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\begin{document}
\title{\bf Some Symmetry Remarks on the Ground State
Configurations of the Two Dimensional Bosonic Falicov-Kimball Model}
\author{\bf J. Rodrigo Parreira\thanks{Supported by CNPq}\\
Department of Physics\\Princeton University\\
Princeton, NJ
\\PO Box 708, 08540-0708\\USA}
\maketitle
\begin{abstract} We show that the ground state configuration of
the nuclei in a
system composed by itinerant bosons in two dimensions described
by the Falicov-Kimball Hamiltonian is composed by symmetric intervals
centered on $45^{\circ}$-degree lines on thje cartesian plane. This adds
up to the previous result
about this problem which asserted symmetry of intervals centered on the
coordinate axii. The combination of these facts strongly suggest the
convexity of the nuclear distributions.
{\bf Keywords:} Falicov-Kimball Model, Itinerant Bosons, Ground State
Nuclear Configurations, Rearrangement Inequalities.
{\bf PACS:} 05.30.Jp; 71.10.Fd
\end{abstract}
\section{Introduction}
The Falicov-Kimball model was introduced in 1969 \cite{FK} as a way to
describe semiconductor-metal phase transitions in transition-metal oxides.
The Hamiltonian which defines the system is given by
\beq
H=-\sum_{x,y\in\Lambda; |x-y|=1}
c_{x}^{\dagger}
c_{y} + U\sum_{x\in\Lambda} W(x)c_{x}^{\dagger}c_{x}\label{H}
\eeq
where, $\Lambda\subset\bf Z^{\rm d}$, and, according to the original purposes
of the authors,
$c_{x}^{\dagger}$ and $c_{x}$ were respectivelly thought to be creation and
annihilation operators for mobile electrons in $d-$shells, while the $W(x)$
indicated the positions of localized $f-$electrons assuming the values
$W(x)=1$ if the site is occupied and $W(x)=0$ otherwise.
The Hamiltonian operator can then be stated as
$$
H=T-V.
$$
Here, $T$ is the operator given by
$$
T=-\Delta - 2d.{\bf 1}.
$$
$\Delta$ is the lattice Laplacean, whose action, in $d$ dimensions is
$$
(-\Delta f)(x) = \sum_{i=1}^{d}[ 2f(x) - f(x+{\bf e}_{i}) - f(x - {\bf
e}_{i})]
$$
where ${\bf e}_{i}$ is a unit vector in the $i$-direction. ${\bf 1}$
represents here the identity operator.
The operator $V$ is just a diagonal operator whose elements of the
principal diagonal are respectively $U$ or $0$ according to the fact that
the corresponding site is occupied or empty.
The same Hamiltonian can be interpreted in several other different
ways. It could just represent a simplification of the Hubbard model
where, say, the spin-down electrons are taken to be infinitely massive.
Or, if $U$ is chosen to be positive, the atractive interaction between
mobile and fixed particles assumes the form of the interaction between
itinerant negative charged electrons and fixed positive nuclei.
Indeed, adopting this last point of view, Kennedy and Lieb \cite{KL} showed
that, in the half-filled case
($\sum_{x}c_{x}^{\dagger}c_{x}=\sum_{x}W(x)=|\Lambda|/2$, where $|\Lambda|$
is the cardinality of the set $\Lambda$), when $\Lambda$ is connected and
can be
decomposed into two sublattices, $\Lambda_{1}$ and $\Lambda_{2}$ such
that $\Lambda= \Lambda_{1} \cup
\Lambda_{2}$, the
ground state of the system is attained when the nuclei arrange
themselves in a checkerboard configuration, occupying all sites of one
of the sublattices. This result shows that the Falicov-Kimball model
can be viewed as a very simple model of matter, leading to the formation of
crystals.
However, an interesting remark in \cite{KL} is related to the fermionic
nature of the mobile particles. It is proved that, when the itinerant
particles happen to be bosons, the ground state changes into a
configuration where all the nuclei are packed together.
In order to appreciate this result in its total extent, some definitions
are necessary.
