\documentstyle[12pt]{article}
\begin{document}
\author{{\bf O. Bolina}\thanks{%
Supported by FAPESP.} \\
%EndAName
Instituto de F\'{\i}sica\\
Universidade de S\~ao Paulo\\
P.O. Box 66318\\
05389-970 S\~ao Paulo, Brazil\\
{\bf e-mail:} {\tt bolina@if.usp.br} \and {\bf J. Rodrigo Parreira}\thanks{%
Supported by CNPq} \\
%EndAName
Department of Physics\\
Princeton University\\
P.O. Box 708\\
08544-0708 Princeton, USA\\
{\bf e-mail: }{\tt parreira@math.princeton.edu}}
\title{{\bf The XY Model in a Transverse Magnetic Field}}
\maketitle
\newpage
\begin{abstract}
We show that the ground state of the XY model (ferromagnetic or
antiferromagnetic) in a transverse magnetic field $h$ (transverse meaning
here perpendicular to the plane where the interactions occur)--- for any
spin value,
in any dimension --- is the state with all spins aligned parallel to the
field, when $h$ is bigger in magnitude than some critical value $h_c$. We
also compute the spatial and time decay of correlation functions
for the spin-$1/2$ linear chain and study the behavior of these
quantities as a function of the external magnetic field.
\end{abstract}
\section{Introduction}
The Hamiltonian for the quantum XY model in the presence of a constant
transverse field is given by
\beq
{\cal H}_\Lambda =\frac J2\sum_{\left\langle x,y\right\rangle \in \Lambda
}[S_1(x)S_1(y)+S_2(x)S_2(y)]+h\sum_{x\in \Lambda }S_3(x). \label{Hsp}
\eeq
As usual, $\Lambda $ is a finite domain in the hypercubic lattice ${\bf Z}^d$
and by $\left\langle x,y\right\rangle $ we mean nearest neighbor sites on
the lattice. At each site $x$ we have a spin angular momentum ${\bf S}(x)$,
such that ${\bf S}^2=S\left( S+1\right) $ is the same for all spins, the
system being conceived with open boundary conditions. The sign of $J$
defines the ferromagnetic ($J<0$) or antiferromagnetic ($J>0$) nature of the
system. Actually, it is a known fact that the XY model can always be
equivalently described as ferromagnetic or antiferromagnetic, since there
exists an unitary transformation (a rotation of spins in a sublattice around
the $z$-axis through an angle of $\pi $) that maps $J\rightarrow -J$. This
property leads to the conclusion that the sign of $J$ is not relevant in
our analysis.
In 1961, Lieb, Schultz and Mattis \cite{Lieb1} solved the one-dimensional
antiferromagnetic model by means of the
Jordan-Wigner \cite{JW}
transformation which maps the system into a free Fermi gas in the presence
of an external potential, thereby giving the complete set of states and
excitation energies through the diagonalization of the Hamiltonian in terms of
fermion operators. Since this transformation is not available in more than
one dimension, in this paper we use a variational principle to determine the
ground state of the model in $d$ dimensions and show that, when $%
|h|>2Sd\left| J\right| $, all spins align antiparallel to the field
(section 2 and the Appendix). For the
antiferromagnetic system, our result is a proof of the so-called ''Jacobsohn
theorem'' \cite{Niem} in a special case and is a simple application of
classical domination to quantum spin chains.
Next we consider the spin-$1/2$ one-dimensional model with open boundary
conditions (section 3). The problem of
determining the eigenstates and eigenvalues for the system with periodic
boundary conditions was solved by Niemeiyer \cite{Th}. In section 4, we
investigate the asymptotic behavior of spatial correlations for large chains
($\Lambda \rightarrow \infty$)
as functions of the external field. We show that the polynomial decay
obtained in \cite{Lieb1} for zero field survives up to the critical value
of the field $h=|J|$. For $h > |J|$ the spins become independent
variables, so that the correlation degenerates into delta functions of
{\it x} and {\it y}. This result should be compared to the exponential
decay obtained by Klein e Perez \cite{Klein} for random fields.
\newline
Finally, in section 5 we
study
the time decay of correlation functions and find out that their behavior
is also controled by the external magnetic field in the following
way
\[ \left \{
\begin{array}{lll}
\delta_{xy}e^{-t(|h|-J)} & {\rm if\ } h>|J| \\
\delta_{xy}\frac{1}{\sqrt{t}} & {\rm if\ } h=|J| \\
\frac{1}{t} & {\rm if\ } h<|J|
\end{array}
\right.
