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\title{GROUND STATE ENERGY OF THE LOW DENSITY BOSE GAS}
\author{Elliott H. Lieb$^{1}$ and Jakob Yngvason$^2$}
\address{$1.$ Department of Physics, Jadwin Hall,
Princeton University, P.~O.~Box 708, Princeton, New Jersey 08544\\
$2.$ Institut f\"ur Theoretische Physik, Universit\"at Wien,
Boltzmanngasse 5, A 1090 Vienna, Austria\\}
\date{October 26, 1997}
\maketitle
\begin{abstract}
Now that the properties of low temperature Bose gases at low density,
$\rho$, can be examined experimentally it is appropriate to revisit
some of the formulas deduced by many authors 4-5 decades ago. One of
these is that the leading term in the energy/particle is $2\pi \hbar^2
\rho a/m$, where $a$ is the scattering length. Owing to the delicate
and peculiar nature of bosonic correlations, four decades of research
have failed to establish this plausible formula rigorously. The only
known lower bound for the energy was found by Dyson in 1957, but it was
14 times too small. The correct bound is proved here.
\end{abstract}
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With the renewed experimental interest in low density, low temperature
Bose gases, some of the formulas posited four and five decades ago have
been dusted off and re-examined. One of these is the leading term in
the ground state energy. In the limit of small particle density, $\rho$,
\begin{equation} e_0(v, \rho) \approx \mu \ 4\pi \rho a,
\label{rho}
\end{equation}
where $e_0(v,\rho)$ is the ground state energy (g.s.e.)
per particle in the thermodynamic limit, $a$ is the scattering length
(assumed positive) of the two-body potential $v$ for bosons of mass $m$,
and $\mu \equiv \hbar^2/2m$.
Is (\ref{rho}) correct? In particular, is it true for the hard-sphere
gas? While there have been many attempts at a rigorous proof of
(\ref{rho}) in the past forty years, none has been found so far. Our aim
here is to supply that proof in many cases. As remarked below,
(\ref{rho}) cannot hold unrestrictedly.
An upper bound for $e_0(v, \rho)$ agreeing with (\ref{rho}) is not easy
to derive, but it was achieved for hard spheres by a variational
calculation \cite {FD}, which can be extended to include general,
positive potentials of finite range. What remained unknown was a good
lower bound. The only one available is Dyson's \cite {FD}, and that is
about {\it fourteen times smaller} than (\ref{rho}). In this paper we
shall provide a lower bound of the desired form, and thus prove
(\ref{rho}). We can also give explicit error bounds for small enough
values of the dimensionless parameter $Y\equiv \rho a^3$:
\begin{equation}
e_0(v,\rho) \geq \mu\ 4\pi \rho a (1- CY^{1/17})
\label{error}
\end{equation}
for some fixed $C$ (which is not evaluated explicitly
because $C$ and the exponent $1/17$ are only of academic interest).
The bound
(\ref{error}) holds for {\it all non-negative, finite range,
spherical, two-body potentials.}
A typical experimental value \cite{TRAP} is $Y=5\times 10^{-6}$.
We conjecture that (\ref{rho}) requires only a positive scattering
length {\it and} the absence of any many-body, negative energy bound
state. If there are such bound states then (\ref {rho}) is certainly
wrong, but this obvious caveat does not seem to have been clearly
emphasized before. There is a `nice' potential with positive scattering
length, no 2-body bound state, but with a 3-body bound state \cite
{BA}.
Our method also applies to the positive temperature free energy (because
Neumann boundary conditions give an upper bound to the solution to the
heat (or Bloch) equation). The obvious details will not be discussed
here. We also give some explicit bounds for {\it finite} systems, which
might be useful for experiments with traps, but we concentrate here on
the thermodynamic limit for simplicity. For traps with slowly varying
confining potentials, $V_{\rm ext}$, our method will prove that the
leading term in the energy is given by the well known local density
approximation \cite{LDA}, which minimizes the gaseous energy
(\ref{rho}) plus the confining energy, i.e., $\int (V_{\rm ext} \rho
+\mu 4 \pi a \rho^2)$.
The fact that Dyson's lower bound was not improved for four decades,
despite many attempts, attests to the fact that bosons are subtle
quantum mechanical objects which can have peculiar correlations unknown
to fermions. {}For example, there is the non-thermodynamic $N^{7/5}$ law
for the charged Bose gas that was discovered by Dyson \cite{FD2},
confirmed only 20 years later \cite {CL}, and which defies any simple
physical interpretation.
