On critical stability of three quantum charges interacting through delta potentials
H.D.\
We consider three one dimensional quantum, charged and spinless particles interacting through delta potentials. We derive sufficient conditions which guarantee the existence of at least one bound state.
duclos.tex
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\def\QED{\mbox{\rule[-1.5pt]{6pt}{10pt}}}
\def\re{{\rm Re\,}}
\def\Ran{{\rm Ran\,}}
\def\spect{{\rm spect\,}}



\def\R{\mathbb{R }}
\def\dsS{\mathbf{S }}




\def\HH{\mathcal{H }}
\def\OO{\mathcal{O }}
\def\SS{\mathcal{S }}

\title{On critical stability of three quantum charges interacting through delta potentials}



\author{H.D.\,Cornean\instnr{1}\thanks{\textit{E-mail address:} 
 cornean@math.aau.dk}, 
 P. Duclos\instnr{2} \thanks{\textit{E-mail address:} 
 duclos@univ-tln.fr}, 
 B. Ricaud
\instnr{2} \thanks{\textit{E-mail address:} 
 ricaud@cpt.univ-mrs.fr} }

\instlist{Dept. of Math., Aalborg University, Fredrik Bajers Vej 7G,  9220 Denmark

\and


Centre de Physique Th\'eorique UMR 6207 - Unit\'e Mixte de Recherche du
CNRS et des Universit\'es Aix-Marseille I, Aix-Marseille II et de
l' Universit\'e du Sud Toulon-Var - Laboratoire affili\'e \`a la
FRUMAM, Luminy Case 907, F-13288 Marseille Cedex 9 France}



\runningauthor{H. Cornean, P. Duclos, B. Ricaud}
\runningtitle{Critical stability of three quantum charges with delta self-interactions}
\sloppy
\setcounter{page}{1}
\begin{document}

\maketitle
\begin{abstract}
We consider three one dimensional quantum, charged and spinless
particles interacting through delta potentials. We derive  sufficient
conditions which guarantee the existence of at least one bound state.

\end{abstract}

%%%%%%%%%%%%%%%
\section{Introduction}%
%%%%%%%%%%%%%%
 Denote by $ x_i, m_i, Z_ie$,  $i=1,2,3$, the position, mass and
 charge of the $i$-th particle. Our system is formally described by the Hamiltonian
$
\sum_{i=1}^3-{\hbar^2\over 2m_i}\partial_{x_i}^2+\sum_{1\le i<j\le 3}Z_iZ_je^2\delta(x_i-x_j)
\quad\mbox{acting in $L^2(\R^3$)}
$ 
which is defined as the unique self-adjoint operator associated to the
quadratic form with domain $\HH^1(\R^3)$:
$$
\sum_{i=1}^3{\hbar^2\over2m_i}\|\partial_{x_i}\psi\|^2+\sum_{1\le
i<j\le3}Z_iZ_je^2\int_{x_i=x_j}|\psi(\sigma_{i,j})|^2d\sigma_{i,j},\quad \psi\in\HH^1(\R^3).
$$
Here $\sigma_{i,j}$ denotes a point in the plane $x_i=x_j$.  
We will consider the cases 
$
m_1=m_2=:m>0,\quad  m_3=:M>0\quad Z_1=Z_2=-1,\quad Z_3=:Z>0
$
and answer to the question: {\sl for what values of $m/M$ and $Z$ does
  this system possess at least one bound state after removing the
  center of the mass?} 