{\bf Definition 1}: A function $f(x):[-L,L] \cap {\bf Z} \rightarrow
{\bf R}^{+}$ is said to be {\it Left Symmetric Decreasing} (LSD) when:
$$
f(0) \geq f(-1) \geq f(1) \geq f(-2) \geq ... \geq f(m)
$$
where we denote by $m$ the biggest integer contained in $[-L,L]$.
{\bf Definition 2}: A function $f(x):[-L,L] \cap {\bf Z} \rightarrow
{\bf R}^{+}$ is said to be {\it Right Symmetric Decreasing} (RSD) when:
$$
f(0) \geq f(1) \geq f(-1) \geq f(2) \geq ... \geq f(-m)
$$
where by $-m$ we denote the smallest integer in $[-L,L]$.
{\bf Definition 3}: A function $f(x):[-L,L] \cap {\bf Z} \rightarrow
{\bf R}^{+}$ is said
to be {\it Symmetric Decreasing} (SD) if it is, at the same time, LSD and
RSD.
{\bf Definition 4}: A {\it coordinate line} in the $y_{1}$ direction is a
set $l \subset {\bf Z}^{d}$ of the form
$$
l=\{(z, y_{2}, y_{3},...,y_{d})|z \in {\bf Z}\}.
$$
Given these four definitions, Kennedy and Lieb \cite{KL} proved that, due
to the bosonic nature of the itinerant particles, the ground state wave
function $\psi$
is, on every coordinate line in a given direction, positive definite and RSD
(or
LSD). It can be also SD, in which case $\psi(x) = \psi(y)$ only if $x$ is
the reflection of $y$ through the midpoint of the line. It is also proved
the existence of a positive threshold $\psi_{0}$ such that
\beq
V(x)=2U \Leftrightarrow {\psi} \geq \psi_{0} \label{A1}
\eeq
and
\beq
V(x)=0 \Leftrightarrow {\psi} < \psi_{0} \label{A2}.
\eeq
The crucial difference between the bosonic and the fermionic cases is
that, in the case of itinerant fermions, to obtain the ground state it is
necessary to calculate the $|\Lambda|/2$ eigenstates corresponding to
the $|\Lambda|/2$ lowest eigenvalues and then putting one electron in
each of these
states. In the bosonic case, it is only necessary to find out
the lowest eigenvalue of the Hamiltonian (unique, as a consequence of
Perron-Frobenius theorem) so that all particles can be simultanously
accomodated in the corresponding eigenstate.
The allowed configurations of nuclei in the ground state, given the above
results, would then be composed by intervals over the coordinate lines
either symmetric to the $x$ or to the $y$ axis (two
dimensional model). Figures like centered squares, rectangles, crosses, and
the
like are good examples. In this note we are going to extend these results to
$45^{\circ}$-lines. Some configurations will then be ruled out, like the
rectangle and the cross. However, despite our result is still not
sufficient to
assert the convexity (in a lattice sense) of the distributions of nuclei,
we think that this is a natural conjecture.
We also remark that, although
our results are obtained for the two dimensional case, the same
techniques used here can be straightforwardly used in higher dimensions.
We obtain our results by exploring the symmetry of the possible
configurations of the ground state according to \cite{KL} and then by
applying rearrangement inequalities
to the ground state eigenfunction along the $45^{\circ}$-lines in order to
minimize its energy.
\section{ Exploring the Symmetry of the Ground State Eigenfunction}
In the two dimensional case, according to \cite{KL}, only a few different
possible configurations of the ground state wave function $\psi$ can
exist:
1) $\psi$ could be LSD in the $x$ direction and RSD in the $y$ direction.
2) $\psi$ could be LSD in the $x$ direction and LSD in the $y$ direction.
3) $\psi$ could be LSD in the $x$ direction and SD in the $y$ direction.
4) $\psi$ could be SD in both directions.
And, of course, all cases obtained by replacing LSD by RSD (and
vice-versa) above. These are trivially equivalent to the ones
presented before.
We now define what is precisely understood as a $45^{\circ}$-line.