\]
An important remark is that, in this kind of system, the time decay of
correlation functions is
a measure of the energy gap of the excited states against the ground state
as a function of the external field.
\section{The Ground State of the $\bf{XY}$ Model
in a Strong Transverse Magnetic Field}
We shall make use of the raising and lowering operators
\begin{eqnarray}
S_{\pm }(x)=S_1(x)\pm iS_2(x), \label{RC}
\end{eqnarray}
which obey the commutation relations
\beq
\left.
\begin{array}{llll}
\left[ S_{+}\left( x\right) ,S_{-}\left( y\right) \right]
=2S_3\left( x\right) \delta _{x,y}, \\
\\
\\
\left[ S_3\left( x\right) ,S_{+}\left( y\right) \right]
=S_{+}\left( x\right) \delta _{x,y}. \label{Komut}
\end{array}
\right.
\eeq
In terms of these operators the Hamiltonian is written as:
\begin{equation}
\label{H}{\cal H}_\Lambda =\frac J2\sum_{x\in \Lambda
}\sum_{k=1}^d[S_{+}(x)S_{-}(x+{\bf e}_k)+S_{+}(x)S_{-}(x+{\bf e}_k)]+h\sum_{x\in \Lambda
}S_3(x)
\end{equation}
with ${\bf e}_k$ being an unit vector in ${\bf Z}^d$. In our perturbative method,
we break up the Hamiltonian operator into two parts:
$$
{\cal H}_\Lambda ={\cal H}_\Lambda ^z+{\cal H}_\Lambda ^{xy},
$$
where
\begin{equation}
\label{Hz}{\cal H}_\Lambda ^z=h\sum_{x\in \Lambda }S_3(x),
\end{equation}
and
\begin{equation}
\label{Hxy}{\cal H}_\Lambda ^{xy}=\frac J2\sum_{x\in \Lambda
}\sum_{k=1}^d[S_{+}(x)S_{-}(x+{\bf e}_k)+S_{+}(x)S_{-}(x+{\bf e}_k)].
\end{equation}
Of these, the external potential ${\cal H}_\Lambda ^z$ is considered the
dominant term, which is perturbed by the $x-y$ planar interaction ${\cal H}%
_\Lambda ^{xy}$.
The partition function of the
unperturbed system, defined by ${\cal H}_\Lambda^z$, factorizes
completely, leading to%
$$
{\cal Z}_\Lambda ^{\left( z\right) }=\left[ 2\sum_{j=1}^S\cosh \left( j\beta
h\right) \right] ^{\left| \Lambda \right| }.
$$
The expectation values of the energy and the magnetization, in the limit $%
\beta \rightarrow \infty $, correspond to the state where all the spins lie
antiparallel to the external field, even when $\left| \Lambda \right|
\rightarrow \infty $, so that the thermodynamical limit is well defined.
We can solve immediately the unperturbed eigenvalue problem by setting up
the
basis that diagonalizes ${\cal H}_\Lambda ^z$, namely
$$
{\otimes }_{\Lambda}|\sigma (x)\rangle
$$
where\
$$
S_3(x)\left| \sigma (x)\right\rangle =\sigma (x)\left| \sigma
(x)\right\rangle , \; {\rm with} \; -S\leq \sigma (x)\leq S, \; \forall
x\in \Lambda ,
$$
As a consequence of the above discussion, ${\cal H}_\Lambda ^z$ has a unique
ground state $|\Omega _\Lambda \rangle $, which is also an eigenstate of the
perturbation ${\cal H}_\Lambda ^{xy}$, with zero eigenvalue. This state is
such that
$$
S_3\left( x\right) |\Omega _\Lambda \rangle =-S|\Omega _\Lambda \rangle ,
$$
for any $x\in \Lambda .$ Moreover, since ${\cal H}_\Lambda ^{xy}$ commute
with $\sum_xS_3(x)$, the total number of particles is conserved. Thus, the
only effect of the perturbation ${\cal H}_\Lambda ^{xy}$ is the mixing of
the degenerate energy levels of ${\cal H}_\Lambda ^z$.