There is a huge amount of literature on 1-$d$ particles interacting through
delta potentials either all repulsive or all attractive, but rather
few papers deal with the mixed case. We mention the work of Rosenthal,
\cite{R},  where he considered $M=\infty$. The aim of this paper is to
make a systematic mathematical study of the Rosenthal results and
extend  them to the case $M<\infty$. It has been shown in \cite{BD}
and \cite{CDP} that these delta models serve as effective Hamiltonians
for atoms in intense magnetic fields or 
quasi-particles in carbon nanotubes. As one can see in (\cite{KB},
\cite{O}, \cite{SPMP},\cite{EI}), they also seem to be relevant for atomic wave guides, nano and leaky wires.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The spectral problem}%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%
\noindent {\bf Removing the center of mass.}%
%%%%%%%%%%%%%%%%%%%%%%
 $\quad$ Using the Jacobi coordinates: $x:=x_2-x_1$, $y:= x_3-(m_1x_1+m_2x_2)/(m_1+m_2)$
and $z:=\sum_{i}m_ix_i/\sum_im_i$ we get the 2-$d$ relative motion
formal Hamiltonian 
$\widetilde H=
-{\hbar^2\over m}\partial_x^2-{2m+M\over
4mM}\hbar^2\partial_y^2+e^2\delta(x)-Ze^2\delta(y-{x\over2}) -Ze^2\delta(y+{x\over
2}).
$
Define $\alpha^2:=(M+2m)/4M$ and
$\nu(\alpha):=\sqrt{1/4+\alpha^2}$. Let $J$ be the Jacobian of 
the coordinate change $(x',y')=\{2\nu(\alpha)\hbar^2/(mZe^2)\}(x,\alpha
y)$, and define the unitary
$(U^{-1}f)(x,y)=\sqrt{J}f(x',y')$. Consider three unit vectors of
$\R^2$ given by 
$A_1:={1\over\nu(\alpha)}\left(\alpha,-{1\over2}\right)$, 
$A_2:={1\over\nu(\alpha)}\left(-\alpha,-{1\over2}\right)$, and 
$A_3:=(0,1)$. Define $A_i^\perp$ as $A_i$ rotated by $\pi/2$ in the positive sense.
Then $U\widetilde{H} U^{-1}=\{mZ^2
e^4\}/\{2\hbar^2\nu(\alpha)^2\}\; H$, where:
$$
H:=-{1\over 2}\partial_x^2-{1\over 2}\partial_y^2-\delta(A_1^\perp.(x,y))
-\delta(A_2^\perp. (x,y))+\lambda\delta(A_3^\perp . (x,y)),\quad \lambda:={\nu(\alpha)\over
  Z}.
$$
We denote by $\theta_{i,j}$ the angle between the vectors $A_i$ and
$A_j$. We give some typical values of all these parameters (see fig. \ref{fig1}).
\begin{figure}
\begin{minipage}{60mm}
\begin{tabular}{|c|c|c|c|c|}
\hline
${M\over m}$ & $\theta_{2,3}$ &  $\theta_{1,2}$ &$\alpha^2$
&$\nu(\alpha)$\cr
\hline
$1$ & $2\pi/3$ & $2\pi/3$& $3/4$ & $1$  \cr
$\infty$ & $3\pi/4$ &$\pi/2$&$1/4$&$1/\sqrt{2}$ \cr
\hline
\end{tabular}
\end{minipage}
\begin{minipage}{60mm}
\centering
\includegraphics[width=5cm]{duclosFig1.eps}
\end{minipage}
\caption{Left: table with corresponding values of angles and masses, right: support of the delta potentials with the unit vectors
$A_i$'s.}\label{fig1}
\end{figure}
\vspace{0.2cm}

%%%%%%%%%%%%%%
\noindent{\bf The skeleton}%
%%%%%%%%%%%%%%
 $\quad$ Let $A$ be unit vector in $\R^2$. If one introduce the "trace" operator  $\tau_{A}:\HH^1(\R^2)\to L^2(\R)$ defined as $(\tau_{A}\psi)(s):=\psi( sA)$ and if we let $\tau:\HH^1(\R^2)\to\bigoplus_{i=1}^3 L^2(\R)$ be defined as $\tau:=(\tau_{A_1},\tau_{A_2},\tau_{A_3})$,
we may rewrite the Hamiltonian $H$ as $H_0+\tau^\star g\tau$ where $2H_0$ stands for the free Laplacian and $g$ is the $3\times3$ diagonal matrix with entries $(-1,-1,\lambda)$. Denoting $R_0(z):=(H_0-z)^{-1}$and $R(z):=(H-z)^{-1}$ the resolvents of $H_0$ and $H$,  one derives at once, with the help of the second resolvent equation, the formula for any $z$ in the resolvent sets of $H_0$ and $H$:
\begin{align}\label{rezform}
R(z)=R_0(z)-R_0(z)\tau^*(g^{-1}+\tau R_0(z)\tau^*)^{-1}\tau
R_0(z).
\end{align}
Using the HVZ theorem (see \cite{S} for the case with form-bounded
interactions), we can easily compute the essential spectrum:
$\sigma_{{\rm ess}}(H)= [-{1\over2},\infty)$. Its bottom is given by the infimum of the spectrum of the subsystem made by the positive charge and one negative charge. 

