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quantum spin systems, quasiparticle states, dispersion relation, transverse Ising system.
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% file name: quasilast.tex
% changes 3-16
% The section on the infinite volume limit has been rewritten.
% I added a title page and made up a title. Feel free to change it
% if you think of a better one. Also, your new address needs to
% be filled in. And the abstract needs to be written.
% p. 2 : first paragraph :
% (with $\beta = 1/k_B T$, $k_B$ being the Boltzmannn constant).
% Do we really need to explain what k_B is ?
% p. 2 : first paragraph : I changed ``Feynman-Kac'' to ``Trotter product''
% since that is what is really used in this sort of work
% Except your papers with Jurg and others where you used something
% a bit different.
% p. 2 : first paragraph : ``expected properties'' has been replaced by
% ``exponential decay of the truncated correlation functions.''
% p.3 : just after eq (1). I changed
% ``anisotropic $XY$- and Heisenberg models'' to ``highly anisotropic...''
% I also added a reference to the Kirkwood Thomas paper and our second
% paper. In the bibliography our second paper needs a title.
% Proof of theorem 1: I made the radius \delta explicit. Hopefully this
% makes things a bit clearer. The changes are minor and are at the very
% start and very end of the proof.
% The references contained the line
% \bibitem{albanese}
% without anything else. Was this a reference we were going to fill
% in latter?
% The book by Malyshev and Minlos, bibitem mm, did not appear to be cited
% anywhere. I added it to the list of papers by the Russians.
% p 11, just after eq (32) I added
% (For $d>1$, $k$ and $l$ are vectors, and $kl$ will denote their dot
% product.)
% The paragraph containing eq (101) has been rewritten a bit.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%% old to do list %%%%%%%%%%%%%%%%%%%%
% * understand what the Russians have done
% * Infinite volume limit details
% DONE
% * Theorem 2 has been rewritten. The problem with the previous version
% was that E_1^\Lambda(k) is defined for a different collection of k
% for each \Lambda. So it doesn't make sense to take the infinte
% volume limit of E_1^\Lambda(k) -E_0^\Lambda.
% New version of the theorem avoids this by working with the Fourier
% coefficients.
% * In statement of theorem 2, orders are only upper bounds
% SEE remark following the theorem
% * Explain when we need perioidicity and when we don't.
% SEE the introduction section
% * I don't see where exactly the inequality (31) and the sentence
% above it (in Theorem 2) has been explicitly proven in the paper.
% \mu is needed to prove inequality (31). We have
% |e_s| \le exp(-\mu |s|) ||e||
% So we choose \mu so that
% exp(-\mu) \le c |\epsilon|
% SEE the very end of the last section
% * The def of the weighted norm has been changed. Actually, the definition
% of w_i(X) has been changed. It is now the number of bonds in the smallest
% connected set of bonds containing X and i. This is the right way
% to do things since the factors of \epsilon are tied to bonds.
% The old way would not give the right orders in \epsilon for various
% bounds.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Using KT equation to study excited states in quantum spin systems %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[12pt]{article}
\baselineskip=4ex
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\newcommand \half {{1 \over 2}}
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\newcommand \e {\epsilon}
\newcommand \g {{\exp^{(1)}}}
\newcommand \h {{\exp^{(2)}}}
\newcommand \f { {1 \over n !}}
% \def \d {\bullet}
\def \d {\triangle}
\def \L {\Lambda}
\def \xx {{\cal X}}
%%%% ?????????????????????????????????
\def \zed {Z}
% following is the function on the Banach space which gives the FPE
\newcommand \fp {F}
% following is the function whose argument is a set of sites which is
% used to write down the FPE for the exicted states in a compact way
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\newtheorem{definition}{Definition}
\newtheorem{proposition}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{theorem}{Theorem}
\begin{document}
\title{A fixed point equation for one quasiparticle states in quantum
spin 1/2 systems}
\title{Expansions for one quasiparticle states in spin 1/2 systems}
\author{Nilanjana Datta
\\Statistical Laboratory
\\Centre for Mathematical Sciences
\\University of Cambridge
\\Wilberforce Road, Cambridge CB30WB
\\ email: n.datta@statslab.cam.ac.uk
\\
\\Tom Kennedy
\\Department of Mathematics
\\University of Arizona
\\Tucson, AZ 85721
\\ email: tgk@math.arizona.edu
\bigskip
}
\maketitle
\begin{abstract}
Convergent expansions of the wavefunctions for the ground state
and low-lying excited states of quantum transverse Ising systems
are obtained. These expansions are employed to prove that
the dispersion relation for a single quasi-particle
has a convergent expansion.
\end{abstract}
\newpage
\section{Introduction}
The ground state and low-lying
excited states of a quantum spin system, governed by
a Hamiltonian $H$, are usually studied by considering a low temperature
expansion of its partition function e.g. \cite{alb, ginibre, ty1, ty2}.
This involves an analysis
of the trace of the
operator, $\exp(-\beta H)$, in the limit of large $\beta$
(with $\beta = 1/k_B T$, $k_B$ being the Boltzmannn constant).
The analysis is usually done by applying the method of cluster
expansions to the Trotter product representation of the operator.
In contrast, Kirkwood and Thomas \cite{kt} studied the ground state
of such systems by working directly with the Schrodinger equation,
$H |\Psi\rangle = E |\Psi\rangle$.
They considered a class of Hamiltonians of quantum spin-$1/2$
systems, which
were small perturbations of classical Hamiltonians.
By making a clever ansatz for the form of the ground state, they
were able to develop a convergent expansion for the ground state which
they used to show that the infinite volume limit of the ground state
exists and has exponential decay of the truncated correlation functions.
The method of Kirkwood and Thomas has been employed, in the $C^*$-algebraic
framework, to establish the uniqueness of translationally invariant ground
states in quantum spin-$1/2$ systems \cite{matsuia}
and has also been extended to study
the ground states of models of arbitrary (though finite) spin,
e.g. the quantum Potts model \cite{matsuib}.
The approach of Kirkwood and Thomas can, however, be applied
only to a restricted
class of Hamiltonians, namely to ones that satisfy a Perron-Frobenius
condition. In this paper, we show how this restriction can be eliminated,
thus extending the range of applicability of their approach.
We also simplify their method -- the simplification
being in the estimates which show that the expansions
converge.
%We also extend the range of applicability of their approach.
%Their treatment required that the Hamiltonian satisfy a Perron-Frobenius
%condition. We give an argument that makes this
%restriction unnecessary.
It is of interest to study, not only the properties of the ground state,
but also the low-lying excited states of quantum spin systems.
There have been a series of papers \cite{zkm,amz, ms, km, pokorny, mm}
devoted to the investigation
of the low-lying spectra of infinite-dimensional quantum lattice systems.
In \cite{zkm} this has been done by reducing the problem to the spectral
analysis of an operator which is unitarily equivalent to the Hamiltonian
governing the lattice system, and is also
the generator of an infinite-dimensional
Markov diffusion process. The method of cluster expansions has been employed
to analyse the low--lying spectral properties. In particular, the spectrum of
the Hamiltonian has been analysed in the one-particle invariant subspaces,
which describe the elementary excitations of the ground state. In \cite{amz,ms}
it has been shown that the one--particle excitations form a segment of
absolutely continuous spectrum, which is separated by energy gaps from the
ground state eigenvalue, and from the rest of the spectrum.
In the second part of this paper, we study one quasi-particle excitations
in a quantum spin-$1/2$
system by employing a method different from the ones mentioned
above. Our method uses techniques, based on the Kirkwood-Thomas
approach, that we develop for the ground state analysis.
The key idea is that the lowest excited states have a definite momentum.
We make an ansatz for the form
of these states based on their momentum, and use it to develop a
convergent expansion for them. As an application of this expansion
we study the dispersion relation for a single quasi-particle and
show that it has a convergent expansion.
In this paper we consider the following formal Hamiltonian
\be
{\widetilde{H}}= - \sum_i \s^z_i + \e \sum_{} \, \s^x_i \s^x_j,
\label{eq_orig_ham}
\ee
The sum over $$ is over all nearest neighbor sites in the lattice.
The technique we use is, however, applicable to a wide class of Hamiltonians,
which include the highly anisotropic $XY$- and Heisenberg models
\cite{dk,kt}.
%Of course, to make sense of the above Hamiltonian we must first restrict
%it to a finite subset $\Lambda$ of $\zed^d$.
We consider the Hamiltonian
to be restricted to finite volumes, $\Lambda$, which are hypercubes,
and impose periodic boundary conditions.
This restriction is essential for our treatment of the one quasi-particle
excited states which relies on the translation symmetry of the model.
However, for our treatment of the ground state we could consider
more general $\Lambda$ and other boundary conditions.