The first understanding of (\ref{rho}) goes back to Bogoliubov
\cite{BO}, who also introduced the notion of `pairing' in Helium (which
resurfaced in the BCS theory for fermions). Later, there were several
derivations of (\ref{rho}) (and higher order) \cite{Lee-Huang-YangEtc},
\cite{EL3a}. The method of the pseudopotential, which is an old idea
of Fermi's, was closest to the Bogoliubov analysis. The `exact'
pseudopotential was constructed in \cite{EL1a}, but it did not help to
make this appealing idea more rigorous. Most of the derivations were in
momentum space, the exception being \cite{EL3a}, which works directly
in physical space and which can handle both long and short
range potentials. See \cite{EL2} for a review. All these methods rely
on special assumptions about the ground state (e.g., selecting special
terms in a perturbation expansion, which likely diverges) and
it is important to derive a fundamental result like (\ref{rho}) without
extra assumptions.
In all of this earlier work one key fact was not understood, or at least not
clearly stated in connection with the derivation of (\ref{rho}).
It is that there are two different regimes, even at
low density, with very different physics, even though the simple
formula
(\ref{rho}) seems to depend only on the scattering length.
Recall that the (two-body) scattering length is defined, for a
spherically symmetric potential, $v$, by
\begin{equation}
-\mu u_0^{\prime\prime}(r) +\frac{1}{2} v(r)u_0(r) =0,
\label{scat}
\end{equation}
with $u_0(0)=0$, $u_0(r)>0$ (which is equivalent to the absence of
negative energy bound states, and which is true for nonnegative $v$).
As $r\to \infty$, $u(r)\approx r-a$. (Note the $v/2$ and not $v$ in
(\ref {scat}) because of the reduced mass.) Thus, $a$ depends on $m$
in a nontrivial way, and there are two extremes:
1. {\it Potential Energy Dominated Region:} The hard sphere ($v(r) =
\infty$ for $r < a$), is the extreme case here; the scattering length
is independent of $m$, and the energy is mostly (entirely) {\it
kinetic}. We see this from (\ref{rho}) because $-m~\partial e_0/
\partial m $ is the kinetic energy (Hellmann-Feynman theorem). In this
regime the potential is so dominant that it forces the energy to be
kinetic. The g.s. wave function is highly correlated.
2. {\it Kinetic Energy Dominated Region:} The typical case is a very
`soft' potential. Then $a\approx (m/\hbar^2)\int_0^{\infty} v(r)r^2
dr$, which implies, from (\ref{rho}) that $e_0$ hardly depends on
$m$. Thus, the energy is almost all {\it potential.} The g.s. wave
function is essentially the noninteracting one in this limit.
In other words, `scattering length' is not a property of $v$ alone, and
the low density gas, viewed from the perspective of the bosons, looks
quite different in the two regimes. Nevertheless, as (\ref{rho}) says,
the energy cannot distinguish the two cases. Whether Bose-Einstein
condensation itself can notice the difference remains to be seen.
Condensation will not be touched upon here, except to note that so far
{\it the only case with 2-body interactions in which B-E condensation has
been rigorously established is lattice bosons with point interaction.}
See \cite{KLS}.
Dyson \cite {FD} effectively converted
region 1 into region 2, and we shall make use of his important idea.
We assume that the $N$ particles are in a $L\times L\times L$ cubic box,
$\Omega$. The particle density is then $\rho = NL^{-3}$. It is well
known that the energy per particle in the thermodynamic
limit, $e_0(v,\rho)$, does not depend on the details of (reasonable)
$\Omega$, so we are free to use a cube and take $N\to \infty$ through
any sequence we please, as far as $e_0(v,\rho)$ is concerned.
We set $N=kM$ with $k$ an integer and $M$ the
cube of an integer, because we shall want to divide up $\Omega$ into $M$
smaller cubes (called {\it cells}) of length $\ell=(k/\rho)^{1/3}$. We
will take $M\to \infty$ with $\ell$ and $k=\rho \ell^3$ fixed, but large.
The N-body Schr\"odinger operator is
\begin{equation}
H= -\mu
\sum_{i=1}^{N} ~ \Delta_i ~ + ~
\sum_{1 \leq i < j \leq N} ~ v(\x_i - \x_j)~.
\label{ham}
%\eqnum{ham}
\end{equation}
{}For boundary conditions we impose Neumann (zero derivative)
boundary conditions on $\Omega$. It is well-known that Neumann boundary
conditions give the lowest possible g.s.e. for $H$, and hence its use
is appropriate for a discussion of a {\it lower} bound for the g.s.e.