From this and formula \eqref{rezform} it is standard to prove the
following lemma:
\vskip2mm
%%%%%%%%%%%%%%%%%%%
\noindent{\bf Lemma~1}.%
%%%%%%%%%%%%%%%%%%%
 $\quad$ {\it Let $k>{1\over\sqrt{2}}$. Define $\dsS:=  k\,g^{-1}+\tau
R_0(-1)\tau^*$. Then $E=-k^{2}<-{1\over2}$ is a discrete eigenvalue
of $H$ if and only if
$
\ker(  g^{-1}+\tau R_0(E)\tau^*)\ne\{0\}$. Note that up to a scaling
this is the same as $\ker\dsS\ne\{0\}$. Moreover, ${\rm mult}(E)={\rm dim}(\ker\dsS)$.}
\vskip2mm

The spectral analysis is thus reduced to the study of $\dsS$, a
$3\times 3$ matrix of integral operators each acting in $L^2(\R)$.  We call $\dsS$ the {\sl skeleton} of $H$.
Let us denote by $T_{A,B}:=\tau_AR_0(-1)\tau_B^\star$,
$T_0:=T_{A,A}$, by $\theta_{A,B}$  the angle between 
two unit vectors $A$ and $B$,  and by $\widehat T_{A,B}$ the Fourier image of $T_{A,B}$.  Then the kernel of $\widehat T_{A,B}$  when
$\theta_{A,B}\not\in \{0,\pi\}$, and of $\widehat T_0$ read as:
\begin{equation}\label{Ttheta}
\widehat T_{A,B}(p,q)=
{1\over2\pi|\sin(\theta_{A,B})|}{1\over
\left({p^2-2\cos(\theta_{A,B}) pq +q^2\over 2\sin^2(\theta_{A,B})}+1\right)},\quad
\widehat T_0(p,q)={\delta(p-q)\over\sqrt{p^2+2}}.
\end{equation}
Then $\widehat T_0$ is a bounded multiplication operator, and
$\widehat T_{A,B}$ only depends on $|\theta_{A,B}|$. Consequently we
denote in the sequel $T_{A_i,A_j}$ by
$T_{\theta_{i,j}}$ or $T_{i,j}$  .


\vspace{0.2cm}

%%%%%%%%%%%%%%
\noindent{\bf Reduction by symmetry.}%
%%%%%%%%%%%%%%
 $\quad$ 
$H$ and $\dsS$ enjoy various symmetry properties which follow from the fact that two particles are identical. Let 
$
\pi:L^2(\R)\to L^2(\R)
$
be the parity operator, i.e. $\{\pi\varphi\}(p)=\varphi(-p)$ and
denote by $\pi_1:=\pi\otimes 1$ and $\pi_2:=1\otimes\pi$ the 
unitary symmetries with respect to the $x$ and $y$ axis. One verifies
that for all $i,j\in\{1,2\}$, we have $[\pi_i,H]=0$ and
$[\pi_i,\pi_j]=0$. 
Thus if we denote by $\pi_i^\alpha$, $\alpha=+,-$ the eigenprojectors of $\pi_i$ on the even, respectively odd functions we may decompose $H$ into the direct sum
$$
H=\bigoplus_{\alpha\in\{\pm\},\,\beta\in\{\pm\}} H^{\alpha,\beta},\quad  H^{\alpha,\beta}:=\pi_1^\alpha\pi_2^\beta H.
$$
Similarly let $\Pi,\sigma:L^2(\R^3)\to L^2(\R^3)$ defined by
$(\Pi\psi)(-p):=\psi(-p)$ and
$\sigma\psi=\sigma(\psi_1,\psi_2,\psi_3):=(\psi_2,\psi_1,\psi_3)$.
They both commute with $\dsS$, 
and also $[\Pi,\sigma]=0$. 

Let $\Pi^\alpha$ and $\sigma^\alpha$, $\alpha=+,-$, denote the eigenprojectors of $\Pi$ and $\sigma$ symmetric and antisymmetric resp..
 Then we can write $
\dsS=\bigoplus_{\alpha\in\{\pm\},\,\beta\in\{\pm\}} \dsS^{\alpha,\beta},\quad  \dsS^{\alpha,\beta}:=
\Pi^\alpha\sigma^\beta\dsS$. 
 