It has been shown \cite{pokorny} that the spectrum of the Hamiltonian
\reff{eq_orig_ham} above the lowest energy level contains a {\em{continuous}}
part and is separated from the lowest level by an energy gap. In this
paper we make no further analysis of the {\em{nature}} of the spectrum
above the ground state eigenvalue.
The first part of this paper is devoted to our version of
the Kirkwood-Thomas approach to the ground state of quantum
spin-1/2 systems. The one-particle excitations of the system
are analysed in the second part of the paper. Unfortunately,
the analysis
of two-particle excitations lies outside the scope of our method.
\section{The ground state}
A classical spin configuration on the lattice is an assignment
of a $+1$ or a $-1$ to each site in the lattice. So for each
$i \in \L$, $\s_i = \pm 1$. We will abbreviate the classical
spin configuration $\{\s_i\}_{i \in \L}$ by $\s$ .
For each such $\s$ we let $|\s>$ be the
state in the Hilbert space which is the tensor product
of a spin up state at each site with $\s_i=+1$ and a spin down state
at each site with $\s_i=-1$.
Thus $|\s>$ is an eigenstate of all the $\s^z_i$ with
$\s^z_i |\s> = \s_i |\s>$.
Any state $|\Psi>$ can be written in terms of this basis:
\be
|\Psi> = \sum_\s \psi(\s) |\s>,
\label{eqa}
\ee
where $\psi(\s)$ is a complex-valued function on the spin configurations
$\s$.
The form of the Hamiltonian in equation \reff{eq_orig_ham}
seems natural for perturbation
theory in $\e$ since the Hamiltonian is diagonal when $\e=0$.
However, following Kirkwood and Thomas, we
study a unitarily equivalent Hamiltonian
\be
H= - \sum_i \s^x_i + \e \sum_{} \, \s^z_i \s^z_j.
\label{eqb}
\ee
For a single site the ground state of $-\s^x$ is the vector
$|+1>+|-1>$. Thus the unnormalized
ground state of the Hamiltonian \reff{eqb} with $\e=0$ is
given by \reff{eqa} with $\psi(\s)=1$ for all $\s$.
For a subset $X$ of the lattice $\L$, let
$$\s(X) = \prod_{i \in X} \s_i.$$
Let $f(\s)$ be a complex-valued function of the $\s$'s.
Then $f(\s)$ can be written in a unique way as
$$f(\s)= \sum_{X} g(X) \s(X),$$
where the $g(X)$ are complex numbers and $X$ is summed over
all subsets of $\L$ (including $\L$ itself and the empty set).
The Fourier coefficients $g(X)$ are given by
$$g(X) = 2^{-|\L|} \, \sum_\s \, \s(X) \, f(\s),$$
where $|\L|$ is the number of sites in $\L$ and
the sum is over all $2^{|\L|}$ spin configurations on $\L$.
The heart of the Kirkwood Thomas method is the following.
We expand the ground state with respect to the basis as
in eq. \reff{eqa} and
assume that $\psi(\s)$ can be written in the form
\be
\psi(\s)= \exp[- {1 \over 2} \sum_X g(X) \s(X)] .
\label{eqc}
\ee
(The factor of $-{1 \over 2}$ is introduced for later convenience.)
Kirkwood and Thomas use a Perron-Frobenius argument to show that
$\psi(\s)>0$ for all $\s$, and so the ground state can be written in
this form with real $g(X)$.
The Perron-Frobenius argument requires that the off-diagonal terms
in the Hamiltonian are all non-positive, and so it appears that
the Kirkwood Thomas approach only applies to a very restricted
class of models. This is not the case. We will show how to
eliminate the Perron-Frobenius argument entirely.
We proceed in two steps. First, we will make the
ansatz \reff{eqc} and prove that there is an eigenstate of this form.
Later we will give a short argument to show that this eigenstate
must in fact be the ground state.
Consider the eigenvalue equation
\be
H |\Psi> = E_0 |\Psi> .
\label{eqbb}
\ee
The operator $\s^z_i \s^z_j$ is diagonal, so
\be
\s^z_i \s^z_j \sum_\s \psi(\s) |\s>
= \sum_\s \s_i \s_j \psi(\s) |\s> .
\ee
The operator $\s^x_i$ just flips the spin at site $i$, i.e.,
$\s^x_i |\s> = |\s^{(i)}>$, where $\s^{(i)}$ is the spin configuration
$\s$ but with $\s_i$ replaced by $-\s_i$.
Hence
\be
\s^x_i \sum_\s \psi(\s) |\s>
= \sum_\s \psi(\s) |\s^{(i)}>
= \sum_\s \psi(\s^{(i)}) |\s> .
\ee
The last equality follows by a change of variables in the sum.
Thus if we
use \reff{eqa} in the Schrodinger equation \reff{eqbb} and pick out the
coefficient of $|\s>$, then for each spin configuration $\s$ we have
\be
- \sum_i \psi(\s^{(i)}) + \e \sum_{} \, \s_i \s_j \psi(\s)
= E_0 \psi(\s).
\ee
Dividing both sides by $\psi(\s)$, we have
\be
- \sum_i { \psi(\s^{(i)}) \over \psi(\s)}
+ \e \sum_{} \, \s_i \s_j = E_0 .
\ee
Now $\s^{(i)}(X)$ is just $\s(X)$ if $i \notin X$ and
$-\s(X)$ if $i \in X$.
Thus
$$\psi(\s^{(i)}) =\exp[- {1 \over 2} \sum_{X: i \notin X} g(X) \s(X)
+ {1 \over 2} \sum_{X: i \in X} g(X) \s(X)],
$$
and so we have
$${\psi(\s^{(i)}) \over \psi(\s)}
= \exp[\sum_{X: i \in X} g(X) \s(X)].
$$
Thus the Schrodinger equation is now
\be
- \sum_i \exp[\sum_{X: i \in X} g(X) \s(X)]
+ \e \sum_{} \, \s_i \s_j = E_0.
\label{eqd}
\ee
This equation must hold for every configuration $\s$.
This is the magic step in the Kirkwood Thomas method because of the
cancellation between terms in $\psi(\s)$ and $\psi(\s^{(i)})$.
As $|\L| \rightarrow \infty$,
the sum $\sum_X g(X) \s(X) $ will diverge,
but the sum $\sum_{X: i \in X} g(X) \s(X)$ that appears in the
above equation will remain bounded.
We will refer to equation \reff{eqd} as the Kirkwood Thomas equation.
We rewrite \reff{eqd} as
\be
- \sum_i \exp^{(2)}\bigg(\sum_{X: i \in X} g(X) \s(X)\bigg)
+ \e \sum_{} \, \s_i \s_j = E_0 + |\L|
+ \sum_i \sum_{X: i \in X} g(X) \s(X),
\ee
where
\be
\h (x) = e^x - 1 - x.
\ee
and $|\L|$ is the number of sites in the lattice $\Lambda$.
Now
\be \sum_i \sum_{X: i \in X} g(X) \s(X)
= \sum_X \sum_{i: i \in X} g(X) \s(X)
= \sum_X |X| g(X) \s(X),
\ee
where $|X|$ is the number of sites in $X$.
So the Kirkwood Thomas equation has become
\be
\sum_X |X| g(X) \s(X) + E_0 + |\L|
=
- \sum_i \h \bigg(\sum_{X: i \in X} g(X) \s(X)\bigg)
+ \e \sum_{} \, \s_i \s_j.
\ee
Expanding $\h$ in a power series we have
\bea
\sum_X |X| g(X) \s(X) + E_0 + |\L|
&=&
- \sum_i \sum_{n=2}^\infty \, \f \bigg(\sum_{X: i \in X} g(X) \s(X)\bigg)^n
+ \e \sum_{} \, \s_i \s_j \nonumber\\
&=&
- \sum_i \sum_{n=2}^\infty \, \f
\sum_{X_1,X_2,\cdots,X_n: i \in X_j}
\prod_{j=1}^n g(X_j) \s(X_j)
+ \e \sum_{} \, \s_i \s_j.\nonumber\\
\label{kt}
\eea
The notation $: i \in X_j$ means that $i$ belongs to $X_j$ for all
$j=1,2,\cdots,n$.
Since $\s_i^2=1$, $\s(X) \s(Y) = \s(X \d Y)$ where the
symmetric difference $X \d Y$ of $X$ and $Y$ is defined by
$X \d Y = (X \cup Y) \setminus (X \cap Y)$.
