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non-relativistic QED, enhanced binding, Pauli-Fierz
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\begin{document}


\markboth{\scriptsize{HVV \version}}{\scriptsize{HVV \version}}
\title{\bf{Enhanced Binding in non-relativistic QED}}
\author{\vspace{5pt} Christian Hainzl$^{1,3}$, Vitali Vougalter$^2$ 
and Semjon A. Vugalter$^1$\\
\vspace{-4pt}\small{$1.$ Mathematisches Institut, LMU M\"unchen}\\ \vspace{-4pt}
\small{Theresienstrasse 39, 80333 M\"unchen, Germany}\\
\small{{\texttt{\{hainzl,wugalter\}@mathematik.uni-muenchen.de}}}\\
\vspace{-4pt}\small{$2.$ Department of Mathematics, University of British Colombia}
\\ \vspace{-4pt} \small{Vancouver, B.C. V6T 1Z2}\\ \small{{\texttt{vitali@math.ubc.ca}}}}


\date{\small\version}
\maketitle

\begin{abstract}
We consider a spinless particle coupled to a photon field
and prove that even if the Schr\"odinger operator $p^2 + V$ does not
have eigenvalues the system can have a ground state. We describe the coupling by means of the 
Pauli-Fierz Hamiltonian and our
result holds in the case where the coupling constant $\al$ is small.
\end{abstract}
\footnotetext[3]{Marie Curie Fellow}


\section{INTRODUCTION}


In the picture of Quantum electrodynamics (QED) atoms consist of charged particles, which are necessarily coupled 
to a photon field. If one neglects the radiation effects one obtains the Schr\"odinger operator. Although the fundamental properties of 
the one-particle and multi-particle Schr\"odinger operators have been successfully studied since the middle 
of the last century, the systematical mathematical study of the non-relativistic QED model was initiated 
by Bach, Fr\"ohlich, and Sigal  in \cite{BFS1,BFS2,BFS3} only a couple of years ago (a comprehensive review of results 
in non-relativistic QED can be found in \cite{GLL}) and some very fundamental problems are remaining still open. One of these problems 
is the question of enhanced binding via interaction with a quantized radiation field.

Consider a three-dimensional particle in a potential well $\be V(x)$ with $V(x) \leq 0$. If the potential well is not deep enough,
i.e. $\be$ is small, the corresponding Schr\"odinger operator does not have a discrete spectrum and binding does not occur. For a negative potential
there exists a critical value $\be_0$ such that for $\be > \be_0$ there is at least one bound state whereas for $\be \leq \be_0$ no particle can be bound.

In a recent paper Griesemer, Lieb, and Loss (\cite{GLL}) proved that a photon field cannot decrease the binding energy. If the 
Schr\"odinger operator with potential $\be V$ has an eigenvalue, the corresponding energy operator in non-relativistic 
QED (Pauli-Fierz Hamiltonian) has a ground state.

However, the physical intuition tells us that interaction with a photon field must increase binding. Leaving the region 
of the potential the particle produces a photon field and consequently it needs more energy to
leave the potential well.

The goal of this paper is to give a mathematical rigorous proof this phenomenon. Previously the enhanced binding 
was studied in the dipole approximation by Hiroshima and Spohn in \cite{HS}. In this approximation it is assumed that the magnetic vector 
potential does not depend on the coordinates of the particle. They proved that, if the potential $\be V$ is fixed, for sufficiently large values of
the coupling parameter $\al$ (which is the fine structure constant, see (\ref{rpf})), binding takes place.

Our approach to this problem is different. On the one hand we study 
the Pauli-Fierz operator without additional restrictions on the magnetic vector potential and on the other hand our results hold for
small values of $\al$ (recall, that the physical value of $\al$ is about $1/137$).
We prove that in case of the Pauli-Fierz operator (and for small $\al$ enough)
binding starts at values of $\be$ strictly less than $\be_0$.




\section{DEFINITIONS AND MAIN RESULTS}

We describe the self-energy of the particle by 
\begin{equation}\label{rpf}
T = (p + \sqrt{\al}A(x))^2 +  H_f.
\end{equation}

The unit of energy is $mc^2/2$, where $m$ is the electron mass, the unit length
is $l_c = 2 \hbar /mc$, twice the Compton wavelength of the particle, and
$\al= e^2/\hbar c$ is the  {\it fine structure constant}. The physical value is $\al \sim 1/137$. In the present paper $\al$
plays the role of a small, dimensionless number. Our results hold for sufficiently small values of $\al$. The charge of the electron is 
$e$.

The operator $p= -i\nabla_x$,
while $A$ is the magnetic vector potential.
The unit of $A(x)^2$ is $mc^2/2l_c$. 
We fix the gauge ${\rm div} A =0$.


The Hilbert space is
\begin{equation}
\Hh =
\Ll^{2}({\mathbb {R}}^{3})\otimes \F,
\end{equation}
where
$\F$ is the symmetric Fock space for the photon field.

Throughout this paper we denote  the position-vector of the particle by $x$
and by $\xi_i$ the position-vectors of the photons. It is also convenient to use
the relative coordinates $x-\xi_i$ for the photons.
The vector potential is
\begin{equation}
A(x) = \sum_{\lambda = 1,2} \int_{\R^3} \frac{\chi(|k|)}{|k|^{1/2}} [\ean \aan(k)
e^{ikx} +
\ec\ac(k) e^{-ikx}] dk,
\end{equation}
where the operators $\aan, \ac$ satisfy the usual commutation relations
\begin{equation}
[a_\nu (k),\ac(q)] = \delta(k-q)\delta_{\lambda,\nu}, \quad [\aan(k), a_\nu(q)] = 0, \ \
{\rm etc.}
\end{equation}
The vectors $\ean(k)$ are the two possible orthonormal polarization vectors perpendicular to $k$.
The units are chosen as in \cite{GLL}.

Obviously,
\begin{equation}
A(x) = D(x) + D^\ast (x),
\end{equation}
where 
\begin{equation}
D(x) = \sum_{\lambda = 1,2} \int_{\R^3} \frac{\chi(|k|)}{|k|^{1/2}} [\ean \aan(k)
e^{ikx}  dk
\end{equation}
and $D^\ast$ is the  operator adjoint to $D$.

For $x\in \R^3$ let
\begin{equation}
\vec G(x) = \F^{-1}[\hat {\vec G} (k)] \equiv \F^{-1}\left[\sum_{\lambda = 1,2} \frac{\ean(k)\chi(|k|)}{|k|^{1/2}}   \right] ,
\end{equation}
where $\F$ is the Fourier transform and let
$\psi_n(x,\xi_1,...,\xi_n)$ be the $n$-photon component of a function $\Psi \in \Hh$.
Obviously, 
\begin{equation}
[D\psi_n]_{n-1} (x,\xi,...,\xi_{n-1}) = \sqrt{n} \int \vec G(x-\xi_n) \psi_n(x,\xi_1,...,\xi_n)d\xi_n.
\end{equation}
We shall also use the following commutation relation \cite{GLL}
\begin{equation}
[D,D^\ast] = 4\pi \Lambda^2.
\end{equation}



The function $\chi(|k|)$ describes the ultraviolet cutoff on the
wavenumbers $k$. 

Concerning this cutoff we do not make further assumptions except those that are needed to apply
\cite{GLL}. It may be a sharp or a smooth cutoff. Only to ease the understanding of the calculations one can suppose that 
$\chi (|k|) $ is  the characteristic
function $\Theta( \Lambda - |k|)$. 


The photon field energy $H_f$ is given by
\begin{equation}
H_f = \sum_{\lambda= 1,2} \int_{\R^3} |k| \ac (k) \aan (k) dk.
\end{equation}

Let again $\psi_n$ be the $n$-photon component of a function $\Psi \in \Hh, \psi_n \in H^{1/2}(\R^{3(n+1)})$.
Then,
\begin{equation}
(\psi_n,H_f \psi_n) = n (\psi_n, |\nabla_{\xi_1}| \psi_n).
\end{equation}
To prove existence of enhanced binding in non-relativistic QED we would like to compare binding in the presence of the photon field and
without it. To this end
let us introduce the  Schr\"odinger operator
\begin{equation}\label{sch}
h_\be = -\Delta  + \be V(x)
\end{equation}
with external potential $\be V(x) \in C(\R^3)$,
which we assume to be radial $V(x)=V(|x|)$, non positive $V(x) \leq 0$, and
with compact support.
It is known that there is a critical value of the parameter $\be_0 > 0$ such that for $\be \leq
\be_0$ there is no ground state and the operator (\ref{sch}) has
only an essential spectrum and at the same time for all $\be > \be_0$ the operator $h_\be$
has at least one eigenvalue.

