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Bose gas, superfluidity
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\begin{document}
\title{Superfluidity in Dilute Trapped Bose Gases}
\author{Elliott H.~Lieb}
\email{lieb@math.princeton.edu}
\affiliation{Department of Physics,
Jadwin Hall, Princeton University,
P.~O.~Box 708, Princeton, New Jersey 08544}
\author{Robert Seiringer}
\altaffiliation{On leave from Institut f\"ur Theoretische Physik,
Universit\"at
Wien, Boltzmanngasse 5, A-1090 Vienna, Austria}
\email{rseiring@math.princeton.edu}
\affiliation{Department of Physics,
Jadwin Hall, Princeton University,
P.~O.~Box 708, Princeton, New Jersey 08544}
\author{Jakob Yngvason}
\email{yngvason@thor.thp.univie.ac.at}
\affiliation{Institut f\"ur Theoretische Physik,
Universit\"at
Wien, Boltzmanngasse 5, A-1090 Vienna, Austria}
\date{\today}
\begin{abstract}
A commonly used theoretical definition of superfluidity in the ground
state of a Bose gas is based on the response of the system to an
imposed velocity field or, equivalently, to twisted boundary
conditions in a box. We are able to carry out this program in the
case of a dilute interacting Bose gas in a trap, and we prove that a gas
with repulsive interactions is 100\% superfluid in the dilute limit in
which the Gross-Pitaevskii equation is exact. This is the first
example in an experimentally realistic continuum model in which
superfluidity is rigorously verified.
\end{abstract}
\pacs{05.30.Jp, 03.75.Fi, 67.40.-w}
\maketitle
\section{Introduction}
The phenomenological two-fluid model of superfluidity (see, e.g., \cite{TT})
is based on the
idea that the particle density $\rho$ is composed of two parts, the
density $\rho_{\rm s}$ of the inviscid superfluid and the normal fluid density
$\rho_{\rm n}$. If an external velocity field is imposed on the fluid
(for instance by moving the walls of the container) only the viscous normal
component responds to the velocity field, while the superfluid
component stays at rest. In accord
with these ideas the superfluid density in the ground state is
often defined as follows \cite{HM}: Let $E_0$ denote the ground
state energy of the system in the rest frame and $E_0'$ the ground
state energy, measured in the
moving frame, when a velocity field ${\bf v}$ is imposed.
Then for small ${\bf v}$
\begin{equation}\label{rhos}
\frac{E_0'}N=\frac {E_0}N+({\rho_{\rm s}}/\rho)\half{ m} {\bf
v}^2+O(|{\bf v}|^4)
\end{equation}
where $N$ is the particle number and $m$ the particle mass. At
positive temperatures the ground state energy should be replaced by
the free energy. (Remark: It is important here that (\ref{rhos}) holds
uniformly for all large $N$; i.e., that the error term $O(|{\bf
v}|^4)$ can be bounded independently of $N$. For fixed $N$ and a
finite box, Eq.\ (\ref{rhos}) with $\rho_{\rm s}/\rho=1$ always holds
for a Bose gas with an arbitrary interaction if ${\bf v}$ is small
enough, owing to the discreteness of the energy spectrum
\cite{remark}.) There are other definitions of the superfluid density
that may lead to slightly different results \cite{PS}, but this is the
one we shall use in this paper. We shall not dwell on this issue here,
since it is not clear that there is a \lq\lq one-size-fits-all\rq\rq\
definition of superfluidity. For instance, in the definition we use
here the ideal Bose gas is a perfect superfluid in its ground state,
whereas the definition of Landau in terms of a linear dispersion
relation of elementary excitations would indicate otherwise. We
emphasize that we are not advocating any particular approach to the
superfluidity question; our contribution here consists in taking one
standard definition and making its consequences explicit.