Thus
$\prod_{j=1}^n \s(X_j) = \s(X_1 \d \cdots \d X_n).$
If we equate the coefficients of $\s (X)$ on both sides of eq. \reff{kt},
then we obtain, for $X \ne \emptyset$,
\be
g(X)
=
- { 1 \over |X|} \sum_i \sum_{n=2}^\infty \, \f
\sum_{X_1,X_2,\cdots,X_n: i \in X_j,
\atop{X_1 \d \cdots \d X_n=X} }
g(X_1) g(X_2) \cdots g(X_n)
\, + \, {1 \over |X|} \e \, 1_{\nn}(X),
\label{fixedpt}
\ee
where $1_{\nn}(X)$ is $1$ if $X$ consists of two nearest neighbor
sites and is $0$ otherwise.
Similarly, for $X=\emptyset$, we obtain the equation
\be
E_0 + |\L| = - \sum_i \sum_{n=2}^\infty \, \f
\sum_{X_1,X_2,\cdots,X_n: i \in X_j,
\atop{X_1 \d \cdots \d X_n= \emptyset} }
g(X_1) g(X_2) \cdots g(X_n).
\ee
We let $g$ denote the collection of coefficients
$\{g(X): X \subset \Lambda, X \ne \emptyset\}$,
and think of eq. \reff{fixedpt} as a fixed point equation, $g=\fp(g)$.
We define a norm by
\be
||g||= \sum_{X \ni i} \, |g(X)| \, |X|\, e^{\mu w_i(X)},
\label{normg}
\ee
where $w_i(X)$ is the number of bonds in the smallest connected set
of bonds that contains $X$ and the site $i$.
(We make the convention that a bond contains its endpoints.)
If $X$ just contains the single site $i$, then $w_i(X)=0$.
With periodic boundary conditions we need only consider
$g(X)$ which are translation invariant, and so
the above quantity does not depend on $i$.
With other boundary conditions we would need to define the norm to
be the supremum over $i$ of the above.
The constant $\mu$ is positive.
The smaller $|\epsilon|$ is, the larger we may take $\mu$. (The product
$e^{\mu} |\epsilon|$ must be small.)
We will prove that the fixed point equation has a solution in the
Banach space defined by this norm if $|\epsilon|$ is sufficiently small
by applying the contraction mapping theorem.
\begin{theorem}
There exists $\eps_0 >0$ such that if $|\eps| e^\mu < \eps_0$,
then the fixed point equation \reff{fixedpt} has a solution $g$,
and $||g|| \le c e^\mu |\epsilon|$ for some constant $c$ which depends
only on the lattice.
\label{thmone}
\end{theorem}
\no {\bf Proof:} We will prove that $F$ is a contraction on
a small ball about the origin, and that it maps this
ball back into itself. The contraction mapping theorem
will then imply that $F$ has a fixed point in this ball.
We denote the radius of the ball by $\delta$ and let
\be
\delta = 4d \, e^\mu \, |\epsilon|.
\ee
(The factor of $4d$ is for the lattice $\zed^d$. For other lattices
this factor would be different.)
For the sake of concreteness, we will prove $F$ is a contraction
with constant $1/2$, but there is nothing special about the choice of
$1/2$. So we need to show
\be
||\fp(g)-\fp(g')|| \le \frac{1}{2} ||g - g'|| \quad {\hbox{for}} \quad
||g||, ||g'|| \le \delta ,
\label{boundone}
\ee
and
\be
||\fp(g)|| \le \delta \quad {\hbox{for}} \quad
||g|| \le \delta .
\label{boundtwo}
\ee
The proof of \reff{boundone} proceeds as follows.
\be
||\fp(g)-\fp(g')|| \le
\sum_i \sum_{n=2}^\infty {1 \over n!}
\sum_{X_1,\cdots,X_n: i \in X_j,
\atop{0 \in X_1 \d \cdots \d X_n} }
|g(X_1) \cdots g(X_n) - g'(X_1) \cdots g'(X_n)|
e^{\mu w_0(X_1 \d \ldots \d X_n)} .
\label{return}
\ee
If $0 \in X_1 \d \ldots \d X_n$, then $0$ is in at least one $X_j$.
Using the symmetry under permutations of the $X_j$, we can
take $0 \in X_1$ at the cost of a factor of $n$.
We claim that if $i \in X_j$ for $j=1,2,\cdots,n$, then
\be
w_0(X_1 \d \cdots \d X_n) \le w_0(X_1) + \sum_{j=2}^n w_i(X_k) .
\label{wtri}
\ee
To prove the claim,
let $C_1$ be a connected set of bonds such that
$X_1 \subset C_1$, $0 \in C_1$, and $|C_1|=w_0(X_1)$.
For $j=2,3,\cdots,n$ let $C_j$ be connected sets of bonds such that
$X_j \subset C_j$, $i \in C_j$ and $|C_j|=w_i(X_j)$.
Define $C=\cup_{j=1}^n C_j$. Since all the $X_j$ contain $i$, all the
$C_j$ contain $i$. So $C$ is connected. Clearly,
$X_1 \d \cdots \d X_n \subset C$, and since $0 \in C_1$, we have $0 \in C$.
So
\be w_0(X_1 \d \cdots \d X_n) \le |C| \le \sum_{j=1}^n |C_j|
= w_0(X_1) + \sum_{j=2}^n w_i(X_j)
\ee
which proves the claim \reff{wtri}.
So
\bea
&& ||\fp(g)-\fp(g')||
\nonumber\\
&& \le
\sum_i \sum_{n=2}^\infty {1 \over (n-1)!}
\sum_{X_1, \cdots, X_n: i \in X_j, 0 \in X_1}
\left|\prod_{j=1}^n g(X_j) -
\prod_{j=1}^n g'(X_j) \right| e^{\mu(w_0(X_1)+w_i(X_2) + \cdots + w_i(X_n))}
\nonumber\\
&& \le
\sum_{n=2}^\infty {1 \over (n-1)!}
\, \sum_{X_1 \ni 0} \, \sum_{i \in X_1} \, \sum_{X_2 \ldots X_n \ni i}
\, \sum_{k=1}^n \bigg( \prod_{j=1}^{k-1} |g(X_j)| \bigg) |g(X_k)-g'(X_k)|
\bigg( \prod_{j=k+1}^n |g'(X_j)| \bigg) \nonumber\\
&& \quad \quad \quad \times \quad
e^{\mu(w_0(X_1)+w_i(X_2) + \cdots + w_i(X_n))} \nonumber\\
&& \le \sum_{n=2}^\infty {1 \over (n-1)!}
\, \sum_{X_1 \ni 0} \, \sum_{i \in X_1} |g(X_1)-g'(X_1)| e^{\mu w_0(X_1)}
\sum_{X_2 \ldots X_n \ni i} \, \bigg( \prod_{j=2}^{n} |g'(X_j)|
e^{\mu w_i(X_j)} \bigg)
\nonumber\\
&& + \sum_{n=2}^\infty {1 \over (n-1)!}
\, \sum_{X_1 \ni 0}
\, \sum_{i \in X_1} |g(X_1)|e^{\mu w_0(X_1)}
\, \sum_{X_2 \ldots X_n \ni i} \, \sum_{k=2}^n
\, \bigg( \prod_{j=2}^{k-1} |g(X_j)|e^{\mu w_i(X_j)} \bigg)
\nonumber\\
&& \quad \quad \times \quad |g(X_k)-g'(X_k)|e^{\mu w_i(X_k)}
\, \bigg( \prod_{j=k+1}^{n} |g'(X_j)| e^{\mu w_i(X_j)} \bigg)
\nonumber\\
&& \le || g - g'|| \sum_{n=2}^\infty {1 \over (n-1)!}
\left[ ||g'||^{n-1} + \sum_{k=2}^n ||g||^{k-1} \, ||g'||^{n-k}\right],
\label{g11}
\eea
where we have used the following
\be
\sum_{X_1 \ni 0} \, \sum_{i \in X_1} |g(X_1)-g'(X_1)|
e^{\mu w_0(X_1)}=\sum_{X_1 \ni 0}
|g(X_1)-g'(X_1)| |X_1| e^{\mu w_0(X_1)} = ||g-g'||,
\ee
\be
\sum_{X_k \ni i} |g(X_k)-g'(X_k)|
e^{\mu w_i(X_k)} \le ||g-g'||,
\ee
\be
\sum_{X_1 \ni 0} \, \sum_{i \in X_1} |g(X_1)|
e^{\mu w_0(X_1)}=\sum_{X_1 \ni 0}
|g(X_1)| |X_1| e^{\mu w_0(X_1)} = ||g||,
\ee
and
\be
\sum_{X_j \ni i} |g(X_j)|e^{\mu w_i(X_j)} \le ||g||.
