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\begin{document}

\author{{\bf J. Rodrigo Parreira}\thanks{%
Supported by CAPES.} \and {\bf O. Bolina}\thanks{%
Supported by FAPESP.} \and {\bf J. Fernando Perez}\thanks{%
Partially supported by CNPq.} \\
%EndAName
Instituto de F\'{\i}sica\\
Universidade de S\~ao Paulo\\
P.O. Box 66318\\
05389-970 S\~ao Paulo, Brazil}
\title{{\bf Energy Gap in Heisenberg Antiferromagnetic Half-Integer Spin Chains
with Long-Range Interactions} }
\date{December, 1995. }
\maketitle

\begin{abstract}
We show that there is no gap in the excitation spectrum of antiferromagnetic
chains with half-integer spins and long-range interactions provided the
exchange function has a sufficiently rapid decay.
\end{abstract}

\section{Introduction}

The ideas and techniques in this paper, on the problem of a gap in the
excitation spectrum of Heisenberg antiferromagnetic chains, share much in
common with those in \cite{Lieb1} and \cite{Lieb2}. In \cite{Lieb1} Lieb,
Schultz and Mattis proved that the ground state of the model with spin $%
\frac 12$ and nearest-neighbor interactions is non-degenerate and has total
spin zero. This result was obtained under the assumption of translational
invariance and the decomposability of the lattice into two sublattices with
antiferromagnetic interactions between them. Under the same assumptions -
plus periodic boundary conditions - it was also proved that the energy gap $%
\Delta _N$ between the ground state and the excited states for a finite
chain of length $N$ goes to zero as $N\rightarrow \infty $. In \cite{Lieb2}
the results about the uniqueness of the ground state were generalized by
Lieb and Mattis: the requirement of translational invariance was dropped,
arbitrary spin values were allowed and ferrimagnetic arrays were considered,
the antiferromagnetic model being treated as a special case.

In this note we show that the results of \cite{Lieb1} and \cite{Lieb2} can
be extended to one- dimensional antiferromagnetic chains with periodic
boundary conditions, half-integer spins and long range interactions without
frustration.

\section{The Model}

We consider antiferromagnetic chains defined by the Hamiltonian 
\begin{equation}
\label{HLA}H_N=-J\sum_{x,y\in \Lambda _N}(-1)^{d_{xy}}\left( d_{xy}\right)
^{-\alpha }{\bf S}_x.{\bf S}_y 
\end{equation}
where $J\geq 0$ , $\Lambda _N=\{1,2,...,N\}$, $N$ even, ${\bf S}_x$ are
usual spin operators at site $x\in \Lambda _N$ with ${\bf S}_x^2=s(s+1)$ and 
$s$ is half-integer. The factor $\left( d_{xy}\right) ^{-\alpha }$ measures
the decay of the interaction with the distance between the spins at sites $x$
and $y$. We also need periodic boundary conditions, so that ${\bf S}_{x+N+1}=%
{\bf S}_x$.

A suitable definition of distance - which takes into account the periodic
boundary conditions and the long-range character of the interactions - is
the one defined by%
$$
d_{xy}=\left\{ 
\begin{array}{l}
\left| x-y\right| , 
\text{ if }\left| x-y\right| \leq \frac N2 \\ N-\left| x-y\right| ,\text{ if 
}\left| x-y\right| >\frac N2 
\end{array}
\right. 
$$

Our result is the following: we prove that the energy gap $\Delta _N$
between the unique ground state and the excited states goes to zero as $%
N\rightarrow \infty $ for any $J>0\,$ and $\alpha >2$.

\section{Proof of the Result}

The model defined by (\ref{HLA}) is a translational invariant system which
can be decomposed into two sublattices with antiferromagnetic interactions
between them. As an immediate consequence we can conclude that the total
spin of the ground state is zero.

We now introduce the state $\left| \Psi _k\right\rangle $ :%
$$
\left| \Psi _k\right\rangle ={\cal O}_k\left| \Psi _0\right\rangle \text{ ,} 
$$
where $\left| \Psi _0\right\rangle $ is the unique ground state of the
system and the operator ${\cal O}_k$ is given by%
$$
{\cal O}_k=\exp \{ik\sum_{x=1}^NxS_x^3\}, 
$$
with $k=\frac{2\pi }N$.