The general $n-$angular momentum state, defined by
$$
\left| \Psi_n\right\rangle =\sum_{x_1,\ldots,x_n}\Phi
(x_1,\ldots,x_n)S_{+}(x_1)\ldots S_{+}(x_n)|\Omega_\Lambda \rangle
$$
satisfies
$$
{\cal H}_\Lambda ^z|\Psi_{n}\rangle =-h(S|\Lambda |-n)|\Psi_{n}\rangle
$$
where $n=0,1,2,...,2S|\Lambda |$ is the change in the value of the system
total angular momentum. Therefore, the spectrum of ${\cal H}_\Lambda ^z$ is
the discrete set of values from $-|\Lambda |Sh$ to $+|\Lambda |Sh$ with
width $h$ between adjacent levels. To verify that the ground state of of the
complete Hamiltonian is also $|\Omega _\Lambda \rangle $, it is sufficient
to show that the expectation value of the perturbation in any unperturbed
eingenstate is never greater than the difference between the energies of the
considered level and of the fundamental one. This situation will be
achieved, as we shall see, with a sufficiently strong magnetic field.
The difference from the case treated in \cite{Oscar} is that, now, the $%
\left| \Psi _n\right\rangle $ state may be generated by the application of $%
n $ raising operators ($n\leq 2S$) on the same point ($x_1=x_2=...=x_n=x$).
The action of the ${\cal H}_\Lambda ^{xy}$ operator on the vector $\left|
\Psi _n\right\rangle $ produces%
$$
{\cal H}_\Lambda ^{xy}\left| \Psi _n\right\rangle =\left| \Psi _n^{\left(
1\right) }\right\rangle +\left| \Psi _n^{\left( 2\right) }\right\rangle
+\left| \Psi _n^{\left( 3\right) }\right\rangle ,
$$
where%
\begin{eqnarray*}
\lefteqn{\left| \Psi _n^{\left( 1\right) }\right\rangle =J\sum_{x\in \Lambda
}\sum_{k=1}^d\sum_{x_1...x_n} \Phi
(x_1,...,x_n) \times} \\ & & \times S_{+}(x)S_{+}(x_1)...S_{+}(x_n)S_{-}(x+{\bf e}_k)|\Omega _\Lambda
\rangle =0, \\
\lefteqn{\left| \Psi _n^{\left( 2\right) }\right\rangle =4S\frac
J2\sum_{k=1}^d\sum_{i=1}^n\sum_{x_1...x_n}\Phi
(x_1,...,x_n) \times} \\ & & \times S_{+}(x_i+{\bf e}_k)
S_{+}(x_1)...S_{+}(x_{i-1})S_{+}(x_{i+1})...S_{+}(x_n)|\Omega
_\Lambda \rangle \\
\lefteqn{\left| \Psi _n^{\left( 3\right) }\right\rangle =-4\frac
J2\sum_{k=1}^d\sum_{i\neq j}\sum_{x_1...,x_i=x_j,...,x_n}\Phi
(x_1,...,x_n)\times} \\ & & \times S_{+}\left( x_i+{\bf e}_k\right)
S_{+}\left( x_1\right) ...S_{+}(x_{i-1})S_{+}(x_{i+1})...S_{+}\left(
x_{j-1}\right)\times \\ & & \hspace{.25in} \times S_{+}\left( x_i\right) S_{+}\left( x_{j+1}\right)
...S_{+}(x_n)|\Omega _\Lambda \rangle .
\end{eqnarray*}
The fact that the coefficients of $\left| \Psi _n^{\left( 3\right)
}\right\rangle $ are less or equal than zero implies that%
$$
\| \ {\cal H}_\Lambda ^{xy}\left| \Psi _n\right\rangle\| \leq 2nSd|J|\left \|
\ | \Psi
_n\right\rangle\| .
$$
To obtain the above results we have used the commutation relations
(\ref{Komut}) together with the characterization of the ground state
of the
unperturbed model as the state where all the spins have a $S_3$ eigenvalue $%
-S$.
Thus, the expectation value of the energy in any subspace defined by $\left|
\Psi _n\right\rangle $ is
$$
-|\Lambda |Sh+n(h-2Sd\left| J\right| )
$$
which, for $|h|>2d\left| J\right| S$, is greater than the energy of the
vacuum, thus proving that the vacuum $|\Omega _\Lambda \rangle $ is the
ground state of the complete system.
