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Dirac-Coulomb operator, Dirac-Fock theory, electron-positron field
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%running head
\markboth{W.H. Aschbacher}{Lowering the
Dirac-Fock minimizer by pair correlation}

%title
\title{Lowering the
Dirac-Fock minimizer by pair correlation}

%authors & affiliations
\author{Walter H. Aschbacher}

\author{Walter H. Aschbacher\footnote{CPT-CNRS Luminy,
 Case 907, 13288 Marseille Cedex 9, France; this author
 is supported by the Swiss National Science Foundation}}


\thispagestyle{empty}
\date{}
\maketitle

%abstract
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{abstract}
We prove by a simple computation that a suitable coupling to the
{\it positronic} sector lowers the energy of the purely electronic
minimizer of the Dirac-Fock functional.
\end{abstract}

\noindent\textbf{Key words.} Dirac-Coulomb operator, Dirac-Fock
theory, electron-positron field.

\noindent\textbf{AMS classification (2000).} 34L40, 47J20.
%KEY WORDS:

%\newpage


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction and motivation}


The energy of a quantum system consisting of massive relativistic
spin-$1/2$ point particles in an exterior field and interacting
via Coulombic (or other) forces may approximately be described by
means of {\it Dirac-Fock theory}: Each particle effectively feels
an additional potential created by the rest of the system.
Although this theory has {\it not} yet been rigorously derived
from QED it is nevertheless used in practice yielding good
agreement with experimental data. Steps in the direction of a
rigorous justification are presently undertaken, cf. \cite{BBHS98,
BBHS99, BES, BFHS} and references therein.

A state of such a system is approximated in this theory to be
quasi-free, generalizing the notion of a Slater determinant. Due
to the fact that a quasi-free state is uniquely determined by its
corresponding one particle density matrix given by a self-adjoint
trace class operator $\gamma$ on the one particle Hilbert space $\hh$, cf. \cite{BBHS99},
we can write the Dirac-Fock energy functional ${\cal E}_{DF}$ as
\begin{eqnarray}
\label{DF}
     {\cal E}_{DF}(\gamma)=\tr(H_c\gamma)+{\cal Q}(\gamma,\gamma)
\end{eqnarray}
where $H_c$ denotes the Dirac-Coulomb operator, and ${\cal Q}$ is
the difference of the {\it direct} and the {\it exchange} terms
well-known from Hartree-Fock theory, cf. \cite{BBHS99} and
Section~\ref{S2} below for precise definitions.

As it has been shown in \cite[Corollary~3.9]{BFHS} there exists a
(not necessarily unique) minimizer $\gamma_0$ of the Dirac-Fock
functional (\ref{DF}) if the minimization procedure is carried out
over the space ${\cal S}^{(\delta)}$ of one particle density
matrices which are, in addition, {\it diagonal}, i.e.,
$(1-\Lambda^{(\delta)})\gamma\Lambda^{(\delta)}=0$,  and have
prescribed charge $\tr\gamma=N\in\N$.
% \begin{eqnarray*}
 %  \begin{array}{ll}
%   {\cal S}^{(\delta)}  := &\{\gamma\in{\cal
%   L}^1(\hh),\gamma^*=\gamma, \tr\gamma = N\,|\,H_c\gamma\in{\cal
%   L}^1(\hh), \\
 %  & \ -(1-\Lambda^{(\delta)})\le
  % \gamma\le\Lambda^{(\delta)}, (1-\Lambda^{(\delta)})
   %\gamma\Lambda^{(\delta)}) = 0\},
   %\end{array}
%\end{eqnarray*}
Here,
$$\Lambda^{(\delta)}:=\chi_{(0,\infty)}(H^{(\delta)})$$
denotes the projection onto the positive spectral part of the
Dirac-Fock operator $H^{(\delta)}=H_c+W^{(\delta)}$ where the
mean-field potential $W^{(\delta)}$ is given by
\begin{eqnarray}
\label{def:W}
   W^{(\delta)}\psi(x)=(\rho_\delta\ast V)(x)
   \,\psi(x)-\int dy \,\,V(x-y)\,\kappa_\delta(x,y)\,\psi(y),
\end{eqnarray}
($\rho_\delta$ and $\kappa_\delta$ are given below in
Section~\ref{S2}, $V$ is the Coulomb potential). Note that
$W^{(\delta)}$ is specified by the {\it electron-positron
distribution} $\delta$ taken from the set $F=\{0\le \delta,\;
  \tr(|H_c|^{1/2}|\delta||H_c|^{1/2})<\infty\}$.
The space $\Lambda^{(\delta)}\hh$ is thus the set of
{\it one-electron states}.

In the case of {\it charge one}, this minimizer $\gamma_0$ is
found to be an orthogonal projection of the form
$\gamma_0=|\psi_0\rangle\langle\psi_0 |$, $\psi_0\in \Ran
\Lambda^{(\delta)}$, which fulfills the Dirac-Fock equations
$\Lambda^{(\delta)} H^{(\gamma_0)}\Lambda^{(\delta)}\psi_0 =
\varepsilon\psi_0$, $\varepsilon\geq 0$ (see \cite[Lemma~4.3,
Corollary~4.5]{BFHS}). Since $\psi_0\in\Ran\Lambda^{(\delta)}$ and
$W^{(\gamma_0)}\psi_0=0$, this is equivalent to
$\Lambda^{(\delta)} H_c \Lambda^{(\delta)} \psi_0 = \varepsilon
\psi_0$. But any eigenfunction $\psi$ of $\Lambda^{(\delta)} H_c
\Lambda^{(\delta)}$ with positive eigenvalue has its corresponding
$\gamma= |\psi\rangle\langle\psi|$ in $S^{(\delta)}$. Hence,
$\gamma_0$ being a minimizer, $\varepsilon$ is the ground state
energy of $\Lambda^{(\delta)} H_c\Lambda^{(\delta)}$, i.e., the
lowest positive eigenvalue of $\Lambda^{(\delta)}
H_c\Lambda^{(\delta)}$.


Moreover, for a system of charge $1$, using boundedness of
$W^{(\delta)}$ \cite[Appendix~A.]{BFHS}, Theorem 3 (i) in
\cite{GS99} yields
\begin{eqnarray*}
  \inf_{\gamma\in S^{(\delta)}}{\cal E}_{DF}(\gamma)&=&{\cal E}_{DF}(\gamma_0)
  = \varepsilon\le\inf\left\{\sigma(H_c)\cap [0,mc²]\right\}
  = \inf_{\gamma\in S^{(\delta=0)}}{\cal E}_{DF}(\gamma)
  ={\cal E}_{DF}(\gamma_c)
\end{eqnarray*}
where $\psi_c$ is the electronic ground state of $H_c$, and
$\gamma_c=|\psi_c\rangle\langle\psi_c|$. This implies that the
``$\sup\inf$''
 \begin{equation}\label{supinf}
  \sup_{\delta\in F}\inf_{\gamma\in S^{(\delta)}}
  \mathcal{E}_{DF}(\gamma)
 \end{equation}
is attained by the pair $\left(\delta=0, \gamma_c\right)$ for a
system of charge $1$: Mittleman in \cite{M} (for the general case
of charge $N$) argued that the right choice of
$\Lambda^{(\delta)}$ that determines the one-electron space should
be given by the stationary points of the $\delta$-dependent
functional
$\mathcal{E}_{DF}(\gamma_0)=\mathcal{E}_{DF}(\gamma_0(\delta))$,
which motivates to investigate the ``$\sup\inf$'' problem
\eqref{supinf}.