 From the expression of $\widehat T_\theta(p,q)$ one also sees that
 $[\pi,T_\theta]=0$ so that $T_\theta$ decomposes into
 $T_\theta^+\oplus T_\theta^-$ where $T_\theta^\pm:=\pi^\pm T_\theta$.   As
 usual we shall consider $T_\theta^\pm$ as operators acting in
 $L^2(\R_+)$. 
Due to these symmetry properties we have
$\ker\dsS=\bigoplus_{\alpha,\beta}\ker\dsS^{\alpha,\beta}$, and each
individual null-space can be expressed as the null-space of a single
operator acting in $L^2(\R_+)$ that we call {\sl effective skeleton}.
We gather in the following table the four effective skeletons we have  to consider with their corresponding subspaces in $L^2(\R^2)$:


\vskip2mm
\begin{minipage}{120mm}
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
$S^{\alpha,\beta}$& effective skeleton & subspace in $L^2(\R^2)$\cr
\hline
$++$    & $k-T_0-T_{1,2}^++2T_{2,3}^+(T_0+k\lambda^{-1})^{-1}T_{2,3}^+$ & $\Ran\pi_1^+\pi_2^+$\cr
$-+$    & $k-T_0-T_{1,2}^-+2T_{2,3}^-(T_0+k\lambda^{-1})^{-1}T_{2,3}^-$ & $\Ran\pi_1^+\pi_2^-$\cr
$+-$    & $k-T_0+T_{1,2}^+$                                   & $\Ran\pi_1^-\pi_2^-$\cr
$--$    & $k-T_0+T_{1,2}^-$                                   & $\Ran\pi_1^-\pi_2^+$\cr
\hline 
\end{tabular}
\vskip1mm
Table~1.
\end{center}
\end{minipage}


%%%%%%%%%%%%%%%%%%%%
\section{Sectors without bound states}%

\vspace{0.2cm}
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent {\bf Properties of the $T_\theta$ operators.}%
%%%%%%%%%%%%%%%%%%%%%%%%%%
 $\quad$ From (\ref{Ttheta}) we get $0\le T_0\le 1/\sqrt{2}$. Then
 $T_\theta$ is self-adjoint and has a finite Hilbert-Schmidt norm. 
The proof of the following lemma in not at all obvious, but will be omitted due to the lack of space.
\vskip2mm
%%%%%%%%%%%%%
\vspace{0.2cm}
\noindent{\bf Lemma~2}.%
%%%%%%%%%%%%%
{\sl\  For all $\theta\in[\pi/2,\pi)$ one has $\pm T_\theta^\pm \ge0$
  and the mapping $[\pi/2,\pi)\ni \theta\mapsto\pm\inf T_\theta^\pm$
  is strictly increasing.}


\vspace{0.2cm}
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent {\bf Absence of bound state in the odd sector with respect to $y$.}%
%%%%%%%%%%%%%%%%%%%%%%%%%%
 $\quad$ We have the following result: 

\vspace{0.2cm}

\noindent{\bf Theorem~3}.%
%%%%%%%%%%%%%
{\sl\  For all $Z>0$ and all $0< M/m\le \infty$, $H$ has no bound
  state in the symmetry sector $\Ran\pi_2^-$.}

\noindent {\it Proof.} The symmetry sector $\Ran\pi_2^-$ corresponds
to the second and third lines in Table~1. For the third line 
one uses that $T_{1,2}^+\ge0$ by Lemma~2, and that $k>1/\sqrt{2}$ since we are
looking for eigenvalues below $\sigma_{\rm ess}(H)=[-{1\over
2},\infty)$. Hence 
$$
k-T_0+T_{1,2}^+\ge k-{1\over\sqrt{2}}>0
$$ 
thus $\ker(k-T_0+T_{1,2}^+)=\{0\}$, and by Lemma~1 this shows that $H$ has no
eigenvalues in $\Ran\pi_1^-\pi_2^-$. By the same type of arguments one has:
$
k-T_0-T_{1,2}^-+2T_{2,3}^-(T_0+k\lambda^{-1})^{-1}T_{2,3}^-\ge k-{1\over\sqrt{2}}>0. \ \QED
$
\vskip2mm
%%%%%%%%%%%%%
\noindent{\bf Remark~4}.%
%%%%%%%%%%%%%
$\quad $ The above theorem has a simple physical interpretation. Wave functions which are antisymmetric in the $y$ variable are those for which the positive charge has a zero probability to be in the middle of the segment joining the negative charges. A situation which is obviously not favorable for having a bound state.



