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\begin{titlepage}
\begin{flushright}
UWThPh-1997-23\\
September 1, 1997
\end{flushright}
\vspace{2cm}
\begin{center}
{\Large \bf Existence of many-particle bound states in spite of a
pair-interaction with positive scattering length}\\[50pt]
Bernhard Baumgartner   \\
Institut f\"ur Theoretische Physik \\ Universit\"at Wien\\
Boltzmanngasse 5, A-1090 Wien 
\vfill
{\bf Abstract} \\
\end{center}

Examples of bound states are presented, for a system of three identical 
particles as well as for a system of particles on a lattice. The 
particles may be bosons as well as fermions. The interaction between them
is given by a pair-potential, which does not allow two-particle bound
states, has positive scattering length and is not too different from
realistic interatomic potentials.

\vfill
\end{titlepage}

\section{Introduction}
Experiments which have produced Bose-Einstein-condensates of several
different elements did spark a renewed interest in the theory of
Bose-Einstein-condensation (C.N. Yang 1997, K. Huang 1996, K. Burnett
1996 and references therein). A great part of theoretical work relies
on the formula
\beq
E/N = 2\pi (\hbar^2/m) \rho a + O(\sqrt{\rho a^3})
\eeq
for the ground-state energy of a non-relativistic Bose gas with low
density $\rho$, where the particles interact pairwise by a potential with
scattering length $a$. The arguments for its validity (Lieb 1965, Lieb
1963 and references therein) rest on the {\em assumption}, that there
exists {\em no many-body bound state}.

A related, less discussed problem would be to find a formula for the ground 
state energy of a non-relativistic low density Fermi gas with pair
interactions. Again it seems reasonable, that the $\rho^{5/3}$-formula for
non-interacting Fermions has to be enlarged by adding low-order terms,
{\em provided} that {\em no many body bound state\/} exists; that
$E > 0$.

Concerning Bosons, a necessary condition for the positivity of the ground
state energy is the non-existence of two-particle bound states, and the
positivity of the scattering length $a$ of the pair-potential $v$, which,
throughout this work, is assumed to be a function of the particle-distance
only. The message of this paper is the demonstration that this condition
is {\em not sufficient}, neither for Bosons nor for Fermions. We present
bound states for systems of three particles and for a system of particles
on a lattice. The family of pair-potentials which are used are not too
different from realistic interatomic potentials: They are thermodynamically
stable, they have a repulsive core and an attractive well. Such examples
have as far as I know never been presented before.

\section{Setup: The interaction potential and statistics}
The interaction potential $v$ is a function of the interparticle distance
$r$ only. The calculations are done with $\hbar^2/m = 1$. But in the
reduced two-particle system a reduced mass equal to $m/2$ has to be used
and the scattering length is to be calculated by solving
\beq
\vp''(r) = v(r) \vp(r)
\eeq
with
$$
\vp(0) = 0 .
$$
Non-existence of a two-particle bound state is equivalent to $\vp(r)$
not changing sign. The scattering length $a$ is
\beq
a := \lim_{r \ra \infty} (r - \vp/\vp').
\eeq
Using parameters $c > 0$, $d \geq 1$ we define the potential $v(r)$ to be
zero for $r > c + d$, and otherwise
\beqa
v(r) &:=& (r - c)^2 - 1 + \mu \delta (c + d - r), \\
\mu &=& d + \frac{1}{c + d}. \no
\eeqa
Absence of bound states and positivity of the scattering length are seen
by comparison of $\vp$ with $f(r) = e^{-(r-c)^2/2}$. The Gaussian $f$
is also a solution of (2) for $0 < r < c + d$. The Wronskian
$w = \vp' f - \vp f'$ is constant in this interval and $\vp$ does not
change sign (Sturm's oscillation theorem). The Wronskian $w$ is strictly
positive, if $\vp$ is chosen as positive. Approaching $c + d$ from the lower
side one finds
$$
\frac{\vp'(c+d-0)}{\vp(c+d)} = \frac{w}{\vp f} + \frac{f'(c+d)}{f(c+d)}
> \frac{f'(c+d)}{f(c+d)} = - d.
$$
The $\delta$-function part of the barrier makes
\beq
\frac{\vp'(c+d+0)}{\vp(c+d)} = \frac{\vp'(c+d-0)}{\vp(c+d)} + \mu
> \frac{1}{c + d}.
\eeq
Outside the barrier $\vp(r)$ is linear, so (5) implies that it does not
change sign and that $a$ is strictly positive.