One of the unresolved issues in the theory of superfluidity is its
relation to Bose-Einstein condensation (BEC). It has been argued that
in general neither condition is necessary for the other (c.f., e.g.,
\cite{huang,giorgini,KT}). A simple example illustrating the fact
that BEC is not necessary for superfluidity is the 1D hard-core Bose
gas. This system is well known to have a spectrum like that of an
ideal Fermi gas \cite{gira}, and it is easy to see that it is
superfluid in its ground state in the sense of (\ref{rhos}). On the
other hand, it has no BEC \cite{Lenard,pita}. The definition of the
superfluid velocity as the gradient of the phase of the condensate
wave function \cite{HM,baym} is clearly not applicable in such cases.
We do not give a historical overview of superfluidity because
excellent review articles are available \cite{leggett, baym}. While
the early investigations of superfluidity and Bose-Einstein
condensation were mostly concerned with liquid Helium 4, it has become
possible in recent years to study these phenomena in dilute trapped
gases of alkali atoms \cite{DGPS}. The experimental success in
realizing BEC in such gases has led to a large number of theoretical
papers on this subject. Most of these works take BEC for granted and
start off with the Gross-Pitaevskii (GP) equation to describe the
condensate wave function. A rigorous justification of
these assumptions is however a difficult task, and only very recently
BEC has been rigorously proved for a physically realistic many-body
Hamiltonian \cite{LS}. It is clearly of interest to show that
superfluidity also holds in this model and this is what we accomplish
here. We prove that the ground state of a Bose
gas with short range, repulsive interaction is 100\% superfluid in the
dilute limit in which the Gross-Pitaevskii description of the gas is
exact. This is the limit in which the particle number tends to
infinity, but the ratio $Na/L$, where $a$ is the scattering length of
the interaction potential and $L$ the box size, is kept fixed. (The
significance of the parameter $Na/L$ is that it is the ratio of the
ground state energy per particle, $\sim Na/L^3$, to the lowest
excitation energy in the box, $\sim 1/L^2$.) In addition we show that
the gas remains 100\% Bose-Einstein condensed in this limit, also for
a finite velocity ${\bf v}$. Both results can be generalized from
periodic boxes to (non-constant) velocity fields in a cylindrical
geometry.
The results of this paper have been conjectured for many years, and it
is gratifying that they can be proved from first
principles. They represent the first example of a rigorous
verification of superfluidity in an experimentally realistic
continuum model.
We wish to emphasize that in this GP limit the fact that there is
100\% condensation does not mean that no significant interactions
occur. The kinetic and potential energies can differ markedly from
that obtained with a simple variational function that is an $N$-fold
product of one-body condensate wave functions. This assertion might
seem paradoxical, and the explanation is that near the GP limit the
region in which the wave function differs from the condensate function
has a tiny volume that goes to zero as $N\to \infty$. Nevertheless,
the interaction energy, which is proportional to $N$, resides in this
tiny volume.
\section{Setting and Main Results}
We consider a Bose gas with the Hamiltonian
\begin{equation}\label{ham}
H_{N}=-\mu\sum_{j=1}^N \nabla_j^2+
\sum_{1\leq i0$, depending only on ${\cal C}$ and
$a(r,z)$, such that for all $|\varphi|<\varphi_0$ there is a unique
minimizer $\phi^{\rm GP}$ of the Gross-Pitaevskii functional
\begin{multline}\label{defgp}
\E^{\rm
GP}[\phi]=\int_{\K}\Big(\mu\big|\big(\nabla+{\rm i}A(\x)\big)
\phi(\x)\big|^2 \\ + V(\x)
|\phi(\x)|^2 + 4\pi\mu N a
|\phi(\x)|^4\Big)d^3\x
\end{multline}
under the normalization condition $\int|\phi|^2=1$. This minimizer
does not depend on $\theta$, and can be chosen to be positive, for the
following reason: The relevant term in the kinetic energy is $T=
-r^{-2}[\partial/\partial \theta + {\rm i}\varphi\, r\, a(r,z)]^2$. If
$|\varphi\, r\, a(r,z)| < 1/2$, it is easy to see that $T\geq \varphi^2
a(r,z)^2$, in which case, without raising the energy, we can replace
$\phi$ by the square root of the $\theta$-average of $|\phi|^2$. This
can only lower the kinetic energy \cite{anal} and, by convexity of
$x\to x^2$, this also lowers the $\phi^4$ term.