\vspace{0.2cm}
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent {\bf Absence of bound state in the odd-even sector with
  respect to $x$ and $y$.}%
%%%%%%%%%%%%%%%%%%%%%%%%%%
 $\quad$ \noindent Looking at the fourth line of Table~1 we have to consider
\begin{equation}\label{Smoinsmoins}
S^{-,-}(k):=k-T_0+T_{1,2}^-=:
\sqrt{k-T_0}\left(1+\widetilde T_{1,2}^-(k)\right)\sqrt{k-T_0}
\end{equation}
where $\widetilde
T_{1,2}^-(k):=(k-T_0)^{-{1\over2}}T_{1,2}^-(k-T_0)^{-{1\over2}}$. Here
we will only consider the case $M\ge m$, i.e. $\pi/2\le \theta_{1,2}\le 2\pi/3$. Assume that we can prove that
$
\widetilde
T_{1,2}^-(2^{-{1\over2}})\ge -1
$
for $\theta_{1,2}=2\pi/3$, this will imply that $S^{-,-}(2^{-{1\over2}})\ge0$ first for $\theta_{1,2}=2\pi/3$ and then for all $\pi/2\le \theta_{1,2}\le 2\pi/3$ by the monotonicity of $\inf T_{1,2}^-$ with respect to $\theta$ as stated in Lemma~2; finally looking at (\ref{Smoinsmoins}) this will show that $S^{-,-}(k)>0$ for all $k>1/\sqrt{2}$ and therefore that $\ker S^{-,-}(k)=\{0\}$.
But 
$
\widetilde
T_{1,2}^-:=T_{1,2}^-(2^{-{1\over2}})
$ ( for $ \theta_{1,2}=2\pi/3$) 
 is Hilbert-Schmidt  since its kernel decay at infinity faster than
 the one of $T_{1,2}^-$ and it has the following 
behavior at  the origin:
$
\widetilde
T_{1,2}^-(p,q)\sim
-{16\sqrt{2}\over3\sqrt{3}\pi}+\OO\left((p^2+q^2)\right).
$
It turns out that 
$-1$ is an eigenvalue of $\widetilde
T_{1,2}^-$ with eigenvector  
$$
\R_+\ni p\mapsto \frac{
\left [\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{p^2+2}}\right ]^{1/2}}{p\left(2 p^2+3
\right) }\quad\buildrel p\to0\over\sim\quad
\frac{1}{3\,2^{\frac{5}{4}}}+\OO(p^2)
$$
and since the Hilbert-Schmidt norm of $\widetilde T_{1,2}$ can be evaluated numerically to $\|\widetilde T_{1,2}^-\|_{HS}\le 1.02$, all the other eigenvalues of $\widetilde T_{1,2}^-$ are above $-1$. Thus we have proved the 
\vskip2mm
%%%%%%%%%%%%%%
\noindent{\bf Theorem~5}.%
%%%%%%%%%%%%%%
{\sl\  For all $Z>0$ and all $1\le  M/m\le \infty$, $H$ has no bound state in the symmetry sector $\Ran\pi_1^-\pi_2^-$.}

%%%%%%%%%%%%%%%%%%%
\section{The fully symmetric sector}%
%%%%%%%%%%%%%%%%%%%
According to Table~1, we need to find under which conditions one has $\ker S^{+,+}(k)\ne\{0\}$ where 
$$ 
S^{+,+}(k):=k-T_0-K(k),\quad{\rm with}\quad K(k):=T_{1,2}^+-2T_{2,3}^+(T_0+k\lambda^{-1})^{-1}T_{2,3}^+.
$$ 
The proof of the following lemma is an easy application of Fredholm 
and analytic perturbation theory:

\vskip2mm
%%%%%%%%%%%%%
\noindent{\bf Lemma~6}.%
%%%%%%%%%%%%%
{\it  {\rm (i)} $\{\re k^2>0\}\ni k\mapsto S^{+,+}(k)$ is a bounded analytic self-adjoint family
of  operators.