\ee
Since $||g||$ and $||g'||$ are both bounded by $\delta$,
it follows from \reff{g11} that
\be
||\fp(g) - \fp(g')||\le K \, ||g - g'|| ,
\label{f1}
\ee
where
\be
K = \sum_{n=2}^\infty {n \over (n-1)!} \delta^{n-1}
= e^{\delta} - 1 + \delta \, e^{\delta}.
\label{kg}
\ee
We have defined $\delta$ to be proportional to $e^\mu |\epsilon|$,
so if $\epsilon_0$ is small enough then $K \le 1/2$.
To prove \reff{boundtwo}, we use \reff{boundone} with $g'=0$.
Using \reff{fixedpt} to compute $\fp(0)$, it follows that
\bea
||\fp(g)|| &\le& ||\fp(g)-\fp(0)|| + ||\fp(0)||
\le {1 \over 2} ||g|| + \sum_{X \ni 0} |\eps| 1_{\nn} (X)
e^{\mu w_0(X)}\nonumber\\
& \le & {1 \over 2} ||g|| + 2 d |\eps| e^\mu \le {\delta \over 2}
+ 2 d |\eps| e^\mu = \delta .
\label{zero}
\eea
\qed
At this point all we have shown is that we can find an eigenfunction of
the form \reff{eqc}.
Kirkwood and Thomas show that this eigenfunction is
the ground state by a Perron-Frobenius argument. This limits their method to
models in which the off-diagonal matrix elements of the Hamiltonian
are non-positive. We give a different argument which does not
impose any restrictions on the signs of the matrix elements of the
Hamiltonian.
When $\eps=0$, we know that our eigenfunction is the ground state.
Since we are working on a finite lattice, we are considering a finite
dimensional eigenvalue problem. So the only way our eigenvalue can cease
being the ground state energy is if it crosses another eigenvalue.
So if we can show that our eigenfunction is nondegenerate for all
$\eps$ with $|\eps|<\eps_0$, then it follows that our eigenfunction
is the ground state for all such $\eps$. Suppose that there is
some $\eps$ such that the degeneracy of the eigenvalue of our eigenfunction
is greater than one. Recall that our eigenfunction is denoted by
$\psi(\s)$. Let $\phi(\s)$ denote a different eigenfunction with the
same eigenvalue. Define
\be
\psi_\alpha(\s)=\psi(\s) + \alpha \phi(\s) .
\ee
So $\psi_\alpha$ is also an eigenfunction with the same eigenvalue.
Since $\psi(\s)>0$ for all $\s$, and there are finitely many
$\s$, if $\alpha$ is sufficiently small, then
$\psi_\alpha(\s)>0$ for all $\s$. Thus it can be written in the form
\be
\psi_\alpha(\s)= \exp[- {1 \over 2} \sum_X g_\alpha(X) \s(X)] .
\ee
Moreover, as $\alpha \rightarrow 0$, $||g-g_\alpha|| \rightarrow 0$.
Since $\psi_\alpha$ is an eigenfunction, $g_\alpha$ satisfies the fixed point
equation.
For sufficiently small $\alpha$, $g_\alpha$ is in the ball of radius
$\delta$ about the origin. But this contradicts the uniqueness of the
solution of the fixed point equation in this ball.
Thus the eigenvalue is nondegenerate, and so the eigenfunction we have
constructed is indeed the ground state.
In the preceding argument, how small $\alpha$ must be depends on
$\Lambda$. This is not a problem. For each $\Lambda$ we need only
find one value of $\alpha$ which leads to a contradiction and so implies
that the eigenfunction is not degenerate.
Finally, we consider the question of convergence to the infinite
volume limit. One approach would be to show that the coefficients
$g(X)$ have convergent expansions in $\epsilon$ and that these
expansions agree up to order $\epsilon^L$ where $L$ is the length
of the smallest side in $\Lambda$.
We will sketch a different approach based on the
fixed point equation. It takes advantage of the fact that if we have a
$g$ which is near the origin and very close to being a fixed point,
then it must be very close to the true fixed point.
We start by asking where the volume dependence is
in the fixed point equation \reff{fixedpt}.
The $X_i$ in this equation must be subsets of the volume $\Lambda$, and
the notion of nearest neighbor in the term $1_{\nn}(X)$ in
\reff{fixedpt} depends on the volume because of the periodic boundary
conditions.
To make the volume dependence of $F(g)$ explicit, we write it as
$F_\Lambda(g)$. We can define a fixed point equation which should give
the infinite volume limit directly.
To do this, in \reff{fixedpt} we let the $X_i$ be subsets of
$\zed^d$ and interpret $1_{\nn}(X)$ to mean nearest neighbors in
$\zed^d$ in the usual sense.
We denote the resulting function by $F_\infty(g)$. Our proof of
theorem \ref{thmone} also shows
that there is a small neighborhood of the origin in which the
fixed point equation $F_\infty(g)=g$ has a unique solution.
Let $g_\Lambda$ and $g_\infty$ denote the fixed points of $F_\Lambda$ and
$F_\infty$. We want to show that as the volume $\Lambda$ converges to
$\zed^d$, $g_\Lambda$ converges to $g_\infty$.
Note that $g_\Lambda$ and $g_\infty$ are defined for different collections
of subsets, so it is not immediately clear in what sense
$g_\Lambda$ should converges to $g_\infty$.
In fact, if a set $X$ is near the boundary, $g_\Lambda(X)$ and $g_\infty(X)$
need not be close.
Consider one dimension and let
$\Lambda=\{1,2,\cdots,L\}$, and let $X=\{1,L\}$. Because of the periodic
boundary conditions, the two sites in $X$ are actually nearest
neighbors. So $g_\Lambda(X)$ is of order $\epsilon$. But $g_\infty(X)$
will be of order $\epsilon^L$.
To compare $g_\Lambda$ and $g_\infty$ correctly, we proceed as follows.
We take our volume to be centered about the origin and let $L$
be the length of the shortest side. So the closest sides are a distance
$L/2$ from the origin. Recall that $w_0(X)$ is the number of bonds
in the smallest set of bonds which contains $X$ and $0$.
We claim that
\be
\sum_{X \ni 0: w_0(X) \le L/4} |g_\Lambda(X) - g_\infty(X)| \label{expfast}
\ee
converges to zero exponentially fast as $\Lambda$ converges to $\zed^d$.
Note that the terms we are neglecting
\be
\sum_{X \subset \Lambda: 0 \in X, w_0(X) > L/4} |g_\Lambda(X)|
\ee
and
\be
\sum_{X \subset \zed^d: 0 \in X, w_0(X) > L/4} |g_\infty(X)|
\ee
are both of order $\exp(-\mu L/4)$ by our bounds on the solution of the
fixed point equations.
Our strategy is to show that $g_\Lambda$ is almost a fixed point
of $F_\infty$. This then implies that $g_\Lambda$ is close to the exact
fixed point $g_\infty$ of $F_\infty$.
To show that $g_\Lambda$ is almost a fixed point
of $F_\infty$ we need to estimate
\be
F_\infty(g_\Lambda) - g_\Lambda = F_\infty(g_\Lambda) - F_\Lambda(g_\Lambda) .
\ee
However, $g_\Lambda$ is only defined on subsets of $\Lambda$,
so we cannot plug $g_\Lambda$ directly into $F_\infty$.
We define a function $g^\prime_\Lambda$ which is defined for all
$X \subset \zed^d$ as follows. For sets $X$ with $0 \in X$ and
$w_0(X) \le L/4$, we let $g^\prime_\Lambda(X)=g_\Lambda(X)$.
We then extend the definition to sets which are translates of such $X$
by taking $g^\prime_\Lambda(X+t)=g^\prime_\Lambda(X)$.
(These translations are done in $\zed^d$.)
Finally we define $g^\prime_\Lambda(X)$ to be zero for the sets for which
it is not yet defined. Note that there is a consistency issue here.
If we have two sets $X$ and $Y$ which both contain $0$, have
$w_0(X),w_0(Y) \le L/4$ and are translates of each other in $\zed^d$, then
we need to be sure $g_\Lambda(X)=g_\Lambda(Y)$. These conditions on $X$ and
$Y$ ensure that they are also translates in $\Lambda$, so this follows
from the translation invariance of $g_\Lambda$.
We will now show that
$g^\prime_\Lambda$ is close to $g_\infty$ by showing that
$g^\prime_\Lambda$ is almost a fixed point of $F_\infty$.
We will estimate
\be
\sum_{X \ni 0} |F_\infty(g^\prime_\Lambda)(X) - g^\prime_\Lambda(X)|
\label{fixdif}
\ee
where $F_\infty(g^\prime_\Lambda)(X)$ is the coefficient of
$\s(X)$ in $F_\infty(g^\prime_\Lambda)$.
Note that we are now using a norm without the weighting factor of
$e^{\mu w_0(X)}$.