A further reason to consider this pair
$\left(\delta=0,\gamma_c\right)$, is that for a system of change
one the mean-field potential is expected to vanish, in which case
the minimizer is a pure electronic ground state of the
relativistic hydrogen atom.

In the present paper, we show that for
$\Lambda^{(\delta=0)}=\chi_{(0,\infty)}(H_c)$ and in the case of
{\it non-vanishing} electron-positron correlations in the one
particle density matrix $\gamma$, i.e., for {\it off-diagonal}
$\gamma$, $(1-\Lambda^{(\delta)} )\gamma \Lambda^{(\delta)} \neq
0$, the energy of the Dirac-Fock functional can be {\it lowered}
compared to the energy of the purely electronic minimizer
$\gamma_c$.


%where $\gamma$ denotes the {\it one particle density matrix}, for $Q:  S^{(\delta_0)}\times  S^{(\delta_0)}\rightarrow \C$, $Q[\gamma,\gamma']:=D[\rho_\gamma,\rho_\gamma']-E[\gamma,\gamma']$, cf. above.



{\bf Acknowledgements} We are very grateful to our collaborator J.-M. Barbaroux for most stimulating exchange. Moreover, we would like to thank B. Helffer, E.
S\'er\'e, and H. Siedentop for useful discussions.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The setting}\label{S2}


We prove in this section that we can lower the Dirac-Fock energy
${\cal E}_{DF}$ on its purely electronic minimizer by adding an
off-diagonal perturbation. In order to give precise definitions of
the ingredients we recall the following facts.\\ The Dirac-Coulomb
operator $H_c$ describes the relativistic motion of a {\it single}
massive spin-$1/2$ point particle in an external electrostatic
Coulomb field $V$ of an infinitely heavy point-like nucleus,
\begin{eqnarray*}
\label{DCop}
  H_c=c\,\alpha \cdot p+mc²\beta+V,
\end{eqnarray*}
with orbital momentum $p=-i\nabla$, mass $m=1$, and Coulomb
potential $V(x):=-\kappa{|x|}^{-1}$ coupled by some
$\kappa\in\R_+$; furthermore, $\alpha$ and $\beta$ denote the
usual Dirac matrices (we will use their standard representation,
cf. appendix \ref{app:DiracEqu}). Our notation hides all the
dependencies on possible parameters but the dependence on the
speed of light, $c$, since we are merely interested in the limit
of large $c$. If $\kappa$ is not too large, $\kappa\in
]0,\sqrt{3}/2\,c[$,  the operator $H_c$ in the Hilbert space
$\hh=L²(\R³)\otimes\C^4=:L²(\R³)^4$ is essentially self-adjoint on
$C_0^\infty(\R³\!\setminus\! \{0\})^4$ and self-adjoint on
$H^1(\R³)^4$, the domain of the free Dirac operator $H_c-V$, cf.
\cite{T92}; set $\kappa=1$.

Now, for a given electron-positron distribution $\delta\in F$ as
above, the Dirac-Fock energy is defined by the functional
 $$
 {\cal E}_{DF}: \begin{array}{lll}
    {\cal S}_{\mathrm{off}}^{(\delta)}&\rightarrow&\C\\
    \gamma&\mapsto& \tr(H_c\gamma)+{\cal Q}(\gamma,\gamma)
  \end{array}
 $$
where the space of one particle density matrices ${\cal
S}_{\mathrm{off}}^{(\delta)}$ of charge one is defined by
\begin{eqnarray*}
   {\cal S}_{\mathrm{off}}^{(\delta)}:=\{\gamma\in{\cal
   L}^1(\hh),\gamma^*=\gamma, \tr\gamma=1\,|\,H_c\gamma\in{\cal
   L}^1(\hh), -(1-\Lambda^{(\delta)})\le
\gamma\le\Lambda^{(\delta)}\},
\end{eqnarray*}
for ${\cal L}^1(\hh)$ the trace class operators on $\hh$. We set
$W_c:=W^{(\gamma_c)}$ from (\ref{def:W}). The functional ${\cal
Q}:({\cal S}_{\mathrm{off}}^{(\delta)})^{\times 2}\rightarrow\C$
is defined to be the difference between the {\it direct} and the
{\it exchange} terms, ${\cal D}$ and ${\cal E}$ respectively,
${\cal Q}(\gamma_1,\gamma_2):={\cal
D}(\rho_{\gamma_1},\rho_{\gamma_2})-{\cal
E}(\kappa_{\gamma_1},\kappa_{\gamma_2})$. The direct term ${\cal
D}$ is given by
\begin{eqnarray*}
{\cal D}(\rho_1,\rho_2):=\frac{1}{2}\,\int_{\R^6}\!d³\!xd³\!y\,\,V(x-y)\,\, \rho_1^\ast(x)\,\rho_2(y)
\end{eqnarray*}
for $\rho_j:\R^3\rightarrow \C$, $j=1,2$, and the exchange term ${\cal E}$ has the form
\begin{eqnarray*}
{\cal E}(\kappa_1,\kappa_2):=\frac{1}{2}\,\int_{\R^6}\!d³\!xd³\!y\,\,V(x-y)\,\, \tr_{\C^4}\left[\kappa_1^\ast(x,y)\,\kappa_2(x,y)\right]
\end{eqnarray*}
where $\kappa_j:\R^3\times\R^3\rightarrow M_4$, $j=1,2$, and $M_4$
denotes the set of complex $4\times 4$ matrices. The definition of
the integral kernel $\kappa_\gamma: \R³\times\R³\rightarrow M_4$
of $\gamma\in {\cal S}_{\mathrm{off}}^{(\delta)}$ reads
$\kappa_\gamma(x,y):=\sum_{i,j}(e_i,\gamma
e_j)_{L²(\R³)^4}\,\,\,e_i(x)\otimes e_j^\ast(y)$ where $\{e_j\}$
is an orthonormal basis of $\hh$, and $e_j^\ast$ denotes the row
covector of the column vector $e_j$ (in $\C^4$). Furthermore,
$\rho_\gamma: \R³\rightarrow\C$ is just the trace of the diagonal
of the integral kernel $\kappa_\gamma$,
$\rho_\gamma(x):=\tr_{\C^4}\,\,\kappa_\gamma(x,x)$. Hence, we have
\begin{eqnarray}
\label{QW}
{\cal Q}(\gamma_1,\gamma_2)=\frac{1}{2}\,\tr\,\,\gamma_1W^{(\gamma_2)}.
\end{eqnarray}
In the following we concentrate on the case $\delta=0$, as it has
been discussed in the introduction. Note that we know from
\cite[Theorem 4]{BBHS99} that ${\cal E}_{DF}$ is {\it nonnegative}
on ${\cal S}^{(\delta=0)}_{\mathrm{off}}$.