We denote the ground state energy of $\E^{\rm GP}$ by
$E^{\rm GP}$, depending on $Na$ and $\varphi$.
The following Theorem \ref{T2}
concerns the ground state energy $E_0$ of
\begin{multline}
H_N^{A}=\sum_{j=1}^N\Big[-\mu \big(\nabla_j+{\rm i}A(\x_j)\big)^2 +
V(\x_j)\Big]
\\ +\sum_{1\leq i0$ and $R>0$ this is
\begin{equation}
\geq \varepsilon T+(1-\varepsilon)(T^{\rm in}+I)+(1-\varepsilon)
T_{\varphi}^{\rm out} \ ,
\end{equation}
with
\begin{equation}\label{15}
T=\mu\int_{\K^{N-1}} d\X\int_\K
d\x_{1}\big|\nabla_1\vert\Psi(\x_{1},\X)
\vert\big|^2 \ ,
\end{equation}
\begin{equation}\label{16}
\quad T^{\rm in}=\mu
\int_{\K^{N-1}} d\X\int_{\Omega^c_{\X}} d\x_{1}\big|\nabla_1
\vert\Psi(\x_{1},\X)\vert\big|^2\ ,
\end{equation}
\begin{equation}
T_{\varphi}^{\rm out}=\mu\int_{\K^{N-1}} d\X\int_{\Omega_{\X}} d\x_{1}
\big\vert\big(\nabla_1+{\rm i}\vp/L \big)
\Psi(\x_{1},\X)\big\vert^2\ ,
\end{equation}
and
\begin{equation}
I=\half\int_{\K^{N-1}}
d\X\int_\K d\x_{1}\sum_{j\geq 2}
v(\vert\x_{1}-\x_{j}\vert)|\Psi(\x_{1},\X)\vert^2 \ .
\end{equation}
Here
\begin{equation}\Omega^c_{\X}=\left\{\x_{1}:\ \vert\x_{1}-
\x_{j}\vert0$ and $C<\infty$ such that for all subsets
$\Omega\subset\K$ and all functions $f$ on the torus $\K$ the
following estimate holds:
\begin{multline}\label{poinc}
\Vert\nabla_\varphi
f\Vert_{L^2(\Omega)}^2\geq \frac{\varphi^2}{L^2}\|f\|_{L^2(\K)}^2
+\frac c{L^2} \Vert
f-L^{-3}\langle 1,f\rangle_{\K}
\Vert_{L^2(\K)}^2\\ -C\left(\|\nabla_\varphi f\|_{L^2(\K)}^2 + \frac
1{L^2}
\|f\|_{L^2(\K)}^2\right) \left(\frac{|\Omega|^c}{|\K|}\right)^{1/2} \
.
\end{multline} Here $\vert\Omega^c\vert$ is the volume of
$\Omega^c=\K\setminus\Omega$, the complement of $\Omega$ in $\K$.