The corresponding operator with a quantized radiation field is
\begin{equation}\label{ham}
{\bf H}_\be = T  + \be V(x).
\end{equation}

Our goal now is to show that the operator 
${\bf H}_{\be}$ has bound states for  values of $\be$ smaller $h_\be$.
To establish the existence of a ground state of ${\bf H}_\be$ we apply the criterion of \cite{GLL}, which says that  ${\bf H}_\be$
has a ground state if
\begin{equation}\label{crite}
\infspec {\bf H}_{\be} <  \infspec T.
\end{equation}
However, in contrast to the Schr\"odinger operator $h_\be$, for which the infimum of the spectrum without potential is always $0$, the 
$\infspec T$ is a complicated function depending on $\al$ and $\Lambda$. To prove the inequality (\ref{crite})
one needs precise estimates on this function. Our first result is the following asymptotic estimate on $\so = \infspec T$.


\begin{thm}[Localization of the spectrum of a free spinless particle]\label{pt2}
Let
\begin{equation}\label{c1}
{{\mathcal E}_0} = 2 \sum_{\mu,\nu =1,2}\int  \frac{ (\fg_\mu (p) \cdot \fg_\nu(q))^2}{[p+q]^2 + |p| + |q|}dpdq.
\footnote{To simplify the formulas we write for $\vec A,\vec B \in \R^3$ $\vec A\cdot \vec B = (\vec A,\vec B)_{\R^3}$}
\end{equation}
Then, for small $\al$,
\begin{equation}\label{coeb}
\Sigma_0 = \infspec T = 4\pi \al \Lambda^2 - \al^2 \Em_0 + o(\al^2).
\end{equation}
\end{thm}

Recall, that the operator $h_\be$ has a critical value $\be_0$ of the parameter $\be$ such that, for $\be\leq \be_0$, 
$h_\be$ does not have a bound state. Using Theorem \ref{pt2} we construct a variational trial function proving for
small values of $\al$ the following:
\begin{thm}[Enhanced binding]\label{mt}
For all   sufficiently small $\al$ there exists a number $\be_1{(\al)} < \be_0$,
such that for all $\be > \be_1{(\al)}$ the operator 
${\bf H}_{\be}$ has a ground state.
\end{thm}




\section{PROOF OF THEOREM \ref{pt2}}

\subsection{Some a-priori estimates} 

We start this section with an a-priori estimate on $\so$.
\begin{lem}\label{al}
For $\al \Lambda$ sufficiently small the inequality
\begin{equation}
\so \geq 4\pi \al \Lambda^2 - \bc\al^2 \Lambda^3,
\end{equation}
holds with a  constant $\bc$ independent of $\al$.
Moreover,
\begin{equation}
T \geq 4\pi\al \Lambda^2 - \bc\al^2 \Lambda^3 + \frac 12 p^2 +\frac 12 H_f
\end{equation}
in the sense of quadratic forms.
\end{lem}
\begin{proof}
For $\Psi $ in the form domain of $T$ we have
\begin{equation}\label{le39}
(\Psi,T\Psi) = \pa p\Psi\pa^2 + 2\sqrt{\al}(p\Psi,A(x) \Psi) + \al \pa A(x) \Psi\pa^2 + (\Psi, H_f 
\Psi).
\end{equation}
Notice that $A(x) = D(x) + D^\ast (x)$ and $[p,A(x)]=0$ imply
\begin{equation}\label{i33}
2\as(p\Psi,A(x)\Psi) = 4\as \Re (p\Psi,D(x)\Psi) \leq  
\sqrt{\al \Lambda}\pa p\Psi\pa^2 + 16 \sqrt{\al/ \Lambda} \pa D(x) \Psi\pa^2.
\end{equation}
Moreover, with the commutation relation
\begin{equation}\label{comm}
 D(x) D^\ast (x) =  D^\ast (x) D(x) + 4\pi\Lambda^2
\end{equation}
we get
\begin{equation}
\al \pa A\Psi\pa^2 = 4\pi \al \Lambda^2 \pa \Psi\pa^2 + 2\al(\Psi,D^\ast D\Psi) + 2\al \Re (\Psi,DD\Psi).
\end{equation}
By Schwarz inequality
\begin{equation}\label{lo1}
2\al\Re D(x) D(x) \leq {C^{-1}} D^\ast (x) D(x) + {C\al^2} D(x) D^\ast (x)\quad {\mbox{for an arbitrary}} \,\, C>0.
\end{equation}
The relations  (\ref{le39})-(\ref{lo1}) together with the estimate 
\cite{GLL}
\begin{equation}\label{conn}
D^\ast (x) D(x) \leq 8\pi \Lambda H_f
\end{equation}
imply
\begin{eqnarray}\label{hugo}
(\Psi, T\Psi) &\geq& (4\pi \al \Lambda^3 -4\pi C\al^2 \Lambda^2)\pa \Psi\pa^2 +\half \{\pa p\Psi \pa^2 +(\Psi,H_f \Psi)\} +
\\ \nonumber
&&  +(\half - \sqrt{\al \Lambda})\pa p\Psi \pa^2 + \left(\half - 8\pi [16 \sqrt{\al \Lambda} + C^{-1} \Lambda + \al^2 C\Lambda] \right) (\Psi,H_f \Psi)  .
\end{eqnarray}
For $C$ large enough and  $\al \Lambda$ sufficiently small the last two terms in (\ref{hugo}) are positive
which completes the proof of the Lemma.
\end{proof}


\begin{rem}
The inequalities (\ref{i33}) and (\ref{conn}) imply also that
\begin{eqnarray}\label{sepp}
&(\Psi, T \Psi)& \geq (1-\sqrt{\al \Lambda})\pa p\Psi \pa^2 + \al \pa A \Psi\pa^2 +
 (1- 128\pi \as \Lambda) (\Psi,H_f\Psi)\\\nonumber
&&\geq  4\pi\al \Lambda^2 + (1-C\as)  \pa p\Psi \pa^2 - 2\al |\Re (\Psi,DD\Psi)| +(1-C \as ) 
(\Psi,H_f\Psi),
\end{eqnarray}
with some constant $C>0$ independent of $\al$. We will use the last relation (\ref{sepp}) for obtaining a sharp lower bound on $\so$.
\end{rem}

\begin{rem}
Obviously $\so \leq 4\pi \al\Lambda^2$. Let $\Psi$ be a state with energy close to $\so$, i.e. $(\Psi,T\Psi) \leq 4\pi\al \Lambda^2 + \bc\al^2 \Lambda^3 $. By 
Lemma \ref{al} we conclude 
\begin{equation}\label{lo2}
\pa \nabla \Psi \pa^2 + (\Psi, H_f \Psi) \leq 4\bc \al^2 \Lambda^3 \pa \Psi\pa^2.
\end{equation}
The last inequality says that for small $\al$ no photon can be concentrated in a small region, in particular in a small region near the particle. 
The state $\Psi$ 
is
mainly supported in the regions where all photons are far away from the particle. This observation 
plays an important role for estimating  $\so$ and we will make it mathematically precise later.
\end{rem}


Another important observation, which we get from the proof of Lemma \ref{al}, is that as long as we consider 
the ground state energy of the \lq\lq free\rq\rq \ Hamiltonian, the term $\as pA$
is irrelevant, because it is relatively bounded by $\eps(p^2 + H_f)$ with some small constant $\eps$.
Consequently, the process of annihilation of {\it one} photon, which is involved in that term,
is not relevant for the estimation of $\so$. 

At the same time the annihilation of two photons, which is represented by the term $2\al \Re (\Psi,DD\Psi)$,
contributes to the next to leading order of $\al^2$. On this basis one can infer that a function $\Psi \in \Hh$ 
containing only two components, a $0$-photon part and $2$-photon part, can be a good candidate
for a minimizer of the quadratic form $(\Psi, T\Psi)$ up to an error of order $o(\al^2)$. Additionally, in order
to satisfy (\ref{lo2}) we shall take the two photon component of $\Psi$ with norm
bounded by $C\al$.

The presence of a negative external potential $V$ changes the role of the term
$\al (p\Psi, A\Psi)$ dramatically. The positive term $\pa p\Psi\pa^2$ can be compensated by the negative term
$(\Psi,V\Psi)$ and the term $\al(p\Psi,A\Psi)$ can be responsible for shifting the bottom of the spectrum.
This observation will be used in the proof of Theorem \ref{mt}.

\subsection{Upper bound for $\Sigma_0$}
Let
\begin{equation}
\Psi_n = \{f_n(x), 0, \al f_n(x) \varphi(\eta_1,\eta_2),0,0...\},
\end{equation}
with 
\begin{equation}\label{chef}
\varphi (\eta_1,\eta_2) = -\sqrt{2} {\mathcal F}^{-1}\left(\frac{\fg(p) \cdot \fg(q)}{[p+q]^2 + |p| 
+|q|}\right).
\end{equation}
We assume $f_n(x)$ to be a real function, with $\pa f_n \pa =1$.

Then 
\begin{eqnarray}\nonumber
(\Psi_n,T\Psi_n) &=& \pa \nabla_x f_n(x) \pa^2 + \al^2 (\varphi, [-i\nabla_{\eta_1} - 
i\nabla_{\eta_2}]^2 \varphi)
+ \al^2 \pa \nabla_x f_n(x) \pa^2 \pa\varphi\pa^2 \\ && + \al^2 \sum_{i=1,2}\pa 
|\nabla_{\eta_i}|^{1/2}\varphi\pa^2 
+ \al (\Psi_n, A^2 \Psi_n).
\end{eqnarray}
Notice, that 
\begin{equation}\label{ub1}
\al (\Psi_n, A^2 \Psi_n) = 4\pi \al\Lambda^2 + 2\al (\Psi_n,D^\ast D \Psi_n) + 2\al^2 \Re 
(f_n(x) , DDf_n(x) \varphi(\eta_1,\eta_2)).
\end{equation}
The first term on the r.h.s. of (\ref{ub1}) can be estimated (\cite{GLL}) as
\begin{equation}
2\al |(\Psi_n,D^\ast D \Psi_n) | \leq 16\pi \al^3 \Lambda \sum_{i=1,2}\pa 
|\nabla_{\eta_i}|^{1/2}\varphi\pa^2,
\end{equation}
and the last term is
\begin{equation}
2\sqrt{2} \al^2 \int \vec G(\eta_1) \cdot \vec G(\eta_2)  \varphi(\eta_1,\eta_2) d\eta_1d\eta_2,
\end{equation}
which implies 
\begin{eqnarray}\nonumber
(\Psi_n, T\Psi_n) &\leq& 4\pi \al \Lambda^2 +(1 + C\al^2) \pa \nabla _x f_n(x)\pa^2 + 
\al^2\{ (\varphi, [(i\nabla_{\eta_1} + i\nabla_{\eta_2})^2 + |\nabla_{\eta_1}| + 
|\nabla_{\eta_2}|]\varphi)\\ &&
+2\sqrt{2} \al^2 \int \vec G(\eta_1) \cdot \vec G(\eta_2)  \varphi(\eta_1,\eta_2) d\eta_1d\eta_2\} + 
O(\al^3),
\end{eqnarray}
where $C$ is independent of $\al, \Lambda, n$.