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Lowering the minimizer}


After having specified the ingredients of the Dirac-Fock
functional ${\cal E}_{DF}$ we now turn to the problem described in
the introduction.\\ Let $\psi_c$ be an electronic ground state of
the Dirac-Coulomb operator $H_c$ (2-fold degenerate), and
$\epsilon\in\R$ a perturbation parameter. We claim that we can
find an off-diagonal perturbation, $\gamma_\epsilon$ say, of the
electronic Dirac-Coulomb ground  state projector
$\gamma_c:=|\psi_c\rangle\langle\psi_c|\in{\cal
S}_{\mathrm{off}}^{(\delta=0)}$ such that we can {\it strictly
lower} the Dirac-Fock energy of $\gamma_c$, i.e.,
\begin{eqnarray}
\label{claim}
{\cal E}_{DF}(\gamma_c+\gamma_\epsilon)<{\cal E}_{DF}(\gamma_c)+{\cal O}(\epsilon²).
\end{eqnarray}

Hence, defining the perturbation $\gamma_\epsilon$ as
\begin{eqnarray*}
\gamma_\epsilon:=\epsilon²\,\,(|\psi_>\rangle\langle \psi_>|-|\psi_<\rangle\langle \psi_<|)+\epsilon (1-\epsilon²)^{1/2} \,\,(|\psi_>\rangle\langle \psi_<|+|\psi_<\rangle\langle \psi_>|),
\end{eqnarray*}
we claim that there exists $\psi_<$ from the negative spectral subspace of $H_c$ and $\psi_>$ in the positive spectral subspace of $H_c$ without the electronic ground state $\psi_c$,
\begin{eqnarray}
\label{conditions}
 \psi_<\in\chi_{(-\infty,0]}(H_c)\hh,\quad \psi_>\in\chi_{(0,\infty)}(H_c)\hh\cap \{\psi_c\}^\bot
\end{eqnarray}
such that the claim (\ref{claim}) is fulfilled. Plugging
$\gamma_c$ and $\gamma_\epsilon$ into (\ref{claim}) and taking
into account the first order in $\epsilon$ only, we find
\begin{eqnarray}
\label{HFper}
{\cal E}_{DF}(\gamma_c+\gamma_\epsilon)-{\cal E}_{DF}(\gamma_c)&=&2\epsilon \, \mbox{Re} \,\,{\cal Q} (|\psi_>\rangle\langle \psi_<|+|\psi_<\rangle\langle \psi_>|,\gamma_c) +{\cal O}(\epsilon²)\nonumber\\
&=&2\epsilon \, \mbox{Re} \,\,\left(\psi_<,W_c \,\psi_>\right)+{\cal O}(\epsilon²),
\end{eqnarray}
where $\gamma_c + \gamma_\epsilon\in {\cal
S}_{\mathrm{off}}^{(\delta=0)}$, $\epsilon\in [0,1]$, and
$W_c=X_c-Y_c$ on $L^2(\R^3)^4$, cf. (\ref{def:W}) and (\ref{QW}),
with
\begin{eqnarray}
\label{def:X,Y}
X_c\psi:=\left(|\psi_c|²_{\C^4}\ast V\right)\,\psi,\quad Y_c\psi:=\left((\psi_c,\psi)_{\C^4}\ast V\right)\,\psi_c.
\end{eqnarray}

Therefore, in order to prove the claim (\ref{claim}), it suffices to show that
\begin{eqnarray}
\label{suff}
\mbox{Re} \,\,\left(\psi_<,W_c \,\psi_>\right)\neq 0
\end{eqnarray}
and to choose the sign of $\epsilon$ appropriately. That's what we
will do next. To this end we introduce the orthogonal projection
$\Lambda_c$ onto the negative spectral part of $H_c$,
$\Lambda_c:=\chi_{(-\infty,0]}(H_c)$.


\vspace{0.5cm}

\begin{theoreme}
\label{theorem} There exists a
$\psi_>\in\chi_{(0,\infty)}(H_c)\hh\cap \{\psi_c\}^\bot$ such that
for $c$ sufficiently large,
\begin{eqnarray}
\label{thm}
\Lambda_cW_c\psi_>\neq 0.
\end{eqnarray}
\end{theoreme}

\vspace{0.5cm}

{\bf Remark} Of course, this now implies (\ref{suff}) for the
choice $\psi_<:=\Lambda_cW_c\psi_>$:
\begin{eqnarray*}
\mbox{Re} \,\left(\psi_<,W_c \,\psi_>\right)=\|\Lambda_cW_c\psi_>\|²\neq 0.
\end{eqnarray*}


In order to prove (\ref{thm}) we extract zeroth and first order
terms from $\Lambda_c$, $W_c$, and $\psi_c$ for large $c$. This
will eventually lead to a first order term on the rhs of
(\ref{thm}) plus {\it remainder terms} of higher order.
\begin{lemma}
\label{Lambda}
\begin{eqnarray}
\label{Lambda_c}
 \Lambda_c=\Lambda^{(0)}+\frac{1}{c} \,\Lambda^{(1)}+\frac{1}{c²}
 \,\Lambda^{(c)},\quad \Lambda^{(0)}=
 \frac{1}{2} (1-\beta),\quad\Lambda^{(1)}=-\frac{1}{2}\alpha\cdot p,
\end{eqnarray}
and $\Lambda^{(c)}$ is ${\cal O}(1)$ on the states appearing in
the remainder terms of (\ref{thm}).
\end{lemma}


