\section{Binding of the lithium anion}\label{sec:06-lithium-binding}

We prove that the lithium anion is strictly bound. To do so, we bound the
anion energy $\AtomE{4}{3}$ from above
(Proposition~\ref{prop:lithium-upper}) and the neutral energy
$\AtomE{3}{3}$ from below (Proposition~\ref{prop:lithium-lower}).

The upper bound is variational. Its trial function is an antisymmetrized
sum of 150 explicitly correlated Gaussians with integer parameters, so all
matrix elements can be enclosed using exact arithmetic.

The lower bound has three steps. First, a scaling argument and the helium
bound of Theorem~\ref{thm:helium-monoanion} compare $\AtomH{3}{3}$ with a
sum of hydrogenic operators. This comparison gives the required bound in
every symmetry sector except two. Second, in each of the two remaining
sectors we add a rank-one projection onto a hydrogenic configuration, which
makes the shifted Hamiltonian positive. Third, we show that the rank-one
term is strictly smaller than the resulting operator, in the manner of the
Birman--Schwinger principle. The third step uses certified inner products
from Appendix~\ref{sec:11-lithium-integrals}.

\begin{proposition}\label{prop:lithium-upper}
Under Set-ups~\ref{setup:atoms} and~\ref{setup:execution},
\[
 \AtomE{4}{3}<-7.497020240.
\]
\end{proposition}

\emph{Idea of the proof.}
The norm and the energy of the trial function are sums of $45300$
closed-form Gaussian integrals. We enclose each term between integers in
exact arithmetic and obtain an upper bound for the Rayleigh quotient. The
full proof is in Appendix~\ref{sec:13-deferred-proofs-upper}.

\begin{proposition}\label{prop:lithium-lower}
Under Set-ups~\ref{setup:atoms} and~\ref{setup:execution},
\[
 \AtomE{3}{3}>-7.493955380.
\]
\end{proposition}

\begin{proof}
We follow the three steps described at the beginning of this section.
All operator comparisons below are comparisons of quadratic forms.

\emph{The comparison operator and the remaining sectors.}
Set
\[
 t=\frac58,\quad a=1-t=\frac38,\quad Q=\frac1a=\frac83,
 \quad C=-2.903724674,\quad z=-7.494,
 \quad\kappa=\frac5{72a}=\frac5{27}.
\]
Write $\mathcal T=-\frac12\sum_{i=1}^3\Delta_i$,
$\mathcal V=\sum_{i<j}d_{ij}^{-1}$ and $\mathcal A=\sum_i r_i^{-1}$.
The unitary dilation $\mathcal D_sf(X)=s^{9/2}f(sX)$, $s>0$, satisfies
\[
 \mathcal D_{1/t}^*(t\mathcal T-2\mathcal A+\mathcal V)\mathcal D_{1/t}
       =t^{-1}\AtomH{3}{2}.
\]
Theorem~\ref{thm:helium-monoanion} consequently gives
\begin{equation}\label{eq:lithium-binding-comparison}
 \AtomH{3}{3}\ge\frac Ct+G_a,\qquad
 G_a=\sum_{i=1}^3\left(-\frac a2\Delta_i-\frac1{r_i}\right).
\end{equation}
By the hydrogenic spectrum theorem (Theorem~\ref{thm:hydrogen};
\cite[Theorem~10.9]{teschl2009}), the one-electron levels in $G_a$ are
$\epsilon_n=-1/(2an^2)$. Let $N_1$ count occupation of its spatial $1s$
orbital, including spin. There are two $1s$ spin orbitals, so $N_1$ has
values $0,1,2$. On the subspace $N_1\le1$,
\begin{equation}\label{eq:lithium-binding-core-hole}
 G_a\ge3\epsilon_2+(\epsilon_1-\epsilon_2)N_1
       \ge\epsilon_1+2\epsilon_2=-\frac3{4a}.
\end{equation}
On $N_1=2$, the core has orbital angular momentum zero, even parity and
spin zero. The valence orbital therefore determines total orbital angular
momentum $L$ and parity $(-1)^L$. If $L\ge2$, its principal index is at
least three, and Theorem~\ref{thm:hydrogen} bounds the entire valence
space below by $\epsilon_3$. In every sector except $L=0$ with even
parity and $L=1$ with odd parity, either this bound or
\eqref{eq:lithium-binding-core-hole} applies. Thus these other sectors obey
\begin{equation}\label{eq:lithium-binding-complement}
 \AtomH{3}{3}\ge\beta I,\qquad
 \beta=\frac Ct-\frac{19}{18a}.
\end{equation}
This includes opposite-parity sectors, in which a filled $1s$ core is
impossible. The spectral complement bounds in Theorem~\ref{thm:hydrogen}
include the continuous subspace.