\end{lem}
\begin{proof}
We shall derive (\ref{poinc}) from a special form of this inequality
that holds for all functions that are orthogonal to the constant
function. Namely, for any positive $\alpha<2/3$ and some constants
$c>0$ and $\widetilde C<\infty$ (depending only on $\alpha$ and
$|\varphi|<\pi$) we claim that
\begin{multline}\label{poinc2}
\|\nabla_\varphi h\|_{L^2(\Omega)}^2 \\ \geq
\frac{\varphi^2+c}{L^2}
\Vert h\Vert_{L^2(\K)}^2-\widetilde
C\left(\frac{|\Omega^c|}{|\K|}\right)^\alpha\Vert\nabla_\varphi
h\Vert_{L^2(\K)}^2 \ ,
\end{multline}
provided
$\langle 1,h\rangle_{\K} =0$. (Remark: Eq.~(\ref{poinc2})
holds also for $\alpha=2/3$, but the proof is slightly more
complicated in that case. See \cite{lsy02}.) If (\ref{poinc2}) is
known
the
derivation of (\ref{poinc}) is easy: For any $f$, the function
$h=f-L^{-3}\langle 1,f\rangle_{\K}$ is orthogonal to
$1$. Moreover,
\begin{eqnarray}\nonumber
&&\Vert\nabla_\varphi h\Vert^2_{L^2(\Omega)}\\ \nonumber &&=
\Vert\nabla_\varphi h\Vert^2_{L^2(\K)}-\Vert\nabla_\varphi
h\Vert^2_{L^2(\Omega^c)}\nonumber\\
&&=
\Vert\nabla_\varphi f\Vert^2_{L^2(\Omega)}-
\frac{\varphi^2}{L^2}\vert\langle L^{-3/2},f\rangle_{\K}\vert^2
\left(1+\frac{|\Omega^c|}{|\K|}\right)
\nonumber \\ &&\quad +2\frac{\varphi}{L}{\rm
Re}\,\langle L^{-3/2},f\rangle_{\K}
\langle\nabla_\varphi f,L^{-3/2}
\rangle_{\Omega^c}\nonumber\\ \nonumber
&&\leq \Vert\nabla_\varphi f\Vert^2_{L^2(\Omega)}-
\frac{\varphi^2}{L^2} \vert\langle L^{-3/2} ,f\rangle_{\K}\vert^2
\\ \nonumber && \quad
+\frac{|\varphi|}{L}
\left(L \Vert\nabla_\varphi f\Vert_{L^2(\K)}^2+\frac 1L \Vert
f\Vert_{L^2(\K)}^2\right) \left(\frac{|\Omega^c|}{|\K|}\right)^{1/2} \\
\label{quadrat1}
\end{eqnarray}
and
\begin{multline}\label{quadrat2}
\frac{\varphi^2+c}{L^2}\Vert
h\Vert_{L^2(\K)}^2=\frac{\varphi^2}{L^2}\left( \Vert
f\Vert_{L^2(\K)}^2- \vert\langle L^{-3/2}
,f\rangle_{\K}\vert^2\right)\\ +
\frac c{L^2} \Vert f-L^{-3}\langle 1 ,f\rangle_{\K}
\Vert_{L^2(\K)}^2 \ .
\end{multline}
Setting $\alpha=\half$, using $\|\nabla_\varphi h\|_{L^2(\K)}\leq
\|\nabla_\varphi f\|_{L^2(\K)}$ in the last term in (\ref{poinc2})
and combining (\ref{poinc2}), (\ref{quadrat1}) and (\ref{quadrat2})
gives
(\ref{poinc}) with $C=|\varphi|+\widetilde C$.
We now turn to the proof of (\ref{poinc2}). For simplicity we set
$L=1$. The general case follows by scaling. Assume that
(\ref{poinc2})
is false. Then there exist sequences of constants $C_{n}\to\infty$,
functions $h_{n}$ with $\Vert h_{n}\Vert_{L^2(\K)}=1$ and
$\langle 1,h_{n}\rangle_{\K} =0$, and domains $\Omega_{n}\subset\K$ such
that
\begin{equation}\label{false}
\lim_{n\to\infty}\left\{
\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\Omega_{n})}^2+C_{n}\vert\Omega_{n}^c\vert^\alpha\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\K)}^2\right\}
\leq \varphi^2 \ .
\end{equation}
We shall show that this leads to a contradiction.