\vspace{0.5cm}


{\bf Proof \ref{Lambda}} Since $0\in\R$  lies in the spectral gap
$(-c²,E_c)$ of $H_c$, with $c²>E_c>0$ the ground state energy of $H_c$ on
its positive spectral subspace, we can write $\Lambda_c$ as (cf.
\cite{K95}, p.359)
\begin{eqnarray*}
\label{U(H)}
\Lambda_c=\slim_{\delta\to 0,\rho\to\infty}\frac{1}{2}\left({\bf 1}-U_{\delta,\rho}(H_c)\right),\quad U_{\delta,\rho}(H_c)&:=&\frac{1}{\pi}\left[\int_{-\rho}^{-\delta}\!+\!\int_\delta^\rho\right]d\eta\,\,(H_c-i\eta)^{-1}.
\end{eqnarray*}
In order to extract the potential $V$ at order $1/c²$ from the resolvent in this principal value like integral we rescale $U_{\delta,\rho}(H_c)=U_{\delta/c²,\rho/c²}(H_c/c²)$ and apply the second resolvent identity:
\begin{eqnarray}
\label{ResExp}
(H_c/c²-i\eta)^{-1}=R^{(0)}+\frac{1}{c} \,R^{(1)}+\frac{1}{c²} \,R^{(c)},
\end{eqnarray}
where we write down the zeroth and first order here,
\begin{eqnarray*}
R^{(0)}:=(\beta-i\eta)^{-1},\quad R^{(1)}:=-(\beta-i\eta)^{-1}
\alpha \cdot p\, (\beta-i\eta)^{-1},
\end{eqnarray*}
%\left(\frac{1}{c}\alpha\cdot  p+m\beta+\frac{1}{c²}V-i\eta\right)^{-1}\\
%R^{(c)}&:=&\sum_{n=2}^\infty\frac{(-1)^n}{c^{n-2}}(m\beta-i\eta)^{-1}\left[\alpha\cdot  p(m\beta-i\eta)^{-1}\right]^n\\
%&&-\left(\frac{1}{c}\alpha\cdot  p+m\beta+\frac{1}{c²}V-i\eta\right)^{-1}V\left(\frac{1}{c}\alpha\cdot  p+m\beta-i\eta\right)^{-1}
whereas $R^{(c)}$ is treated in appendix \ref{app:estimates}. Note
that the application of the resolvent identity is well-defined on
the domain of $p$, $D(p)=H^1(\R³)$. Defining now for $j=0,1,c$,
\begin{eqnarray}
\label{def:Lambda}
\Lambda^{(j)}:=\slim_{\delta\to 0,\rho\to\infty} \frac{1}{2}\left({\bf 1}-\frac{1}{\pi}\left[\int_{-\rho/c²}^{-\delta/c²}+\int_{\delta/c²}^{\rho/c²}\right]d\eta\,\,R^{(j)}\right),
\end{eqnarray}
and performing some elementary integration we find $\Lambda^{(0)}$ and $\Lambda^{(1)}$ in (\ref{Lambda_c}). The estimates involving $\Lambda^{(c)}$ are carried out in appendix \ref{app:estimates}.
%&:=&\slim \frac{1}{2}\left({\bf 1}-U^{(j)}_{\delta/c²,\rho/c²}[D_c/c²]\right)\\
$\Box$

\vspace{0.5cm}

The expansion of a bound state $\psi_c\in\hh_{pp}(H_c)$ is given in \cite{T92}.
\begin{lemma}
\label{psic}
Let $\psi_c\in \hh_{pp}(H_c)$ with eigenvalue $E_c$.
\begin{eqnarray}
\label{psidec}
\psi_c=\psi_c^{(0,1)}+\frac{1}{c²}\,\psi^{(c)},\quad
\psi_c^{(0,1)}:={{\phi}\choose {\frac{1}{2c}\sigma\cdot   p\,
\phi}}, \qquad \|\psi^{(c)}\|_{H^1(\R³)^4}={\cal O}(1),
\end{eqnarray}
and $\phi\in \hh_{pp}((-\Delta/2+V){\bf 1}_2)$ with eigenvalue $\,\,\lim_{c\to\infty} (E_c-c²)$.
\end{lemma}

\vspace{0.5cm}

In the next lemma we extract the low order terms of $W_c=X_c-Y_c$ by computation using the bispinor notation $\psi=:(\psi_+,\psi_-)^T$. The expressions $\phi$, $\psi^{(0,1)}_c$, $\psi^{(c)}$ stem from (\ref{psidec}). The higher order terms are estimated in appendix \ref{app:estimates}.
\begin{lemma}
\label{W}
The direct part decomposes as
\begin{eqnarray}
\label{X_c}
X_c=X^{(0)}+\frac{1}{c²}X^{(c)},\quad X^{(0)}=|\phi|²_{\C²}\ast V,
\end{eqnarray}
and the exchange part reads
\begin{eqnarray*}
\label{Y_c} Y_c\psi&=&\left(Y^{(0)}\psi_++\frac{1}{c} \,
Y^{(1)}\psi_-+\frac{1}{c²}\,Y^{(c)}\psi\right)
\,\psi^{(0,1)}_c+\frac{1}{c²}\,Y_c\psi^{(c)}\\
Y^{(0)}&=&(\phi,\cdot)_{\C²}\ast V,\quad
Y^{(1)}=\frac{1}{2}(\sigma\cdot   p\,\phi,\cdot)_{\C²}\ast V.
\end{eqnarray*}
The contribution $Y^{(c)}$ is ${\cal O}(1)$ on the states
appearing in the remainder terms of (\ref{thm}).
\end{lemma}

\vspace{0.5cm}

Now we can prove Theorem \ref{theorem} by assembling the forgoing
large $c$ expansions.

{\bf Proof \ref{theorem}}  Denote again by $E_c$ the electronic
ground state energy of $H_c$. Choose
$\psi_>:=\varphi_c:=\psi_{-1/2}$ and $\psi_c:=\psi_{1/2}$ the
corresponding spinors with quantum numbers $(n,j,m)=(1,1/2,-1/2)$
and $(1,1/2,1/2)$, respectively,
$\psi_c,\varphi_c\in\mbox{ker}(H_c\!-\!E_c)$.  Note that the
condition (\ref{conditions}) on $\psi_>$ is satisfied since
$\psi_{-1/2}$ is $L^2$ orthogonal to $\psi_{1/2}$ since even
$(\psi_{-1/2},\psi_{1/2})_{\C^4}=0$. This implies that the
exchange energy on this state vanishes $Y_c\varphi_c=0$,  leaving
us with the direct term $X_c$ in $W_c$ only. Until the final
formula (\ref{xi}), we carry out the computation for a general
${\cal J}\psi_c$ where ${\cal J}$ is a general $SU(2)$ rotation in
$\hh$, cf. appendix \ref{app:DiracEqu}.