It suffices to examine orbital magnetic number zero and spin magnetic
number $1/2$. Orbital and spin rotations commute with the Hamiltonian and
its form: every integer-$L$ orbital representation contains magnetic
number zero, and each of the three-electron spin representations
$1/2,3/2$ contains spin magnetic number $1/2$. Bounds on these components
therefore hold on their full magnetic multiplets. For $T\in\{S,P\}$ put
$\ell_S=0$, $\ell_P=1$, and let $\mathcal B_T$ denote the sector with
$L=\ell_T$, parity $(-1)^{\ell_T}$ and those two magnetic numbers,
without restricting total spin. Write $H_T=\AtomH{3}{3}|_{\mathcal B_T}$.

\emph{The hydrogenic configurations and an inverse bound.}
Use the trial functions $u_T$ and the vectors $p_{T,n}$ of
Definition~\ref{def:li-trials}. We call the vectors $p_{T,n}$ sources. By
\eqref{eq:lithium-binding-sources} below, they are the antisymmetrized
hydrogenic configurations with a filled $1s$ shell and the third electron
in the level $n$. The spatial representation of the trial functions and
sources is
antisymmetric under the exchange $P_{13}$ of the positions $1$ and $3$.
Extending its
$\uparrow\downarrow\uparrow$ component, divided by $\sqrt3$, to the other
two spin strings by signed permutations is well-defined and isometric:
the stabilizer is generated by $(13)$ and the three component norms are
equal. This identifies the spatial and physical scalar products, without
a further spin factor.

To fix the source phases, let $\phi_{100}(x)=(Q^3/\pi)^{1/2}e^{-Q|x|}$,
and choose the real, normalized charge-$Q$ orbitals $\phi_{nL0}$ with
angular factors $Y_{00}=1/\sqrt{4\pi}$ and
$Y_{10}(\widehat x)=\sqrt{3/(4\pi)}x_3/|x|$, and positive leading radial
coefficient at the origin. The Laguerre polynomials and positive constants
in Definition~\ref{def:li-trials} give, with these phases,
\begin{equation}\label{eq:lithium-binding-sources}
 p_{T,n}=\frac{1-P_{13}}{\sqrt2}
       [\phi_{100}(x_1)\phi_{100}(x_2)\phi_{n\ell_T0}(x_3)].
\end{equation}
For $S$ this identity uses
$L_{n-1}^{(1)}(y)=\sum_{j=0}^{n-1}(-1)^j\binom{n}{n-1-j}y^j/j!$;
for $P$ the valence radial polynomials are $1$ and $1-Qr/6$.
Orthogonality of the valence orbitals to the core and to each other makes
these sources orthonormal. They span the indicated discrete levels of
the filled-core space. Lemma~\ref{lem:li-physical-moments} places both
trials and sources in $D(\AtomH{3}{3})$. Their stated symmetries and the
reducing character of $\mathcal B_T$ therefore place them in $D(H_T)$.