Thermodynamic stability does obviously hold only if $c$ is not too small.
We prove it for $c \geq 5$ by bounding $v$ from below by a positive
definite function. First, for any $c$, $d$ under consideration,
$$
v(r) \geq u_c(r) := [(r-c)^2 - 1] \theta(c+1-r),
$$
bounding $v$ from below by removing all of the barrier. For $c = 5$ it
is easy to check numerically that
$$
u_5(r) > g(r) := 20 \; e^{-0.8 \times r} - 4 \; e^{-0.2 \times r}.
$$
This function is positive definite. Its Fourier transform as a function
of $\vec x$ in ${\bf R}^3$, with $r = |\vec x|$, is a positive function
of $\vec k$ with $|\vec k | = k$:
$$
\wt g(k) = \frac{2 \sqrt{2/\pi}}{(0.8^2+k^2)^2(0.2^2+k^2)^2}
[ 0.128 + 15.2 k^2] > 0.
$$
For $c > 5$ we make a comparison by scaling:
$$
u_c(r) \geq (c/5)^2 u_5 (5r/c).
$$
So each $v(r)$ with $c \geq 5$ is a sum of a positive definite and a 
positive function, which implies thermodynamic stability (Ginibre 1968).

Concerning the statistics it will not be necessary to take special care
of the Bosons. The true ground state wave function for distinguishable
but equal particles is automatically symmetric under exchange. So any
proof for the existence of a bound state of distinguishable particles
is also a proof valid for Bosons.

The Fermions on the lattice will be satisfied with a product of one-particle
wave functions which do not overlap. Such a product can be antisymmetrized
without any change in energy. For three Fermions the antisymmetry is 
achieved by choosing the right angular momenta.

\section{Existence of a three-particle bound state}
The Hamiltonian
\beq
- \frac{1}{2} \sum_{i=1}^3 \Delta_{x_i} + \sum_{i < j} v(|\vec x_i - \vec x_j|)
\eeq
acts on $\cL^2({\bf R}^9)$.

The free movement of the center of mass with the coordinates
$\vec x_{CM} = (\vec x_1 + \vec x_2 + \vec x_3)/3$ can be separated off.
Since the transformation in ${\bf R}^9$ to the system of coordinates
$$
\sqrt{3} \; \vec x_{CM}
$$
\beqa
\vec y_1 &=& \frac{1}{\sqrt{6}}(2 \vec x_1 - \vec x_2 - \vec x_3) \\
\vec y_2 &=& \frac{1}{\sqrt{2}}(\vec x_2 - \vec x_3) \no
\eeqa
is orthogonal, the kinetic energy of inner rotations and oscillations is
represented by
\beq
- \frac{1}{2} \Delta_{y_1} - \frac{1}{2} \Delta_{y_2}.
\eeq
The interaction energy depends only on the $r_i = |\vec y_i|$ and the
angle $\vt \in [0,\pi]$ defined by
\beq
\vec y_1 \cdot \vec y_2 = r_1 r_2 \cos \vt,
\eeq
since
\beqa
| \vec x_1 - \vec x_{2(3)}| &=& (3r_1^2/2 + r^2_2/2 - (+) \; \sqrt{3}\;
r_1 r_2 \cos \vt)^{1/2} \no \\
| \vec x_2 - \vec x_3| &=& \sqrt{2} \; r_2.
\eeqa
For $r_1,r_2,\vt$ fixed there remain the inner rotations, which may be
expressed using three Euler angles.

The mathematical situation is thus analogous to the one in the treatment
of the Helium atom by (Hylleraas 1928) and (Breit 1930). Restriction to
$s$-waves of the rotation gives as the operator for the kinetic energy of
oscillations:
\beq
- \frac{1}{2} \left[
\frac{1}{r^2_1} \frac{\partial}{\partial r_1} r^2_1 
\frac{\partial}{\partial r_1}
+ \frac{1}{r^2_2} \frac{\partial}{\partial r_2} r^2_2 
\frac{\partial}{\partial r_2} +
\left( \frac{1}{r^2_1} + \frac{1}{r^2_2} \right)
\frac{1}{\sin \vt} \frac{\partial}{\partial \vt} \sin \vt
\frac{\partial}{\partial \vt} \right]
\eeq
acting on the Hilbert space
$$
\cL^2({\bf R}_+ \times {\bf R}_+ \times [0,\pi], r^2_1 r^2_2
\sin \vt dr_1 dr_2 d \vt).
$$
With a unitary transformation to the $\cL^2$ on the same set, but with
Lebesgue-measure, multiplying the wave functions by 
$r_1 r_2 \sqrt{\sin \vt}$, the kinetic energy operator becomes
\beq
- \frac{1}{2} \left[ \frac{\partial^2}{\partial r^2_1} +
\frac{\partial^2}{\partial r^2_2} +
\left( \frac{1}{r^2_1} + \frac{1}{r^2_2} \right)
\left( \frac{\partial^2}{\partial \vt^2} + \frac{1}{4}
\left( 1 + \frac{1}{\sin^2 \vt} \right) \right) \right].
\eeq
We dont have to take care of the boundary conditions, since we will
finally apply this operator to functions with compact support in the
interior only. 

