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\topmatter
\title Intermediate Phases in mixed Nematic/Heisenberg Spin-Models
\endtitle
\bigskip
\bigskip
\leftheadtext {M. Campbell and L. Chayes}
\rightheadtext\nofrills {Mixed Nematic/Heisenberg Models}
\author
\hbox{\hsize=2.8in
\vtop{\centerline{M. Campbell}
\centerline{{\it Department of Mathematics}}
\centerline{{\it University of California, Los Angeles}}}
\vtop{\centerline{L. Chayes}
\centerline{{\it Department of Mathematics}}
\centerline{{\it University of California, Los Angeles }}}}
\endauthor
%
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\address
M. Campbell
\hfill\newline
Department of Mathematics
\hfill\newline
University of California
\hfill\newline
Los Angeles, California 90095-1555
\endaddress
\email
mcampbe\@math.ucla.edu
\endemail
\address
L. Chayes
\hfill\newline
Department of Mathematics
\hfill\newline
University of California
\hfill\newline
Los Angeles, California 90095-1555
\endaddress
\email
lchayes\@math.ucla.edu
\endemail
\keywords
Liquid crystal model, Classical Heisenberg model
\endkeywords
\bigskip
\abstract
We prove that in three or higher dimensions, an isotropic,
ferromagnetic, nearest-neighbor O(N) perturbation of a nematic liquid crystal
model (also having isotropic, nearest-neighbor
interactions) has an intermediate phase. In this intermediate phase,
the liquid crystal order parameter is non-zero but the spontaneous
magnetization is zero while at low temperatures, these systems exhibit
magnetization.
\endabstract
\endtopmatter
\document
\baselineskip = 15pt
This paper concerns the (lattice) nematic liquid crystal model. Here, at
each site
$i$ of a regular lattice, resides an $O(N)$ spin--variable, $\vec s_i$.
(We will
take $|\vec s_i| = 1$; for physical considerations, one focuses on $N =
2,3$.) This
variable, in some approximation, usually represents a rod--like molecule --
or density thereof
-- that is symmetric about its center. Thus the spin--spin interaction
boils down to
the square of the usual Heisenberg term.
Such models have been investigated mathematically in \cite{AZ}, \cite{Z}
where, in
dimension $d \geq 3$, a low temperature phase was established. This phase,
called the
nematic phase, is characterized by the non--vanishing of the nematic order
parameter,
i.e. the matrix
$$
\Bbb Q^{\alpha, \gamma}_i = s_i^\alpha s_j^\gamma - \frac 1 N
\delta^{\alpha\gamma}
\tag 1
$$
does not average to zero. Needless to say in these models the spins themselves
have zero average at all temperatures.
Here we wish to consider a situation where the liquid crystal
molecules have
some asymmetry along the axis of the rod, that will become physically
relevant at
sufficiently low temperatures. For example, the ``head'' and ``tail'' ends
of the
molecules may have different interactions that tend to align the heads of the
molecules, as was discussed in \cite{LG}. Thus the natural model to
consider is the
standard liquid crystal model (e.g. as written down in \cite{AZ}) with the
Hamiltonian perturbed by a Heisenberg term. This is precisely what was done in
\cite{LG} for the case of the two-dimensional $O(2)$ model.
Here, mostly for technical reasons, we will focus on $d \geq 3$.
The techniques of reflection positivity
and infrared bounds \cite{FSS}, \cite{FILS} tell
us that both the liquid crystal and the magnetic
order parameters have associated temperatures, below which they are non-zero.
The purpose of this paper is to show that these temperatures can be
distinct, and
consequently the existence of an intermediate phase with nematic order and no
magnetic order. We will use a generalized graphical representation of the type
discussed in \cite{CM} to distinguish these temperatures.
The setup for this model in finite volume consists of a box
$\Lambda \subset {\Bbb Z}^d$, $d \geq 3$, with bonds ${\Bbb B}$, and a ``liquid
crystal rod'' (i.e., an $O(N)$ variable) centered at each site. The ends
of the rod
are antipodal points on the $(N -1)$--dimensional unit sphere $S^{N-1}$ and the
(matrix) observable for this molecule is $\Bbb Q_i^{\alpha, \gamma}$ as
defined in
Equ. (1). The unperturbed nematic interaction between two molecules at
neighboring
sites $i$ and $j$ is given by $-J\text{Tr}[\Bbb Q_i\Bbb Q_j]$ with $J > 0$.