\section{\sloppy{The Spin-$\bf{1/2}$ Linear Chain with Open Boundary
Conditions}}
We now consider a linear chain
of $2L+1$ (L even) spins $1/2$ in an external transverse magnetic field $ h$,
with free boundary conditions. The system, still defined by (\ref{Hsp}), is
restricted to the box $\Lambda={\bf Z} \cap [-L,L]$ and
${\bf S }^{2}=\frac{3}{4}$.
The problem can be exactly solved by first introducing the raising and
lowering operators given by (\ref{RC}) and the fermion creation and annihilation
operators defined by the Jordan-Wigner transformation \cite{JW}. For
$-L < x \leq L$, let
\beq
\left.
\begin{array}{llll}
a^{\dag}(x)=S_{+}(x)\exp[i\pi\sum_{-L\leq z < x} S_{+}(z)S_{-}(z)]
\\
\\
\\
a(x)=\exp[-i\pi\sum_{-L\leq z < x} S_{+}(z)S_{-}(z)]
S_{-}(x) \label{JWtrans}
\end{array}
\right.
\eeq
The Hamiltonian (\ref{Hsp}) is then written as a quadratic form
\[
{\cal H}_L =-(J/2)\sum_{-L\leq x \leq L-1}[a^{\dag}(x)a(x+1) +
a^{\dag}(x+1)a(x)] +
\]
\[+ h\sum_{-L\leq x \leq L} a^{\dag}(x)a(x).
\]
The canonical transformation defined by
\beq
\left.
\begin{array}{lll}
a^{\dag}_{k} & = \sum_{-L\leq x \leq L}\varphi_{k}(x)a^{\dag}(x) \\
\\
a_{k} & = \sum_{-L\leq x \leq L}\varphi_{k}(x)a(x) \label{kanon}
\end{array} \label{canon}
\right.
\eeq
sends ${\cal H}_{L}$ into the form
\[
{\cal H}_L =\sum_{k}\Gamma_{k}a^{\dag}_{k}a_{k}.
\]
In (\ref{canon}), the coefficients $\varphi_{k}(x)$ ($x=-L,...,L$) are the
$2L+1$ components of the vector $\Phi_{ k}$ in the matrix equation
\begin{equation}
A\Phi_{k} = \Gamma^{2}_{k}\Phi_{k} \label{autov}
\end{equation}
where
\[
A= \frac{1}{2}\left [
\begin{array}{cccccc}
2h & -J & 0 & ... & 0 & 0 \\
-J & 2h & -J & ... & 0 & 0 \\
0 & -J & 2h & ... & 0 & 0 \\
. & . & . & . & . & . \\
0 & 0 & . & -J & 2h & -J \\
0 & 0 & ...& 0 & -J & 2h \\
\end{array}
\right ]
\]
The set of solutions of (\ref{autov}) is then
\beq
\varphi_{k}(x)= \left \{
\begin{array}{lll}
A_{+}\cos{kx},\:\:\:\: k \in I_{1}=\{ \frac{(2l+1)\pi}{2L+1},\:\:\:\:
\: \: l=0,1,\,...\, ,L \} \\
\\
A_{-}\sin{kx}, \:\:\:\: k \in I_{2}= \{ \frac{l\pi}{L+1}, \:\:\:\:\:\:
l=1,2 \, ... \, ,L \}, \label{solut}
\end{array}
\right.
\eeq
which belong to the set of eigenvalues $\Gamma_{k}=h - J\cos{k}$. It is also
usefull to define $\Lambda^{*}=I_{1} \cup I_{2}$.
The normalization constants are given by
\beq
A_{\pm}=\frac {1}{\sqrt{\frac {2L+1}{2} \pm
\frac{\sin{k(2L+1)}}{2\sin{k}}}} \label{norm}
\eeq
The eigenvectors of the Hamiltonian operator, when indexed by the set of
energy levels, can be written as
\[
|\Psi_{I} \rangle =\prod_{k \in I}a^{\dag}_{k}|\Omega \rangle
\]
with eigenvalues
\[
E_{I}=\sum_{k \in I}\Gamma_{k}, \: \: \: \: \: \: \: \: \:
I\subset \Lambda^{*},
\]
where $|\Omega \rangle$ is the Fock vaccum
\[
a(x)|\Omega \rangle =0, \: \: \: \: \: \forall \: x \in \Lambda,
\:\:\:\:
{\rm or} \:\:\:\: a_{k}|\Omega \rangle=0, \: \: \: \: \: \forall \: k
\in \Lambda^{*}.