In order to proof the claim we just assemble the expansions given in Lemma \ref{Lambda}, \ref{psic}, \ref{W}.  Let us first compute $W_c{\cal J}\psi_c$:
\begin{eqnarray}
\label{WJpsi}
W_c{\cal J}\psi_c=X^{(0)}{\cal J}\psi^{(0,1)}_c-(Y^{(0)}A\phi\!\circ\! R^{-1})\,\psi^{(0,1)}_c+\frac{1}{c²}\sum_{j=1}^5 S_j.
\end{eqnarray}
All the remainder terms $S_j$, $j=1,2,...$, and $S$ below are defined and estimated in appendix \ref{app:estimates}. We apply now $\Lambda_c$ on (\ref{WJpsi}),
\begin{eqnarray*}
\Lambda_cW_c{\cal
J}\psi_c&=&\left(\Lambda^{(0)}+\frac{1}{c}\Lambda^{(1)}\right)\,W_c{\cal
J}\psi_c+\frac{1}{c²}S_6,
\end{eqnarray*}
and compute the zeroth order contribution,
\begin{eqnarray*}
\Lambda^{(0)}\,W_c{\cal J}\psi_c&=&\frac{1}{2}(1-\beta)\left(X^{(0)}{\cal J}\psi^{(0,1)}_c-(Y^{(0)}A\phi\!\circ\! R^{-1})\,\psi^{(0,1)}_c\right)+\frac{1}{c²}\sum_{j=1}^5 \Lambda^{(0)}S_j.
\end{eqnarray*}
For the action in first order $\Lambda^{(1)}$ we find that
\begin{eqnarray*}
\frac{1}{c}\Lambda^{(1)}\,W_c{\cal J}\psi_c&=&%-\frac{1}{2m}\,\alpha\cdot  p \left(X^{(0)}{\bf A}\varphi^{(0,1)}_c-(Y^{(0)}A\phi\circ R)\,\varphi^{(0,1)}_c\right)\\
%&&+\frac{1}{c²}\sum_{j=1}^5 \Lambda^{(1)} S_j\\
- \,\Lambda^{(0)}\,W_c{\cal J}\psi_c\nonumber\\
&&+\frac{1}{2}\left((\phi,A\phi\!\circ\! R^{-1})_{\C²}\ast p_j V\right)\,{{0\choose \sigma_j\phi}}-\frac{1}{2}\left(|\phi|^2_{\C²}\ast p_j V\right)\,{{0\choose \sigma_jA\phi\!\circ\! R^{-1}}}\nonumber\\
&&+\frac{1}{c²}\sum_{j=1}^5 \left(\Lambda^{(0)}+\frac{1}{c}\Lambda^{(1)}\right)S_j+\frac{1}{c²}\sum_{j=7}^{10} S_j.
\end{eqnarray*}
Putting it all together we finally arrive at
\begin{eqnarray}
\label{FinalDec}
\Lambda_cW_c{\cal J}\psi_c&=&\frac{i}{2c}{{0\choose \xi}}+\frac{1}{c²}S,
\end{eqnarray}
where we used the relation $A\,z\cdot\sigma A^\ast=
(Rz)\cdot\sigma$ from appendix \ref{app:DiracEqu}, and $\xi$ is
given by
\begin{eqnarray}
\label{xi} \xi:=\left ((\phi,A\phi\!\circ\! R^{-1}))_{\C²}\ast
(-\nabla V)\cdot\sigma\right)\phi-\left(|\phi|²_{\C²}\ast (-\nabla
V)\cdot\sigma\right)\,A\phi\!\circ\! R^{-1}.
\end{eqnarray}
In order to prove that $\xi$ is {\it non-zero} we use the fact
stated in Lemma \ref{psic} that for our choice of
$\psi_c=\psi_{1/2}$ the wavefunction $\phi$ in (\ref{xi}) is the
ground state $\psi_0$ of the (Pauli) hydrogen atom,
\begin{eqnarray}
\label{phi}
\phi(x)=\psi_0(|x|)\,\chi_+,
\end{eqnarray}
where $\chi_+:=(1,0)^T$, and $\chi_-:=(0,1)^T$ (we could have
chosen $\chi_-$ instead of $\chi_+$ in (\ref{phi})). Plugging
$A=-i\sigma_2$ into (\ref{xi}) (the rotation ${\cal J}$ with this
$A$ and the corresponding $R$ being a rotation around the 2-axis
by the angle $\pi$ maps $\psi_{1/2}$ onto $\psi_{-1/2}$) and using
the fact that $\psi_0$ has vanishing angular momentum (\ref{phi})
we find the lower component  $\xi_-$ of $\xi\in L²(\R³)²$ to be
\begin{eqnarray*}
\xi_-=\left(|\psi_0|²\ast (-\nabla V)\right)\,\psi_0.
\end{eqnarray*}
But the factor $E:=|\psi_0|²\ast (-\nabla V)$ is just the
electrostatic Coulomb field generated by the spherically symmetric
charge distribution $|\psi_0|²$. Hence, Newton's theorem implies
that $E$ is a central field generated by the {\it point charge}
$Q(|x|)$ centered at the origin composed of all the charge inside
the ball $B_{|x|}(0)$ of radius $|x|$ around $0$, and $Q(|x|)$ is
{\it strictly positive} due to the nature of $\psi_0$,
\begin{eqnarray*}
E(x)=Q(|x|)\, \frac{x}{|x|},\quad Q(|x|)=\|\psi_0\|^2_{L²(B_{|x|}(0))}>0.
\end{eqnarray*}
This concludes the proof of Theorem \ref{theorem}. $\Box$

\vspace{1cm}



%\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%appendix

\begin{appendix}
\label{appendix}

\section{Ingredients for  the Dirac equation}
\label{app:DiracEqu}

\subsection{$SU(2)$ rotation in $L^2(\R³)\otimes\C^4$}

We denote by ${\cal J}: L²(\R³)^4\rightarrow L²(\R³)^4$ the
$SU(2)$ rotations on $\hh=L²(\R³)^4$,
\begin{eqnarray*}
\label{J}
{\cal J}\psi(x):=e^{-i\varphi nJ}\psi(x)=e^{-i\varphi nS}\psi(R^{-1}(n,\varphi)\,x)
\end{eqnarray*}
for the spin angular momentum $S=-i/4\alpha\wedge\alpha$, the
orbital angular momentum $L=x\wedge p$, and the total angular
momentum $J=L+S$; $R(\varphi,n)$ denotes the $SO(3)$ rotation by
the  angle $\varphi\in [0,4\pi)$ around the axis $n\in\R³$,
$|n|_{\R³}=1$. Using the standard representation
(\ref{def:StdRep}) and the bispinor notation
$\psi=:(\psi_+,\psi_-)^T$, we can write the action of ${\cal J}$
as
\begin{eqnarray*}
\label{rot}
{\cal J}\psi=\left(\begin{array}{c}A\psi_+\!\circ\! R^{-1}\\
 A\psi_-\!\circ R^{-1}\!\end{array}\right),
\end{eqnarray*}
where $A\in SU(2)$ and $R\in SO(3)$ are connected by the above
$SU(2)/_{ \{ \pm {\bf 1} \}}\simeq SO(3)$ epimorphism stemming
from the defining relation $A\,z\cdot\sigma A^\ast=
(Rz)\cdot\sigma$, where again
$\sigma=(\sigma_1,\sigma_2,\sigma_3)$, and  $z\in\C³$.