{\bf Definition 5}: A $45^{\circ}${\it -degree line} which contains the
point $(x,y)$
and goes from the second to the fourth quadrants is a set of points
$\overline{l}^{(2,4)} \subset {\bf Z}^{2}$ of the form
$$
\overline{l}^{(2,4)} = \{(x,y)+n(1,-1)|n\in{\bf Z} \}.
$$
In a similar way one can define a $45^{\circ}$-degree line that goes
from the third to the first quadrants by
$$
\o{l}^{(3,1)}= \{ (x,y)+n(1,1)|n\in{\bf Z} \}.
$$
We will also be calling
the $45^{\circ}$-lines that contain the origin as {\it central} and
indicating them as $\o{l}^{(2,4)}_{c}$ (and, of course $\o{l}^{(3,1)}_{c}$).
Let us take, as an example, the ground
state wave function $\psi$ corresponding to the first case above. Now,
imagine points located over $\o{l}^{(2,4)}_{c}$. Since,
as a hypothesis, $\psi$ is LSD in the $x$
direction and RSD in the $y$ direction, we have that
$$
\psi(0,0) \geq \psi(-1,1) \geq \psi(-1,-1)\geq \psi(1,-1) \geq \psi(1,2)
\geq \psi(-2,2) \geq ...
$$
so that
$$
\psi(0,0) \geq \psi(-1,1) \geq \psi(1,-1) \geq \psi(-2,2) \geq \psi(2,-2)
\geq ...
$$
and we have then that the values assumed by $\psi$ over
$\o{l}^{(2,4)}_{c}$ are also ordered in a symmetric decreasing
way. If it is LSD or RSD it depends only on the chosen orientation of the
line. Let us call it LSD. It is easy to discover that, for
each of the above cases, 1)-4), at least
one of the
distributions of $\psi$ over central $45^{\circ}$-lines is ordered in a LSD
or RSD way. Both of them
can also be ordered, corresponding to the case when the distribution of
$\psi$
is SD along one of the coordinate lines. Finally, if the
distribution is SD in both directions one is able to find out that both
of these central lines are also SD.
\section{Rearrangement Inequalities}
Rearrangement inequalities will be used below in order to minimize the
energy of the ground state. The following
straightforward inequality (for a proof, see \cite{HLP}) is a basic Lemma.
{\bf Lemma 1}: For both $f:{\bf Z} \rightarrow {\bf R}^{+}$ and $g:
{\bf Z} \rightarrow {\bf R}^{+}$ we have that
$$
\langle f | g \rangle \leq
\langle f^{*} | g^{*} \rangle = \langle f^{**} | g^{**}
\rangle.
$$
where by $f^{*}$ and $f^{**}$ we respectively indicate the LSD and
RSD rearrangements of the function $f$, defined as follows:
{\bf Definition 6}: Given $f:{\bf Z} \rightarrow {\bf R}^{+}$, it is always
possible to arrange the values of $f$ in a strictly decreasing order:
$$
f_{1} \geq f_{2} \geq f_{3} \geq ...
$$
We then define the
{\it LSD rearrangement of f}, indicated by $f^{*}$ as $f(0)=f_{1}$,
$f(-1)=f_{2}$, $f(1)=f_{3}$, and so on. In a similar way the {\it RSD
rearrangement of f}, indicated by $f^{**}$ can be constructed.
We now introduce the following operator $\Upsilon$:
$$
[\Upsilon f](x)=f(x) + f(x+1),
$$
whose adjoint is easily verified to be
$$
[\Upsilon^{\dagger}f](x)=f(x-1)+f(x).
$$
We then have
{\bf Lemma 2}: Given $f:{\bf Z} \rightarrow {\bf R}^{+}$, a LSD
function, and $g :{\bf Z} \rightarrow {\bf R}^{+}$, we have that
$$
\langle f^{*} | \Upsilon g \rangle \leq \langle f^{*} | \Upsilon
g^{**} \rangle.
$$
{\bf Proof}:
Since $f$ is a LSD function we have that $f^{*}=f$.