We divide the terms in this sum into two classes, those with
$w_0(X) \le L/4$ and those with $w_0(X) > L/4$.
For the latter, $g^\prime_\Lambda(X)=0$ and
\be
\sum_{X \ni 0: w_0(X)>L/4}
|F_\infty(g^\prime_\Lambda)(X)| \le
e^{-\mu L/4} \sum_{X \ni 0: w_0(X)>L/4}
|F_\infty(g^\prime_\Lambda)(X)| e^{\mu w_0(X)}
= e^{-\mu L/4} ||F_\infty(g^\prime_\Lambda)||
\label{latter}
\ee
where $||F_\infty(g^\prime_\Lambda)||$ is the original norm with the
exponential weight.
Now consider an $X$ containing $0$ for which $w_0(X) \le L/4$. Then
$g^\prime_\Lambda(X)=g_\Lambda(X)=F_\Lambda(g_\Lambda)(X)$.
So we need to compare $F_\infty(g^\prime_\Lambda)(X)$ and
$F_\Lambda(g_\Lambda)(X)$.
Consider a term $X_1,\cdots,X_n$ in $F_\infty$ with $\sum_i w_0(X_i) \le L/4$
and $X_1 \d \cdots \d X_n = X$, {with $X$ containing the origin}.
If $w_0(X_i)\le L/4$, then $X_i$ has a translate $X_i+t$ with $0 \in X_i+t$
and $w_0(X_i+t)\le L/4$.
So we have $g_\Lambda^\prime(X_i)=g_\Lambda(X_i)$,
and the contribution of this term to
$F_\infty(g^\prime_\Lambda)(X)$
will be the same
as the contribution of this term to $F_\Lambda(g_\Lambda)(X)$.
So the difference between $F_\infty(g^\prime_\Lambda)(X)$ and
$F_\Lambda(g_\Lambda)(X)$ comes from terms with $\sum_i w_0(X_i) > L/4$.
%$\sum_i w_0(X_i) \ge L/4$.
Their total contribution to \reff{fixdif} is of order $\exp(-\mu L/4)$.
Combining this with \reff{latter}, we have shown
\reff{fixdif} is of order $\exp(-\mu L/4)$.
If we apply the contraction mapping theorem
with $\mu=0$ in the definition of the norm, this
implies
\be
\sum_{X \ni 0} |g^\prime_\Lambda(X) - g_\infty(X) |
\ee is of order $\exp(-\mu L/4)$.
And so \reff{expfast} converges to zero like $\exp(-\mu L/4)$ in the
infinite volume limit.
% For later use, we note the following property of the $g(X)$.
% Let $R$ be reflection about the origin, i.e., $R j = -j$ for a site $j$.
% For a set of sites $X$,
% $R X$ is defined in the obvious way. We claim that for the solution
% of the fixed point equation, $g(X)=g(RX)$. To prove this, define
% $g'$ by $g'(X)=g(RX)$. Then it is easy to see that $\fp(g)=g$
% implies $\fp(g')=g'$. Since the solution of the fixed point equation
% is unique in the small ball about the origin, this implies
% $g=g'$. So $g(X)=g(R X)$.
\section{Excited states : transverse field plus perturbation}
In this section we study the low-lying excited
states with one quasiparticle for the model defined by \reff{eqb}
when $\eps$ is small.
When $\epsilon$ is $0$, these states for the Hamiltonian
\reff{eq_orig_ham} have all spins pointing in the $z$-direction except
for one spin pointing in the opposite direction.
Thus the energy of the first excited state is $|\L|$-fold degenerate
($|\L|$ is the number of sites). When $\epsilon$ is small but not zero,
this degenerate eigenvalue should open up into a band of continuous
spectrum in the infinite volume limit. The goal is to get a convergent
expansion for the eigenfunctions corresponding to this continuous
spectrum. This looks impossible since there is no gap around
these eigenvalues. However, these excited states are uniquely
determined by their momentum. If $T_l$ denotes translation
by $l$, then an eigenstate with momentum $k$ satisfies
\be
T_l | \Psi > = e^{-ikl} | \Psi > .
\label{momentum}
\ee
(For $d>1$, $k$ and $l$ are vectors, and $kl$ will denote their dot
product.)
If you look in the subspace with momentum $k$, then the first excited state
has an isolated eigenvalue. We will show that by building into the
wave function the constraint that it has momentum $k$, one
can get a convergent expansion for the eigenfunction.
One of the consequences of our expansion is the following
theorem about the energy dispersion relation.
We will always work on a hyper-rectangle:
\be
\Lambda=\{(i_1,\cdots,i_d): 1 \le i_j \le L_j, j=1,2,\cdots d\}
\ee
where $L_i$ is the length in the $i$th direction.
The possible momenta are given by
$k=(k_1,\cdots,k_d)$
where $k_i=\frac{2\pi n_i}{L_i}$ and $n_i=0,1,....,L_i-1$.
We denote the ground state energy for a volume $\L$ by $E_0$.
We let $E_1(k)$ denote the
lowest eigenvalue in the subspace of momentum $k$ if $k \ne 0$ and
the next to lowest eigenvalue if $k = 0$.
(We think of this as the energy of the state with one quasiparticle
with momentum $k$.)
These eigenvalues depend on the volume $\L$. When we wish to emphasize
this dependence, we will write them as
$E_0^\L$ and $E_1^\L(k)$.
The dispersion relation is
\be
\Delta^\Lambda(k)=E_1^\L(k)-E_0^\L .
\ee
When $\epsilon=0$, $\Delta^\Lambda(k)=2$ for all $k$.
For small nonzero $\epsilon$ we expect $\Delta^\Lambda(k)$ to equal
$2$ plus a function that is of order $\epsilon$.
The following theorem essentially says that this function has an
infinite volume limit, and this limit has a convergent expansion.
We state the theorem in terms of the Fourier coefficients of
$\Delta^\Lambda(k)$ rather than the function itself since
$\Delta^\Lambda(k)$ is defined for different values of $k$
for different $\Lambda$.
\begin{theorem}
There exists an $\eps_0 >0$ such that for all $|\eps| < \eps_0$
the following is true.
Write $\Delta^\Lambda(k)$ as a Fourier series:
\be
\Delta^\Lambda(k)= 2 + \sum_{s \in \Lambda} e^\Lambda_s \, e^{iks} .
\label{eq_quasi}
\ee
There are coefficients $e_s$, defined for all $s \in \zed$, such that
for any sequence of finite volumes which are hyper-rectangles and
converge to $\zed^d$,
the coefficients $e^\Lambda_s$ converge uniformly to $e_s$.
The coefficients $e^\Lambda_0$ and $e_0$ are at least first order
in $\epsilon$, and $e^\Lambda_s$ and $e_s$ are at least of
order $\epsilon^{|s|}$ in the sense that there is a constant $c$ such that
\be
|e_s| \le (c |\eps|)^{|s|} .
\ee
Thus the infinite volume dispersion relation may be defined by
\be
\Delta(k)= 2 + \sum_{s \in \zed^d} e_s e^{iks} .
\ee
\label{thm_excited}
\end{theorem}
\bigskip
\no {\bf Remark:} The bounds on the $e_s$ are only upper bounds.
Some of the $e_s$ may be higher order in $\epsilon$ than the
theorem indicates.
\bigskip
We start by considering what states with momentum $k$ look like.
Let $\s-l$ be the configuration whose value at site $i$ is $\s_{i+l}$.
A function $\psi(\s)$ has momentum $k$ if
\be
\psi(\s-j)=e^{-ikj} \psi(\s) .
\label{basisk}
\ee
If $X$ is any subset of the lattice, then we can form such a
function by letting
\be
\phi_{X,k}(\s)=\sum_l e^{ikl} \s(X+l) ,
\ee
where $X+l$ is the set of sites $\{i+l:i \in X\}$.
The above sum is over sites in $\Lambda$. All subsequent sums in
this section will be over sites in $\Lambda$ unless otherwise indicated.
We claim that these functions span the subspace of momentum $k$.
It suffices to show that if we consider all $k$, then these
functions span the entire state space. Since
\be
\sum_k \phi_{X,k}(\s)= |\Lambda| \s(X) ,
\ee
the span of the functions $\phi_{X,k}(\s)$ includes the
functions $\s(X)$ for all $X$. The latter form a
basis for the state space, so the $\phi_{X,k}$ span the
state space.
While these functions span the subspace of momentum $k$, they are
not linearly independent.
For some choices of $X$ and $k$,
$\phi_{X,k}(\s)$ will be zero. For example, if $X=\emptyset$ or
$\Lambda$, then the function is zero unless $k=0$.