Set $m_S=6$, $m_P=3$, and let
$P_{T,n}=|p_{T,n}\rangle\langle p_{T,n}|$ for $2\le n\le m_T$.
For every integer $n\ge2$ define
\[
 g_n=\frac Ct-\frac{1+1/(2\max(3,n)^2)}a-z
     =\frac{170037989}{937500000}-\frac4{3\max(3,n)^2}.
\]
Then $g_2=g_3=\beta-z>0.0332257$, and $g_n$ is nondecreasing.
Put $K_T=H_T-zI+\kappa P_{T,2}$. Since $\kappa=\epsilon_3-\epsilon_2$,
the rank-one term $\kappa P_{T,2}$ raises the comparison level of the
source with $n=2$ to the level $n=3$.
On the orthogonal complement of $p_{T,2},\ldots,p_{T,m_T}$,
Theorem~\ref{thm:hydrogen} gives the lower bound
$2\epsilon_1+\epsilon_{m_T+1}$ on the part with $N_1=2$, and
\eqref{eq:lithium-binding-core-hole} is stronger on the part with
$N_1\le1$. Hence
\begin{equation}\label{eq:lithium-binding-lift}
 K_T\ge g_{m_T+1}I+
       \sum_{n=2}^{m_T}(g_n-g_{m_T+1})P_{T,n}\ge g_2I>0.
\end{equation}
The variational identity
\[
 \langle f,K_T^{-1}f\rangle
 =\sup_{v\in D(K_T^{1/2})}
       \{2\operatorname{Re}\langle f,v\rangle-\|K_T^{1/2}v\|^2\}
\]
then yields
\begin{equation}\label{eq:lithium-binding-inverse}
 K_T^{-1}\le\frac1{g_{m_T+1}}I+
       \sum_{n=2}^{m_T}\left(\frac1{g_n}-\frac1{g_{m_T+1}}\right)P_{T,n}.
\end{equation}
This argument does not assume that $H_T$ commutes with the projections
$P_{T,n}$.

\emph{Bound on the rank-one term.}
Set $c_S=1.026$, $c_P=1.007$, $w_T=c_Tu_T$, and define
\[
\begin{gathered}
 \eta_T=(H_T-zI)w_T,\quad b_T=\langle p_{T,2},w_T\rangle,
 \quad e_T=\langle w_T,\eta_T\rangle,\quad d_T=\|\eta_T\|^2,\\
 v_{T,n}=\langle p_{T,n},\eta_T\rangle,\qquad j_T=1-\kappa b_T,
 \qquad \rho_T=p_{T,2}-K_Tw_T=j_Tp_{T,2}-\eta_T.
\end{gathered}
\]
All these scalar products are real, and $w_T\in D(K_T)$ by the domain
assertion above. Expanding the residual gives the exact identity
\[
 \langle p_{T,2},K_T^{-1}p_{T,2}\rangle
 =2b_T-e_T-\kappa b_T^2+
       \langle\rho_T,K_T^{-1}\rho_T\rangle.
\]
Using \eqref{eq:lithium-binding-inverse}, define the upper bound
\begin{equation}\label{eq:lithium-binding-score}
\begin{split}
 \mathcal S_T=\kappa\bigg[&2b_T-e_T-\kappa b_T^2
 +\frac{j_T^2-2j_Tv_{T,2}+d_T}{g_{m_T+1}}\\
 &+\sum_{n=2}^{m_T}\left(\frac1{g_n}-\frac1{g_{m_T+1}}\right)
                  (j_T\mathbf1_{n=2}-v_{T,n})^2\bigg].
\end{split}
\end{equation}
Thus $\kappa\langle p_{T,2},K_T^{-1}p_{T,2}\rangle\le\mathcal S_T$.
If $\mathcal S_T\le s_T<1$, Cauchy--Schwarz applied to
$K_T^{-1/2}p_{T,2}$ and $K_T^{1/2}f$ gives
$\kappa|\langle p_{T,2},f\rangle|^2\le s_T\langle f,K_Tf\rangle$
for every form-domain vector $f$. Subtraction of the rank-one term and
\eqref{eq:lithium-binding-lift} imply
\begin{equation}\label{eq:lithium-binding-sector-floor}
 H_T\ge\bigl[z+(1-s_T)g_2\bigr]I.
\end{equation}

\emph{Evaluation from the certified intervals.}
Lemma~\ref{lem:li-physical-moments} encloses in rational intervals the
inner products among the trial functions, the sources and the images of
the trial functions under the Hamiltonian. We call these inner products
moments. The lemma uses $h_T=(\AtomH{3}{3}+15/2)u_T$. Consequently, at
$z=-7.494$,
\[
 \eta_T=c_T(h_T-\delta u_T),\qquad\delta=0.006.
\]
Denote the moments that enter by
\[
 A_T=\langle u_T,h_T\rangle,\quad D_T=\|h_T\|^2,
 \quad N_T=\|u_T\|^2,\quad B_{T,n}=\langle p_{T,n},u_T\rangle,
 \quad V_{T,n}=\langle p_{T,n},h_T\rangle.
\]
They enter through the exact identities
\begin{equation}\label{eq:lithium-binding-recentering}
\begin{aligned}
 b_T&=c_TB_{T,2},&
 e_T&=c_T^2(A_T-\delta N_T),\\
 d_T&=c_T^2(D_T-2\delta A_T+\delta^2N_T),&
 v_{T,n}&=c_T(V_{T,n}-\delta B_{T,n}).
\end{aligned}
\end{equation}
No normalization of $u_T$ is introduced.

