\section{The helium dianion}\label{sec:04-helium-dianion}

We prove that a helium nucleus does not bind four electrons. The proof
applies Proposition~\ref{prop:quadratic-removal} with the weight $r^2$.
Section~\ref{sec:07-lithium-dianion} applies the same proposition to the
lithium dianion. The proposition requires four inputs. The first is an
upper bound $U_0$ for the neutral helium energy
(Lemma~\ref{lem:helium-neutral-cap}). The second is a pair of exterior
gaps (Lemma~\ref{lem:helium-dianion-exterior-gaps}). The third is a
pointwise lower bound for the weighted electron repulsion by a sum of
one-electron potentials (Lemma~\ref{lem:he4-quadratic-screen}). The fourth
is a pair of bounds on the number of negative eigenvalues of a one-electron
operator (Lemma~\ref{lem:he4-quadratic-levels}). The last two inputs are
computer-assisted and are proved in
Appendix~\ref{sec:12-helium-quadratic-certificates}.

\begin{lemma}\label{lem:helium-neutral-cap}
The neutral helium energy satisfies
\[
 \AtomE{2}{2}\leq-\frac{1406349}{484600}<-\frac{29}{10}.
\]
\end{lemma}

\begin{proof}
Let $\alpha=9/5$, $s=r_1+r_2$, $t=r_1-r_2$, and $\rho=d_{12}$,
and let $\chi\in\mathbb C^2\otimes\mathbb C^2$ be a normalized
antisymmetric spin vector. Set
\[
 P(s,t,\rho)=1+\frac{\alpha\rho}{6}+\frac{\alpha^2t^2}{24},
 \qquad \Psi(x_1,x_2)=e^{-\alpha s}P(s,t,\rho)\chi.
\]
The spatial factor is symmetric. The distance functions are locally
Lipschitz with bounded gradients, and the exponential controls the
polynomial and its derivatives at infinity. Thus $\Psi$ belongs to the
fermionic form domain.

After angular integration, the volume element is
$\pi^2(s^2-t^2)\rho\,ds\,dt\,d\rho$ on
\[
 \mathcal D=\{(s,t,\rho):s\geq0,\ 0\leq\rho\leq s,
                              \ -\rho\leq t\leq\rho\}.
\]
For nonnegative integers $i,j,k$, successive integration in $t,\rho,s$
gives
\[
 \int_{\mathcal D}e^{-2\alpha s}s^it^j\rho^k\,ds\,dt\,d\rho
 =\frac{2(i+j+k+2)!}
 {(j+1)(j+k+2)(2\alpha)^{i+j+k+3}}
 \quad (j\ \text{even});
\]
the integral vanishes for odd $j$. Expanding $P$ and differentiating
$\Psi$, this identity yields
\begin{align*}
 \|\Psi\|^2&=\pi^2\alpha^{-6}\frac{2423}{1152},\\
 \mathfrak q_{2,2}[\Psi]
 &=\pi^2\alpha^{-6}
   \left(\frac{709}{384}\alpha^2-\frac{20627}{3072}\alpha\right).
\end{align*}
The variational principle therefore gives
\[
 \AtomE{2}{2}\leq
 \frac{\mathfrak q_{2,2}[\Psi]}{\|\Psi\|^2}
 =\frac{2127\alpha^2-(61881/8)\alpha}{2423}
 =-\frac{1406349}{484600}
 =-\frac{29}{10}-\frac{1009}{484600}.
\]
\end{proof}

Throughout this section
\begin{equation}\label{eq:helium-dianion-cap}
 U_0=-\frac{1406349}{484600},\qquad
 (b_1,\delta_1)=\Bigl(2,\frac{1629}{1000}\Bigr),\qquad
 (b_2,\delta_2)=\Bigl(4,\frac38\Bigr).
\end{equation}
By HVZ (Theorem~\ref{thm:hvz}) and Lemma~\ref{lem:helium-neutral-cap},
$\AtomE{3}{2}\le\AtomE{2}{2}\le U_0$. Hence the choice $U=U_0$ is
admissible in Proposition~\ref{prop:quadratic-removal} with $N=4$, $Z=2$.
Nothing from Section~\ref{sec:05-helium-monoanion} is used.

\begin{lemma}[Exterior gaps]\label{lem:helium-dianion-exterior-gaps}
Let $a\in\{1,2\}$ and let $u$ belong to the fermionic form domain of
$\AtomH{4}{2}$. If $u$ vanishes almost everywhere on $\{q_a\le b_a\}$, then
\[
 \mathfrak q_{4,2}[u]-U_0\|u\|^2\ge\delta_a\|u\|^2.
\]
\end{lemma}