Denote this Neumann g.s.e. by $E_0(v,N,L)$.
Now divide $\Omega$ into $M$ cells and impose Neumann conditions on
each cell, which, as stated before, lowers the energy further. We also
neglect the interaction between particles in different cells; this,
too, can only lower the energy because $v\ge 0$.
A lower bound for $E_0 (v, N,L)$ is obtained by distributing
the $N$ particles in the $M$ cells and then finding a lower bound for
the energy in these cells, which are now independent. We then add these
$M$ energies. {}Finally, we minimize the total energy over {\it all
choices} of the particle number in each cell (subject to the total
number being $N$). Despite the independence of the cells, the latter
problem is not easy. In particular, something has to be invoked to
make sure that we do not end up with some cells having too large a
number of particles and some cells having too few.
With $L,N$ and $M=N/(\rho \ell^3)$ fixed, let $Mc_n$, for $n=0,1,2,
\ldots$ denote the number of cells containing exactly $n$ particles.
Then the particle number and cell number constraints are
\begin{equation}
\sum_{n \geq 0} ~ n c_n ~=~ k=\rho \ell^{3}, \qquad\quad
\sum_{n \geq 0} ~ c_n ~ =1
\label{con}
\end{equation}
and our energy bound is
\begin{equation}
E_0 (v, N, L ) ~ \geq~ M~\min~\sum_{n \geq 0} ~ c_n ~ E_0 (v, n, \ell),
\label{lb}
\end{equation}
where the minimum is over all $c_n \ge 0$ satisfying (\ref{con}).
The minimization would be easy if we knew that $E_0
(v, n, \ell)$ (or a good lower bound for it) is convex in $n$,
for then the optimum would be $c_n =
\delta_{n,k}$. This convexity is very plausible, but we cannot
prove it (except in the thermodynamic limit, where it amounts to
thermodynamic stability). What we do know instead is {\it
superadditivity:}
\begin{equation}
E_0(v, {n+n'}, \ell)\geq
E_0(v, n, \ell)+E_0(v, n', \ell)
\label{superadd}
\end{equation}
for all $n$, $n'$, and this
turns out to be an adequate substitute for controlling
the large $n$ terms in (\ref{lb}).
Eq. (\ref{superadd}) is
an immediate consequence of the positivity of the potential
and it is used as follows.
Suppose, provisionally, that we have a lower bound of the form
\begin{equation}
E_0(v, n, \ell) \ge K(v, \ell)~ n (n-1) \qquad
{\rm for}\ 0\le n \leq 4k,
\label{provis}
\end{equation}
with $K(v,\ell)$ independent of $n$ for $0\le n \leq 4k$.
In fact, we shall later prove that for small enough $\rho$ (and
hence small enough $k$) (\ref{provis}) holds with
\begin{equation}
K(v,\ell) \ge \mu~4\pi a~\ell^{-3} (1-C'~Y^{1/17}),
\label{kay}
\end{equation}
with $C'$ some constant. [However, the analysis we give now, leading to
(\ref{bound}), does not depend on this particular form of $K(v,\ell)$.]
Split the sum in (\ref{lb}) into two pieces:
$0\le n <4k$ and $4k \le n$. Let $t\equiv \sum_{n <4k} nc_n \le k$,
so that $k-t = \sum_{n \ge 4k} nc_n $.
{}From (\ref{provis})
and Cauchy's inequality (and $\sum_{n <4k} c_n \le 1$)
\begin{equation}
\sum_{n <4k} c_n E_0(v, n, \ell) \ge K(v,\ell)~
t(t-1).
\label{lowersum}
\end{equation}
On the other hand, if $n\ge 4k$ then, by (\ref{superadd}),
$E_0(v, n, \ell) \ge (n/8k )E_0(v, 4k, \ell)$, so
\begin{equation}
\sum_{n \ge 4k} c_n E_0(v, n, \ell) \ge \frac{k-t}{2}K(v,\ell)
(4k-1).
\label{uppersum}
\end{equation}
Upon adding (\ref{lowersum}) and (\ref{uppersum}) the factor
$t(t-1)+(k-t)(2k-1/2)$ is obtained. Although the number $t$ is unknown,
we note that
this factor is monotone decreasing
in $t$ in the interval $0\le t\le k$ ( which is where $t$ lies,
by (\ref{con})). Thus, we can set $t=k$ and obtain the same bound
as if we had convexity, i.e.,
\begin{equation}
E_0(v,N,L) \ge N~K(v,\ell)~(\rho \ell^3-1).