The full
Hamiltonian, including the Heisenberg perturbation that we consider is given by
$$
{\Cal H}_{\Lambda}({\vec h}, {\Bbb K}) =
-\sum_{\langle i,j \rangle \in {\Bbb B}}
\left(
J\text{Tr}[\Bbb Q_i\Bbb Q_j] + \epsilon(\vec s_i\cdot \vec s_j)
\right)
- \sum_{i\in\Lambda} ( \text{Tr}[\Bbb K_i \Bbb Q_i] +
\vec h_i\cdot \vec s_i)
\tag 2
$$
with ${\vec s}_{i} \in S^{N-1}$, ${\vec h} \in {\Bbb R}^N$, and
$\epsilon > 0$. The objects $\Bbb K_i$ are traceless $N{\times}N$ self-adjoint
matrices:
$$
\Bbb K_i^{\alpha\gamma} = k_i^{\alpha} k_i^{\gamma} - \frac 1 N |\vec
k_i | \delta^{\alpha \gamma}
$$
where $\vec k_i$ is a vector like $\vec h_i$, with components $k_i^\alpha,
\alpha = 1,\dots,N$.
The external fields, $\Bbb K \equiv {\Bbb K_i}$ and $\vec h \equiv {\vec
h_i}$ are for
auxiliary purposes only and will not enter into the final results.
The partition function is given by
$$
{\Cal Z}_{\Lambda}({\vec h}, {\Bbb K}) =
\partition{_{\Lambda}({\vec h}, {\Bbb K})}
\tag 3
$$
where $ds$ denotes the Haar measure on $S^{N-1}$. We let $\langle \cdot
{\rangle}_{\Lambda; {\vec h},\Bbb K}$ denote the Gibbs measure corresponding to
${\Cal Z}_{\Lambda}(\vec h, {\Bbb K})$.
Since we will be using reflection positivity to establish phase
transitions,
we first consider only periodic boundary conditions (cf.
\cite{FILS}). Thus, in the absence of spatially dependent external fields,
the states
we consider are translation invariant so the magnetization observable is
equal to
${\vec s}_{0}$ and the nematic observable equal to $\Bbb Q_0$.
The order parameters that we will consider are the infinite volume quantities
$\langle {\vec s}_{0} \rangle_{0^+,0}$ and
$\langle {\Bbb Q}_{0} \rangle_{0,0^+}$. In both cases, the zero field
limit is taken
with uniform external fields pointing along a particular axis.
Using the techniques in \cite {FSS} and \cite{FILS}, we have first order phase
transitions with respect to ${\vec h}$ and ${\Bbb K}$. Furthermore we have
lower
bounds on the transition temperatures for each order parameter. We will denote the
lower bounds on the transition temperatures for
$\langle {\vec s}_{0} \rangle_{0^+,0}$ and
$\langle {\Bbb Q}_{0} \rangle_{0,0^+}$
by $T_*^{spin}$ and
$T_*^{lc}$, respectively. The actual transition temperatures
(below which $\langle {\vec s}_{0} \rangle_{0^+,0}$ and
$\langle {\Bbb Q}_{0} \rangle_{0,0^+}$ are non zero and above
which they vanish) will be denoted by $T_c^{spin}$ and $T_c^{lc}$. We summarize
the current status in the following:
\proclaim{Proposition 1}
There is a first order phase transition in each
variable ${\vec h}$ and ${\Bbb K}$ with
$\langle {\vec s}_{0} \rangle_{0^+,0}$ and
$\langle {\Bbb Q}_{0} \rangle_{0,0^+}$
the respective order parameters for these phase transitions.
These quantities are non--zero
when $T < T_*^{spin}$ and/or
$T < T_*^{lc}$; here the
lower bounds on the transition temperatures are related to the interaction
constants by $T_*^{spin} \equiv 4\epsilon /[N D]$ and
$T_*^{lc} \equiv 8J(N-1)/[N^2(N+1)D]$, where $D$ is the constant in
\cite{FILS}:
$$
D \equiv (2\pi)^{-d} \negthickspace \int\limits\Sb
|p_m| \leq \pi \\ m =1,\ldots,d \endSb
\biggl( \sum\limits_{m=1}^d 1 - \cos
p_m \biggr)^{-1} d^d p \, \text{,}
$$
and $d$ is the lattice dimension. Furthermore, the temperatures
$T_c^{spin}$ and $T_c^{lc}$ are finite.