We now bound $\mathcal S_S$ and $\mathcal S_P$ using only the rational
endpoints of these intervals. Temporarily suppress $T$, and let subscripts
$-$ and $+$ on $A,D,N,B_n,V_n$ denote the endpoints printed in
Lemma~\ref{lem:li-physical-moments}. Set
\[
\begin{aligned}
 b_\pm&=cB_{2,\pm},& e_-&=c^2(A_--\delta N_+),\\
 d_+&=c^2(D_+-2\delta A_-+\delta^2N_+),&
 v_{n,\pm}&=c(V_{n,\pm}-\delta B_{n,\mp}),\\
 j_-&=1-\kappa b_+,& j_+&=1-\kappa b_-.
\end{aligned}
\]
These formulas give outward bounds by their coefficient signs. The
specified endpoints satisfy $b_->0$, $j_->0$, $v_{2,-}>0$ and
$j_--v_{2,+}>0$. Define
\[
\begin{gathered}
 F_T=2b_+-e_--\kappa b_-^2,\qquad
 R_T=j_+^2-2j_-v_{2,-}+d_+,\\
 a_{T,2}=j_+-v_{2,-},\qquad
 a_{T,n}=\max(|v_{n,-}|,|v_{n,+}|)\quad(3\le n\le m_T).
\end{gathered}
\]
Positivity of the coefficients in \eqref{eq:lithium-binding-score} gives
\[
 \mathcal S_T\le\kappa\left[F_T+\frac{R_T}{g_{m_T+1}}
 +\sum_{n=2}^{m_T}\left(\frac1{g_n}-\frac1{g_{m_T+1}}\right)a_{T,n}^2\right].
\]
Table~\ref{tab:lithium-binding-score} gives integer upper bounds for
$10^{15}$ times each term in these brackets. Each comparison uses just
the rational endpoints above and the displayed formula for $g_n$.
Equivalently, it is verified by multiplying by the positive denominators.

\begin{table}[ht]
\centering
\caption{Integer upper bounds for $10^{15}$ times the terms in the bound
for $\mathcal S_T$.
The row indexed by $n$ bounds
$(g_n^{-1}-g_{m_T+1}^{-1})a_{T,n}^2$.}
\label{tab:lithium-binding-score}
\begin{tabular}{lrr}
\hline
Term & $S$ & $P$\\
\hline
$F_T$ & 4740317261675618 & 2838404429685468\\
$R_T/g_{m_T+1}$ & 567896606970620 & 1791632727708472\\
$n=2$ & 20913699346983 & 123610754064044\\
$n=3$ & 62462513402733 & 139024913705375\\
$n=4$ & 1036687051373 & --\\
$n=5$ & 105796401819 & --\\
$n=6$ & 14991159233 & --\\
\hline
Sum & 5392747556008379 & 4892672825163359\\
\hline
\end{tabular}
\end{table}

The final strict comparisons are the integer identities
\[
\begin{aligned}
 27\cdot998656955\cdot10^6-5\cdot5392747556008379&=4958105>0,\\
 27\cdot906050524\cdot10^6-5\cdot4892672825163359&=22183205>0.
\end{aligned}
\]
Hence $\mathcal S_S<0.998656955$ and $\mathcal S_P<0.906050524$.
This endpoint calculation covers every combination of values in the
certified intervals. It requires no assumption on correlations between
them.

Take $s_* =0.998656955$, the larger of the two bounds.
Equation~\eqref{eq:lithium-binding-sector-floor} controls both remaining
sectors, and \eqref{eq:lithium-binding-complement} controls all other
orbital and parity sectors. The rotation argument includes every spin and
magnetic sector. Since $z+(1-s_*)g_2<z+g_2=\beta$, they give on the full
fermionic space
\[
 \AtomH{3}{3}\ge\bigl[z+(1-s_*)g_2\bigr]I.
\]
Finally,
\[
 (1-s_*)g_2>
 0.001343045\cdot0.0332257
 =0.0000446236102565>0.000044620,
\]
so $\AtomE{3}{3}>-7.493955380$, as asserted.
\end{proof}