\noindent{\rm (ii)} If $\inf \sigma\left (S^{+,+}(2^{-{1\over2}})\right )<0$, then there exists  $k>1/\sqrt{2}$ so that
$\ker S^{+,+}(k)\ne \{0\}$.
}

\vskip2mm


Denote by $\mathcal{K}(p,q)$ the integral kernel of
$K(2^{-{1\over2}})$. Our last result is:
\vskip2mm
%%%%%%%%%%%%%%%%%%%%%
\noindent{\bf Theorem~7}.%
%%%%%%%%%%%%%%%%%%%%%
{\sl\ For all  $0< M/m\le \infty$, $H$ has at least one bound state in the symmetry sector $\Ran\pi_1^+\pi_2^+$ if $Z$ is such that
$
\mathcal{K}(0,0)>0.
$
}
\vskip2mm

\noindent {\bf Proof.} We will now look for an upper bound on $\inf S^{+,+}(2^{-{1\over2}})$ by the variational method. Let 
$
j\in C_0^\infty(\R_+,\R_+)
$
so that $\int_{\R_+} j(x) dx=1$ and define two families of  functions: 
$
\forall \epsilon>0,\
\psi_\epsilon(p):=\epsilon^{-1}j(p\epsilon^{-1}),\
\phi_\epsilon := \frac{\sqrt{\epsilon}}{||j||}\psi_\epsilon,\
\|\phi_\epsilon\|=1.
$
We know that $\psi_\epsilon$ converges as $\epsilon\to0$ to the Dirac distribution. First one has
\begin{align}
((2^{-{1\over2}}-T_0)\phi_\epsilon,\phi_\epsilon)&=
{1\over\epsilon\sqrt{2}\|j\|^2}\int_{\R_+} 
%:
[1-(1+p^2/2)^{-1/2}]j^2({p/\epsilon})dp\nonumber \\
&\le{\epsilon^2\over 2\sqrt{2}\|j\|^2}\int_{\R_+} p^2j(p)^2dp.\end{align}
Then 
one has
$
(K(2^{-{1\over2}})\phi_\epsilon,\phi_\epsilon)={\epsilon\over\|j\|^2}(K(2^{-{1\over2}})\psi_\epsilon,\psi_\epsilon)=
{\epsilon\over\|j\|^2}\left(\mathcal{K}(0,0)+\OO(\epsilon)\right)
$
so that
$(S^{+,+}(2^{-{1\over2}})\phi_\epsilon,\phi_\epsilon)=2^{-{1\over2}}-(T_0\phi_\epsilon,\phi_\epsilon)-(K(2^{-{1\over2}})\phi_\epsilon,\phi_\epsilon)$
will be negative for $\epsilon>0$ small enough, provided
$\mathcal{K}(0,0)>0$. \QED

\vspace{0.2cm}

 It is possible to compute $\mathcal{K}(0,0)$
analytically. It can be shown that there exists $Z_c^{\rm ub}(M/m)$
such that for any $Z$ larger than this value, we have
$\mathcal{K}(0,0)>0$. If we now define the critical $Z$ as $
Z_c(M/m):=\inf \{ Z>0,\, H=H(Z,M/m)\  \mbox{has at least one bound state}\},
$ it follows from our last theorem that $Z_c(M/m)\leq Z_c^{\rm ub}(M/m)$.

The curve $Z_c^{\rm ub}(M/m)$ is plotted on figure~\ref{fig2}, where
we used $\theta_{1,2}$ instead of the ratio $M/m$.
\begin{figure}
\centering
\includegraphics[width=8cm]{duclosFig2.eps}
\caption{Graph of $Z_c^{\rm ub}$.}\label{fig2}
\end{figure}
\vskip2mm
%%%%%%%%%%%%%%%%%%%%%
\noindent{\bf Remarks~8}.%
%%%%%%%%%%%%%%%%%%%%%
\ (a) Rosenthal found numerically $Z_c^{\rm ub}(\frac{\pi}{2})$, i.e. $Z_c^{\rm ub}$ for $M=\infty$ to be $0.374903$. With our analytical expression of $\mathcal{K}(0,0)$ we know this value to any arbitrary accuracy: $Z_c^{\rm ub}(\frac{\pi}{2})=0.37490347747000593278...$
\vskip1mm
\noindent (b) The above curve shows that an arbitrarily small positive charge of mass $M<0.48m$ can bind two electrons. However we believe that the exact critical curve will show that $M<m$ and $Z>0$ is sufficient to bind these two electrons.