As a consequence of translational invariance, we have 
\begin{equation}
\label{f2}\left\langle \Psi _0\mid \Psi _k\right\rangle =\left\langle \Psi
_0\mid {\cal O}_k\exp \{ikNS_1^3\}\exp \{-ik\sum_{x=1}^NS_x^3\}\mid \Psi
_0\right\rangle \text{ .} 
\end{equation}

>From the fact that the ground state is a singlet, it follows that%
$$
\exp \{-ik\sum_{x=1}^NS_x^3\}\left| \Psi _0\right\rangle =\left| \Psi
_0\right\rangle , 
$$
and since for half-integer spins, in the basis that diagonalize $S^3$, which
shall be used throughout in this paper,%
$$
\exp \{ikNS_1^3\}=-I, 
$$
we have the orthogonality relation$\left\langle \Psi _0\mid \Psi
_k\right\rangle =-\left\langle \Psi _0\mid \Psi _k\right\rangle =0.$

\begin{theorem}
If $\alpha >2$ the energy gap $\Delta _N$ between the ground state and the
excited states goes to zero as $N\rightarrow \infty $.
\end{theorem}

\TeXButton{Proof}{\proof}

We begin by rewritting the Hamiltonian (\ref{HLA}) as 
\begin{equation}
\label{f1}H_N=-J\sum_{x\in \Lambda _N}\sum_{n=1}^{\frac N2-1}\left[
(-1)^nn^{-\alpha }{\bf S}_x.{\bf S}_{x+n}+\frac{\left( -1\right) ^{\frac N2}}%
2\left( \frac N2\right) ^{-\alpha }{\bf S}_x.{\bf S}_{x+\frac N2}\right] , 
\end{equation}
and then compute the expected value of the energy in the state $\mid \Psi
_k\rangle $ :%
$$
E_k\equiv \langle \Psi _k\mid H_N\mid \Psi _k\rangle =\langle \Psi _0\mid 
{\cal O}_k^{-1}H_N{\cal \ O}_k\mid \Psi _0\rangle . 
$$
\ Now we point out that%
$$
{\cal O}_k^{-1}H_N{\cal O}_k=H_N+J\sum_{x\in \Lambda _N}\sum_{n=1}^{\frac
N2-1}\{\left( -1\right) ^nn^{-\alpha }[(S_x^1S_{x+n}^2-S_x^2S_{x+n}^1)\sin
kn+ 
$$
$$
+(S_x^1S_{x+n}^1+S_x^2S_{x+n}^2)\left( \cos kn-1\right) ]+ 
$$
$$
+\left( -1\right) ^{\frac N2}\left( \frac N2\right) ^{-\alpha
}[(S_x^1S_{x+\frac N2}^2-S_x^2S_{x+\frac N2}^1)\sin \frac{kN}2+ 
$$
\begin{equation}
\label{E}+(S_x^1S_{x+\frac N2}^1+S_x^2S_{x+\frac N2}^2)\left( \cos \frac{kN}%
2-1\right) ]\}, 
\end{equation}
where we have used that: 
\begin{equation}
\label{PropOk1}{\cal O}_k^{-1}S_x^1{\cal O}_k=S_x^1\cos kx+S_x^2\sin kx\text{
,} 
\end{equation}
\begin{equation}
\label{PropOk2}{\cal O}_k^{-1}S_x^2{\cal O}_k=S_x^2\cos kx-S_x^1\sin kx\text{
,} 
\end{equation}
\begin{equation}
\label{PropOk3}{\cal O}_k^{-1}S_x^3{\cal O}_k=S_x^3\text{,} 
\end{equation}
It is easy to see that the last two terms on the r. h. s. of (\ref{E}) equal
zero. We also have that%
$$
J\sum_{x\in \Lambda _N}\sum_{n=1}^{\frac N2-1}\left( -1\right) ^nn^{-\alpha
}\left( S_x^2S_{x+n}^1-S_x^1S_{x+n}^2\right) \sin kn=i\sum_{y\in \Lambda
_N}\left[ yS_y^3,\widetilde{H}\right] , 
$$
with%
$$
\widetilde{H}=-J\sum_{x\in \Lambda _N}\sum_{n=1}^{\frac
N2-1}(-1)^nn^{-\alpha -1}\sin kn{\bf S}_x.{\bf S}_{x+n}, 
$$
so that%
$$
\langle \Psi _0\mid i\sum_{y\in \Lambda _N}\left[ yS_y^3,\widetilde{H}%
\right] \mid \Psi _0\rangle =0, 
$$
where the self-adjointness of the operator $S^3$ was taken into account. The
expression in (\ref{E}) is then reduced to%
$$
\langle \Psi _0\mid {\cal O}_k^{-1}H_N{\cal \ O}_k\mid \Psi _0\rangle = 
$$
$$
=\langle \Psi _0\mid H_N+J\sum_{x\in \Lambda _N}\sum_{n=1}^{\frac
N2-1}\left( -1\right) ^nn^{-\alpha }(S_x^1S_{x+n}^1+S_x^2S_{x+n}^2)[\cos
kn-1]\mid \Psi _0\rangle . 
$$