Now, since
$$
[\Upsilon^{\dagger} f^{*}](x) = f^{*}(x-1) + f^{*}(x)
$$
and
$$
[\Upsilon^{\dagger} f^{*}](-x) = f^{*}(-x-1) + f^{*}(-x)
$$
for any $x \in {\bf Z}$, we have that $\Upsilon^{\dagger}f^{*}$ is a
RSD function. This is easy to see, because, since $f^{*}$ is LSD, we
have that
$$
f^{*}(x-1) \geq f^{*}(-x)
$$
and
$$
f^{*}(x) \geq f^{*}(-x-1).
$$
Implying $[\Upsilon^{\dagger} f^{*}](x) \geq
[\Upsilon^{\dagger}f^{*}](-x)$. In a similar fashion one is also able
to prove that $[\Upsilon^{\dagger}f^{*}](-x) \geq [\Upsilon^{\dagger}
f^{*}](x+1)$.
So, by using Lemma 1, we have that
$$
\langle f^{*} | \Upsilon g \rangle =
\langle \Upsilon^{\dagger}f^{*} | g \rangle \leq
\langle \Upsilon^{\dagger} f^{*} | g^{**} \rangle = \langle
f^{*} | \Upsilon^{\dagger} g^{**} \rangle.
$$
This completes the proof.
\section{Minimizing the Ground State Energy}
Let ${\cal F}$ be the functional
\beq
{\cal F}=-\langle \o{\psi} | V \o{\psi} \rangle - \langle \o{\psi} |
\w{\Upsilon}_{1}
Q_{1} \rangle - \langle \o{\psi} | \w{\Upsilon}_{2} Q_{2} \rangle
\label{F}
\eeq
so that we are able to define the number
\beq
E(V,Q_{1},Q_{2})=\min_{\o{\psi}}\{ {\cal
F}(\o{\psi};V,Q_{1},Q_{2})|\langle \o{\psi} | \o{\psi} \rangle = c\}
\label{E}
\eeq
The notation $\o{\psi}$ is here being used in order to indicate the fact
that these functions have support over $45^{\circ}$-lines. The operators
$\w{\Upsilon}_{1,2}$ are then respectively defined as
$$
[\w{\Upsilon}_{1}\o{\psi}](x,y)=\o{\psi}(x+1,y) +\o{\psi}(x,y \pm 1)
$$
and
$$
[\w{\Upsilon}_{2}\o{\psi}](x,y)=\o{\psi}(x-1,y) + \o{\psi}(x,y \mp 1)
$$
the upper sign corresponding to the situation where the support of $\psi$
goes from the second to the fourth quadrants, the
lower sign when it goes from the third to the first quadrants.
We shall now assume that both (\ref{A1}) and (\ref{A2}) are valid
in order to prove the following Lemma.
{\bf Lemma 3}: Given the functional
${\cal F}$ defined by (\ref{F}), we have that, under the assumptions
(\ref{A1}) and (\ref{A2}), whenever $\o{\psi}=\o{\psi}^{*}$:
$$
E(V,Q_{1},Q_{2}) \geq
E(V^{*},Q_{1}^{**},Q_{2}^{**})=E(V^{**},Q_{1}^{*},
Q_{2}^{*}).
$$
{\bf Proof}:
Follows immediately after Lemma 2 and by noting that, since
$\o{\psi}=\o{\psi}^{*}$, the assumptions (\ref{A1}) and (\ref{A2}) imply
$V=V^{*}$ so that
$$
\langle \o{\psi} | V \o{\psi} \rangle = \langle \o{\psi}^{*} | V^{*}
\o{\psi}^{*} \rangle = \langle \o{\psi}^{**} | V^{**} \o{\psi}^{**} \rangle.
$$
Now, in order to simplify the proof of our main result we recall a
particular case, namely the one discussed in section 2. This will be
done without any loss of generality. The reader will easily find out that
a similar version of the theorem below is available for every possible
distribution of $\psi$ along the coordinate line directions.