\label{bound}
\end{equation}
In summary, if we can show (\ref{provis}) for a box of a {\it fixed}
size $\ell$, for all particle numbers up to $n=4\rho \ell^{3}$, then we
will have obtained our goal, (\ref{error}), in the thermodynamic limit
{\it provided} we can show that the $K$ in (\ref{provis}) satisfies
(\ref{kay}) with the constant $C'$ when $\ell$ is large compared to the
mean particle spacing, i.e., $\rho ~\ell^3 > C''~Y^{-1/17}$. Then the
$C$ in (\ref{error}) is equals $C'+C''$.
We now focus on a single cell and denote the $n$ coordinates
$(\x_1,...,\x_n)$ collectively by $\X$.
The first step in proving (\ref{kay}) is to replace the
total potential, $\sum_{iR_0$. Let $U(r)\geq 0$ be any
function
satisfying $ \int U(r)r^2dr \leq 1$ and $U(r)=0$ for $rR_0$, we find that (\ref{oned}) is true if $a\leq R$,
which is true since $u_0\geq0$. {}Finally, by linearity and the fact
that
$U(r)=\int r^{-2}\delta (r-s)U(s) s^2 ds$, the $\delta$-function case
implies the general case. Q.E.D.
We select our $U$ by picking some $R \gg R_0$ and setting
\begin{equation}
U(r) ~= ~3(R^3-R_0^3)^{-1} \qquad {\rm for}~ R_0\widetilde H> \mu a\ {\cal W}_U(\X),
\label{lower}
\end{equation}
where ${\cal W}_U$ is as in (\ref {nnpot}), with $v$ replaced by $U$.
Dyson estimates the minimum (over all $\X$) of
${\cal W}_U(\X)$, for a $U$ similar to (\ref{yooo}), and ends up with a
lower bound for all $\rho$, but 14 times smaller than (\ref{rho}). We
follow another route.
A quantity of central importance for us will be the average
value of ${\cal W}_U(\X)$ in a cell,
denoted by $\langle ~{\cal W}_U ~\rangle$.
To compute $ \langle ~ U (\x_1 - \x_{j(1)})
~ \rangle$, for example it is easiest to do the $\x_2 ,
\ldots , \x_n$ integrations over the cell first and then the $\x_1$
integration. Provided $\x_1$ is in the smaller cube which is a distance
$R$ from the cell boundary (whose volume is $(\ell-2R)^3$),
the probability that $R_0 < |\x_j-\x_1| < R$
is $4 \pi (R^3 -R_0^3)/3\ell^{3} $.
Thus, performing the $\x_1$ integration
over the smaller cube, and then adding similar contributions from
$U(\x_2-\x_{j(2)})$, etc., and using (\ref{yooo}), we get
\begin{eqnarray}
\langle ~{\cal W}_U ~ \rangle
& \geq & \frac{3n(\ell -2R)^3}{(R^{3}-R_0^3)\ell^3}\left[ 1-\left( 1-
Q \right)^{n-1} \right]
\label{avone} \\
& \geq & \frac{4\pi}{\ell^3} n (n-1)
\left(1 -\frac{2R}{\ell} \right)^3 \frac{1}{1+ Q(n-1)} .
\label{avtwo}
\end{eqnarray}
with
\begin{equation}
Q= 4 \pi (R^3 -R_0^3) /3\ell^{3}\ll 1.
\end{equation}
In (\ref{avtwo}) we
used $[1-x]^{n-1} \leq [1+(n-1)x]^{-1}$ for
$0\leq x\leq 1$. Note that (\ref{avtwo}) is of the form (\ref{provis}).
By similar reasoning, we obtain the upper bound
\begin{eqnarray}
\langle ~{\cal W}_U ~ \rangle
& \leq & \frac{3n}{R^{3}-R_0^3}\left[ 1-\left( 1-
Q \right)^{n-1} \right]
\label{avoneupper} \\
& \leq & \frac{4\pi }{\ell^3}n (n-1).
\label{avtwoupper}
\end{eqnarray}
Since $U(r)^2 = 4\pi (Q\ell^3)^{-1} U(r)$, we easily obtain
another useful fact:
\begin{equation}
\langle~{\cal W}_U^2 ~ \rangle \le 4\pi n~ (Q\ell^3)^{-1}
\langle~{\cal W}_U ~ \rangle
\label{square}
\end{equation}
With this preparation, we show now how to use Lemma 1 and these
averages to obtain (\ref{provis}) and (\ref{kay}). Instead of using
(\ref{lower}) alone, we
pick some $0<\varepsilon \ll 1$ and, borrowing a bit of kinetic energy,
define
\begin{equation}
\widehat H \equiv \varepsilon {\cal T} +
( 1 - \varepsilon)~ \mu a {\cal W}_U (\X) ~ .