The $s$-waves of Euler angles are not allowed for Fermions.
They require higher angular momentum introducing a centrifugal force.
More details of the necessary modifications for their treatment follow
at the end of this section.

We restrict our attention to ``small'' oscillations near the minimum of
the potential, assuming the parameters $c$ and $d$ to be large. The final
change of coordinates is therefore
\beqa
u &:=& \frac{1}{\sqrt{2}}\; (r_1 + r_2) - c \no \\
v &:=& \frac{1}{\sqrt{2}}\; (r_1 - r_2) \\
w &:=& \frac{1}{2}\; c\left(\vt - \frac{\pi}{2}\right). \no
\eeqa
These coordinates are chosen according to symmetries: $u$ is the 
coordinate for a ``breathing mode'', where $\vt$ and $r_1/r_2 = 1$
remain fixed. Because of the symmetry under particle exchange, the
other oscillations should be degenerate, with equal frequencies. 

Expanding the potential in powers of $u,v$ and $\vt - \pi/2$ about
its minimum at $r_1 = r_2 = c/\sqrt{2}$, $\vt = \pi/2$ gives an
expansion in powers of $1/c$.
This expansion of the interaction energies involves
\beqa
(| \vec x_1 - \vec x_{2(3)}| - c)^2 &=& 
\left( u + \frac{v}{2} - (+)\;\frac{\sqrt{3}}{2} w\right)^2 + 
O\left(\frac{1}{c}\right) \no \\
(| \vec x_2 - \vec x_3| - c)^2 &=& (u - v)^2.
\eeqa
Summing up gives the potential energy
\beq
V = - 3 + 3 u^2 + \frac{3}{2} \; v^2 + \frac{3}{2} \; w^2 +
O \left( \frac{1}{c} \right).
\eeq
The operator for the kinetic energy is now
\beq
- \frac{1}{2} \Delta_{u,v,w} + \frac{1}{2} \left[ 1 -
\frac{c^2}{4} \left( \frac{1}{r^2_1} + \frac{1}{r^2_2} \right) \right]
\frac{\partial^2}{\partial w^2} - \frac{1}{8}
\left( 1 + \frac{1}{\sin^2 \vt}\right)
\left( \frac{1}{r^2_1} + \frac{1}{r^2_2} \right),
\eeq
$$
\Delta_{u,v,w} := \frac{\partial^2}{\partial u^2} +
\frac{\partial^2}{\partial v^2} + \frac{\partial^2}{\partial w^2}.
$$
The harmonic-oscillator Hamiltonian made from the leading terms,
\beq
- \frac{1}{2} \; \Delta_{u,v,w} + 3u^2 + \frac{3}{2} v^2 +
\frac{3}{2} w^2 - 3
\eeq
has the negative ground state energy
\beq
\frac{1}{2} \; \sqrt{6} + \sqrt{3} - 3 = - 0.043...
\eeq
It remains to discuss the corrections from the anharmonic terms in (15)
and (16), and also from the restriction to a compact support. All these
corrections can be bounded and the bounds can be controlled in the limit
of large $c$, since they fall off at least as $1/c$, as is shown in the
appendix. Hence it is proven, that there exists a bound state, a molecule
of three Bosonic atoms arranged in an equilateral triangle.

When treating the model with Fermions, we introduce the antisymmetrizing
factor
$$
\frac{(\vec y_1 \times \vec y_2) \cdot \vec a}{|\vec y_1 \times \vec y_2|}
$$
where $\vec a$ is a constant vector. This is a function of the Euler
angles, determining the rotational state of the molecule. It is
completely antisymmetric under particle exchange. Its contribution to the
kinetic energy gives the angular momentum barrier
$$
\left( \frac{1}{r^2_1} + \frac{1}{r^2_2} \right) \frac{1}{\sin^2 \vt},
$$
which has to be added to the Hamiltonian of the oscillations, in formulas
(11), (12) and (16).