Since the sequence $h_{n}$ is bounded in $L^2(\K)$ it has a
subsequence, denoted again by $h_{n}$, that converges weakly to some
$h\in L^2(\K)$ (i.e., $\langle g,h_{n}\rangle_{\K}\to \langle
g,h\rangle_{\K}$ for all $g\in L^2(\K)$). Moreover, by H\"older's
inequality the $L^p(\Omega_{n}^c)$ norm $\Vert \nabla_\varphi
h_{n}\Vert_{L^p(\Omega_{n}^c)}=(\int_{\Omega^c_{n}} \vert
h(\x)\vert^pd\x)^{1/p}$ is bounded by
$\vert\Omega_{n}^c\vert^{\alpha/2}\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\K)}$ for $p=2/(\alpha+1)$. From (\ref{false}) we
conclude that $\Vert \nabla_\varphi h_{n}\Vert_{L^p(\Omega_{n}^c)}$ is
bounded and also that $\Vert \nabla_\varphi
h_{n}\Vert_{L^p(\Omega_{n})}\leq\Vert \nabla_\varphi
h_{n}\Vert_{L^2(\Omega_{n})}$ is bounded. Altogether, $\nabla_\varphi
h_{n}$ is bounded in $L^p(\K)$, and by passing to a further
subsequence if necessary, we can therefore assume that $\nabla_\varphi
h_{n}$ converges weakly in $L^p(\K)$. The same applies to $\nabla
h_{n}$. Since $p=2/(\alpha+1)$ with $\alpha<2/3$ the hypotheses of the
Rellich-Kondrashov Theorem \cite[Thm~8.9]{anal} are fulfilled and
consequently $h_{n}$ converges {\it strongly} in $L^2(\K)$ to $h$
(i.e., $\Vert h-h_{n}\Vert_{L^2(\K)}\to 0$). We shall now show that
\begin{equation}\label{lowersemi}
\liminf_{n\to\infty}\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\Omega_{n})}^2\geq \Vert\nabla_\varphi
h\Vert_{L^2(\K)}^2 \ .
\end{equation}
This will complete the proof because the $h_{n}$ are normalized and
orthogonal to $1$ and the same holds for $h$ by strong
convergence. Hence the right side of (\ref{lowersemi}) is necessarily
$>\varphi^2$, since for $|\varphi|<\pi$ the lowest eigenvalue of
$-\nabla_\varphi^2$, with constant eigenfunction, is
non-degenerate. This contradicts (\ref{false}).
Eq.~(\ref{lowersemi}) is essentially a consequence of the weak lower
semicontinuity of the $L^2$ norm, but the dependence on $\Omega_{n}$
leads to a slight complication. First, Eq.~(\ref{false}) and
$C_{n}\to
\infty$ clearly imply that $\vert\Omega_{n}^c\vert\to 0$, because
$\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\K)}^2>\varphi^2$. By choosing
a subsequence we may assume that
$\sum_{n}\vert\Omega_{n}^c\vert<\infty$. For some fixed $N$ let
$\widetilde\Omega_{N}=\K\setminus\cup_{n\geq N}\Omega_{n}^c$. Then
$\tilde\Omega_{N}\subset\Omega_{n}$ for $n\geq N$.
Since $\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\Omega_{n})}^2$ is bounded, $\nabla_\varphi
h_{n}$ is also bounded in $L^2(\widetilde\Omega_{N})$ and a
subsequence
of it converges weakly in $L^2(\widetilde\Omega_{N})$ to
$\nabla_\varphi
h$. Hence
\begin{multline}\label{lowersemi2}
\liminf_{n\to\infty}\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\Omega_{n})}^2 \\ \geq
\liminf_{n\to\infty}\Vert\nabla_\varphi
h_{n}\Vert_{L^2(\widetilde \Omega_{N})}^2\geq\Vert\nabla_\varphi
h\Vert_{L^2(\widetilde\Omega_{N})}^2 \ .
\end{multline}
Since
$\widetilde\Omega_{N}\subset \widetilde\Omega_{N+1}$ and
$\cup_{N}\widetilde\Omega_{N}=\K$ (up to a set of measure zero), we
can
now let $N\to\infty$ on the right side of (\ref{lowersemi2}). By
monotone convergence this converges to $\Vert\nabla_\varphi
h\Vert_{L^2(\K)}^2$. This proves (\ref{lowersemi}) which, as remarked
above, contradicts
(\ref{false}).