\endproclaim
\demo{Proof}
The temperatures $T_*^{spin}$ and $T_*^{lc}$ are obtained by the
methods in section four of \cite{FILS}; we will outline the
derivation in \cite{AZ} for $T_*^{lc}$.
We use matrices that have a single non--zero entry
on or above the diagonal at ($\alpha,\gamma$) to obtain all of
the mixed-term bounds for Gaussian domination:
${\Cal Z}_{\Lambda}(0, {\Bbb H}^{\alpha \gamma}) \leq {\Cal
Z}_{\Lambda}( 0, 0)$. We have N(N+1)/2 bounds (since
there are that many elements on or above the diagonal of an $N{\times}N$
matrix). By summing these bounds and following the derivation in
\cite{FSS} and \cite{FILS},
we see that if
$$
\langle Tr( {\Bbb Q}_{0}^2 ) \rangle -
\frac{1}{4\beta} \frac{N(N+1)}{2} J D \geq 0
\tag 4
$$
there will be multiple phases, and thus for some ergodic state $\mu$,
$\mu({\Bbb Q}_{0}) \neq 0$. This will happen if $T <
[N(N+1)D]^{-1} 8J \langle \text{Tr}({\Bbb Q}_{0}^2) \rangle =
[N(N+1)D]^{-1} 8J(N-1)/N$. The classic derivation (in \cite{FSS}) results
in the
value for $T_*^{spin}$.
Standard high temperature techniques (\cite{S, V.1.3}, Dobrushin's uniqueness
theorem, or the method in \cite{AC}) shows that the temperatures
$T_c^{spin}$ and
$T_c^{lc}$ are finite.
\qed
\enddemo
We will now employ a primitive graphical representation of our
model to obtain an upper bound on the critical temperature $T_c^{spin}$
for the magnetic order parameter. The goal of this paper is achieved if
our upper bound lies below $T_*^{lc}$ (which will be satisfied if $\epsilon$ is
small).
Let us start this with some preliminary
notation. As before, Let $\Lambda \subset {\Bbb Z}^d$ be a finite box with
bonds
${\Bbb B}$. We define
$$
\matrix \format \l & \l \\
{\Cal H}_{\Lambda}^{lc} &= -\sum\limits_{\langle i,j
\rangle \in {\Bbb B}}
J \thinspace Tr({\Bbb Q}_{i} {\Bbb
Q}_{j}) \, \text{,} \\ \vspace{1\jot}
{\Cal H}_{\Lambda}^{spin} &= -\spinHamiltonian \,
\text{,} \\ \vspace{1\jot}
{\Cal H}_{\Lambda} &= {\Cal H}_{\Lambda}^{lc} +
{\Cal H}_{\Lambda}^{spin} \, \text{,}
\endmatrix
\tag 5
$$
Here we allow arbitrary boundary conditions on $\partial \Lambda$ -- but
these will
not enter into the notation. Observe that in the magnetic term, we have added a
constant (which of course, has no physical consequences). To get our
graphical representation, we write
$$
{\Cal Z}_{\Lambda} = \text{Tr}[e^{-\beta\Cal H_{\Lambda}}] =
\text{Tr}[e^{-\beta\Cal H_\Lambda^{lc}}
\prod_{\langle i,j \rangle \in \Bbb B}
e^{\beta\epsilon (\vec s_i\cdot \vec s_j + 1)}]
\tag 6
$$
where the operation of ``Tr'' includes any special considerations that take
place at
the boundary. We expand the product in Equ.(6) associating each term in the expansion
with a bond configuration $\omega \in \{0,1\}^{\Bbb B}$. In particular,
for each
$\langle i,j
\rangle$ we must ``choose'' the term $(e^{\beta \spinHamiltonianTerm} - 1)$
or $+1$
which corresponds to $\omega_{\langle i,j \rangle} = 1$ (occupied) or
$\omega_{\langle i,j \rangle} = 0$ (vacant) respectively. It is noted that the
contribution from each term in the sum is positive. Thus the weights
$$
V_{\Lambda}(\omega) = \partition{_{\Lambda}^{lc}} \prod\limits_{\langle i,j
\rangle \in \omega}
\left( e^{\beta
\spinHamiltonianTerm} - 1 \right)
\tag 7
$$
when divided by the partition function may be interpreted as the probability of
observing the configuration $\omega$. We denote the associated probability
measure
by $v_\Lambda(-)$.