\]
The ground state, in particular, is given by
\begin{equation}
|\Psi_{0} \rangle = |\Psi_{I_{0}} \rangle =\prod_{k \in I_{0}}
a^{\dag}_{k}| \Omega \rangle \label{GS}
\end{equation}
where
\[
I_{0}= \{k\subset \Lambda^{*}:\Gamma_{k} < 0 \}.
\]
There are three regions of localization of the ground state
as a function of the external field:
\[
\left \{
\begin{array}{lll}
{\rm if} \:\:\:\: |h|J \:\:\:\: \Rightarrow E_{I}>0 \:\:\:\:
\forall \:\:\:\: I\subset \Lambda^{*} \:\:\:\: {\rm then} \:\:\:\: I_{0}=
\emptyset \\
{\rm if} \:\:\:\: h< -J \:\:\:\: \Rightarrow E_{I}<0 \:\:\:\: \forall
\:\:\:\: I\subset \Lambda^{*} \:\:\:\: {\rm then} \:\:\:\: I_{0}=\Lambda^{*}
\end{array}
\right.
\]
In the first case we will prove that the correlation function between the
first
and last spins of the chain exhibits a polynomial decay. On the other
hand, whether $h>J$ (so that $I_{0}=\emptyset$), or $h<-J$ (so that
$I_{0}=\Lambda^{*}$), the correlation length will be shown to be zero.
\section{Correlation Functions in One Dimension}
We define the correlation function between spins at sites $x$ and
$y$ by
\[
\langle \Psi|[S_{+}(x)S_{-}(y) +
S_{-}(x)S_{+}(y)]|\Psi \rangle
\]
where $ | \Psi \rangle$ is an eigenvector of the total particle number
operator
\[
\sum n(x)|\Psi \rangle = N|\Psi \rangle
\]
with $n(x)=S_{+}(x)S_{-}(x)$, for some integer $N\geq 0$.\\
\noindent
To prove that for $h<|J|$, the spin-spin correlation function between the
first and
last spins decays polynomially we are going, first of all, to write the
correlation function in terms of Fermi operators, by inverting the system
(\ref{JWtrans})
\[
S_{+}(x)S_{-}(y)=a^{\dag}(x)\prod_{x< z < y}e^{i\pi n(z)}a(y)
\]
\[
S_{-}(x)S_{+}(y)=-a(x)\prod_{x< z < y}e^{i\pi n(z)}a^{\dag}(y).
\]
for $x < y $.
\newline
We then obtain (again for $x|J|$, just note that
since $S_{+}(x) | \Psi \rangle $ creates a particle at $x$, and
$S_{+}(y) | \Psi \rangle $ creates, independently of $x$, another particle
at $y$, we have simply
\[
\langle \Psi | S_{+}(y)\sigma_{+}(x) | \Psi \rangle=\delta_{xy}
\]
which proves that the spin variables become independent when the
field is sufficiently strong.
\section{Time Decay of Correlation Functions}
The time decay of correlation functions, for $h< |J|$, is given by
\begin{eqnarray}
\langle \Psi_{I} | a^{\dag}(x)e^{-t({\cal
H}-E_{I})}a(y) | \Psi_{I} \rangle &=& \sum_{l,m}\varphi_{l}
(x)\varphi_{m}(y) \times \nonumber \\ &\times&\langle \Psi_{I}
| a^{\dag}_{l}e^{-t(\sum_{k}\Gamma_{k}a^{\dag}_{k}a_{
k} - E_{I})}a_{m} | \Psi_{I} \rangle \nonumber = \\
&=&\sum_{l,m}\varphi_{l}(x)\varphi_{m}(y)
\langle \Psi_{I} | a^{\dag}_{l}e^{t\Gamma_{m}} a_{m} | \Psi_{I}
\rangle = \nonumber \\
&=&\sum_{l\in I}\varphi_{l}(x)\varphi_{l}(y)e^{t\Gamma_{l}}.\nonumber
\end{eqnarray}
Substituting for $\varphi_{k}$ the expressions given in (8),
we get
\[
\int_{-k_{c}}^{k_{c}}dk\,\cos{k(x-y)}e^{t\Gamma_{k}}.