\begin{acknowledge}
H.C. acknowledges support from the Statens Naturvidenskabelige
Forskningsr{\aa}d grant {\it Topics in Rigorous
  Mathematical Physics}, and partial support through the European Union's IHP
network Analysis $\&$ Quantum HPRN-CT-2002-00277. P.D. and B.R. thanks the Universit\'e du Sud Toulon-Var for its support from its mobility program.
\end{acknowledge}
\begin{thebibliography}{99}
%
\bibitem{BD}
R. Brummelhuis, P. Duclos:  Few-Body Systems {\bf 31},  1-6, 2002
%
\bibitem{CDP}
Cornean H., Duclos P., Pedersen Th. G:  Few-Body Systems {\bf 34}, 155-161, 2004
%
\bibitem{EI} P. Exner, T. Ichinose: J. Phys. {\bf A34}(2001), 1439-1450
%
\bibitem{KB} K. Kanjilal, D. Blume: Phys. Rev. A {\bf
70} (2004), 042709
%
\bibitem{O} Maxim Olshanii: Phys. Rev. Letters {\bf 81}(5) 1998,
938-941
%
\bibitem{SPMP} 
A.F. Slachmuylders, B. Partoens, W. Magnus, F.M. Peeters:  Preprint cond-mat/0512385
%
\bibitem{R} C.M.~Rosenthal: Journ. Chem. Phys. {\bf 55}(5) 2474-2483 (1971)
%
\bibitem{S} B.~Simon: {\em Quantum mechanics for Hamiltonians defined as
quadratic forms}. Princeton University press, New Jersey 1971
\end{thebibliography}


\end{document}


\noindent Proof of lemma~6. Statement (i) is clear. To show (ii) write $S:=S^{+,+}(2^{-{1\over2}})=:2^{-{1\over2}}-T_0-K$ where $K:=K({2^{-{1\over2}}})$.
Clearly $K(k)$ is compact so that $\spect\!_{\rm ess}S(k)=\spect_{\rm ess}(k-T_0)=[k-2^{-{1\over2}},k]$. Thus if  $\inf S<0$ it  is a discrete eigenvalue of $S$. 
Since $S$ is analytic in the vicinity of $k={1\over\sqrt{2}}$ one has ( by regular
perturbation theory) that
$\inf S(k)$ is still in $\spect\!_d(S(k))\cap\R_-^\star)$ for $k$ in a small enough
neighbourhood of ${1\over\sqrt{2}}$.  Let $k_\star$ be defined as
$$
k_\star=\inf\left\{k\ge {1\over\sqrt{2}},\, \inf S(k)<\inf\spect\!_{\rm ess}S(k)\right\}
$$
then $k_\star>{1\over\sqrt{2}}$ by what as been said above. Thus either $k_\star=\infty$ and
in that case $\inf S(k)$ which is continuous and takes a negative value for
$k={1\over\sqrt{2}}$ and positive values for $k$ close to $\infty$ must go through 0 by the
intermediate value theorem. Or $k_\star$ is finite; then necessarily 
$\inf S(k_\star)=\inf\spect\!_{\rm ess}S(k_\star)>0$ and again by the continuity of $\inf
S(k)$ there must be a value ${1\over\sqrt{2}}<k_0<k_\star$ so that $\inf
S(k_0)=0\in\spect_{\rm d}S(k_0)$. \QED

\end{document}



math-ph0604003.extracted_bib
\begin{thebibliography}{99}
%
\bibitem{BD}
R. Brummelhuis, P. Duclos:  Few-Body Systems {\bf 31},  1-6, 2002
%
\bibitem{CDP}
Cornean H., Duclos P., Pedersen Th. G:  Few-Body Systems {\bf 34}, 155-161, 2004
%
\bibitem{EI} P. Exner, T. Ichinose: J. Phys. {\bf A34}(2001), 1439-1450
%
\bibitem{KB} K. Kanjilal, D. Blume: Phys. Rev. A {\bf
70} (2004), 042709
%
\bibitem{O} Maxim Olshanii: Phys. Rev. Letters {\bf 81}(5) 1998,
938-941
%
\bibitem{SPMP} 
A.F. Slachmuylders, B. Partoens, W. Magnus, F.M. Peeters:  Preprint cond-mat/0512385
%
\bibitem{R} C.M.~Rosenthal: Journ. Chem. Phys. {\bf 55}(5) 2474-2483 (1971)
%
\bibitem{S} B.~Simon: {\em Quantum mechanics for Hamiltonians defined as
quadratic forms}. Princeton University press, New Jersey 1971
\end{thebibliography}