\begin{proof}
Let $\mathfrak h$ be the Coulomb integral of Set-up~\ref{setup:atoms}
with $N=4$, $Z=2$ on the full space $H^1((\mathbb R^3)^4;\mathbb C^{16})$.
It agrees with $\mathfrak q_{4,2}$ on fermionic vectors. We apply
Lemma~\ref{lem:regional-bound} with $N=4$, $Z=2$, so that a core electron
contributes $-2$ and $A_j=2-\kappa-jR/(R+b)$. The three parameter choices
used are
\begin{equation}\label{eq:helium-dianion-regional-values}
 \begin{aligned}
 M_0\bigl(2,2,\tfrac{21}{50},\tfrac{37}{20}\bigr)&=\frac{5089}{4000}
   &&(\kappa=\tfrac{28749}{20000},\ C=\tfrac{9597}{40000},\ A_0=\tfrac{11251}{20000}),\\
 M_0\bigl(\tfrac45,4,\tfrac35,\tfrac{31}{20}\bigr)&=\frac{19190}{8000}
   &&(\kappa=\tfrac{2883}{2000},\ C=\tfrac{651}{1600},\ A_0=\tfrac{1117}{2000}),\\
 M_1\bigl(4,4,\tfrac14,2\bigr)&=\frac{77}{32}
   &&(\kappa=1,\ C=\tfrac{3}{32},\ A_1=\tfrac12).
 \end{aligned}
\end{equation}
In each case $\frac12\le A_j\le\frac{308}{250}$, so the maximum in $M_j$
is attained at $h=2$ (in the third case also at $h=1$), and the values
follow by rational arithmetic.

\emph{The first gap.} If $u=0$ on $\{q_1\le2\}$, then all four radii exceed
$2$ on the support of $u$, and the first value gives
\[
 \mathfrak q_{4,2}[u]-U_0\|u\|^2
 \ge\Bigl(\frac{1406349}{484600}-\frac{5089}{4000}\Bigr)\|u\|^2
 =\frac{15796333}{9692000}\|u\|^2>\frac{1629}{1000}\|u\|^2 .
\]

\emph{The second gap.} Suppose $u=0$ on $\{q_2\le4\}$. Let $\theta(r)$
be zero for $r\le4/5$, affine from zero to $\pi/2$ on $[4/5,4]$, and
$\pi/2$ for $r\ge4$. For $S\subset\{1,\ldots,4\}$ let
$\chi_S=\prod_{i\in S}\cos\theta(r_i)\prod_{i\notin S}\sin\theta(r_i)$. Then
$\sum_S\chi_S^2=1$ and
$\mathfrak h[u]=\sum_S\mathfrak h[\chi_Su]-\frac12\int\sum_i|\theta'(r_i)|^2|u|^2$.
On the support of $u$ at most one radius lies below $4$. Hence
$\chi_Su=0$ for $|S|\ge2$, and the localization error is at most
$\frac12|\theta'|^2=\pi^2/(8\cdot(16/5)^2)=\pi^2/81.92<\frac{987}{8192}$,
using $\pi^2<9.87$. The term with $S=\varnothing$ has all radii above
$4/5$ and is bounded by the second value in
\eqref{eq:helium-dianion-regional-values}. A term with $S=\{i\}$ has
$r_i<4$ and the other three radii above $4$, and is bounded by the third
value. The localized spinors are global $H^1$ functions. Their vanishing
conditions, sliced in the other variables, give the Dirichlet traces
required by Lemma~\ref{lem:regional-bound}. No symmetry is needed.
Since $\frac{19190}{8000}<\frac{77}{32}$,
\[
 \mathfrak q_{4,2}[u]-U_0\|u\|^2
 \ge\Bigl(\frac{1406349}{484600}-\frac{77}{32}-\frac{987}{8192}\Bigr)\|u\|^2
 >\Bigl(2.90208-2.40625-0.12049\Bigr)\|u\|^2>\frac38\|u\|^2 .
\]
The displayed decimals are outward roundings of the three rationals. The
exact difference is $0.37534\ldots>0.375$. All arithmetic in this proof
is rational and is replayed by \texttt{certify\_gaps.py} with the cap
$U_0$ as its argument.
\end{proof}

We next compare the weighted form $M_{\beta,U}$ of
Proposition~\ref{prop:quadratic-removal} with a sum of one-electron
operators. Let $g$ be the piecewise affine function of
Definition~\ref{def:he4-quadratic-data}, which we call the screen. Let $h$
be the scalar spatial operator with closed form
\begin{equation}\label{eq:helium-dianion-onebody}
 q_h[v]=\frac12\int_{\mathbb R^3}(2+|x|^2)|\nabla v|^2\,dx
 +\int_{\mathbb R^3}\Bigl(-\frac4{|x|}-\frac32+g(|x|)\Bigr)|v|^2\,dx
\end{equation}
on $\{v\in H^1(\mathbb R^3):\int|x|^2|\nabla v|^2<\infty\}$. Let
$h_i$ act as $h\otimes I_{\mathbb C^2}$ on electron $i$.