\subsection{The standard representation of the Dirac matrices}

The standard representation for the Dirac matrices reads as follows.
\begin{eqnarray}
\label{def:StdRep}
\beta=\left(\begin{array}{cc}1 & 0\\ 0 & -1 \end{array}\right),\quad
\alpha_j=\left(\begin{array}{cc}0 & \sigma_j\\ \sigma_j& 0 \end{array}\right),
\end{eqnarray}
for $j=1,2,3$, $\alpha=(\alpha_1,\alpha_2,\alpha_3)$, with the Pauli matrices
\begin{eqnarray*}
  \sigma_1=\left(\begin{array}{cc}0 & 1\\ 1 & 0 \end{array}\right),\quad
  \sigma_2=\left(\begin{array}{cc}0 & -i\\ i & 0 \end{array}\right),\quad
  \sigma_3=\left(\begin{array}{cc}1 & 0\\ 0 & -1 \end{array}\right).
\end{eqnarray*}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Estimates on higher order terms}
\label{app:estimates}


In the following are collected all the higher order terms from the
large $c$  expansions used in the proof of Theorem \ref{theorem}.
By definition, all the $S_j$ are multiplied by the factors $1/c²$.
Hence, given (\ref{FinalDec}), it suffices to show that for $c$
sufficiently large,
\begin{eqnarray*}
\|S_j\|_{L²(\R³)^4}={\cal O}(1).
\end{eqnarray*}
Since in our case at hand the {\it exchange} energy vanishes,
$Y_c\psi_{-1/2}=0$, we will restrict ourselves to the discussion
of the terms involving the {\it direct} operator $X_c$ only. The
estimates involving $Y_c$ are similar.

\subsection{Direct remainder terms}


In the sequel, all the norms and scalar products {\it without}
index act on $\hh=L²(\R³)^4$. A norm $\|\cdot\|_{m,p}$ stands for
the norm in the Sobolev space $W^{m,p}(\R³)$, $m=0,1,2,...$, $1\le
p\le \infty$, and we denote the $L^p$ norms $\|\cdot\|_{0,p}$
simply as $\|\cdot\|_p$. Furthermore, we write $L²$ instead of
$L²(\R³)^4$ and similarly for the other spaces ($H^1\equiv
W^{1,2}$).  Finally, let us write $\varphi_c$ instead of
$\psi_{-1/2}={\cal J}\psi_c$.
\begin{lemma}
\label{est:Xrem} The following remainder terms are uniformly
bounded in $c$ as soon as $c$ is sufficiently large,
\begin{eqnarray*}
\|S_j\|={\cal O}(1),\quad j=1,2,6,7,8.
\end{eqnarray*}
\end{lemma}
The remainder terms stemming from the direct part are given by
\begin{eqnarray*}
  S_1&:=&W_c{\cal J}\psi^{(c)},\\
  S_2&:=&X^{(c)}{\cal J}\psi^{(0,1)}_c,\\
  S_6&:=&\Lambda^{(c)}\,W_c{\cal J}\psi_c,\\
  S_7&:=&-\frac{1}{4}\left(|\phi|²_{\C²}\ast
     p_jV\right)\,{{\sigma_jA(\sigma\cdot   p\phi)\circ R^{-1} \choose 0}},\\
  S_8&:=&-\frac{1}{4}\left(|\phi|²_{\C²}\ast V\right)
     \,{{\sigma_jA(R^{-1}(p(\sigma\cdot   p\phi))\circ R^{-1})_j\choose 0}},\\
  S&:=&\sum_{j=1}^2 \left(\Lambda^{(0)}+\frac{1}{c}
     \Lambda^{(1)}\right)S_j+\sum_{j=6}^{8} S_j.
\end{eqnarray*}
{\bf Proof \ref{est:Xrem}}\\ We use the decomposition of the
Coulomb potential as $V=-1/|x|=V_0+V_\infty$ with
$V_0:=V\chi_{B}\in L¹\cap L^{3-\varepsilon}$ and
$V_\infty:=V\chi_{\R³\setminus B}\in L^{3+\varepsilon}\cap
L^\infty$ where $\epsilon>0$, $\chi_A$ is  the characteristic
function $\chi$ of the subset $A\in\R³$, and $B:=B_R(0)$ denotes
the ball of finite radius $R<\infty$ centered at the the origin
$0\in\R³$.

$\bullet$  $S_1$\\ We start with an $L^\infty$ estimate on the
direct term $X_c=|\psi_c|²\ast V$ where $\psi_c\in W^{1,2}$.
%The following estimates use H\"older, Young, and the Sobolev continuous imbedding $W^{1,2}(\R³)\hookrightarrow L^p(\R³)$ for $2\le p\le 6$ to show that $X_c\in L^\infty(\R³)$
%\begin{eqnarray}
%\||\psi_c|²\ast V_0\|_\infty\le \|\psi_c\|^2_4\,\|V_0\|,\quad  \||\psi_c|²\ast% V_\infty\|_\infty\le \|\psi_c\|²\,\|V_\infty\|_\infty.
%\end{eqnarray}
Using Kato's inequality \cite[p.307]{K95},
$\|\psi/|x|^{1/2}\|^2\le \pi/2\,(\psi,|\nabla|\psi)$, we
immediately have
\begin{eqnarray*}
   \|X_c\|_\infty\le\frac{\pi}{2}\|\psi_c\|\,\|\nabla\psi_c\|<\infty,
\end{eqnarray*}
and hence, due to  Lemma \ref{psic}, uniformly  in the speed of
light $c$, $\|S_1\|=\|X_c \varphi_c\|\le \|X_c\|_\infty
\|\varphi_c\|$.
%\begin{eqnarray}
%\|S_1\|&=&\|X_c \varphi_c\|\le \|X_c\|_\infty \|\varphi_c\|.
%\end{eqnarray}
%Uniformity in the speed of light $c$ is guaranteed by Lemma \ref{psic} and the Sobolev imbedding.
%\begin{eqnarray}
%\label{est:X,Y}
%\|X_c\psi\| \,\,\mbox{and}\,\,\|Y_c\psi\|\le 2\,\|p\psi\|
%\end{eqnarray}
%\begin{eqnarray}
%\|S_1\|=\|(X_c-Y_c){\cal J}\psi^{(c)}\|\le 4\,\|\psi^{(c)}\|_{H¹(\R³)^4}={\cal O}(1),
%\end{eqnarray}

%\vspace{0.5cm}

$\bullet$  $S_2$\\ All the terms in $S_2$ have the form $((fg)\ast
V)\varphi_c^{(0,1)}$ for $f,g\in W^{1,2}$ and
$\varphi_c^{(0,1)}\in L^1\cap L^\infty$ due to the nature of the
ground state wave function $\phi(x)=e^{-1/2\,|x|}$. Using H\"older
and Young we have
\begin{eqnarray*}
\|((fg)\ast V)\varphi_c^{(0,1)} \|\le \|f\|\,\|g\|\,\|V_0\|\, \|\varphi_c^{(0,1)}\|_\infty+\|f\|\,\|g\|\,\|V_\infty\|_\infty\, \|\varphi_c^{(0,1)}\|.
% \|((fg)\ast V_0) h\|+\|((fg)\ast V_\infty) h\|\\
%&\le& \|(fg)\ast V_0\|\, \|h\|_\infty+\|(fg)\ast V_\infty\|_\infty\, \|h\|\\
%&\le& \|f\|\,\|g\|\,\|V_0\|\, \|h\|_\infty+\|f\|\,\|g\|\,\|V_\infty\|_\infty\, \|h\|
\end{eqnarray*}
Uniformity in $c$ follows again from Lemma \ref{psic}.