The inequality
\begin{equation}
\pa F_1\pa^2 + 2 (F_1,F_2) \geq - \pa F_2\pa^2,
\end{equation}
with 
\begin{equation}
F_1 = [(i\nabla_{\eta_1} + i\nabla_{\eta_2})^2 + |\nabla_{\eta_1}| + 
|\nabla_{\eta_2}|]\varphi(\eta_1,\eta_2) \, \, \in \Ll^2(\R^3)
\end{equation}
and
\begin{equation}
F_2 = \sqrt{2} [(i\nabla_{\eta_1} + i\nabla_{\eta_2})^2 + |\nabla_{\eta_1}| + |\nabla_{\eta_2}|]^{1/2} 
\vec G(\eta_1) \cdot \vec G(\eta_2)
\end{equation}
implies now that
\begin{equation}\label{ub32}
(\Psi_n, T\Psi_n) \leq 4\pi \al \Lambda^2 - \al^2\Em_0 +  (1 + C\al^2) \pa \nabla _x f_n(x)\pa^2 + 
O(\al^3),
\end{equation}
with 
$
\Em_0$ is given by (\ref{c1}).
The upper bound on $\so$ follows from (\ref{ub32}) if we take a sequence $f_n(x)$ with 
$ \pa \nabla _x f_n(x)\pa \to 0$.



\subsection{Lower bound on $\so$.}



\subsubsection{Partition of unity for the photon space and field energy estimates}


Let $\psi_n$ be the $n$-photon component of the state $\Psi$ and let $a_1$ and $\eps >0$ be some fixed 
but arbitrary numbers. 
Let $u_1(\eta)$ and $u_2(\eta)$ be the localization functions defined in Theorem \ref{loclem} with the above chosen 
$a_1 > 0$ and $\eps >0$. 
Obviously the products 
\begin{equation}
\ukn(\eta_1,...,\eta_n) := \Pi_{i=1}^{n}u_{k_i}(\eta_i), \quad k_i=1,2
\end{equation}
make the partition of unity in $\R^{3n}$ in the sense that $\sum_{k_1,...,k_n =1,2} \ukn^2 =1$,
where the sum is taken over all possible combinations of $k_i, k_i= 1,2$; $i=1,...,n$.

Applying Theorem \ref{loclem} we immediately arrive at 
\begin{equation}\label{lo3}
(\psi_n, H_f\psi_n) \geq (1-\eps) \sum_{k_1,...,k_n =1,2} (\ukn \psi_n, H_f \ukn \psi_n)
\end{equation}
The inequality (\ref{lo3}) has two important consequences. 
First, notice that 
the sum on the r.h.s. of (\ref{lo3}) contains $\frac{n(n-1)}2 $ terms corresponding to the products
$\ukn$ in which the function $u_1(\eta_i)$ appears exactly twice. Using the symmetry of $\psi_n$ 
we derive that for $n\geq 2$
\begin{equation}\label{lo4}
(\psi_n, H_f\psi_n) \geq (1-\eps)\frac{n(n-1)}2 ( \psi_n\unz, H_f  \psi_n\unz),
\end{equation}
where 
\begin{equation}
\unz = u_1(\eta_1)u_1(\eta_2)u_2(\eta_3)...u_2(\eta_n)
\end{equation}
and
\begin{equation}
(\Psi,H_f \Psi) \geq (1 -\eps) \sum_{n\geq 0} \frac{(n+2)(n+1)}2 ( \psi_{n+2}\uz, H_f  \psi_{n+2}\uz).
\end{equation}
Second, denote by $\gamma_0$ the lowest eigenvalue of the 
operator $|\nabla_\eta|^{1/2}$ with the Dirichlet boundary condition at $|\eta|= b_1$. 
(Recall that $u_1(\eta) = 0$ for $|\eta|\geq b_1$.)
Then for all $\ukn$ with $\sum_{i=1}^n k_i < 2n$ (at least one photon is localized near the electron) 
holds:
\begin{equation}
(\psi_n \ukn, H_f \psi_n \ukn) \geq \gamma_0 \pa \psi_n \ukn\pa^2,
\end{equation}
which implies
\begin{equation}\label{lo7}
(\psi_n, H_f\psi_n) \geq \gamma_0 (\pa \psi_n \pa^2 - \pa \psi^0_n\pa^2),
\end{equation}
with $\psi_n^0 = \psi_n \Pi_{i=1}^n u_2(\eta_i)$. Let $\Psi^0$ be the state with components 
$\{\psi_0^0, \psi_1^0,...,\psi_n^0,...\}$.
Comparing (\ref{lo7}) with (\ref{lo2}) we immediately conclude that for all states $\Psi$, with energy 
close to $\so$,
holds
\begin{equation}\label{lo8}
\pa\Psi^0\pa^2 \geq (1-4\bc\gamma_0^{-1} \al^2 \Lambda^3) \pa \Psi\pa^2,
\end{equation}
which means that for small $\al$ the difference between $\Psi_0$ and $\Psi$ is small. As the last 
remark for this section notice,
that for the operator $p_n^2 = - (\nabla_x + \sum_{i=1}^n \nabla_{\eta_i} )^2 $ the localization error 
corresponding to the localization with the functions $\ukn$ is the same as for the operator 
$-\sum \Delta_{\eta_i}$ and applying Remark \ref{locrem} to this operator in an analogical way as to  
(\ref{lo4}) we get 
\begin{equation}
(\psi_n,p_n^2\psi_n) \geq (1-\eps) \frac{n(n-1)}2 \pa (\nabla_x + \sum_{i=1}^n 
\nabla_{\eta_i} ) \psi_n\unz \pa^2
+(1-\eps)\pa (\nabla_x + \sum_{i=1}^n \nabla_{\eta_i} )\psi_n^0\pa^2,
\end{equation}
which implies 
\begin{equation}\label{lo10}
(\Psi,p^2\Psi) \geq (1-\eps) \sum_{n=0}^\infty \left\{ \frac{(n+1)(n+2)}2 \pa 
(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )  \psi_{n+2}\uz\pa^2+ \pa (\nabla_x + \sum_{i=1}^n \nabla_{\eta_i} )\psi_n^0\pa^2 \right\}.
\end{equation}
Summarizing the inequalities (\ref{sepp}), (\ref{lo2}) and (\ref{lo10}) we conclude that for small 
$\al$
\begin{eqnarray}\nonumber
(\Psi,T\Psi) &\geq& \al \pa A\Psi\pa^2 + \sum_{n=0}^\infty (1-2\eps) \left\{ 
\frac{(n+1)(n+2)}2 (\tps, H_f\tps) +  \right.\\ \label{lo11} &&
\left. +\pa (\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} ) \tps\pa^2 + \pa (\nabla_x + \sum_{i=1}^{n}
\nabla_{\eta_i} ) \psi^0_n\pa^2 \right\},
\end{eqnarray}
with 
\begin{equation}
\tps= \psi_{n+2} u_1(\eta_1) u_1(\eta_2) \prod_{i=3}^{n+2} u_2(\eta_i).
\end{equation}

\subsubsection{Estimate on the $A^2(x)$ term}


The inequality (\ref{sepp}) shows that the only term which can make the value of the quadratic form $(\Psi,T\Psi)$
less that $4\pi \al \Lambda^2$ is $2\al \Re (\Psi, DD\Psi)$ coming from
$ A(x)^2$. Our next goal is to estimate this term precisely.