\begin{lemma}[Occupation]\label{lem:helium-dianion-occupation}
Suppose $t_0<t_1$ satisfy $\NegIndex(h-t_0)=0$ and $\NegIndex(h-t_1)\le1$,
spin excluded. Then every compactly supported fermionic $u\in H^1$
satisfies $\sum_{i=1}^4\langle u,h_iu\rangle\ge(2t_0+2t_1)\|u\|^2$.
\end{lemma}

\begin{proof}
Let $\Pi=\mathbf 1_{(-\infty,t_1)}(h)$, of rank at most one. The index
hypotheses give $h\ge t_1I-(t_1-t_0)\Pi$ in the form sense, and
$\Pi\otimes I_{\mathbb C^2}$ has rank at most two. In an orthonormal wedge
basis adapted to its range, each spin orbital occurs at most once, so the
occupation of that range is at most $\min\{4,2\}=2$ for every fermionic
state. Hence $\sum_i\langle u,h_iu\rangle\ge[4t_1-2(t_1-t_0)]\|u\|^2$.
\end{proof}

\begin{theorem}\label{thm:helium-dianion}
Under Set-ups~\ref{setup:atoms} and~\ref{setup:execution},
$\AtomE{4}{2}=\AtomE{3}{2}$, and $\AtomH{4}{2}$ has no nonzero $L^2$
eigenfunction at any energy at or below $\AtomE{3}{2}$.
\end{theorem}

\begin{proof}
We verify the hypotheses of Proposition~\ref{prop:quadratic-removal} with
$N=4$, $Z=2$, $\beta=2$, $U=U_0$, $J=\{1,2\}$, $d_a=\delta_a$ and
\[
 \lambda_1=\frac{22}{25},\qquad\lambda_2=\frac{19}{100}.
\]
The inequality $U_0\ge\AtomE{3}{2}$ was noted after
\eqref{eq:helium-dianion-cap}, and
Lemma~\ref{lem:helium-dianion-exterior-gaps} supplies the exterior
hypotheses. Let $B_a$ and $M_{2,U_0}$ be the moment functions and the
weighted form of Proposition~\ref{prop:quadratic-removal} for these
constants. With $f(r)=2r+g(r)$ they are precisely the functions in
Definition~\ref{def:he4-quadratic-data}.

\emph{Lower bound for the repulsion.} Lemma~\ref{lem:he4-quadratic-screen} gives, almost
everywhere on $(\mathbb R^3)^4$,
\[
 \sum_{i<j}\frac{r_i^2+r_j^2+2}{d_{ij}}
 \ge\sum_{i=1}^4f(r_i)-\lambda_1B_1(q_1)-\lambda_2B_2(q_2)-1 .
\]
Inserting this into $M_{2,U_0}$ and using
$-2(r_i+2/r_i)+f(r_i)=-4/r_i+g(r_i)$ and $3N/2=6=4\cdot\frac32$,
\[
 M_{2,U_0}[u]+\sum_{a=1}^2\lambda_a\int B_a(q_a)|u|^2
 \ \ge\ \sum_{i=1}^4\langle u,h_iu\rangle-\|u\|^2-2U_0\|u\|^2
\]
for every compactly supported fermionic $u\in H^1$.

\emph{Lower bound for the one-electron operators.} Lemma~\ref{lem:he4-quadratic-levels} gives
$\NegIndex(h+\frac{197}{100})=0$ and $\NegIndex(h+\frac{3}{25})\le1$,
spin excluded. Lemma~\ref{lem:helium-dianion-occupation} with
$t_0=-\frac{197}{100}$, $t_1=-\frac{3}{25}$ gives
$\sum_i\langle u,h_iu\rangle\ge-\frac{209}{50}\|u\|^2$.

\emph{Conclusion.} Consequently
\[
 M_{2,U_0}[u]+\sum_{a=1}^2\lambda_a\int B_a(q_a)|u|^2
 \ge\Bigl(-\frac{209}{50}-1+\frac{1406349}{242300}\Bigr)\|u\|^2
 =\frac{151235}{242300}\|u\|^2>\frac{624}{1000}\|u\|^2 .
\]
All hypotheses of Proposition~\ref{prop:quadratic-removal} hold with
$\varepsilon=\frac{624}{1000}$. Its conclusion gives the energy equality
and excludes every nonzero $L^2$ eigenfunction at or below
$\AtomE{3}{2}$, including the threshold.
\end{proof}

\begin{remark}[Comparison with the lithium dianion]
Theorems~\ref{thm:helium-dianion} and~\ref{thm:lithium-dianion} are two
instances of Proposition~\ref{prop:quadratic-removal} with the same
weight $r^2$, the same constant $\beta=2$ and two ordered-radius moments.
The helium instance is simpler. It does not reserve part of the angular
kinetic energy, and it needs no fixed radial fields such as those of
Definition~\ref{def:geometric-certificates-lithium-fields}. A single
screen $g$ with five knots suffices, and the geometric certificate covers
the four ordered radii directly. The margin $0.624$ hartree is large
compared with the $10^{-5}$ margin of the lithium instance. A distant
electron sees the net charge $N-1-Z=1$ in both cases. The
difference is the size of the core loss, which grows with $Z$.
\end{remark}