We write
$i \leftrightarrow \partial {\Lambda}$
if there is a path of occupied bonds in ${\omega}$
connecting the site $i$ to $\partial {\Lambda}$, and $i
\not\leftrightarrow \partial {\Lambda}$ if there is no such path
connecting $i$ to $\partial\Lambda$.
Although the weights appear rather unwieldy it is straightforward to obtain {\it
bounds} on certain probabilities. For example, we have
\proclaim{Lemma 2}
Let $\{e_1, e_2, \ldots, e_n\}$ denote a given set of edges.
Then in any boundary condition, the probability
that all of these
edges are occupied,
$v_{\Lambda}(e_1\!\!=\!\!1,e_2\!\!=\!\!1,\ldots,e_n\!\!=\!\!1)$,
is bounded above by
$(e^{2\beta\epsilon } -1)^n$.
\endproclaim
\demo{Proof}
This is an elementary observation:
$$
\aligned
{v}_\Lambda(e_1\!\!=\!\!1,e_2\!\!=\!\!1,\ldots,e_n\!\!=\!\!1) &= \!\!\!
\sum\limits\Sb \omega : \omega(e_m)=1 \\
m=1,\ldots,n \endSb {\Cal Z}_{\Lambda}^{-1}
\partition{_{\Lambda}^{lc}} \prod\limits_{\langle i,j \rangle \in {\Bbb
B}}
\left( e^{\beta \spinHamiltonianTerm} - 1
\right)^{{\omega}\langle i,j \rangle} \\
&\leq \left( e^{2 \beta \epsilon} - 1 \right)^n
v_{\Lambda}(e_1\!=\!0,e_2\!=\!0,\ldots,e_n\!=\!0) \\
&\leq \left( e^{2 \beta \epsilon} - 1 \right)^n.
\endaligned
\tag{8}
$$
It is noted that the inequality in Equ.(8) is independent of
boundary conditions.
\qed
\enddemo
\proclaim{Theorem 3} Let $T_0^{spin}$ be defined by
$\exp\{2\epsilon/T_0^{spin}\} - 1 = 1/[2d -1]$. Then for
$T > T_0^{spin}$, the spontaneous magnetization for the system defined by the
Hamiltonian in Equ.(2) (at zero external fields) is zero.
\endproclaim
\demo{Proof}
Let $\Lambda$ be a finite box as above and let
$\langle \cdot \rangle_\Lambda^*$ be the finite-volume Gibbs measure
using the Hamiltonian ${\Cal H}_{\Lambda}$ in Equ.(2) (at zero external field)
with boundary conditions $*$ and the temperature dependence suppressed.
Then, by well
known arguments (cf.
\cite{S, ch\. III\.3}) the spontaneous
magnetization is bounded above by
$\sup_{*} \langle s_{0}^{1}
\rangle_\Lambda^*$. Thus it is sufficient to show that for large
$\Lambda$, this
quantity is small in any boundary condition.
Let us write, according to the preliminary steps of the graphical
representation, the
identity
$$
\langle s_{0}^{1} \rangle_{\Lambda, \, \beta}^* =
\sum\limits_{{\omega} \in \{0,1\}^{{\Bbb B}}}
{\Cal Z}_{\Lambda}^{-1}
\partition{_{\Lambda}^{lc}} \, s_{0}^{1}
\prod\limits_{\langle i,j \rangle \in \omega}
\left( e^{\beta \spinHamiltonianTerm} - 1 \right).
\tag 9a
$$
We claim that for any configuration ${\omega}$ such that $0 \not\leftrightarrow
\partial{\Lambda}$, the corresponding term in Equ.(9)
vanishes.
To see this let $C_{0}({\omega})$ denote the connected cluster of the
origin in the
configuration $\omega$. If $C_{0}({\omega})$ does not touch the boundary,
then the
integrals of all of the $\vec s_k,\ k \in C_{0}({\omega})$ are performed
with respect
to the full Haar measure. Now consider the change of variables ${\vec
s}_{k} \mapsto
-{\vec s}_{k}, k \in C_{0}({\omega})$. Obviously if $i$ and $j$ are both in
$C_{0}({\omega})$, then the term $\vec s_i \cdot \vec s_j$ is unaffected.
Further,
since each spin component $s_i^\alpha$ appears quadratically in $\Cal
H_{\Lambda}^{lc}$ these terms are also invariant (regardless of whether or
not $i\in
C_{0}({\omega})$). On the other hand, this transformation negates the
$s_0^1$ that sits
in front and hence the whole term cancels out.