Obviously
\begin{eqnarray}\nonumber
\al(\Psi, DD\Psi) &=& \al \sum_n \sqrt{(n+2)(n+1)} \int \overline{\vec G(\eta_1) \cdot \vec G(\eta_2) 
\psi_{n+2}
(x,\eta_1,...,\eta_{n+2})} \times\\
&&\times \psi_n(x, \eta_3,...,\eta_{n+2})dx \prod_{i=1}^{n+2} d\eta_i= \al \sum_n(I_{1,n} + I_{2,n}),
\end{eqnarray}
with
\begin{eqnarray}\nonumber
&&I_{1,n} = \sqrt{(n+2)(n+1)} \int \overline{\vec G(\eta_1) \cdot \vec G(\eta_2) \psi_{n+2}
(x,\eta_1,...,\eta_{n+2})} \times \\ && \qquad\qquad \times \psi_n(x, \eta_3,...,\eta_{n+2}) 
\prod_{i=3}^{n+2} u_2(\eta_i)^2 dx  d\eta
\end{eqnarray}
and
\begin{eqnarray}\nonumber
&&I_{2,n} = \sqrt{(n+2)(n+1)} \int \overline{\vec G(\eta_1) \cdot \vec G(\eta_2) \psi_{n+2}
(x,\eta_1,...,\eta_{n+2})} \times \\ && \qquad\qquad \times\psi_n(x, \eta_3,...,\eta_{n+2})(1- 
\prod_{i=3}^{n+2} u_2(\eta_i)^2) dx  d\eta,
\end{eqnarray}
where we used the notation $d\eta =\prod_{i=1}^{n+2} d\eta_i$.
Similarly to (\ref{lo1}) the bound
\begin{eqnarray}\nonumber
|\al I_{2,n}| &\leq& \al \pa D\psi_{n+2} \pa \pa D^\ast \psi_n (1- \prod_{i=1}^{n} u_2(\eta_i)^2)\pa 
\\
\nonumber &\leq& \al\sqrt{8\pi\Lambda} (\psi_{n+2},H_f\psi_{n+2})^{1/2} \left\{4\pi \Lambda^2 \pa 
\psi_n(1- \prod_{i=1}^{n} u_2(\eta_i)^2)
\pa^2 +\right.\\ \label{lo12} &&\left.+ 
\pa D \psi_n(1- \prod_{i=1}^{n} u_2(\eta_i)^2)
\pa^2\right\}^{1/2}
\end{eqnarray}
holds.
To estimate the last term on the r.h.s. of (\ref{lo12}) in terms of $(\psi_n, H_f\psi_n) $
we cannot use directly the inequality \cite{GLL}
\begin{equation}
D^\ast (x) D (x) \leq 8\pi  \Lambda H_f.
\end{equation}
However, a very similar estimate, given by Lemma \ref{applem} in the Appendix,
holds for some constant $C >0$ independent of $\psi_n$. Namely
\begin{equation}\label{lo13}
\pa D \psi_n(1- \prod_{i=1}^{n} u_2(\eta_i)^2)
\pa^2 \leq C(\psi_n,H_f\psi_n),
\end{equation}
which implies 
\begin{equation}\label{lo15}
|\al I_{2,n}| \leq \al 8\pi \Lambda [(\psi_{n+2} , H_f \psi_{n+2}) + (\psi_n, H_f,\psi_n)] + \al 4\pi \Lambda^2 C\pa\psi_n(1- 
\prod_{i=1}^{n} u_2(\eta_i)^2)
\pa^2 ,
\end{equation}
where $C$ is independent of $\al$.
Notice, that 
\begin{equation}
\pa\psi_n(1- \prod_{i=1}^{n} u_2(\eta_i)^2)
\pa^2 \leq \pa \psi_n(1- \prod_{i=1}^{n} u_2(\eta_i)^2)^{1/2}
\pa^2 = \pa \psi_n\pa^2 - \pa \psi_n^0\pa^2.
\end{equation}
Applying (\ref{lo8}) and summing over all $n$ we arrive at 
\begin{eqnarray}\nonumber
2\sum_n |\al I_{2,n}| &\leq& 16\pi \Lambda\al (\Psi,H_f\Psi) + 4C\bc \gamma_0^{-1} \al^3 \Lambda^3 \pa \Psi\pa^2\\ 
\label{lo16} &\leq& 16\pi \Lambda\al (\Psi,H_f \Psi) + O(\al^3) \pa \Psi\pa^2. 
\end{eqnarray}
Let us now turn to estimating the terms $I_{1,n}$. In these terms we will separate three parts.
The first two parts $\bar I_{1,n}$ and $\tilde I_{1,n}$ correspond to the process of annihilation 
of two photons very far from the particle or one photon near the particle and the second one far away 
from the particle. 
It is intuitively clear that these two parts are small if we compare them to the process of the 
annihilation of two photons near the
particle. We will prove it rigorously and after that we will analyze the main part $I^0_{1,n}$.
Let
\begin{eqnarray}\label{lo81}
&&\bar I_{1,n} = \sqrt{(n+2)(n+1)} \int \overline{\vec G(\eta_1) \cdot \vec G(\eta_2) \psi_{n+2}
(x,\eta_1,...,\eta_{n+2}) } \times \\ \nonumber &&\qquad\qquad \times\psi_n(x, 
\eta_3,...,\eta_{n+2})\prod_{i=1}^{n+2} u_2(\eta_i)^2 dx  d\eta,
\end{eqnarray}
\begin{eqnarray}\label{lo82}
&&\tilde I_{1,n} = \sqrt{(n+2)(n+1)} \int \overline{\vec G(\eta_1) \cdot \vec G(\eta_2) \psi_{n+2}
(x,\eta_1,...,\eta_{n+2})} \times \\ \nonumber &&\qquad\qquad \times\psi_n(x, 
\eta_3,...,\eta_{n+2})\prod_{i=2}^{n+2} u_2(\eta_i)^2 u_1(\eta_1)^2 dx  d\eta
\end{eqnarray}
and
\begin{eqnarray}\label{lo83}
&&I_{1,n}^0  = \sqrt{(n+2)(n+1)} \int \overline{\vec G(\eta_1) \cdot \vec G(\eta_2) \psi_{n+2}
(x,\eta_1,...,\eta_{n+2})} \times \\ \nonumber &&\qquad\qquad \times\psi_n(x, 
\eta_3,...,\eta_{n+2})\prod_{i=3}^{n+2} u_2(\eta_i)^2 u_1(\eta_1)^2 u_1(\eta_2)^2dx  d\eta.
\end{eqnarray}
Obviously
\begin{equation}\label{lo84}
I_{1,n}=\bar I_{1,n}+ 2\tilde I_{1,n}+I^0_{1,n} .
\end{equation}
Let us get the bound for  $\tilde I_{1,n}$, the term $\bar I_{1,n}$ can be evaluated in a similar  and 
simpler way.
Define the operators $D_1$ and $D_2$ analogously to $D$ replacing the vector function $\vec G(\eta)$ by 
$\vec G(\eta) u_1(\eta)^2$ for $D_1$ and by $\vec G(\eta) u_2(\eta)^2$ for $D_2$. Similar to 
(\ref{comm}) we have
\begin{equation}\label{lo85}
[D_2,D^\ast_2] = 2\pa \vec G(\eta) u_2(\eta)\pa^2,
\end{equation}
where the coefficient $2$ describes the $2$ possible photon polarizations.
Notice that if the number $a_1$ in the definition of the function $u_2$ grows,
\begin{equation}
\pa \vec G(\eta) u_2(\eta)^2\pa \to 0.
\end{equation}
Hence
\begin{eqnarray}\nonumber
\al |\tilde I_{1,n}| &\leq& \al \pa D_1\psi_{n+2}\pa \pa D_2^\ast \psi_n \prod_{i=1}^{n} 
u_2(\eta_i)^2\pa \\ \nonumber
&\leq& \al \pa D_1\psi_{n+2}\pa\left(2\pa \vec G(\eta) u_2(\eta)\pa^2\pa\psi_n\pa^2 +  \pa D_2 \psi_n 
\prod_{i=1}^{n} u_2(\eta_i)^2\pa^2\right)^{1/2}\\
\label{lo92}
&\leq& \al^2 \eps(a_1) \nu \pa\psi_n\pa^2 + \nu^{-1} C(\psi_{n+2}, H_f \psi_{n+2})+ \al^2 C\nu (\psi_n, 
H_f \psi_n),
\end{eqnarray}
where $\nu$ is an arbitrary number and $C$ is a constant, which does not depend $\al$ and $\Psi$ (see 
Lemma \ref{alem}).
For fixed $\eps_0$ we first choose $\nu$ and $\al$ such that $\nu^{-1} C + \nu\al^2C < \frac 12 \eps_0$ 
and then 
choose $a_1$ so large that $\eps(a_1) \nu < \frac 12 \eps_0$.
Thus we arrive at
\begin{equation}\label{lo910}
2\sum_n \al |\tilde I_{1,n}| \leq \eps_0 \al^2 \pa \Psi\pa^2 + \eps_0 (\Psi,H_f \Psi),
\end{equation}
which together with the similar estimate for $\bar I_{1,n}$ implies
\begin{equation}\label{lo101}
2\Re \sum_n \al  I_{1,n} \geq -2\eps_0 \al^2 \pa \Psi\pa^2-2 \eps_0 (\Psi,H_f \Psi) + 2\Re \sum_n \al  
I^0_{1,n}.
\end{equation}
Finally for $\pa A\Psi\pa^2$ we have
\begin{equation}\label{lo102}
\al \pa A\Psi\pa^2 \geq 4\pi\al^2 \Lambda^2 \pa\Psi\pa - 2 \eps_0 (\Psi,H_f \Psi) + 2\Re \sum_n \al  
I^0_{1,n}
+ (2\eps_0 \al^2 + O(\al^3))\pa \Psi\pa^2.
\end{equation}
Applying (\ref{lo102}) to (\ref{lo11}) we get
\begin{eqnarray}\nonumber
(\Psi,T\Psi) &\geq& 4\pi \al \Lambda^2 \pa \Psi \pa^2 + \sum_{n=0}^\infty 
\left[(1-\eps_1(a_1,\al))\left(\frac{(n+1)(n+2)}2 \left\{
(\tps,H_f \tps) + \right.\right. \right.\\ \nonumber &&\left.\left. \left.+\pa (\nabla_x + 
\sum_{i=1}^{n+2}
\nabla_{\eta_i} ) \tps\pa^2\right\}  + \pa (\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} ) \psi^0_n\pa^2\right) +  2\Re \sum_n \al  I^0_{1,n}\right] \\ \label{lo103} &&- 
\eps_1(a_1,\al)\al^2 \pa \Psi\pa^2,
\end{eqnarray}
where $\eps_1(a_1,\al) \to 0$ if $a_1 \to \infty$ and $\al \to 0$.
For the integral $\al  I^0_{1,n}$ the a-priori estimate 
\begin{equation}
|\al  I^0_{1,n}| \leq \frac 14(\psi_{n+2} , H_f \psi_{n+2}) + C\al^2\pa \psi^0_n\pa^2
\end{equation}
holds, with some constant $C$ independent of $\al$.
Assume, that for some $n= \bn $ 
\begin{equation}\label{lo112}
\frac{(\bn+1)(\bn+2)}2 
(\widetilde \psi_{\bn +2},H_f \widetilde \psi_{\bn +2}) + \pa(\nabla_x + \sum_{i=1}^{n}
\nabla_{\eta_i} ) \psi_{\bn}^0\pa^2 \geq C\al^2 \pa \psi_{\bn}^0\pa^2.
\end{equation}
Then the term in the sum (\ref{lo103}) corresponding to $\bn$ is positive, thus we can  omit it.
We replace this sum by the sum $\sum^\ast$ over all $n$, for which
(\ref{lo112}) does not hold.

Notice, that the inequality
\begin{equation}
\frac{(n+1)(n+2)}4 (\tps,H_f \tps) \leq C\al^2 \pa \psi_n^0\pa^2
\end{equation}
implies
\begin{equation}\label{lo114}
\pa \tps \pa^2 \leq \frac{C_1(a_1) \al^2 4}{(n+1)(n+2)} \pa \psi_n^0\pa^2.
\end{equation}
Let us project the function $\tps$ on $\psi_n^0$:
\begin{equation}
\tps(x, \eta_1,...,\eta_{n+2}) = \al \gamma_n(\eta_1,\eta_2) \psi_n^0 (x,\eta_3,...,\eta_{n+2}) + 
g_n(\eta_1,...,\eta_{n+2}),
\end{equation}
where $g_n \perp \psi_n^0$ for all $\eta_1,\eta_2$.