It is possible to estimate the sum $(S_x^1S_{x+n}^1+S_x^2S_{x+n}^2)$ by a
positive number $\Theta (s)$, which depends only on the value of the spin at
each site, $s(s+1)=$ ${\bf S}_x^2$, so that:%
$$
E_k\leq E_0+J\sum_{x\in \Lambda _N}\sum_{n=1}^{\frac N2-1}\left| \Theta
(s)\left( -1\right) ^nn^{-\alpha }[\cos kn-1]\right| . 
$$
We have, therefore, for $\Delta E_k\equiv E_k-E_0$, the bound:%
$$
\Delta E_k\leq J\Theta (s)\sum_{n=1}^{\frac N2-1}n^{-\alpha }[1-\cos kn]%
{\cal N}(n). 
$$
Here ${\cal N}(n)$ is the number of pairs of sites at distance $n$. In the
periodic lattice, ${\cal N}(n)=N$, so that:%
$$
\Delta E_k\leq J\Theta (s)\sum_{n=1}^{\frac N2-1}n^{-\alpha }[1-\cos kx]N. 
$$

The sum on the r. h. s. can be estimated for some $c>0$, by parts, for $%
1\leq n\leq \left[ \frac ck\right] $, and $\left[ \frac ck\right] <n\leq N$
- where by $\left[ \frac ck\right] $ we denote the largest integer less or
equal than $\frac ck$ - as follows:%
$$
J\Theta (s)N\sum_{n<\left[ \frac ck\right] }n^{-\alpha }[1-\cos kn]\leq 
\frac{4\pi ^2}NJ\Theta (s)\sum_{n<\left[ \frac ck\right] }n^{2-\alpha }, 
$$
while%
$$
J\Theta (s)\sum_{N\geq n>\left[ \frac ck\right] }n^{-\alpha }[1-\cos
kn]N\leq 2J\Theta (s)N\sum_{n>\left[ \frac ck\right] }n^{-\alpha }\leq 
$$
$$
\leq 2J\Theta (s)N^2\left( \frac ck\right) ^{-\alpha }, 
$$
and we obtain for the energy gap $\Delta _N$ the following upper bound: 
$$
\Delta _N\leq \Delta E_{\frac{2\pi }N}\leq 2J\Theta (s)\left[ N^2\left( 
\frac{cN}{2\pi }\right) ^{-\alpha }+\frac{\pi ^2}N\sum_{n<\frac
ck}n^{2-\alpha }\right] , 
$$
which goes to zero if $\alpha >2$ when $N\rightarrow \infty $.%
\TeXButton{End Proof}{\endproof}

\begin{thebibliography}{9}
\bibitem{Lieb1}  Lieb, E.; Schultz, T.; Mattis, D.; {\it Ann. Phys.}, {\bf 16%
}, 407 (1961).

\bibitem{Lieb2}  Lieb, E.; Mattis, D.; {\it J. Math. Phys.},{\it \ }{\bf 3},
n. 4, 749 (1961).
\end{thebibliography}

\end{document}