$\bullet$  $S_6$\\ We use the expansions of $\Lambda_c$ from
(\ref{Lambda_c}), of $X_c$ from (\ref{X_c}), and the one of
$\varphi_c$ from (\ref{psidec}),
\begin{eqnarray*}
   S_6&=&\Lambda^{(c)}X_c\varphi_c
   =\Lambda^{(c)}X^{(0)}\varphi_c^{(0,1)}+\frac{1}{c²}
      \Lambda^{(c)}X^{(0)}\varphi^{(c)}+\frac{1}{c²}\Lambda^{(c)}X^{(c)}\varphi_c\\
   &=&\Lambda^{(c)}X^{(0)}\varphi_c^{(0,1)}+\left(\Lambda_c -\Lambda^{(0)} -\frac{1}{c}\Lambda^{(1)}\right)
      X^{(0)}\varphi^{(c)}+\left(\Lambda_c -\Lambda^{(0)}
      -\frac{1}{c}\Lambda^{(1)}\right) X^{(c)} \varphi_c,
\end{eqnarray*}
and write $S_6=:\sum_{j=1}^3 S_{6,j}$.\\ $\bullet$  $S_{6,1}$\\
Using the definition of $\Lambda^{(c)}$ given in
(\ref{def:Lambda}) we decompose $S_{6,1}$ into the following
parts, $S_{6,1}=S_{6,1,1}+S_{6,1,2}+S_{6,1,3}$,
\begin{eqnarray}
S_{6,1,1}&:=&\frac{1}{2}X^{(0)}\varphi_c^{(0,1)},\nonumber\\
S_{6,1,2}&:=&\lim_{\delta\to 0,\rho\to\infty} \frac{1}{\pi}\left[\int_{-\rho/c²}^{-\delta/c²}+\int_{\delta/c²}^{\rho/c²}\right]d\eta\,\,R_1^{(c)}(\eta)X^{(0)}\varphi_c^{(0,1)},\nonumber\\
\label{def:S613}
S_{6,1,3}&:=&\lim_{\delta\to 0,\rho\to\infty} \frac{1}{\pi}\left[\int_{-\rho/c²}^{-\delta/c²}+\int_{\delta/c²}^{\rho/c²}\right]d\eta\,\,R_2^{(c)}(\eta)X^{(0)}\varphi_c^{(0,1)},
\end{eqnarray}
for the ingredients $R^{(c)}=R^{(c)}_1+R^{(c)}_2$ of
$\Lambda^{(c)}$ in (\ref{ResExp}) which are well-defined on the
domain of $p$, $D(p)=W^{1,2}$:
\begin{eqnarray*}
R^{(c)}_1&:=&(\beta-i\eta)^{-1}\alpha \cdot
p\,\left(\frac{1}{c}\alpha \cdot p+\beta-i\eta\right)^{-1}\alpha
\cdot p\,(\beta-i\eta)^{-1},\\
R^{(c)}_2&:=&-\left(\frac{1}{c}\alpha \cdot
p+\beta+\frac{1}{c²}V-i\eta\right)^{-1}V\,\left(\frac{1}{c}\alpha
\cdot p+\beta-i\eta\right)^{-1}.
\end{eqnarray*}
Since $X^{(0)}\in L^\infty$ we can immediately estimate $S_{6,1,1}$ as $\|S_{6,1,1}\|\le1/2\,\|X^{(0)}\|_\infty\|\varphi_c^{(0,1)}\|$. In order to estimate $S_{6,1,2}$ we  split $\phi_c^{(0,1)}$ into its upper part $\varphi_0\in W^{2,2}$ and its lower part $\varphi_1\in W^{1,2}$,
\begin{eqnarray*}
\varphi_c^{(0,1)}=\varphi_0+\frac{1}{c}\varphi_1,\quad
\varphi_0:={\phi\choose 0},\quad \varphi_1:=\frac{1}{2}{0\choose
\sigma\cdot   p\, \phi}.
\end{eqnarray*}
We will treat the action of $R^{(c)}_1$ on $\varphi_0$ and on
$\varphi_1$ differently. This leads us to the further
decomposition $S_{6,1,2}=:S_{6,1,2,1}+S_{6,1,2,2}$. Consider first
$S_{6,1,2,1}$, i.e., the action on $\varphi_0$. In order to act on
the domain of the Laplacian $W^{2,2}$ we can write
\begin{eqnarray*}
R_1^{(c)}&=&\left(\frac{1}{c}\alpha \cdot
p+\beta-i\eta\right)^{-1}\alpha\cdot
p\,(\beta-i\eta)^{-1}\alpha\cdot p\,(\beta-i\eta)^{-1}\\
&=&-\frac{1}{1+\eta²}\left(\frac{1}{c}\alpha\cdot
p+\beta-i\eta\right)^{-1}\left(\begin{array}{cc}-\Delta & 0\\ 0 &
-\Delta \end{array}\right),
\end{eqnarray*}
where we used $(\beta-i\eta)^{-1} \alpha\cdot
p\,(\beta-i\eta)^{-1}=-1/(1+\eta²)\,\alpha\cdot  p$ and
$\alpha\cdot p\,\alpha\cdot  p=\mbox{diag}(-\Delta,-\Delta)$. Now,
since $\Delta (X^{(0)}\phi)=(\Delta X^{(0)})\phi+2(\nabla
X^{(0)})(\nabla \phi)+X^{(0)}\Delta \phi$ and $\Delta
X^{(0)}=|\phi|²$, we estimate the integrand of $S_{6,1,2,1}$ as
follows,
\begin{eqnarray*}
\|R_1^{(c)}X^{(0)}\varphi_0\|\le\frac{1}{1+\eta²}\,\left(\|\phi\|_\infty^2+\|X^{(0)}\|_\infty+\|\nabla X^{(0)}\|_\infty\right)\|\phi\|_{2,2},
\end{eqnarray*}
using $\nabla X^{(0)}\in L^\infty$, and we bounded the resolvent
by $1$ uniformly in $c$. $S_{6,1,2,1}$ is now uniformly bounded in
$c$ since the integrand decays sufficiently fast in $\eta$.\\ We
now turn to the estimate on $S_{6,1,2,2}$, i.e., to the action of
$R_1^{(c)}$ on $\varphi_2$. To this end, we rewrite $R_1^{(c)}$ by
inserting $\beta$ in order to bound $1/c\,\alpha\cdot  p+\beta$ by
its resolvent. Using the formula $H(H-z)^{-1}=1+z(H-z)^{-1}$ leads
to
\begin{eqnarray}
\label{R1c}
R^{(c)}_1&=&%-\frac{c}{1+\eta²}\,(\beta-i\eta)^{-1}\left(\frac{1}{c}\alpha\cdot  p+\beta\right)\,\left(\frac{1}{c}\alpha\cdot  p+\beta-i\eta\right)^{-1}(\beta-i\eta)\,\alpha\cdot  p\nonumber\\
%\frac{c}{1+\eta²}\,(\beta-i\eta)^{-1}\beta\,\left(\frac{1}{c}\alpha\cdot  p+\beta-i\eta\right)^{-1}(\beta-i\eta)\,\alpha\cdot  p\\
-\frac{c}{1+\eta²}\,\alpha\cdot
p-\frac{c}{1+\eta²}\,(\beta-i\eta)^{-1}i\eta\,\left(\frac{1}{c}\alpha\cdot
p+\beta-i\eta\right)^{-1}(\beta-i\eta)\,\alpha\cdot  p\nonumber\\
&&+\frac{c}{1+\eta²}\,(\beta-i\eta)^{-1}\beta\,\left(\frac{1}{c}\alpha\cdot
p+\beta-i\eta\right)^{-1}(\beta-i\eta)\,\alpha\cdot  p.
\end{eqnarray}
Similarly to $S_{6,1,2,1}$ acting on $1/c \,\varphi_1$ yields the estimate
\begin{eqnarray*}
   \frac{1}{c}\|R^{(c)}_1\varphi_1\|\le \frac{2}{(1+\eta²)}\,
   \left(\|X^{(0)}\|_\infty+\|\nabla X^{(0)}\|_\infty\right)\|\phi\|_{2,2}
\end{eqnarray*}
where, this time, we used the decay of the resolvents in $\eta$ to
compensate the additional factor $\eta^2$ in (\ref{R1c}).
Collecting all the estimates yields a finite bound on $S_{6,1,2}$
which is uniform in $c$. In order to come to an end with the
desired bound on $S_{6,1}$ it remains to estimate $S_{6,1,3}$
given in (\ref{def:S613}). This time, we estimate the integrand by
using the fact that the Coulomb potential $V$ is relatively
bounded with respect to the momentum $p$, i.e., $\|V\psi\|\le 2
\|p\psi\|$ (Hardy's inequality, uncertainty principle).
\begin{eqnarray*}
   \|R^{(c)}_2\varphi_c^{(0,1)}\|\le \frac{1}{\sqrt{((1/2)²+\eta²)
   (1+\eta²)}}\,\|V(p+i)^{-1}\|\,\|(p+i)X^{(0)}\varphi_c^{(0,1)}\|
\end{eqnarray*}
Here, we estimated the resolvents by extracting each time a
decaying factor ($1/2$ for the ground state energy of $1/c²\,H_c$
on the electronic sector), and we introduced $1=(p+i)^{-1}(p+i)$
on $W^{1,2}$ which bounds $V$. Concluding the estimate on
$S_{6,1}$ we have
\begin{eqnarray*}
\|R^{(c)}_2\varphi_c^{(0,1)}\|\le \frac{8}{1+\eta²}\,\left( \|X^{(0)}\|_\infty+\|\nabla X^{(0)}\|_\infty\right)\|\varphi_c^{(0,1)}\|_{1,2}.
\end{eqnarray*}
$\bullet$ $S_{6,2}$\\ Due to the fact that $\Lambda_c$ is a
projection we immediately have
$\|S_{6,2,1}\|=\|\Lambda_cX^{(0)}\varphi^{(c)}\|\le
\|X^{(0)}\|_\infty\|\varphi^{(c)}\|_{1,2}$. The same estimate
holds for $S_{6,2,2}:=\Lambda^{(0)}X^{(0)}\varphi^{(c)}$ and
$S_{6,2,3}:=\Lambda^{(1)}X^{(0)}\varphi^{(c)}$ due to
(\ref{Lambda_c}) and $X^{(0)}\in W^{1,\infty}$, $\varphi^{(c)}\in
W^{1,2}$, cf. above.