By (\ref{lo114})
\begin{equation}
\pa g_n\pa^2 \leq \frac{C \al^2}{(n+1)(n+2)} \pa \psi_n^0\pa^2
\end{equation}
and
\begin{equation}\label{lo122}
\pa \gamma_n\pa^2 \leq \frac{C }{(n+1)(n+2)}.
\end{equation}
Our goal is to estimate the term
\begin{eqnarray}\nonumber
&&\pa(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} ) \tps\pa^2 = \\ \nonumber && \qquad= \pa \al  (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \gamma_n + \al \psi_n^0(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n +
(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n\pa^2 \\ \nonumber
&&\qquad =\pa \al  (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \gamma_n + \al \psi_n^0(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2 + \pa 
(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n\pa^2\\ \label{lo113} &&
\qquad \quad+2\Re (\al  (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \gamma_n + \al \psi_n^0(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n, (\nabla_x 
+ \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n)
\end{eqnarray}
For the first part of the cross term we have 
\begin{eqnarray}\nonumber
&&|((\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n,\al  (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \gamma_n )| \leq \\ \label{lo131}
&& \qquad\quad \leq \al \pa (\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n \pa^2 + \al \pa(\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0\pa^2 \pa \gamma_n\pa^2
\end{eqnarray}
and the inequality (\ref{lo122}) together with the condition
\begin{equation}\label{132}
\pa (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0\pa^2 \leq C\al^2 \pa\psi_n^0\pa^2
\end{equation}
implies that the last term on the r.h.s. of (\ref{lo131}) is less than $\frac{C\al^3}{(n+1)(n+2)}\pa \psi_n^0\pa^2$.
The second part of the cross term can be evaluated as
\begin{eqnarray}\nonumber
&&|(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n,\al \psi_n^0(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n)|= \\ \nonumber
&&\quad= |(g_n, -(\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \al (\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n))|\leq \\ \nonumber
&&\qquad \quad\leq  \{ \al \pa g_n\pa^2 + C\al^3 \pa \psi_n^0 \pa^2 \pa 
(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2\}\\ \label{lo133}
&&\qquad \quad\leq  \frac{C\al^3}{(n+1)(n+2)}\pa \psi_n^0 \pa^2 + C\al^3 \pa \psi_n^0 \pa^2 \pa 
(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2.
\end{eqnarray}

Summing the inequalities (\ref{lo131}) and (\ref{lo133}) we obtain
\begin{eqnarray}\nonumber
&&2\left |\Re \left(\al  (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \gamma_n + \al \psi_n^0(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n, (\nabla_x 
+ \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n\right)\right| \leq \\
\label{sampras} && \leq 2\al \pa 
(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )g_n\pa^2 + \pa \psi_n^0\pa^2 \pa (\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2+ 
\frac{C\al^3}{(n+1)(n+2)}\pa \psi_n^0 \pa^2.
\end{eqnarray}

Let us turn now to the first term on the r.h.s. of (\ref{lo113})
\begin{eqnarray}\nonumber
&&\pa \al  (\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \gamma_n + \al \psi_n^0(\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2 \geq\\ 
\nonumber
&&\qquad \quad\geq (1-\al) \al^2 \pa \psi_n^0 \pa^2 \pa (\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2 
- 
\al \pa(\nabla_x + \sum_{i=3}^{n+2}
\nabla_{\eta_i} )\psi_n^0 \pa^2\pa\gamma_n\pa^2\geq \\ \label{lo141}&&
\qquad \quad\geq
(1-\al) \al^2 \pa \psi_n^0 \pa^2 \pa (\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2 - 
\frac{C\al^3 }{(n+1)(n+2)}\pa \psi_n^0 \pa^2.
\end{eqnarray}
Applying (\ref{sampras}) and (\ref{lo141})  to (\ref{lo113}) we arrive
at 
\begin{eqnarray}\nonumber
&&\frac{(n+1)(n+2)}2\pa(\nabla_x + \sum_{i=1}^{n+2}
\nabla_{\eta_i} )\tps\pa^2  \geq\\ \label{lo142} &&\qquad 
\frac{(n+1)(n+2)}2\al^2 \pa \psi_n^0 \pa^2 \pa (\nabla_{\eta_1}+\nabla_{\eta_2})\gamma_n\pa^2(1-C\al)
- C\al^3 \pa \psi_n^0 \pa^2,
\end{eqnarray}
which implies 
\begin{eqnarray}\nonumber
(\Psi,T\Psi) &\geq& 4\pi \al \Lambda^2 \pa \Psi \pa^2 + \mbox{ $\sum^\ast_{n}$} 
\left[(1-\eps_1(a_1,\al))\al^2\pa\psi_n^0\pa^2\frac{(n+1)(n+2)}2\times
\right.\\ \nonumber &&\left.
\times \left\{\pa (\nabla_{\eta_1} + \nabla_{\eta_2}) \gamma_n\pa^2 + \pa |\nabla_{\eta_1}|^{1/2} 
\gamma_n\pa^2 + 
\pa |\nabla_{\eta_2}|^{1/2} \gamma_n\pa^2\right\}
+  2\Re \sum_n \al  I^0_{1,n}\right] \\ \label{lo143} &&- \eps_1(a_1,\al)\al^2 \pa \Psi\pa^2,
\end{eqnarray}
Finally, we turn to the estimation of $ I^0_{1,n}$.

Notice, that $\sqrt{(n+1)(n+2)}\pa\gamma_n(\eta_1,\eta_2)\pa \leq C$ which implies
\begin{eqnarray}\nonumber
\al\Re    I^0_{1,n} &=& \al^2 \sqrt{(n+1)(n+2)} \pa \psi_n^0\pa^2 \Re \int \overline{\vec G(\eta_1)\cdot 
\vec G(\eta_2)}
u_1(\eta_1) u_1(\eta_2) \overline{\gamma_n(\eta_1,\eta_2)}d\eta_1d\eta_2 \geq\\ \nonumber &\geq&
\al^2 \sqrt{(n+1)(n+2)} \pa \psi_n^0\pa^2 \Re \int \overline{\vec G(\eta_1)\cdot \vec G(\eta_2)
\gamma_n(\eta_1,\eta_2)}d\eta_1d\eta_2 -\\ \label{lo161} && \qquad \qquad-\al^2 C \pa \psi_n^0\pa^2\pa\vec G(\eta_1)\cdot 
\vec G(\eta_2)\sqrt{1-u_1(\eta_1) u_1(\eta_2)}\pa^2
\end{eqnarray}
The last term in (\ref{lo161}) can be estimated as $\eps(a_1,\al) \al^2$ as $a_1 \to \infty$.
Finally, using the inequality
\begin{equation}
\pa f\pa^2 + 2 \Re (f,h) \geq -\pa h\pa^2 ,\qquad f,h \in \Ll^2,
\end{equation}
we arrive at 
\begin{eqnarray}\nonumber
&&(1- \eps_1(a_1,\al))\frac{(n+1)(n+2)}2\left\{\pa (\nabla_{\eta_1} + \nabla_{\eta_2}) \gamma_n\pa^2 + \pa 
|\nabla_{\eta_1}|^{1/2} \gamma_n\pa^2 + 
\pa |\nabla_{\eta_2}|^{1/2} \gamma_n\pa^2\right\} -\\ \nonumber && \qquad  - \sqrt{(n+1)(n+2)} 2 \Re \int 
\overline{\vec G(\eta_1)\cdot \vec G(\eta_2)
\gamma_n(\eta_1,\eta_2)}d\eta_1d\eta_2 \geq \\ \label{lo162}&& \qquad\qquad\qquad \geq - \frac 
1{1-\eps_1(a_1,\al)} \Em_0,
\end{eqnarray}
with $\eps_1(a_1,\al)$ arbitrarily small,
which completes the proof of the Theorem.



\section{PROOF OF THEOREM \ref{mt}}


\begin{proof}[Proof of Theorem \ref{mt}]