Furthermore,
\be
\phi_{X+t,k}(\s)= \sum_l e^{ikl} \s(X+t+l) = \sum_l e^{ik(l-t)} \s(X+l)
= e^{-ikt} \phi_{X,k}(\s) .
\label{eqnotind}
\ee
So if two $X$'s are translates of each other, then the
corresponding functions will be the same up to a multiplicative constant.
Define $X$ and $Y$ to be translationally equivalent if $X=Y+n$ for
some $n \in \Lambda$. Then the subsets of $\Lambda$ can be partitioned
into equivalence classes. Pick one set from each equivalence class
and let $\xx$ be the resulting collection of sets. Every subset of $\Lambda$
can be written in the form $X + n $ for some $X \in \xx$.
While $X$ is uniquely determined, $n$ is not. (For example, consider
$X=\emptyset$ or $\Lambda$.)
If we consider the collection
$\phi_{X,k}$ for $X \in \xx$,
then this collection spans the subspace of momentum $k$.
Now suppose that for each $k$ we have an eigenstate $\psi_k(\s)$
with momentum $k$.
Then $\psi_k(\s)/\Omega(\s)$ also has momentum $k$. So it can be
written in the form
\be
{ \psi_k(\s) \over \Omega(\s)}
= \sum_{X \in \xx} c(X,k) \, \phi_{X,k}(\s)
\ee
for some coefficients $c(X,k)$, which depend on $k$.
We rewrite this as
\be
\psi_k(\s) = \Omega(\s) \, \sum_{X \in \xx} c(X,k) \, \phi_{X,k}(\s) .
\ee
For each $X$, $c(X,k)$ has a Fourier series
\be
c(X,k) = \sum_n e^{-ikn} \, e(X,n) .
\ee
(The sum over $n$ is over the sites in $\Lambda$.)
Then
\be
\psi_k(\s) = \Omega(\s) \, \sum_{X \in \xx}
\sum_n e^{-ikn} \, e(X,n) \, \phi_{X,k}(\s) .
\ee
Using \reff{eqnotind} this becomes
\be
\psi_k(\s) = \Omega(\s) \, \sum_{X \in \xx}
\sum_n e(X,n) \, \phi_{X+n,k}(\s)
= \Omega(\s) \, \sum_X e(X) \, \phi_{X,k}(\s) ,
\ee
where the coefficients $e(X)$ are given by
\be
e(X)= \sum_{Y \in \xx,n: Y+n=X} e(Y,n) .
\ee
We now have
\be
\psi_k(\s)= \Omega(\s) \, \sum_l e^{ikl} \, \sum_X e(X) \, \s(X+l) .
\label{eqwave}
\ee
If we define
\be
\phi(\s) = \sum_X e(X) \s(X)
\ee
then
\be
\phi(\s-l) = \sum_X e(X) \s(X+l)
\ee
and so
\be
\psi_k(\s) = \Omega(\s) \sum_l e^{ikl} \, \phi(\s-l) .
\label{ansatz}
\ee
Note that the only momentum dependence in the above is in the
explicit factor of $e^{ikl}$.
When $\epsilon=0$, the lowest excited states with momentum $k$
of the Hamiltonian $H$ defined in \reff{eqb} are
\be
\psi_k(\s) = \sum_l e^{ikl} \, \sigma_l .
\ee
In the above formulation this corresponds to
$c(\{0\},k)=1$ for all $k$ and
$c(X,k)=0$ for all $k$ and all $X \in \xx$ other than $\{0\}$.
This is equivalent to $e(\{0\})=1$ and $e(X)=0$ for
all $X$ other than $\{0\}$.
For nonzero $\epsilon$ we can normalize the eigenstates so that
$c(\{0\},k)=1$ for all $k$. This is equivalent to taking
$e(\{0\})=1$ and $e(\{l\})=0$ for $l \ne 0$.
We will show that the other $e(X)$ are small and localized
near $0$ when $\epsilon$ is small.
As we showed in the last section, the ground state wave function
$\Omega(\s)$ can be written as
$$\Omega(\s) = \exp[-\half \sum_X g(X) \s(X)]$$
The ground state is translationally invariant, so $g(X+l)=g(X)$.
The action of the Hamiltonian on the wavefunction \reff{eqwave} is
\begin{eqnarray*}
(H \psi_k)(\s) &&= \Omega(\s) \sum_l e^{ikl} \\
&& \biggl[ - \sum_j \exp(\sum_{Y \ni j} g(Y) \s(Y))
\Bigl[ \sum_{X \not\ni j-l} e(X) \s(X+l)
- \sum_{X \ni j-l} e(X) \s(X+l) \Bigr] \\
&& + \epsilon \sum_{} \, \s_i \s_j
\sum_X e(X) \s(X+l) \biggr] .\\
\end{eqnarray*}
The above needs to equal
$$E_1(k) \psi_k(\s) = E_1(k) \Omega(\s) \sum_l e^{ikl}
\sum_X e(X) \s(X+l) .
$$
Cancelling the common factor of $\Omega(\s)$, the Schrodinger
equation for the excited state becomes
\bea
\sum_l e^{ikl}
&& \biggl[ - \sum_j \exp(\sum_{Y \ni j} g(Y) \s(Y))
[ \sum_{X \not\ni j-l} e(X) \s(X+l)
- \sum_{X \ni j-l} e(X) \s(X+l)] \nonumber \\
&& + \epsilon \sum_{} \, \s_i \s_j
\sum_X e(X) \s(X+l)
- E_1(k) \sum_X e(X) \s(X+l) \biggr] =0 .
\label{excitedse}
\eea
Recall that the ground state wave function satisfies
\be
- \sum_j \exp(\sum_{Y \ni j} g(Y) \s(Y))
+ \epsilon \sum_{} \, \s_i \s_j = E_0 .
\ee
Multiplying this equation by $\sum_l e^{ikl} \sum_X e(X) \s(X+l)$ and
subtracting the result from eq.
\reff{excitedse}, we have
\bea
&& 2 \sum_l e^{ikl} \sum_j \exp(\sum_{Y \ni j} g(Y) \s(Y))
\sum_{X \ni j-l} e(X) \s(X+l) \nonumber \\
= && \sum_l e^{ikl} [E_1(k)-E_0] \sum_X e(X) \s(X+l) .
\label{exse}
\eea
Define $h(Y)$ by
\be
\exp(\sum_{Y \ni 0} g(Y) \s(Y)) = 1+ \sum_Y h(Y) \s(Y) .
\ee
Explicitly,
\be
h(Y) = \sum_{n=1}^\infty {1 \over n !}
\sum_{Y_1,\cdots,Y_n: 0 \in Y_j,
\atop{Y_1 \d \cdots \d Y_n=Y} }
g(Y_1) \cdots g(Y_n) .
\label{hdef}
\ee
Note that $h(Y)$ is localized near the origin.
Using the translation invariance of the $g(Y)$, we have
\bea
&& \exp(\sum_{Y \ni j} g(Y) \s(Y)) =
\exp(\sum_{Y \ni 0} g(Y+j) \s(Y+j)) \nonumber \\
&& = \exp(\sum_{Y \ni 0} g(Y) \s(Y+j)) =
1+ \sum_Y h(Y) \s(Y+j) .
\eea
Inserting this in \reff{exse}
\bea
&& 2 \sum_l e^{ikl}
\sum_j [1+ \sum_Y h(Y) \s(Y+j)]
\sum_{X \ni j-l} e(X) \s(X+l) \nonumber \\
= && \sum_l e^{ikl} [E_1(k)-E_0] \sum_X e(X) \s(X+l).
\label{exseb}
\eea
The change of variables $j \rightarrow j+l$ changes this to
\bea
&& 2 \sum_l e^{ikl} \sum_j
[1+ \sum_Y h(Y) \s(Y+j+l)]
\sum_{X \ni j} e(X) \s(X+l) \nonumber \\
= && \sum_l e^{ikl} [E_1(k)-E_0] \sum_X e(X) \s(X+l).
\eea
Using
\be
\sum_j \sum_{X \ni j} e(X) \s(X+l)
= \sum_X |X| e(X) \s(X+l)
\ee
this becomes
\bea
&&
2 \sum_l e^{ikl} \sum_X |X| e(X) \s(X+l)
+ 2 \sum_l e^{ikl} \sum_Y \sum_X \sum_{j \in X}
h(Y) e(X) \s([(Y+j) \d X]+l)
\nonumber \\
= && \sum_l e^{ikl} [E_1(k)-E_0] \sum_X e(X) \s(X+l).
\label{exsec}
\eea
We can write $E_1(k)-E_0$ as a Fourier series in $k$. Since this
quantity equals $2$ when $\epsilon=0$, it is natural to write it in
the form
\be
E_1(k)-E_0 = 2 + \sum_m e^{-ikm} e_m .