\]
The major contribution to the integral comes from the behavior of
$\Gamma_{k}$ in the neighborhood of the extremes of integration. So
for an arbitrary $0< \epsilon < k_{c}$ we divide the integral into
three parts:
\[
\int_{-k_{c}}^{-k_{c}+\epsilon} dk \cos{k(x-y)}e^{t\Gamma_{k}}
+\int_{|k| < k_{c}-\epsilon} dk \cos{k(x-y)}e^{t\Gamma_{k}}+
\]
\[
+\int_{k_{c}-\epsilon}^{k_{c}}dk \cos{k(x-y)}e^{t\Gamma_{k}}
\]
For $|k| < k_{c}-\epsilon$ the following estimate holds
\[
\int_{|k_{c}| |J|$, the time decay of the correlation function is given by
\begin{eqnarray}
\langle \Psi_{I} | a(x)e^{-t({\cal
H}-E_{I})}a^{\dag}(y) | \Psi_{I} \rangle &=&\sum_{l,m}\varphi_{l}
(x)\varphi_{m}(y) \times \nonumber \\ &\times& \langle
\Psi_{I} | a_{l}e^{-t(\sum_{k}\Gamma_{k}a^{\dag}_{k}a_{k}-
E_{I})}a^{\dag}_{m} | \Psi_{I} \rangle= \nonumber \\
&=&\sum_{l,m}\varphi_{l}(x)\varphi_{m}(y)\langle
\Psi_{I} | a_{l}e^{-t\Gamma_{k}}
a^{\dag}_{m} | \Psi_{I} \rangle= \nonumber \\
&=&\sum_{l}\varphi_{l}(x)\varphi_{l}(y)e^{-t\Gamma_{l}}.\nonumber
\end{eqnarray}
The behavior of the correlation is determined by the energy gap between
the ground state and the excited ones. This gap is just
$\Gamma_{0}=|h|-J$. Since $\Gamma_{0} \leq \Gamma_{k}$,
$\forall \, k \in \Lambda^{*}$, we have
\begin{eqnarray}
\int_{-\pi}^{\pi}dk\,\varphi_{k}(x)\varphi_{k}(y)e^{-t\Gamma_{k}}& \leq &
\int_{-\pi}^{\pi}dk\,|\varphi_{k}(x)| \, |\varphi_{k}(y)|
e^{-t(|h|-J)}\nonumber \\
&=& C\delta_{xy}e^{-t(|h|-J)}.\nonumber
\end{eqnarray}
For $h=|J|$, we expand $\Gamma_{k}$ around $k=0$,
\begin{eqnarray}
\int_{-\pi}^{\pi}dk\,\varphi_{k}(x)\varphi_{k}(y)e^{-t\Gamma_{k}}& = &
\int_{-\pi}^{\pi}dk\, \varphi_{k}(x) \, \varphi_{k}(y)
e^{-th_{c}(\frac{k^{2}}{2!} - \frac{k^{4}}{4!} + ...)} \nonumber \\
\end{eqnarray}
and make the change of variable $u=k\sqrt{\frac{th_{c}}{2}}$, to obtain
\[
\frac{1}{\sqrt{t}}\int_{-\infty}^{+\infty}du\,\varphi_{u}(x)\varphi_{u}(y)
e^{-u^{2} + o(1/t)}
\]
\section{Appendix}
It is also possible to obtain the result of section 2 by formally
identiffying the effect of the Hamiltonian over the state $\left| \Psi
_n\right\rangle $ with the difference Laplacean in $d$ dimensions,
$-{\Delta }_d$, acting on the function $\Phi(x_1,...,x_n)$, which
we define in the standard way, as
\begin{eqnarray*}
\lefteqn{-\Delta _d\Phi (x_1,...,x_n)=\sum_{i=1}^n\sum_{k=1}^d[2\Phi
(x_1,...,x_i,...,x_n)-} \\
& & -\Phi (x_1,...,x_i+{\bf e}_k,...,x_n)-\Phi (x_1,...,x_i-{\bf e}_k,...,x_n)].