To prove the Theorem we will check the binding condition of \cite{GLL} for
$\be = \be_0$. Namely, we will show that 
\begin{equation} \label{bindcon}
\infspec {\bf H}_{\be_0} < \so - \delta \al^2 + o(\al^2).
\end{equation}
The binding for all $\be \in (\be_1,\be_0] $ with some $\be_1 > \be_0$ follows from (\ref{bindcon})
and the continuity of the quadratic form in $\be$.
In the proof of  Theorem \ref{pt2} we have seen that the sequence 
\begin{equation}
\Psi_n = \{f_n(x),0, \al f_n (x)\varphi(x-\xi_1,x-\xi_2),0,0,..\},
\end{equation}
with $\pa f_n(x)\pa =1, \lim_{n \to \infty} \pa \nabla f_n(x) \pa = 0$ and $\varphi$ is the function given by 
(\ref{chef}), is a minimizing
sequence up to the order $\al^2$ for the quadratic form $(\Psi_n,T\Psi_n)$.
More precisely, according to (\ref{ub32}),  $\pa\nabla f_n\pa^2 \leq O(\al) \pa f_n\pa^2$ yields 
\begin{equation}
(\Psi_n,T\Psi_n) \leq \pa \nabla f_n\pa^2 + (\so + o(\al^2))\pa \Psi_n\pa^2 .
\end{equation}
Our next goal is to modify this sequence in such a way that for the modified function
$\Psi_0$
\begin{equation}
(\Psi^0 , {\bf H}_{\be_0} \Psi^0) \leq (\so - \delta \al^2 + o(\al^2))\pa \Psi_0\pa^2,
\end{equation}
with some $\delta > 0$. To this end we will make a special choice  of the function $f(x)$ and 
additionally
add a one photon term, which leads to
\begin{equation}\label{psal}
\Psi^0 = \{f(x),i\as d\nabla f(x)\cdot \vec G(\eta), \al f (x)\varphi(x-\xi_1,x-\xi_2),0,0,...\},
\end{equation}
where $d$ is a parameter to be chosen later. 
We assume  $f(x)$ $\in C^2_0(\R^3)$ to be a real, spherically symmetric function
and to fulfill the condition
\begin{equation}\label{cond}
\pa \Delta f(x)\pa \leq C_1 \pa \nabla f(x)\pa \leq C_2 \as \pa f(x) \pa,
\end{equation}
with some constants $C_{1,2}$.
Next, let us turn to the estimate of the quadratic form $(\Psi^0,{\bf H}_{\be_0}\Psi^0)$.
The negativity of the terms $(\psi_1,V\psi_1) $ and $(\psi_2, V \psi_2)$ together with 
(\ref{ub32}) implies
\begin{eqnarray}\label{2.5}
(\Psi^0 , {\bf H}_{\be_0} \Psi^0) &\leq& (\so  + o(\al^2))\pa \Psi_0\pa^2 + \pa \nabla f(x)\pa^2 + 
\be_0 (f,V(x) f) \\ \nonumber
&& + 4\as \Re \{(p\psi_1, D\psi_2) + (p\psi_0,D\psi_1)\} + \pa (\nabla_x + \nabla_\eta)\psi_1\pa^2 + 
\pa |\nabla_\eta|^{1/2} \psi_1\pa^2,
\end{eqnarray}
where $\psi_0 = f(x), \psi_1, \psi_2$ are the three non vanishing components of $\Psi^0$. 
Since we assumed $f(x)$ to be real and spherically symmetric we derive
\begin{eqnarray}\nonumber
\pa (\nabla_x + \nabla_\eta)\psi_1\pa^2 &=& -(\psi_1, \Delta_x \psi_1)-(\psi_1, \Delta_\eta \psi_1)\\ 
\nonumber
&\leq& \al d^2 \left\{ \int |\vec G(\eta)|^2 d\eta \int \left [ \left (\frac{\partial^2 f}{\partial 
x_1^2}\right )^2 + 2\left(
\frac{\partial^2 f}{\partial x_1\partial x_2}\right )^2\right]dx \right. \\ && \left. + \int \left 
(\frac{\partial f}{\partial x_1}\right)^2dx \int \sum_{i,k=1}^3
\left (\frac{\partial G_i(\eta)}{\partial \eta_k}\right)^2 d\eta \right\},
\end{eqnarray}
where $G_i$ are the components of $\vec G$. Applying (\ref{cond}) we arrive at
\begin{equation}\label{99}
\pa (\nabla_x + \nabla_\eta)\psi_1\pa^2 \leq C\al d^2 \pa \nabla f\pa^2,
\end{equation}
with $C$ depending only on $C_{1}$ of (\ref{cond}).

Obviously
\begin{equation}\label{100}
\pa |\nabla_\eta|^{1/2} \psi_1\pa^2 = \al d^2 \sum_{i=1}^3 \int \left(\frac{\partial f}{\partial 
x_i}\right)^2 dx \pa |\nabla_\eta |^{1/2} G_i\pa^2
\leq C\al d^2 \pa \nabla f(x)\pa^2.
\end{equation}
Substituting (\ref{99}) and (\ref{100}) into (\ref{2.5})
we arrive at
\begin{eqnarray}\label{2.99}
(\Psi^0 , {\bf H}_{\be_0} \Psi^0) &\leq& (\so  + o(\al^2))\pa \Psi_0\pa^2 + \pa \nabla f(x)\pa^2 + 
\be_0 (f,V(x) f) \\ \nonumber
&& + 4\as \Re \{(p\psi_1, D\psi_2) + (p\psi_0,D\psi_1)\} +C\al d^2 \pa \nabla f\pa^2.
\end{eqnarray}

Next, we estimate the term $\as (p\psi_1,D\psi_2)$.
By Schwarz inequality   and condition (\ref{cond})
\begin{equation}\label{2.9}
|\as ( p\psi_1,D\psi_2)| \leq \as d \pa D\psi_2\pa^2 + \as d^{-1}\pa p \psi_1\pa^2 \leq Cd[\al^{5/2} \pa 
f\pa^2 + \al^{1/2}d^{-1}\pa p\psi_1\pa^2] 
\leq C_1\al^{5/2} d \pa f\pa^2.
\end{equation}
By (\ref{2.99}) and (\ref{2.9}) we get
\begin{eqnarray}\nonumber
(\Psi^0 , {\bf H}_{\be_0} \Psi^0) &\leq& (\so  + o(\al^2))\pa \Psi_0\pa^2 + \pa \nabla f(x)\pa^2 + 
\be_0 (f,V(x) f)\\
&& +  4\as \Re  (p\psi_0,D\psi_1)+C\al d^2 \pa \nabla f(x)\pa^2,
\end{eqnarray}
where $C$ is independent of $\al$.

Notice that 
\begin{equation}
\as \Re (p \psi_0,D\psi_1) = -  \al d \int\left( \frac{\partial f}{\partial x_1}\right)^2 dx \int 
|\vec G(\eta)|^2 d\eta =
- \frac 13 \al d \pa \nabla f\pa^2 4\pi \Lambda^2 = - \frac {4\pi}3 \al\Lambda^2 d \pa \nabla f\pa^2,
\end{equation}
which implies 
\begin{equation}
(\Psi^0 , {\bf H}_{\be_0} \Psi^0) \leq (1 - \frac {16\pi}3 \al\Lambda^2 d + C\al d^2) \pa \nabla f\pa^2 + 
\be_0 (f,Vf) + (\so  + o(\al^2))\pa \Psi_0\pa^2.
\end{equation}

As the next step
we choose $d < \frac{8\pi \Lambda^2}{3C}$ which gives
\begin{equation}\label{2.13}
(\Psi^0 , {\bf H}_{\be_0} \Psi^0) \leq (1 -\nu \al) \pa \nabla f\pa^2 + 
\be_0 (f,Vf) + (\so  + o(\al^2))\pa \Psi_0\pa^2,
\end{equation}
where 
\begin{equation}
\nu = \frac{64}{3C} \Lambda^2 \pi^2.
\end{equation}
Due to our choice of $\be_0$, obviously the operator
\begin{equation}\label{2.14}
-(1-\nu \al)\Delta + \be_0 V(x)
\end{equation}
has at least one negative eigenvalue. However, the r.h.s. of (\ref{2.13}) contains the terms which are 
of order $o(\al^2)$
and to prove Theorem \ref{mt} we have to provide more precise estimates on the negative eigenvalues of (\ref{2.14}).

The following Lemma  completes the proof of  Theorem \ref{pt2}.

\begin{lem}\label{alem}
Let $\be_0$ be the critical value (the maximal value of the constant $\be$, for which the Schr\"odinger operator
with the potential $\be V$ does not have a discrete spectrum). Then 
\begin{equation}
\infspec \{-(1-\nu \al)\Delta + \be_0 V(|x|)\} < - \delta \al^2, 
\end{equation}
for some $\delta >0$ and $\al$ small enough.

Moreover,
there exists a function $f_\al(x)$, real, spherical symmetric and satisfying condition (\ref{cond}),
with constants $C_{1,2}$ independent of $\al$, such that 
\begin{equation}
(1-\nu \al)\pa \nabla f_\al\pa^2 + \be_0 (f_\al,  V(|x|)f_\al) < - \delta \al^2\pa f_\al \pa^2. 
\end{equation}

\end{lem}
\end{proof}
The proof of Lemma \ref{alem} is given in Appendix B.



\begin{appendix}
\section{APPENDIX: Localization error for the relativistic kinetic energy}

\begin{thm} \label{loclem}
For arbitrary numbers $a > 0, \eps > 0$ there exists a parameter $b>0$ and functions 
$u_1(|x|), u_2(|x|) \in C^2(\R^3)$ such that
\begin{eqnarray}\nonumber
i)&& u_1(|x|)^2 + u_2(|x|)^2 = 1;\\ \nonumber
ii)&& u_1(|x|)=1 \,\, {\rm for} \,\, |x| \leq a \,\, {\rm and} \\ \nonumber
&&  u_1(|x|)=0 \,\, {\rm for} \,\, |x| \geq b;\\\label{6.1}
iii)&& |(\psi,|\nabla| \psi) - (\psi u_1,|\nabla| \psi u_1)- (\psi u_2,|\nabla| \psi u_2)|
\leq \eps \pa |x|^{-1/2}\psi\pa^2, \,\, \psi \in W^{1,2}(\R^3).
\end{eqnarray}
\end{thm}
\begin{proof}
According to \cite{LY} the localization error for the operator
$|\nabla|$ is
\begin{equation}\label{6.2}
\frac 1{4\pi^2} \sum_{i=1}^2 \int \frac{\psi(x)\psi(y)}{|x-y|^4}(u_i(x) - u_i(y))^2 dxdy.
\end{equation}
Our goal is to show that the functions $u_1(x), u_2(x)$ and the number $b$ can be chosen such that the r.h.s. of 
(\ref{6.1}) is greater than (\ref{6.2}) for all $\psi \in W^{1,2}(\R^3)$.
Therefore we take $u_1 (|x|)$ and choose it on some interval $[a \leq |x| \leq a + \delta]$,  $\delta > 0$,
strictly decreasing and such that 
\begin{equation}
\lim_{|x| \to a + 0} \frac{u_1'(x)^2}{1-u_1(x)^2} = 0.
\end{equation}
For fixed $\eps_1$ we pick a number $\delta > 0$ so small, that 
for $|x| \in [a,a+\delta]$
\begin{equation}\label{621}
u_1'(x)^2 + u_2'(x)^2\leq u_1'(x)^2 + \frac{u_1'(x)^2}{1-u_1(x)^2} < \eps_1.
\end{equation}
Because of symmetry in $x$ and $y$ we can write
\begin{equation}\label{622}
\sum_{i=1}^2 \int \frac{\psi(x)\psi(y)}{|x-y|^4}(u_i(x) - u_i(y))^2 dxdy
= 2\sum_{i=1}^2 \int_{|x| \leq |y|} \frac{\psi(x)\psi(y)}{|x-y|^4}(u_i(x) - u_i(y))^2 dxdy
\leq 2(I_1 + I_2),
\end{equation}
where $I_1$ is the integral over the region $\{ (x,y) \in \R^3 \times \R^3 | |x| \leq |y|, |y| \leq a+\delta\}$
and $I_2$ over the region  $\{ (x,y) \in \R^3 \times \R^3 | |x| \leq |y|, |y| > a+\delta\}$.