{\bf Theorem}: Let $U<0$ and take $V$ in order to minimize the ground
state energy of the operator $H=T-V$ in a rectangle $\Lambda \subset {\bf
Z}^{2}$ under the
condition $\sum_{x}V(x)=2UN$. Let $\psi$ be the ground state wave function
corresponding to the operator $H$. According to \cite{KL} the only possible
distributions of $\psi$ along the coordinate lines are those presented in
section 2. Take the first case. We then have that:
{\bf i)} The distribution of $\psi$ along the central $45^{\circ}$-line
that goes from the second to the fourth quadrant is LSD in the sense that
$$
\psi(0,0) \geq \psi(-1,1) \geq \psi(1,-1) \geq \psi(-2,2) \geq ...
$$
{\bf ii)} The distribution of $\psi$ along any other $45^{\circ}$-line
that goes from the second to the fourth quadrant will be LSD (RSD)
according to the fact that the number of $45^{\circ}$-lines between it
and the parallel central line is an odd (even) number.
{\bf iii)} Along each $45^{\circ}$-degree line that goes from the second
to the fourth quadrants $\psi$ attains its maximum over points
on the central $45^{\circ}$-degree line that goes from the third
to the first quadrants.
As a consequence we have that the support of the function $V(x)$ is
composed by
symmetric intervals with respect to the ce
{\bf Proof}:
{\bf i)} This assertion is already proved in section 2.
{\bf ii)} Take one configuration of $\psi(x)$ that is at the sime time
LSD along all coordinate lines in the horizontal direction and RSD along all
coordinate lines in the vertical direction. Now, look at the distribution
of values of $\psi(x)$ over $\o{l}^{(2,4)}_{c}$. Since this distribution is LSD
in the sense of section 2, in order to minimize the functional ${\cal F}$
in (\ref{F}), both $Q_{1}$ and $Q_{2}$ (the two adjacent
$45^{\circ}$-lines to $\o{l}^{(2,4)}_{c}$, that will be respectively
indicated as $\o{l}^{(2,4)}_{-1}$ and $\o{l}^{(2,4)}_{1}$) should be RSD.
If they are not, rearrange them in this way, so that by Lemma 3 the
energy is lowered. Now, move to the $45^{\circ}$ degree line
$\o{l}^{(2,4)}_{1}$, whose neighbors are $\o{l}^{(2,4)}_{c}$ and
$\o{l}^{(2,4)}_{2}$ and, since $\o{l}^{(2,4)}_{c}$ is already arranged in
the right way, rearrange $\o{l}^{(2,4)}_{2}$ in a LSD way. When all
$45^{\circ}$-degree lines are rearranged following this recipe, it may
be possible that some of the LSD (RSD) arrangements in the horizontal
(vertical) direction are spoiled. It is then necessary to rearrange the
coordinate lines in
the horizontal (vertical) direction again. This will lower the energy
even more. The above procedure can then be repeatedly applied untill no
more rearrangements are possible. This will then be the configuration of
$\psi$ with the lowest energy.
{\bf iii)} By inspection, after Lemma 2.
As a consequence of the theorem above, together with the fact that
(\ref{A1}) and \ref{A2} were proved to be valid in the ground state in
\cite{KL} we have that the support of the function $V(x)$, identical to
the distribution of nuclei, will be composed by symmetric intervals with
respect to the $45^{\circ}$-lines. This , together with the previous
result that states symmetry of intervals where $W(x)\neq 0$ with respect to
coordinate axii strongly suggest that the real configuration is some kind
of lattice approximation of a circle.
\begin{thebibliography}{9}
\bibitem{FK} Falicov, L. M.; Kimball, J. C.: {\it Phys. Rev. Lett.}, {\bf
22}, 997 (1969).
\bibitem{KL} Kennedy, T.; Lieb, E. H.: {\it Physica}, {\bf 138A}, 320 (1986).
\bibitem{HLP} Hardy, G. H.; Littlewood, J. E.; Polya, G.: {\it
Inequalities}, second edition,
Cambridge University Press, Cambridge, UK (1959). For a proof of Lemma 1,
see \S 10.2, Theorem 368.
\end{thebibliography}
\end{document}