\end{eqnarray*}
The argument of the function $\Phi(x_{1},x_{2},\ldots,x_{n}$) has now the
following characteristics:
\begin{eqnarray*}
x_{1}=x_{2}=\ldots=x_{m_{1}}=y_{1}, \\
x_{m_{1}+1}=x_{m_{1}+2}=\ldots=x_{m_{1}+m_{2}}=y_{2}, \\
\vdots \\
x_{(m_{i-1})+1}=x_{(m_{i-1})+2}=\ldots=x_{(m_{i})}=y_{i},
\end{eqnarray*}
where we have introduced the notation
$x_{(m_{p})}=x_{m_{1}+\ldots+m_{p}}$. We also notice that
\begin{eqnarray*}
\lefteqn{i) \; i \leq n, \; m_{p} \leq 2S \Rightarrow \sum_{p=1}^{i} m_p \leq 2nS} \\
\lefteqn{ii) \; \Phi(x_{1},\ldots,x_{(m_{p})+1}+a,x_{(m_{p}+2},\ldots,x_{(m_{p})+m_{p+1}},\ldots,x_{n})=} \\
& &=\Phi(x_{1},\ldots,x_{(m_{p})+1},x_{(m_{p})+2}+a,x_{(m_{p})+3},\ldots,x_{(m_{p})+m_{p+1}},\ldots,x_{n})= \\
& &\vdots \\
& &=\Phi(x_{1},\ldots,x_{(m_{p})+m_{p+1}-1},x_{(m_{p})+m_{p+1}}+a,\ldots,x_{n}),
\end{eqnarray*}
for any $a\in{\bf Z}^{d}$, $1 \leq p \leq i$.
So that the action of the lattice Laplacean over the function $\Phi(x_{1},\ldots,x_{n})$ produces
\begin{eqnarray*}
\lefteqn{-\Delta_{d} \Phi(x_{1},\ldots,x_{n})=\sum_{p=1}^{i}m_{p}\sum_{j=1}^{d}[2\Phi(x_{1},\dots,x_{n})-} \\
& &- \Phi(x_{1},\ldots,x_{(m_{p})+1}+{\bf e}_{j},\ldots,x_{n}) -
\Phi(x_{1},\ldots,x_{(m_{p})+1}-{\bf e}_{j},\ldots,x_{n})].
\end{eqnarray*}
For a $n$-angular momentum state we are able to estimate the expectation
value of the perturbation to be always less than the norm of the Laplacean
operator. Using the Schwartz inequality we have
$$
|\left\langle \Psi _n\left| {\cal H}_\Lambda ^{xy}\right| \Psi
_n\right\rangle |\leq \frac {|J|}{2} \| \left\langle \Psi \right| \| \ \| \ \Delta_{d}\left|\Psi \right\rangle \| \leq 2nSd|J|
$$
Note that the proof requires the total number of particles to be
conserved, and the ground state of ${\cal H}_{z}$ to be the same as
that of ${\cal H}_{xy}$.
\begin{thebibliography}{9}
\bibitem{Lieb1} Lieb, E. H.; Schultz, T. D.; Mattis, D. C.: {\it Ann. Phys.
}{\bf 16}, 407 (1961)
\bibitem{JW} Jordan, P.; Wigner, E.: {\it Z. Physik} {\bf 37}, 631 (1928)
\bibitem{Niem} Falk, H.: {\it Phys. Rev.} {\bf 133}, A1382 (1964). The
''Jacobsohn Theorem'' states that the ground state of the complete
Heisenberg antiferromagnet subject to a strong magnetic field $h$ is
ferromagnetic (in the sense that all spins lie parallel to the external
field, despite the absence of long range correlations), provided $h$ is
bigger than a critical field $h_c$. The proof, as far as we know,
has never been published. See also reference [6] below.
\bibitem{Oscar} Bolina, O.: M. Sc. Dissertation, University of S\~ao Paulo,
Brazil (1992), unpublished. See also: Wreszinski, W. F.; Salinas, S. R. A.:
''Disorder and Competition in Soluble Lattice Models'', World Scientific
(1993).
\bibitem{Klein} Klein, A., Perez, J.F.: Commun. Math. Phys.128,
99-108(1990).
\bibitem{Th} Niemeiyer, Th.: Physica {\bf 36}, 377 (1967).
\end{thebibliography}
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