\label{eq_fourier}
\ee
Thus the right side of \reff{exsec} becomes
\bea
&& \sum_l e^{ikl} [E_1(k)-E_0] \sum_X e(X) \s(X+l)
= \sum_l e^{ikl}
[2 + \sum_m e^{-ikm} e_m]
\sum_X e(X) \s(X+l)
\nonumber \\
&& = 2 \sum_l e^{ikl} \sum_X e(X) \s(X+l)
+ \sum_{l,m} e^{ik(l-m)} e_m \sum_X e(X) \s(X+l)
\nonumber \\
&& = 2 \sum_l e^{ikl} \sum_X e(X) \s(X+l)
+ \sum_{l,m} e^{ikl} e_m \sum_X e(X) \s(X+l+m)
\nonumber \\
&& = 2 \sum_l e^{ikl} \sum_X e(X) \s(X+l)
+ \sum_{l,m} e^{ikl} e_m \sum_X e(X-m) \s(X+l)
\eea
where the last two equalities made use of the change of variables
$l \ra l+m$ and then $X \ra X-m$.
Thus eq. \reff{exsec} becomes
\bea
&&
2 \sum_l e^{ikl} \sum_X |X| e(X) \s(X+l)
+ 2 \sum_l e^{ikl} \sum_Y \sum_X \sum_{j \in X}
h(Y) e(X) \s([(Y+j) \d X]+l)
\nonumber \\
&& = 2 \sum_l e^{ikl} \sum_X e(X) \s(X+l)
+ \sum_{l,m} e^{ikl} e_m \sum_X e(X-m) \s(X+l) .
\label{exsed}
\eea
This equation is of the form
\be
\sum_l e^{ikl} \sum_X f(X) \s(X+l) =0
\label{eqgen}
\ee
where
\be
f(X) = 2 |X| e(X)
+ 2 \sum_{Y,Z,j \in Z:(Y+j) \d Z=X} h(Y) e(Z)
- 2 e(X) - \sum_m e_m e(X-m) .
\label{eqallk}
\ee
Obviously, if $f(X)=0$ for all $X$, then eq. \reff{eqgen} holds.
We will argue that the converse is true.
We rewrite \reff{eqgen} as
\be
\sum_l e^{ikl} \sum_X f(X-l) \s(X) =0 .
\label{eqgenb}
\ee
This must hold for all $\s$, so for all $X$ and $k$
we must have
\be
\sum_l e^{ikl} f(X-l) =0 .
\ee
Fix an $X$ and regard $f(X-l)$ as a function of $l$.
Since the above equation must hold for all $k$,
it says that the Fourier transform of this function is zero. So the
function is zero, i.e., $f(X-l)=0$ for all $l$.
This proves the converse, and so
eq. \reff{exsed} is equivalent to $f(X)=0$ for all $X$.
We now assume that $e(\{0\})=1$ and $e(\{l\})=0$ for all $l \ne 0$.
A priori there is no reason that a solution with these properties must
exist, but we will show that it does.
As we saw before this is essentially a normalization condition on
the eigenstates.
If $X=\{s\}$, then the $2 |X| e(X)$ and $2 e(X)$ terms in $f(X)$
cancel. For such an $X$, $e(X-m)$ is
zero except when $m=s$, in which case it
is $1$. So for such $X$, $f(X)=0$ becomes
\be
e_s = 2 \sum_{Y,Z,j \in Z: (Y+j) \d Z = \{s\}} h(Y) e(Z) .
\label{es}
\ee
For $X$ with $|X| \ne 1$ (including the case of $X= \emptyset$),
the equation $f(X)=0$ can be written as
\be
e(X) = { 1 \over 2(|X|-1)}
\left[ - 2 \sum_{Y,Z,j \in Z: (Y+j) \d Z = X} h(Y) e(Z)
+ \sum_m e_m e(X-m) \right] .
\label{ex}
\ee
We will show these equations have a solution by writing them as a
fixed point equation.
Consider the set of variables
\be e := \{e(X): |X| \ne 1\} \cup \{e_s: s \in \L \} .
\ee
Equations \reff{es} and \reff{ex} form a fixed point equation for $e$
\be
\fp(e)=e .
\label{map}
\ee
Let us introduce the norm
\be
||e|| := \sum_s |e_s|\,e^{\mu |s|} + 2 \sum_{X: |X| \ne 1} |e(X)| ||X|-1|
e^{\mu w(X)} .
\label{norm}
\ee
The factor of 2 is included merely for convenience.
Here $w(X)$ denotes the number of bonds in the smallest
connected set of bonds containing $X$ and the origin, and
$|s|$ is the $l^1$ norm of the site $s$. Note that $|s|=w(\{s\})$.
If $X = X_1 \d \cdots \d X_n$, then
\be
w(X) \le w(X_1) + \cdots + w(X_n) .
\label{wx}
\ee
We prove that the fixed point equation for $e$ has a solution
by using the contraction mapping theorem as we did in the previous
section. We must show that there is a $\delta>0$ such that
\be
||\fp(e)-\fp(e')|| \le \frac{1}{2} ||e - e'|| \quad {\hbox{for}} \quad
||e||, ||e'|| \le \delta ,
\label{exboundone}
\ee
\be
||\fp(e)|| \le \delta \quad {\hbox{for}} \quad
||e|| \le \delta .
\label{exboundtwo}
\ee
To verify \reff{exboundone}, we use
\reff{es} and \reff{ex} to see that
\bea
&& ||\fp(e) - \fp(e')||
\nonumber \\
&& \le \sum_{X:|X| \ne 1} e^{\mu w(X)}
\left[ 2 \sum_{Y,Z,j \in Z: (Y+j) \d Z = X} | h(Y) [e(Z)-e^\prime(Z)] |
+ \sum_s | e_s e(X-s) - e^\prime_s e^\prime(X-s)| \right]
\nonumber \\
&& + 2 \sum_l e^{\mu |l|} \sum_{Y,Z,j \in Z: (Y+j) \d Z = \{l\}}
| h(Y) [e(Z)-e^\prime(Z)] | .
\eea
For a site $l$, we have $w(\{l\})=|l|$, so the above
\bea
&& = 2 \sum_{Y,Z,j \in Z} e^{\mu w((Y+j) \d Z)}
| h(Y) [e(Z)-e^\prime(Z)]|
\nonumber \\
+ && \sum_s \sum_{X:|X| \ne 1} e^{\mu w(X)}
| e_s e(X-s) - e^\prime_s e^\prime(X-s)| .
\label{exboundonea}
\eea
We claim that for $j \in Z$ we have $w((Y+j) \d Z) \le w(Y) + w(Z)$.
To see this, let $A$ and $B$ be connected sets of bonds containing the
origin such that $Y \subset A$, $Z \subset B$
and $|A|=w(Y)$ and $|B|=w(Z)$.
Since the origin is in $A$, the site $j$ is in $A+j$.
The site $j$ is also in $Z$ and so is in $B$.
Since each of $A+j$ and $B$ is connected and they both contain $j$,
their union is connected.
So if we define $C=(A+j) \cup B$,
then $C$ is a connected set of bonds which contains the origin and
$(Y+j) \d Z$. So
\be
w((Y+j) \d Z) \le |C| \le |A+j| + |B| = |A| + |B| = w(Y) + w(Z) .
\ee
We also have $w(X) \le w(X-s) +|s|$. The sum over $j$ in \reff{exboundonea}
then produces a factor of $|Z|$. So \reff{exboundonea} is
\bea
&& \le 2 \sum_{Y,Z} e^{\mu(w(Y)+w(Z))} |Z| | h(Y) [e(Z) - e^\prime(Z)] |
\nonumber \\
&& + \sum_s \sum_{X:|X| \ne 1} e^{\mu(w(X-s)+|s|)}
| e_s e(X-s) - e^\prime_s e^\prime(X-s)| .
\label{eqcbounda}
\eea
Since $e(Z)=e^\prime(Z)$ if $|Z|=1$, and $|Z| \le 2 ||Z|-1|$
if $|Z| \ne 1$, we have
\be
\sum_Z e^{\mu w(Z)} |Z| | e(Z) - e^\prime(Z)|
\le \sum_{Z:|Z| \ne 1} e^{\mu w(Z)} 2 ||Z|-1| | e(Z) - e^\prime(Z)|
\le ||e-e^\prime|| .