The trial wave functions for oscillations have to be completely symmetric
under particle exchange. We may use the same functions as in the case of the
Bosons. Once their support is sufficiently restricted, the symmetrization
has no effect on kinetic or potential energy.

And the contribution from the angular momentum is of the order of $1/c^2$.

\section{A manybody bound state in the form of a lattice}
The setup is the same as above, only at the end some variations on
defining the interaction potential will be discussed. The number of
particles is very large or infinity. We will calculate the energy of one
particle in the interior. As the state of the many particle system we
choose a product of independent one-particle states. Each one-particle
state is defined by the Gaussian wave function
\beq
\psi(\vec x) = (\lambda/\pi)^{3/4} \; e^{-\lambda\; \vec x^2/2}
\eeq
translated to be centered at a lattice point of a fcc or hcp lattice (close 
packing with 12 nearest neighbours). The distance of neighbours is chosen
as $c$, the parameter which defines the distance of two particles at the
minimum of interaction potential. (This choice is not optimal in minimizing
the energy, but it simplifies the calculations.)

The kinetic energy per particle is
\beq
E_{\rm kin} = 3 \lambda/4.
\eeq
The potential energy of a pair of particles, belonging to a pair of 
lattice points in the distance $b = |\vec b|$, is
\beqa
E_{{\rm pair},b} &=& \int\!\!\int v(|\vec z - \vec y|) \psi^2(\vec z)
\psi^2(\vec y - \vec b) d^3 z d^3y = \no \\
&=& \int v(|\vec x|) \left( \int \psi^2(\vec x + \vec y)
\psi^2(\vec y - \vec b) d^3y\right) d^3x.
\eeqa
Straightforward this can be reduced to
\beq
E_{{\rm pair},b} = \left( \frac{\lambda}{2\pi}\right)^{1/2}
\frac{1}{b} \int_0^\infty v(r) \left[ e^{-\lambda(r-b)^2/2} -
e^{-\lambda(r+b)^2/2} \right] r dr.
\eeq
In order to get a low energy, we have to place the cut-off distance
$c + d$ of the interaction potential between the distances to nearest
and next nearest neighbours. For example at half way:
$$
d = \frac{\sqrt{2} - 1}{2} \; c.
$$
Then the contributions of the next nearest and all the other far away
located particles to the energy are small, according to the fall-off of the
Gaussians. The nearest neighbours are
located at distance $b = c$, where the bottom of $v(r)$ lies. The
harmonic part of $v(r)$ near its bottom gives thus the main contribution
to the integral (22). The negative Gaussian and the boundaries to the
harmonic part at $r = 0$ and $r = c + d$ have little influence, again
according to the fall-off of the Gaussians. The approximate main
contribution to the potential energy per particle, half of the pair
interactions with 12 nearest neighbours, can thus be calculated
analytically as
\beq
6 \left(\frac{\lambda}{2\pi}\right)^{1/2} \frac{1}{c} 
\int_{-\infty}^{+\infty} [(r-c)^2 - 1] e^{-\lambda(r-c)^2/2} r dr =
6 \left( \frac{1}{\lambda} - 1 \right).
\eeq
The energy per particle, the sum of (20) and (23)
\beq
\frac{3 \lambda}{4} + \frac{6}{\lambda} - 6
\eeq
is minimized with $\lambda = 2 \sqrt{2}$:
\beq
E = 3 \sqrt{2} - 6 \approx - 1.757.
\eeq
This model can also be used for Fermions: One may proceed by restricting
the one-particle wave functions to a compact support, so that they do not
overlap. This procedure will decrease the potential energy and increase
the kinetic energy. The correction is always to be measured by the fall-off
of the Gaussian. (See the Appendix.) Antisymmetrization has then no
effect on the energy.

\section{Discussion}
Having established the existence of many body bound states we would like
to identify the reason why the ``repulsive effect'' of the positive
scattering length breaks down: It is essentially the dominance of
many-particle-correlations, when they cannot be decomposed into
pair-correlations. For three particles this correlation has to be
fine tuned, determining the wave functions of relative motion, and it
leads to weak binding. For many particles much cruder correlations,
determining only the relative mean positions, give stronger binding.
(One may then consider pair potentials which are more realistic, with
reduced repulsive barrier outside the attractive well.)

