

\begin{proof}
The proof has three steps. We first add the interior and exterior bounds,
whose boundary terms cancel. We then combine the bounds for all symmetry
sectors. Finally we show that a threshold eigenfunction vanishes on $K_7$,
and we extend this vanishing to the whole space by analyticity.

\emph{Matching at the exact threshold.}
Put $E=\AtomE{2}{2}$. Let $K_7,\Omega_7,q_>$ be as in
Lemma~\ref{lem:helium-monoanion-exterior}. Let the full sectors
$S,P$, the form $q_<$, and the constants $z,\eta_S,\eta_P$ be as in
Lemma~\ref{lem:he3-interior} in
Appendix~\ref{sec:10-helium-interior-certificate}.
The execution hypothesis of that lemma is Set-up~\ref{setup:execution}.
Its volume form has the same Hartree normalization and spin summation as
$q_>$.
For $J\in\{S,P\}$ and a fermionic $H^1$ vector $u$ in the full
sector $J$, restriction preserves antisymmetry and sector membership,
and the two restrictions have the same trace.
Adding Lemmas~\ref{lem:he3-interior}
and~\ref{lem:helium-monoanion-exterior} therefore gives
\begin{equation}\label{eq:helium-monoanion-matching}
 \mathfrak q_{3,2}[u]-E\|u\|^2
       \geq(z+\eta_J-E)\|u\|_{L^2(K_7)}^2 .
\end{equation}
The two boundary coefficients are both $1/16$ with the same
Euclidean surface measure. Lemma~\ref{lem:helium-monoanion-neutral}
gives $E<z$, so both coefficients on the right are strictly positive.
The numerical bounds and the completeness of the interior spaces,
including every magnetic component, are precisely those of
Lemma~\ref{lem:he3-interior}. No boundary condition is imposed on
the restrictions of $u$.

\emph{Exhaustion and equality of energies.}
Spatial rotations, spin rotations, and inversion reduce the Coulomb
operator. Three spin-$1/2$ particles have total spin $1/2$ or $3/2$.
Their orbital angular momenta are the integers $L\geq0$ and their
parities are $\pm1$.
Lemma~\ref{lem:helium-monoanion-complement} covers the quartets,
the doublets with parity different from $(-1)^L$, and every $L\geq2$.
The remaining sectors are exactly $S$ and $P$, controlled by
\eqref{eq:helium-monoanion-matching}.
Orthogonal summation yields $\AtomH{3}{2}\geq EI$ on the unrestricted
fermionic space. Theorem~\ref{thm:hvz} places $E$ in its essential
spectrum, so $\AtomE{3}{2}\leq E$ as well.
This proves equality. The lower neutral endpoint in
Lemma~\ref{lem:helium-monoanion-neutral} gives the stated numerical
operator bound.

\emph{Exclusion at threshold.}
Suppose $\Psi\in D(\AtomH{3}{2})$ and $\AtomH{3}{2}\Psi=E\Psi$.
Every symmetry projection satisfies the same eigen-equation.
The first bound in Lemma~\ref{lem:helium-monoanion-complement}
annihilates the quartet and the indicated parity projections.
Its second bound annihilates every remaining projection with
$L\geq2$, since $\rho^{-2}>0$ almost everywhere.
Equation~\eqref{eq:helium-monoanion-matching} makes the $S$ and $P$
projections vanish on $K_7$. Thus the entire eight-component spinor
$\Psi$ vanishes almost everywhere on $K_7$.

To apply Theorem~\ref{thm:analyticity} componentwise, let
$\mathcal A$ be the orthogonal antisymmetrizer on the full spin space
and $\widetilde{\mathfrak q}$ the polarized Coulomb form on its full
$H^1$ domain. Permutation invariance implies
$\widetilde{\mathfrak q}(\Psi,\varphi)
=\widetilde{\mathfrak q}(\Psi,\mathcal A\varphi)$ and
$\langle\Psi,\varphi\rangle=\langle\Psi,\mathcal A\varphi\rangle$.
The fermionic eigen-equation consequently holds against every
smooth compactly supported spinor test. Each scalar spin component
solves the Coulomb equation distributionally and is real analytic
on $\mathbb R^9\setminus\Sigma$, with $\Sigma$ as in
Theorem~\ref{thm:analyticity}.

This open set is connected. In fact its excluded sets are six
linear subspaces $V_j$ of dimension six. For two points $a,b$
outside them choose $c$ outside the finite union of
$V_j+\operatorname{span}\{a\}$ and
$V_j+\operatorname{span}\{b\}$.
These spaces have dimension at most seven, and the two segments
from $a$ to $c$ to $b$ avoid every $V_j$.
The nonempty open set $K_7\setminus\Sigma$ contains, for example,
$((1,0,0),(2,0,0),(3,0,0))$.
Almost-everywhere vanishing there implies pointwise vanishing of
each analytic representative by continuity. The identity theorem
for real analytic functions gives vanishing throughout the connected
collision-free set. Finally $\Sigma$ has measure zero, so $\Psi=0$
in the Hilbert space. The lower form bound already excludes every
$\lambda<E$, completing the assertion at and below threshold.
\end{proof}