\ee
So the first of the two terms in \reff{eqcbounda} is bounded by
\be
2 \sum_Y e^{\mu w(Y)} \, |h(Y)| \, ||e-e^\prime|| . \label{eqboundz}
\ee
For the second term in \reff{eqcbounda} we do a change of variables
and then use the triangle inequality
\bea
&& \sum_s \sum_{X:|X| \ne 1} e^{\mu(w(X-s)+|s|)}
| e_s e(X-s) - e^\prime_s e^\prime(X-s)|
\nonumber \\
&& = \sum_s \sum_{X:|X| \ne 1} e^{\mu(w(X)+|s|)}
| e_s e(X) - e^\prime_s e^\prime(X)|
\nonumber \\
&& \le
\sum_s \sum_{X:|X| \ne 1} e^{\mu(w(X)+|s|)}
[ | e_s e(X) - e_s e^\prime(X)|
+ | e_s e^\prime(X) - e^\prime_s e^\prime(X)| ]
\nonumber \\
&& \le
\sum_s e^{\mu |s|} |e_s| \sum_{X:|X| \ne 1}
e^{\mu w(X)} | e(X) - e^\prime(X)|
+ \sum_s e^{\mu |s|} | e_s - e^\prime_s|
\sum_{X:|X| \ne 1} e^{\mu w(X)} |e^\prime(X)|
\nonumber \\
\nonumber \\
&& \le
{ 1 \over 2} ||e|| \, ||e-e^\prime||
+ { 1 \over 2} ||e-e^\prime|| \, ||e^\prime||
\le \delta \, ||e-e^\prime|| ,
\label{eqcboundb}
\eea
since $||e||$ and $||e'||$ are no greater than $\delta$.
Combining \reff{eqboundz} and \reff{eqcboundb}, we have
\be
||\fp(e) - \fp(e')||\le K \, ||e - e'|| ,
\label{f1e}
\ee
where
\be
K = 2 \sum_Y e^{\mu w(Y)} |h(Y)| + \delta .
\label{k1}
\ee
{From} \reff{hdef} and \reff{wx}
\be
\sum_Y e^{\mu w(Y)} |h(Y)| \le \sum_{n=1}^\infty {1 \over n !}
\sum_{Y_1 \ni 0,\cdots,Y_n \ni 0}
e^{\mu (w(Y_1) + \cdots + w(Y_n))}
|g(Y_1) \cdots g(Y_n)|
= e^{||g||}-1 .
\ee
So
\be
K \le 2(e^{||g||} -1) + \delta .
\ee
If $\delta$ and $\epsilon$ are small enough, then $K \le 1/2$.
To prove \reff{exboundtwo}, we use \reff{es} and \reff{ex} to compute
$\fp(0)$. Note that $e=0$ means that $e(X)=0$ for all $X$ except
$X=\{0\}$. We always have $e(\{0\})=1$.
Letting $e^\prime$ denote $F(0)$, we have
\be
e^\prime_l = 2 \sum_{Y: Y \d \{0\} = \{l\}} h(Y)
\ee
and for $X$ with $|X| \ne 1$ (including the case of $X= \emptyset$),
\be
e^\prime(X) = - { 1 \over (|X|-1)}
\sum_{Y: Y \d \{0\} = X} h(Y) .
\ee
Thus
\be
||\fp(0)|| \le 2 \sum_Y e^{\mu w(Y)} |h(Y)| \le 2(e^{||g||}-1) .
\ee
Hence, for $||e|| \le \delta$
and $\epsilon$ small enough that
$ 2(e^{||g||}-1) < \delta/2 $,
\be
||\fp(e)|| \le ||\fp(e)-\fp(0)|| + ||\fp(0)||
\le {1 \over 2} ||e|| + ||\fp(0)|| \le \delta .
\ee
This completes the proof that the fixed point equation has a solution.
% To show that the solution satisfies $e_s=e_{-s}$, we argue as
% we did in the ground state case.
% Recall that $R$ is reflection about the origin, and we
% showed $g(X)=g(RX)$.
% Define $e'$ by $e'(X)=e(RX)$ and $e'_s=e_{-s}$.
% Then it is easy to see that $\fp(e)=e$
% implies $\fp(e')=e'$. The uniqueness of the solution of the fixed
% point equation in the small ball about the origin then implies
% $e=e'$. In particular, $e_s=e_{-s}$.
We still must prove
that for each $k$, the eigenstate we have constructed has the
lowest eigenvalue in the subspace of momentum $k$ if $k \ne 0$ and
has the next to lowest eigenvalue if $k = 0$. When $\epsilon=0$ we
know that this is true. So if we can prove that within the subspace
of momentum $k$, our eigenstate is never degenerate for
$|\eps| \le \eps_0$, then we are done. We will use the local
uniqueness of solutions to the fixed point equation to prove this.
Let $\psi_k(\s)$ continue to denote
the eigenfunctions constructed above with eigenvalues $E_1(k)$.
Suppose that for some $k_0$ there is another eigenfunction
$\psi^\prime(\s)$ with momentum $k_0$ and eigenvalue $E_1(k_0)$.
We can write it in the form
\be
\psi^\prime(\s) = \Omega(\s) \, \sum_{X \in \xx} c^\prime(X) \,
\phi_{X,k_0}(\s) .
\ee
For small $\alpha$ we define
\be
\psi^\alpha_k(\s) = \psi_k(\s)
\ee
for $k \ne k_0$ and
\be
\psi^\alpha_{k_0}(\s) = {\psi_{k_0}(\s) + \alpha \psi^\prime(\s)
\over 1 + \alpha c^\prime(\{0\})} .
\ee
(We need only consider small $\alpha$, so we are not dividing by zero in
the above.)
Then for all small $\alpha$, $\psi^\alpha_k(\s)$ is an eigenstate of $H$
with eigenvalue $E_1(k)$ and momentum $k$ for all $k$.
These states can be written in the form
\be
\psi_k^\alpha(\s) = \Omega(\s) \, \sum_{X \in \xx} c_\alpha(X,k)
\, \phi_{X,k}(\s) ,
\ee
where $c_\alpha(X,k)$ is just $c(X,k)$ for $k \ne k_0$ and
\be
c_\alpha(X,k_0) = {c(X,k_0) + \alpha c^\prime(X,k_0)
\over 1 + \alpha c^\prime(\{0\})} .
\ee
As before we can write this in the form
\be
\psi_k^\alpha(\s) = \Omega(\s) \, \sum_X e_\alpha(X) \, \phi_{X,k}(\s) ,
\ee
where
\be
c_\alpha(X,k) = \sum_n e^{-ikn} \, e_\alpha(X,n)
\ee
and
\be
e_\alpha(X)= \sum_{Y \in \xx,n: Y+n=X} e_\alpha(Y,n) .
\ee
We have normalized things so that $e_\alpha(\{0\})=1$
and $e_\alpha(\{l\})=0$ for $l \ne 0$.
It follows from what we have done before that for small $\alpha$,
$e_\alpha(X)$ is a
solution of the fixed point equation \reff{eqallk}. As $\alpha \ra 0$,
$e_\alpha(X) \ra e(X)$. However, our solution of the fixed point equation
is locally unique by the contraction mapping theorem.
Thus $e_\alpha(X)=e(X)$ for small $\alpha$. This implies that
the second eigenfunction $\psi^\prime$ is in fact just a multiple
of $\psi_{k_0}$. This completes the proof that the eigenfunctions
we have constructed are indeed the lowest excited states.
The existence of the infinite volume limit and the convergence of
$e^\Lambda_s$ to $e_s$ follow from the estimates of this section and
the arguments of the previous section.
In the fixed point equation, \reff{es} and \reff{ex}, $h(Y)$ depends on the
volume. However, the difference between it and its infinite volume
limit is exponentially small in $L$. The extension of $e(X)$ to a
function $e^\prime(X)$ that is defined for all $X$ is simpler
than the definition of $g^\prime(X)$ in the previous section.
We can simply define $e^\prime(X)$ to be $e(X)$ if $X$ is in $\Lambda$
and $0$ otherwise.
The bounds on the $e_s$ follow from
our choice of norm \reff{norm} which implies that
\be
|e_s| \le \exp(-\mu |s|) ||e||
\ee
Our arguments only require that $e^\mu |\epsilon|$ be small, so we can
choose $\mu$ by setting $e^{-\mu}$ equal to a large constant times
$|\epsilon|$.
Thus the above bound implies $e_s$ is of order $\epsilon^{|s|}$.
\bigskip
\bigskip
\bigskip
\bigskip
\medskip
\noindent {\bf Acknowledgements:}
ND thanks C. Gruber, N. Macris and M. Salmhofer for
interesting discussions. TK thanks the D\'epartment de Physique
and the Institut de Physique Th\'eorique of the Ecole Polytechnique
F\'ed\'erale de Lausanne for their hospitality during a visit when
much of this work was done. He also acknowledges the support of the
National Science Foundation (DMS-9970608).
\bigskip
\bigskip
\bigskip
\bigskip
\medskip
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