$\bullet$ $S_{6,3}$\\
Since $X^{(c)}\in L^\infty$ we immediately have for $S_{6,3,1}:=\Lambda_cX^{(c)}\varphi_c$ and $S_{6,3,2}:=\Lambda^{(0)}X^{(c)}\varphi_c$ that $\|S_{6,3,1}\|,\, \|S_{6,3,2}\| \le \|X^{(c)}\|_\infty\|\varphi_c\|$ which is again uniform in $c$. It remains to estimate $S_{6,3,3}:=\Lambda^{(1)}X^{(c)}\varphi_c$ which is of the form $((fg)\ast\nabla V)\varphi_c$ for $f,g\in W^{1,2}$. Using the Sobolev imbedding $W^{1,2}(\R^3)\hookrightarrow (L^2\cap L^6)(\R^3)$ we finally find
\begin{eqnarray*}
       \|((fg)\ast\nabla V_0)\varphi_c\|\le\|f\|_6\|g\|_6\|
       \varphi_c\|_6\|V_0\|_1,\quad\|((fg)\ast\nabla V_\infty)
       \varphi_c\|\le \|f\|\,\|g\|\,\|\varphi_c\|\,\|V_\infty\|_\infty.
\end{eqnarray*}
$\bullet$ $S_{7}$, $S_{8}$, $S$\\ The estimates of $S_{7}$,
$S_{8}$ are carried out as above, only using the properties of $V$
and its gradient $\nabla V$. The estimates in $S$ correspond to
the estimates already performed in $S_{6,2}$ and $S_{6,3}$. $\Box$

\subsection{Exchange remainder terms}
For the sake of completeness we list he exchange remainder terms.
The estimates on these terms may be performed following similar
lines as for the direct part.
\begin{eqnarray*}
S_3&:=&-\frac{1}{2}(Y^{(1)}A(\sigma\cdot   p \phi)\circ
R^{-1})\,\psi^{(0,1)}_c\\ S_4&:=&-(Y^{(c)}{\cal
J}\psi^{(0,1)}_c)\,\psi^{(0,1)}_c \\ S_5&:=&-Y_c\psi^{(c)}\\
S_9&:=&\frac{1}{4}\left((\phi,A\phi\circ R^{-1})_{\C²}\ast
p_jV\right)\,{{\sigma_j(\sigma\cdot   p\phi)\choose 0}}\\
S_{10}&:=&\frac{1}{4}\left((\phi,A\phi\circ R^{-1})_{\C²}\ast
V\right)\,{{\sigma\cdot   p(\sigma\cdot   p\phi)\choose 0}}\\
S&:=&\sum_{j=3}^5
\left(\Lambda^{(0)}+\frac{1}{c}\Lambda^{(1)}\right)S_j+\sum_{j=9}^{10}
S_j
\end{eqnarray*}

\end{appendix}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%appendix

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%references


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\end{thebibliography}

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\end{document}
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%X^{(c)}&:=&\left[\frac{1}{4} |\sigma\cdot   p \phi|_{\C²}+ 2\,\mbox{Re}\,\,(\varphi^{(0,1)}_c,\psi^{(c)})_{\C^4}+\frac{1}{c²}(\psi^{(c)},\psi^{(c)})_{\C^4}\right]\ast V

%Y^{(c)}&:=&(\psi^{(c)},\cdot)_{\C^4}\ast V
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