Let us estimate $I_1$. For $|x| < |y| \leq a+\delta$
\begin{equation}
\sum_{i=1}^2 (u_i(x) - u_i(y))^2 \leq \max_{|y| \leq a +\delta} \{u_1'(y)^2 + \frac{u_1'(y)^2}{1-u_1(y)^2}\} |x-y|^2,
\end{equation}
which implies 
\begin{eqnarray}\nonumber
|I_1| &\leq& \eps_1 \int_{|y| \leq a +\delta, |x| \leq |y|} \frac{|\psi(x)||\psi(y)|}{|x-y|^2} dxdy \leq
2\eps_1 \int_{|x| \leq 2a} |\psi(x)|^2 dx \int_{|x-y|\leq 3a} |x-y|^{-2} d(x-y)\\ \label{631}
&& \leq 2\eps_1 \pa \psi(x)\pa^2_{|x| \leq 2a} 4\pi 3a \leq 48\pi \eps_1 a^2 \pa\psi (x) |x|^{-1/2}\pa^2.
\end{eqnarray}

Taking $\eps < \eps_1 (48\pi a^2)^{-1}$ we arrive at 
\begin{equation}\label{632}
|I_1| < \eps \pa \psi(x) |x|^{-1/2}\pa^2.
\end{equation}
Notice that $u_1(a+\delta) <1 $ and for $|y| \geq a +\delta, |y| >|x|$ 
\begin{eqnarray}\nonumber
&& (u_2(x) - u_2(y))^2 = \left((1-u_1(x)^2)^{1/2} - ((1-u_1(y)^2)^{1/2}\right)^2 =\\ \nonumber
&&  = \frac{(u_1(x) -u_1(y))^2(u_1(x) +u_1(y))^2}{\left((1-u_1(x)^2)^{1/2} + ((1-u_1(y)^2)^{1/2}\right)^2}\leq
\\ \label{633} &&\qquad \leq
4 (u_1(x) -u_1(y))^2(1-u_1(a+\delta)^2)^{-1}= C_0(u_1(x) -u_1(y))^2
\end{eqnarray}
holds.
The last inequality implies 
\begin{equation}\label{634}
|I_2| \leq (C_0 +1) \int_{|x| < |y|,|y| \geq a +\delta} |\psi(x)||\psi(y)||x-y|^{-4}((u_1(x) -u_1(y))^2 dxdy.
\end{equation}
Now we define for $|x| \in [a+\delta, b]$ 
\begin{equation}\label{635}
u_1(|x|) = \ln\left[\frac{|x|}b\right] \ln^{-1}\left[\frac{a+\delta}b\right] u_1(a+\delta).
\end{equation}

It is easy to see that 
\begin{equation}\label{636}
|u_1(x) - u_1(y)| \leq \ln^{-1}\left[\frac b{a+\delta}\right]|\ln\left[\frac{|x|}{|y|}\right]|\leq \ln^{-1}\left[\frac b{a+\delta}\right]
\frac{|y-x|}{|x|}.
\end{equation}
Let $C > 0$ be a constant, which we specify later. First, we estimate the part of
$I_2$ corresponding to the region where $|x-y| > C|x|$.
Let
\begin{equation}
\Omega_1 = \{(x,y)| |x| < |y|,\, |y| > a +\delta, \,  |x-y| > C|x|\},
\end{equation}
then
\begin{eqnarray}\nonumber
&& \int_{\Omega_1} \frac{|\psi(x)||\psi(y)|}{|x-y|^4}(u_1(x) - u_1(y))^2 dxdy \leq 2 \int_{\Omega_1} \frac{\psi(x)\psi(y)}{|x-y|^4}dxdy \leq
\\ \label{641} && \leq 2 \left(\int \frac{|\psi(x)|^2}{|x-y|^4}dxdy\right)^{1/2}\left( \int\frac{|\psi(y)|^2}{|x-y|^4}dxdy\right)^{1/2}.
\end{eqnarray}
Notice that 
\begin{equation}
\int_{\Omega_1}\frac{|\psi(x)|^2}{|x-y|^4}dxdy \leq \int_{\R^3} |\psi(x)|^2 dx \int_{|x-y| > C|x|, |x| < |y|} |x-y|^{-4} dy
\leq \frac{4\pi}C \pa \psi(x) |x|^{-1/2}\pa^2.
\end{equation}
The second integral on the r.h.s. of (\ref{641}) can be estimated as  
\begin{equation}
\int_{\Omega_1}\frac{|\psi(y)|^2}{|x-y|^4}dxdy \leq \int_{\R^3} |\psi(y)|^2 dy \int_{|x-y| \geq C|x|, |x| < |y|} |x-y|^{-4} dx
\leq 4\pi\sqrt{\frac{(C-1)}{C(C-2)}}\pa \psi(x) |x|^{-1/2}\pa^2.
\end{equation}
For the region 
\begin{equation}
\Omega_2 = \{(x,y)| |x| < |y|,\, |y| > a +\delta, \,  |x-y| < C|x|\}
\end{equation}
we have 
\begin{eqnarray}\nonumber
\int_{\Omega_2} \frac{|\psi(x)||\psi(y)|}{|x-y|^4} dxdy &\leq& \ln^{-2}\left[\frac{b}{a+\delta}\right] \left(
\int_{\Omega_2} \frac{|\psi(x)|^2}{|x-y|^2|x|^2}dxdy\right)^{1/2} \left(
\int_{\Omega_2} \frac{|\psi(y)|^2}{|x-y|^2|x|^2}dxdy\right)^{1/2}\\ \nonumber & \leq& \ln^{-2}\left[\frac{b}{a+\delta}\right]
(4\pi C)^{1/2} \pa \psi(x) |x|^{-1/2}\pa(4\pi C)^{1/2}(C+1) \pa \psi(y) |y|^{-1/2}\pa\\
& \leq& 4\pi C(C+1)\ln^{-2}\left[\frac{b}{a+\delta}\right]
\pa \psi(x) |x|^{-1/2}\pa^2.
\end{eqnarray}
Now we take $C$ large enough to fulfill the inequality 
\begin{equation}
[4(1-u_1(a+\delta)^2)^{-1} + 1]4\pi (C-1)^{1/2}[C(C-2)]^{-1/2} < \eps
\end{equation}
and then for fixed $C$ we choose $b$ so large that 
\begin{equation}
4\pi C(C+1)\ln^{-2}\left[\frac{b}{a+\delta}\right] [4(1-u_1(a+\delta)^2)^{-1} + 1] < \eps
\end{equation}
which together with (\ref{6.2}) and (\ref{632}) implies $iii)$. To complete the proof of the Theorem
it suffices to notice that the functions $u_1$ and $u_2$ can easily be approximated with some function
in $C^2(\R^3)$ satisfying $iii)$.
\end{proof}
\begin{rem}\label{locrem}
With the same functions  $u_1$ and $u_2$ as in Theorem \ref{loclem} one has (see \cite{VZ2})
\begin{equation}
|(\psi,-\Delta \psi) - (\psi u_1,-\Delta \psi u_1)- (\psi u_2,-\Delta \psi u_2)|
\leq \eps \pa |x|^{-1}\psi\pa^2.
\end{equation}
\end{rem}

The last inequality will be used in the proof of Theorem \ref{pt2}



\section{APPENDIX: Proof of Lemma \ref{alem}}

\begin{proof}[Proof of Lemma \ref{alem}]
Let us start recalling some properties of the operator
\begin{equation}
h_\be = -\Delta + \be V(x),
\end{equation}
$V(x)$ negative, radial and compactly supported, with critical value $\be =\be_0$. For $\be =\be_0$ the 
operator $h_\be$ has a so-called virtual level or zero-resonance. It means that the equation
\begin{equation}\label{app1}
-\Delta \psi + \be_0 V(x) \psi = 0
\end{equation}
has a generalized spherical symmetric solution $\tilde \psi$ with the following properties \cite{VZ}:

\begin{enumerate}
\item[(i)]
Let ${\mathcal B}$ be a closure of the space $C_0^\infty (\R^3)$ in the norm $\pa \psi \pa_B = \pa 
\nabla
\psi \pa $. Then $\tilde \psi \in {\mathcal B}$. From this point we assume that $\tilde \psi$ is a 
normalized
solution in the sense that $\pa \tilde \psi\pa_B =1$. Notice, that $\tilde \psi \in \Ll^2_{\rm 
loc}(\R^3)$,
but $\tilde \psi \notin \Ll^2(\R^3)$.

\item[(ii)]
$-\Delta \tilde \psi \in \Ll^2(\R^3)$ and $V(x)\tilde \psi \in \Ll^2(\R^3)$.

\item[(iii)]
Outside the support of $V(x)$ holds 
\begin{equation}\label{a44}
\tilde \psi(x) = C|x|^{-1}.
\end{equation}
\end{enumerate}
The last property follows immediately from the fact that outside the support of 
$V(x)$  the solutions of (\ref{app1}) can be written as $C_1 |x|^{-1} + C_2$,
and $\tilde \psi \in {\mathcal B}$ implies  $C_2 =0$.