Thus we are left with the more restrictive sum
$$
\langle s_{0}^{1} \rangle_{\Lambda, \, \beta}^* =
{\Cal Z}_{\Lambda}^{-1} \sum\limits_{{\omega} : 0
\leftrightarrow \partial{\Lambda}}
\partition{_{\Lambda}^{lc}} \, s_{0}^{1}
\prod\limits_{\langle i,j \rangle \in {\Bbb B}}
\left( e^{\beta \spinHamiltonianTerm} - 1
\right).
\tag 9b
$$
Since $|s_{0}^{1}| \leq 1$, we see that
$$
|\langle s_{0}^{1} \rangle_{\Lambda, \, \beta}^*| \leq
v_\Lambda(0 \leftrightarrow
\partial{\Lambda})
\tag 10
$$
(where on the right hand side we suppress in our notation any dependence on
boundary
conditions).
At this point all we need is a Peierls' argument. Any configuration
${\omega}$ with $0 \leftrightarrow \partial{\Lambda}$ must have a
connected path of occupied bonds (a self--avoiding {\it walk}) between the
origin and
the boundary. Let $R$ denote the distance between the origin and the
boundary. If
$\{e_1, e_2,\ldots,e_S\}$ (with $S \geq R$) is one such walk, let
${\Cal W}(e_1,e_2,\ldots,e_S)$ denote the event that all these edges are
occupied.
Using the result of Lemma 2 we have
$$
\aligned
v_\Lambda^*(0 \leftrightarrow \partial{\Lambda})
&\leq
\sum\limits\Sb \text{all walks},\ S \geq R \\ {\Cal
W}(e_1,e_2,\ldots,e_S)\endSb
v_\Lambda^*(
{\Cal W}(e_1,e_2,\ldots,e_S) ) \\
&\leq \sum\limits\Sb \text{all walks},\ S \geq R \\ {\Cal
W}(e_1,e_2,\ldots,e_S)\endSb
\left( e^{2 \beta \epsilon} - 1 \right)^S \\
&\leq C \left( e^{2 \beta \epsilon} - 1
\right)^R
\left(2d - 1 \right)^R
\endaligned
\tag 11
$$
for some constant $C$ independent of $R$ that is finite if
$(e^{2 \beta \epsilon} - 1)(2d - 1) < 1$.
Since these arguments are uniform in $|\Lambda|$ and independent of
boundary conditions it
follows immediately that when $(e^{2 \beta \epsilon} - 1)(2d - 1) < 1$, the
spontaneous
magnetization vanishes.
\qed
\enddemo
\remark{Remark}
Improved values for $T_0^{spin}$ can be obtained in a couple of ways.
First, rather
trivially, quantity $(2d - 1)$ can be replaced by the connectivity constant
for the
$d$--dimensional square lattice. On the basis of a little more effort, it
can be
established that the measures $v_{\Lambda}(-)$ are dominated by independent
(Bernoulli)
measures with parameter $p = 1 - e^{-2 \beta \epsilon}$. This may be
derived by going to a
full blown Edwards-Sokal measure \cite{ES} along with an associated cluster
algorithm.
Then, at each bond step, regardless of the spin configuration, one notices
that the maximum
bond probability can never exceed $1 - e^{-2 \beta \epsilon}$. The upshot
is that the
improved $T_0^{spin}$ satisfies $1 - \exp\{-2\epsilon/T_0^{spin}\} =
p_c(d)$ where
$p_c(d)$ is the bond percolation threshold for the $d$--dimensional lattice.
\endremark
\proclaim{Corollary}
If $d \geq 3$ and $\epsilon/J$ is sufficiently small (depending on $N$ and
$d$) there is an
intermediate phase where the magnetization is zero and the nematic order
parameter is
positive.
\endproclaim
\demo{Proof}
To demonstrate the above, we need only find a ``window'' between $T_*^{lc}$ and
$T_0^{spin}$. Since, (for fixed $N$ and $d$) the former scales with $J$
and the latter with
$\epsilon$ (for $\epsilon \ll 1$) the desired result follows.
\qed
\enddemo
\remark{Remark}
A detailed analysis of the minimal requirements on $\epsilon/J$ will appear
in the
forthcoming \cite{Ca}.
\endremark
\newpage
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\tenpoint
%\baselineskip = 20pt
\baselineskip = 14pt
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\medskip
\bigskip
\endRefs
\enddocument