\end{proof}
We now are able to finish the proof of Theorem~\ref{T1}. From
Lemmas~\ref{L1} and~\ref{L2} we infer that, for any symmetric $\Psi$ with
$\langle \Psi,\Psi\rangle=1$ and for $N$ large enough,
\begin{eqnarray} \nonumber
&&\frac1N\langle\Psi,H_{N}'\Psi\rangle \big(1-\const
Y^{1/17}\big)^{-1}\\ \nonumber && \geq 4\pi\mu\rho a + \mu
\frac{\varphi^2}{L^2} \\ \nonumber &&
\quad - C Y^{1/17} \Big(\frac 1{L^2} + \frac 1N
\big\langle\Psi,\mbox{$ \sum_j$} (\nabla_j+{\rm i}\vp) \Psi\big\rangle\Big)
\\ \nonumber && \quad + \frac c{L^2}
\int_{\K^{N-1}}
d\X \int_{\K} d\x_{1}\Big|
\Psi(\x_1,\X)\\ &&\qquad\qquad -L^{-3}\big[\mbox{$ \int_\K$} d\x \Psi(\x,\X)\big]
\Big|^2 \ ,\label{lowerbd2}
\end{eqnarray}
where we used that $|\Omega^c|\leq \mbox{$\frac{4\pi}3$} N R^3= \const L^3 Y^{2/17}$. From this
we can infer two things. First, since the kinetic energy, divided by
$N$, is certainly bounded independent of $N$, as the upper bound
shows, we get that
\begin{equation}
\liminf_{N\to\infty} \frac{E_0(N,a,\varphi)}N \geq 4\pi \mu \rho a +
\mu \frac{\varphi^2}{L^2}
\end{equation}
for any $|\varphi|<\pi$. By continuity this holds also for
$|\varphi|=\pi$, proving (\ref{i}). (To be precise,
$E_0/N-\mu\varphi^2L^{-2}$ is concave in $\varphi$, and therefore stays
concave, and in particular continuous, in the limit $N\to\infty$.)
Secondly, since the upper and the lower bound to $E_0$ agree in the
limit considered, the positive last term in (\ref{lowerbd2}) has to vanish in the limit. I.e., we get that for the ground state wave function
$\Psi_0$ of $ H_N'$
\begin{multline}
\lim_{N\to\infty} \int_{\K^{N-1}}
d\X \int_{\K} d\x_{1}\Big|
\Psi_0(\x_1,\X)\\ -L^{-3}\big[\mbox{$ \int_\K$} d\x \Psi_0(\x,\X)\big]
\Big|^2 = 0 \ .
\end{multline}
This proves (\ref{ii}), since
\begin{multline}
\int_{\K^{N-1}}
d\X \int_{\K} d\x_{1}\Big|
\Psi_0(\x_1,\X)-L^{-3}\big[\mbox{$ \int_\K$} d\x \Psi_0(\x,\X)\big]
\Big|^2 \\ = 1- \frac 1{NL^3} \int_{\K\times\K} \gamma(\x,\x') d\x d\x'
\ ,
\end{multline}
and therefore $N^{-1}\langle L^{-3/2}|\gamma_N|L^{-3/2}\rangle \to 1$. As explained in \cite{LS,LSSY} this suffices for the convergence $N^{-1}\gamma_N \to |L^{-3/2}\rangle \langle L^{-3/2}|$ in trace class norm.
\end{proof}
\section{Conclusions}
We have shown that a Bose gas
with short range, repulsive interactions is both a 100\% superfluid and
also 100\% Bose-Einstein condensed in its ground state in the
Gross-Pitaevskii limit where the parameter $Na/L$ is kept fixed as $N\to
\infty$. This is a simultaneous large $N$ and low density limit,
because the dimensionless density parameter $\rho a^3$ is here
proportional to $1/N^2$. If $\rho a^3$ is not zero, but small, a
depletion of the Bose-Einstein condensate of the order $(\rho
a^3)^{1/2}$ is expected (see, e.g., \cite{PeS}). Nevertheless,
complete superfluidity in the ground state, e.g. of Helium 4, is
experimentally observed. It is an interesting open problem to deduce
this property rigorously from first principles. In the case of a
one-dimensional
hard-core Bose gas superfluidity in the ground state is easy to show, but nevertheless there is no Bose-Einstein condensation at all, not even in the ground state \cite{Lenard,pita}.
\begin{acknowledgments}
E.H.L. was partially
supported by the U.S. National Science Foundation
grant PHY 98-20650.
R.S. was supported by the Austrian Science Fund in the from of an Erwin Schr\"odinger fellowship.
\end{acknowledgments}
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