Now we proceed directly to the proof Lemma \ref{alem}.
Let 
\begin{equation}
u \in C_0^2 (\R^3), \ u(x) \leq 1, \ u(x) =1 \, \, {\rm for} \,\,|x|\leq 1, \,\, u(x) =0  
\, \, {\rm for} \,\,|x|\geq 2
\end{equation}
and set
\begin{equation}
f_n(x) = \tilde \psi (x) u(|x|  \al n^{-1}) \pa \tilde\psi (x)u(|x|  \al n^{-1})\pa^{-1}.
\end{equation}
Obviously $\pa f_n (x)\pa =1$ and for large $n$ 
\begin{eqnarray}\nonumber
\pa \nabla f_n(x)\pa &\leq& \pa \tilde\psi (x)u(|x|  \al n^{-1})\pa^{-1}\{\pa \nabla \tilde \psi\pa_{|x|\leq 
2\al^{-1} n} +
\pa C|x|^{-1}\pa_{\al^{-1}n\leq |x| \leq 2\al^{-1} n} \max [|\nabla u(|x|\al n^{-1})|]\}\\ \label{app2}
& \leq &\pa \nabla \tilde\psi (x)u(|x|  \al n^{-1})\pa^{-1}\{\pa \nabla \tilde \psi\pa + C_1 (\al^{-1} n)^{1/2}\al 
n^{-1}\} 
\leq 2 \pa\tilde \psi(x)u(|x|  \al n^{-1})\pa^{-1}.
\end{eqnarray}

Assume $V(x)$ is supported in a ball of radius $a_0$. Then $\tilde \psi(x) = C|x|^{-1}$ for $ |x| \geq 
a_0$ and
\begin{equation}\label{app3}
\pa \tilde \psi (x)u(|x|  \al n^{-1})\pa^2 \geq 4\pi \int_{a_0 \leq |x| \leq 2\al^{-1}n} |x|^{-2} dx = 4\pi(\al^{-1} n - a_0) \geq 
2\pi \al^{-1} n
\end{equation}
for $n \geq \frac{2a_0}\al$.

The inequalities (\ref{app2}) and (\ref{app3}) imply the second relation 
in (\ref{cond}). To check the first inequality in (\ref{cond}) let us estimate 
\begin{equation}\label{app4}
\pa \Delta  \left(\tilde \psi (x)u(|x|  \al n^{-1})\right)\pa \leq \pa \Delta \tilde \psi\pa + \left( \sum_{i=1}^3 
\int \left (
\frac{\partial \tilde \psi}{\partial x_i}\right)^2 \left (
\frac{\partial u}{\partial x_i}\right)^2 dx \right)^{\frac 12} + \left(\int \frac{C}{|x|^2} \Delta u(|x|\al 
n^{-1}) dx\right)^{\frac 12}.
\end{equation}
According to (ii) the first term on the r.h.s. of (\ref{app4}) is bounded. The second term is also 
bounded, 
since $|\nabla u(|x|\al n^{-1})| \leq$ const., (recall that $\pa \nabla \tilde \psi\pa =1$) and the 
last term
is also bounded by a constant for $n \geq \frac{2a_0}\al$. Finally we arrive at 
\begin{equation}
\pa \Delta \left(\tilde \psi (x) u(|x|\al n^{-1})\right)\pa \leq C\pa \nabla \tilde \psi (x)\pa,
\end{equation}
which implies 
\begin{equation}
\pa \Delta f_n (x) \pa \leq C\pa \nabla f_n(x)\pa .
\end{equation}

To prove the Lemma it suffices now to show that for large $n$
\begin{equation}
(1-\nu \al)\pa \nabla f_n\pa^2 + \be_0(f_n,Vf_n)\leq - \delta \al^2\pa f_n\pa^2,
\end{equation}
with some $\delta >0$ independent of $\al$. This is equivalent to 
\begin{equation}
(1-\nu \al)\pa \nabla (\tilde \psi (x)u(|x|  \al n^{-1}))\pa^2 + \be_0(\tilde \psi,V\tilde \psi)\leq - 
\delta 
\al^2\pa \tilde \psi (x)u(|x|  \al n^{-1})\pa^2.
\end{equation}
Recall that 
\begin{equation}\label{sex}
\pa \tilde \psi u(|x|\al n^{-1})\pa^2 \leq C_1 + 8\pi C^2 \int_{a_0}^{2\al^{-1}n} d|x| \leq 2C_2 \al^{-1}n,
\end{equation}
for large $n$, and 
\begin{equation}
\pa \nabla \tilde \psi\pa^2 + \be_0(\tilde \psi,V\tilde\psi) = 0,
\end{equation}
which implies 
\begin{eqnarray}\nonumber
&&(1-\nu \al)\pa \nabla (\tilde \psi (x)u(|x|  \al n^{-1}))\pa^2 + \be_0(\tilde \psi,V\tilde \psi)\leq \\ 
\label{app13}
&& \quad \leq -\nu \al [\pa (\nabla \tilde \psi)u\pa - \pa \tilde \psi \nabla u \pa]^2 + 
[\pa\nabla \tilde \psi\pa^2 - \pa \nabla (\tilde \psi u)\pa^2]\\\nonumber 
&& \quad\leq -\nu \al \left[ \frac 12 \pa \nabla \tilde \psi\pa - C\al^{1/2} n^{-1/2} \right]^2 + 
2\pa \nabla \tilde \psi\pa^2_{|x| \geq \al^{-1} n} + \pa \tilde \psi |\nabla u|\pa^2_{\al^{-1} n \leq 
|x| \leq 2 \al^{-1} n}.
\end{eqnarray}
For $n$ large we have
\begin{eqnarray}\nonumber
&i)& \pa \nabla \tilde \psi\pa^2_{|x| \geq \al^{-1} n} =4\pi C^2 \int_{\al^{-1} n}^\infty |x|^{-2} dx = C^2 
\al n^{-1},\\ 
&ii)& \pa \tilde \psi |\nabla u|\pa^2_{\al^{-1} n\leq |x| \leq 2 \al^{-1} n} \leq C_1 \al^2 n^{-2} 
4\pi\int_{\al^{-1} n}^{2\al^{-1} n}  dr
= C_3 \al n^{-1},\\ \label{app14}
&iii)& C_3\al^{1/2} n^{-1/2} < \frac 14 = \frac 14 \pa \nabla \tilde \psi\pa,
\end{eqnarray}
which implies together with (\ref{app13}) 
\begin{equation}
- (1-\nu \al)\pa \nabla (\tilde \psi (x)u(|x|  \al n^{-1}))\pa^2 + \be_0(\tilde \psi,V\tilde \psi)\leq
- \nu\al/16\leq -\frac {\nu \al^2}{32C_2 n} \pa \tilde \psi u(|x|  \al n^{-1}))\pa^2.
\end{equation}
To complete the proof of the Lemma it suffices now to choose $n$ so large that (\ref{app14}) holds 
(notice, that it can be done uniformly in 
$\al$ for $\al \leq 1$) and for this $n$ take $\delta =\frac 1{32} \nu C_2^{-1} n^{-1}$, where $C_2$ is the 
constant in (\ref{sex}),
which depends on the zero-resonance solution $\tilde\psi$ only.
\end{proof}



\begin{lem}\label{applem}
Let $D_1$ and $D_2$ be the operators defined in Section 5.3. Then
\begin{equation}
\pa D_1\Psi\pa^2 \leq \frac{8\pi \Lambda^2}{(1-\eps)^2} (\Psi, H_f\Psi), \quad 
\pa D_2\Psi\pa^2 \leq \frac{8\pi \Lambda^2}{(1-\eps)^2} (\Psi, H_f\Psi),
\end{equation}
where $\eps$ is chosen as in Lemma \ref{loclem}, corresponding to
the functions $u_1$ and $u_2$ in the definition of $D_1$ and $D_2$.
\end{lem}
\begin{proof}
Our goal is to estimate the term
\begin{equation}\label{al1}
\pa D_1\psi_n\pa^2 = n\int |\int\vec G(\eta_1) u_1(\eta_1)^2 \psi_n(\eta_1,...,\eta_n)d\eta_1|^2 d\eta_2...d\eta_n.
\end{equation}
For fixed values of $\eta_2,...,\eta_n$ we get according to \cite{GLL} Lemma A.4
\begin{eqnarray}\nonumber
&&|\int \vec G(\eta_1) u_1(\eta_1)^2 \psi_n(\eta_1,...,\eta_n)d\eta_1|^2 \leq \\\label{al2} &&\qquad\qquad
8\pi\Lambda^2 \int ||\nabla_{\eta_1}|^{1/2} u_1(\eta_1)^2 \psi_n(\eta_1,...,\eta_n)|^2 d\eta_1.
\end{eqnarray}
By the Localization Theorem 
\begin{equation}\label{al3}
\pa |\nabla|^{1/2}\varphi\pa^2 \geq (1-\eps)\pa |\nabla|^{1/2}\varphi u_1\pa^2 +  (1-\eps)\pa |\nabla|^{1/2}\varphi u_2\pa^2 \geq
(1-\eps) \pa |\nabla|^{1/2}\varphi u_1\pa^2
\end{equation}
for $\varphi \in C_0^2(\R^3)$. Applying (\ref{al3}) twice, first with $\varphi = \psi_n u_1$ and then with $\varphi = \psi_n$ we arrive at
\begin{equation}
\pa D_1\Psi\pa^2 \leq \frac{8\pi \Lambda^2 n}{(1-\eps)^2}\pa |\nabla_{\eta}|^{1/2} \psi_n\pa^2 = 
\frac{8\pi \Lambda^2}{(1-\eps)^2} (\Psi, H_f\Psi).
\end{equation}
An analogical estimate holds for $D_2$.
\end{proof}




\end{appendix}



\bigskip
\noindent {\it Acknowledgment:} The work was partially supported by
the European Union through its Training, Research, and Mobility program FMRX-CT 96-0001. C. Hainzl has been supported by a Marie Curie Fellowship
of the European Community programme \lq\lq Improving Human Research Potential and the
Socio-economic Knowledge Base\rq\rq\ under contract number HPMFCT-2000-00660. C. H. and S.-A. V.  thank 
Robert Seiringer for many valuable comments. 





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