\label{hamtwo}
\end{equation}
By Lemma 1 and $v\ge 0$, we have
\begin{equation}
H > \widetilde H > \widehat H
\end{equation}
We shall derive (\ref{provis}) and (\ref{kay}) by obtaining a lower
bound to $ \widehat H$.
Although $\varepsilon$ is small, we regard
$H_0\equiv \varepsilon {\cal T}$ as our unperturbed Hamiltonian and
$V\equiv ( 1 - \varepsilon) \mu a {\cal W}_U (\X)$ as a
perturbation of $H_0$.
The ground state wave function for $H_0$ is $\Psi_0(\X)
= \ell^{-3n/2}$ and $H_0\Psi_0 = \lambda_0 \Psi_0=0$
(Neumann conditions). The second eigenvalue of $H_0$
is $\lambda_1 =\varepsilon \mu
\pi/ \ell^2$.
Note that the ground state expectation, $\langle \Psi_0 |
{\cal W}_U |\Psi_0 \rangle$, is precisely the average
$\langle~{\cal W}_U ~ \rangle$ mentioned in (\ref{avone})--(\ref
{square}).
Temple's inequality \cite{TE} states that when a perturbation $V$ is
non-negative (as here) and when $\lambda_1-\lambda_0 \ge \langle \Psi_0
|V|\Psi_0 \rangle$ then the g.s.e., $E_0$, of the perturbed Hamiltonian
$H=H_0+V$ satisfies
\begin{equation}
\!E_0 \ge \lambda_0\! +\! \langle \Psi_0|V|\Psi_0 \rangle \!-
\frac{ \langle \Psi_0
|V^2|\Psi_0 \rangle \!-\!\langle \Psi_0
|V|\Psi_0 \rangle^2 }{\lambda_1-\lambda_0 -\langle \Psi_0
|V|\Psi_0 \rangle }.
\label{temple}
\end{equation}
We apply this to our case with $\lambda_1-\lambda_0 = \varepsilon \mu
\pi/ \ell^2$ and $V= (1-\varepsilon) \mu a {\cal W}_U$. We neglect the
(positive) term $\langle \Psi_0 |V|\Psi_0 \rangle^2$ in (\ref{temple})
and we use (\ref{avtwo}), (\ref{avtwoupper}) and (\ref{square}).
We also use $1-\varepsilon < 1$ in two appropriate places and find
\begin{equation}
\! \frac{E_0(v,n,\ell)}{\mu a \langle~{\cal W}_U~\rangle } \ge
(\!1-\! \varepsilon)\!
\left( 1- \frac{4\pi a n}
{Q~\ell} \frac{ 1}{ \varepsilon \pi - a\ell^2\langle~{\cal W}_U~\rangle}
\right).
\label{tempbound}
\end{equation}
Apart from some higher order errors, (\ref{tempbound}) is just what we
need in (\ref{provis}), (\ref{kay}). Let us denote the order of the
main error by $Y^{\alpha}$, and we would like to show that $\alpha =
1/17$ suffices. The errors are the following:
{}From the $(1-\varepsilon)$ factor, we need $\varepsilon
\le O(Y^{\alpha})$.
{}From the $Q(n+1)$ error in (\ref{avtwo}) we need $Q\rho \ell^3
\le O(Y^{\alpha})$.
{}From the $R/\ell$ error in (\ref{avtwo}) we need $R/\ell \le
O(Y^{\alpha})$.
{}From (\ref{tempbound}) we need
$a\ell^{2} \langle~{\cal W}_U~\rangle /\varepsilon \le O(Y^{\alpha})$
and
$ \rho \ell^5 a/(R^3
\varepsilon) \le O(Y^{\alpha})$.
All these desiderata can be met with $\varepsilon=Y^{\alpha}$,
$R/\ell = Y^{\alpha}$, $Q=O(Y^{\alpha})$,
$\rho R^3 =Y^{2\alpha}$
and $\alpha = 1/17$ --- as claimed.
The partial support of U.S. National Science Foundation grant
PHY95-13072A01 (EHL) and the Adalsteinn Kristjansson Foundation of the
University of Iceland (JY) is gratefully acknowledged.
%%%%%%%%%%%%%%%%%%%%
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\end{document}