Dropping the requirement for thermodynamic stability, there are much
simpler arguments against the sufficiency of the positive scattering
length as a condition for the absence of bound states: Consider a pair
potential 
which is negative for small $r$ in a neighbourhood of the origin
$(v(r) = (r^2-3)\theta(d-r)+d\delta(r-d)$ for example).
Then, $N$
particles in a cluster with such a small diameter will have a negative
potential energy, proportional to $N^2$. But kinetic energies need not
increase faster than $N$ in the case of Bosons, $N^{5/3}$ in the case
of Fermions.
(Considering Bosons, already three of them can have a bound state with
the given example potential.)

In a low-density gas the occurrence of special structures has low
probability, so one might philosophize, that a gaseous state without
bounds persists some time as a metastable state. Such considerations
have already been stated in connection with the existence of a 
Bose-condensed gas in spite of a negative scattering length
(R.J. Dodd 1996, B.D. Esry 1996).

\section*{Appendix: Bounds to the effects of localization of the wave
functions and of anharmonic terms}
We use a continuous cut-off approximation to the ground state wave function
$\vp(x)$ of a harmonic oscillator (1-dimensional, centered at $x = 0$):
$$
\psi(x) = \gamma \left( \frac{\omega^2}{\pi}\right)^{1/4}
\left[ e^{-x^2 \omega^2/2} - e^{-\ell^2 \omega^2/2}\right]
\theta(\ell - |x|).
$$
The normalization factor $\gamma$ is greater than 1. It can be bounded by
$$
\gamma < \left( 1 - \frac{2}{\sqrt{\pi}\;\omega\ell} \;
e^{-\ell^2\omega^2} - 4 e^{-\ell^2\omega^2/2}\right)^{-1/2}.
$$
(Derived by finding lower bounds to the integrals which determine
$\|\psi\|$.)

The expectation value of any increasing positive $V(|x|)$ (the interaction
with a partner at the right distance) gets smaller
by this cut-off. The line of reasoning is:

$\gamma > 1 \Ra \psi'(x) < \vp'(x)$ for $x \in (0,\ell) \Ra \psi - \vp$
is decreasing on $(0,\ell) \Ra \psi^2 - \vp^2$ is decreasing $\Ra$ 
(by Chebyshev's inequality)
$$
\int_0^\infty V(x) \psi^2(x) dx = \int_0^\ell V \psi^2 <
\int_0^\ell V \vp^2 < \int_0^\infty V \vp^2.
$$
The kinetic energy increases in the procedure of cutting-off. This
increase is bounded:
$$
\int_{-\infty}^{+\infty} |\psi'(x)|^2 dx = \gamma^2 \int_{-\ell}^{+\ell}
|\vp'|^2 < \gamma^2 \int_{-\infty}^{+\infty} |\vp'|^2.
$$
For the three particle model, $\ell < c/2$ will guarantee that $\psi$
is in the Hilbert space. Also $\ell < d/2$ is useful, in order not to
involve the $\delta$-barrier as contributing to the energy. (Otherwise,
the fall-off of the Gaussian would make good bounds to these
contributions.) Finally, $\ell/c$ shall be small.

The trial function $\psi_1(u) \psi_2(v) \psi_3(w)$ (with the appropriate
$\omega_i$) is used to estimate the ground state energy of the
Hamiltonian (16) + (15). For the harmonic-oscillator part we get (18)
plus the changes discussed above. In the anharmonic part,
$$
f(r_1,r_2) \frac{\partial^2}{\partial w^2} + g(r_1,r_2,\vt) +
V_{\rm anh}(u,v,w),
$$
$g$ is of no concern for Bosons, since it is negative, see (16). For 
Fermions it is positive (the angular momentum barrier), of the order 
of $1/c^2$. The anharmonic part of the
potential can be studied by expanding the interactions as Taylor series
in $u,v,w$, see (14). That the corrections to the harmonic quadratic terms 
are of the order $1/c$ can be understood as a geometric effect.

The operator $f \dfrac{\partial^2}{\partial w^2}$ contributes the
kinetic energy of $\psi_3(w)$, multiplied by the expectation of
$f = \dfrac{c^2}{4} \left( \dfrac{1}{r^2_1} + \dfrac{1}{r^2_2}\right)-1$.
On the support of $\psi_1 \psi_2$ this function is bounded by
$\dfrac{\ell}{c} + \dfrac{2\ell^2}{(c - 2\ell)^2}$.

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