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\centerline{\bf UNIFORM DENSITY THEOREM FOR THE HUBBARD MODEL}
\bigskip
\bigskip
\centerline{Elliott H. Lieb}
\centerline{Departments of Physics and Mathematics}
\centerline{Princeton University}
\centerline{P.O. Box 708, Princeton, NJ 08544-0708}
\bigskip
\centerline{Michael Loss}
\centerline{School of Mathematics}
\centerline{Georgia Institute of Technology}
\centerline{Atlanta, GA 30332-0160}
\bigskip
\bigskip\noindent
{\bf Abstract:}
A general class of hopping models on a bipartite lattice is considered,
including the Hubbard model and the Falicov-Kimball model. For the
half-filled band, the single-particle density matrix $\uprho (x,y)$ in
the ground state and in the canonical and grand canonical ensembles is
shown to be constant on the diagonal $x=y$, and to vanish if $x \not=
y$ and if $x$ and $y$ are on the same sublattice. Physical
implications of this result are discussed.
\bigskip
\bigskip
\bigskip
\bigskip
\noindent
PACS numbers: 75.10.Lp, 71.10+x
\vfill\eject
The one-particle reduced density matrix, $\uprho (x,y)$, of a
many-electron quantum system can reveal a good deal about the presence
or absence of spatial uniformity. We have uncovered a curious --- and,
to us, surprising --- fact about $\uprho$ in the case of a half-filled
band in certain tight binding models, including the Hubbard model and
the Falicov-Kimball model. Our result, which is applicable to a
bipartite lattice, is that $\uprho (x,x)$ is always exactly equal to
one for all $x$ on a finite lattice, even though the hopping matrix
elements and interaction are nonuniform, random and uncorrelated.
Furthermore $\uprho (x,y) = 0$ when $x \not= y$ but $x$ and $y$ are in
the same sublattice. This result applies not only to ground states but
also to positive temperature states. (The infinite volume state will be
discussed at the end of this paper.) This striking result --- the
persistence of uniformity in the face of randomness --- presumably has
physical consequences. For instance, it hints at the stability of some
periodic structures in solids, which is to say that the occurrence of
periodic structures in a system might be insensitive to some of the
details of the Hamiltonian of the system. It is noteworthy that this
interesting result has a simple proof --- essentially based on
hole-particle symmetry.
To establish some notation, we consider a finite graph (lattice)
consisting of sites labeled by $x, y$, etc. and edges (or bonds)
connecting certain pairs of sites. We assume that the graph is
bipartite, i.e., the vertices can be divided into two disjoint subsets
$A$ and $B$ such that there is no edge connecting $x$ and $y$ if $x$
and $y$ are both in $A$ or both in $B$. The total number of sites in
$\Lambda, A$ or $B$ is denoted by $\vert \Lambda \vert, \vert A \vert$
and $\vert B \vert$. We assume that $\vert A \vert \geq \vert B
\vert$. We are given a hermitian $\vert \Lambda \vert \times \vert
\Lambda \vert$ hopping matrix \ $T$ with elements $t_{xy}^{\phantom{*}}
= t_{yx}^*$. These elements are nonzero only if $x$ and $y$ are
connected by an edge; in particular the elements $t_{xy}$ are zero
whenever $x$ and $y$ are both in $A$ or both in $B$. Physically, $T$
is real in the absence of magnetic fields that interact with the
electron orbital motion.
It is easy to see that the nonzero eigenvalues of $T$ come in opposite
pairs, i.e. for every eigenvalue $\lambda$ there is an eigenvalue
$-\lambda$. The two corresponding eigenvectors $\phi^\lambda$ and
$\phi^{-\lambda}$ are simply related: if $\phi^\lambda = (f^\lambda,
g^\lambda)$ with $f^\lambda$ being the $A$-part of $\phi^\lambda$ and
$g^\lambda$ being the $B$-part of $\phi^\lambda$ then $\phi^{-\lambda}
= (f^\lambda, - g^\lambda)$. Alternatively, if $V =
\pmatrix{1&0\cr0&-1\cr} = V^\dagger$ is the diagonal unitary matrix
that multiplies $\phi$ by $-1$ on the $B$-sites, then $VTV = -T$. It
is possible that $T$ has zero-modes; indeed if $\vert A \vert > \vert B
\vert$ then $T$ will have at least $\vert A \vert - \vert B \vert$
zero-modes and all these $\lambda = 0$ eigenvectors have the form $\phi
= (f,0)$.
Suppose now that we have a half-filled band, i.e., $N = \vert \Lambda
\vert$ electrons. By virtue of the two spin states for each electron
we have that the ground state energy of $H_0 = \sum\nolimits_{x,y}
t_{xy} c^\dagger_x c_y$ is $E = 2 \sum\nolimits_{\lambda < 0}
\lambda$. (Note: If $\vert \Lambda \vert$ is odd there is at least
one zero-mode, and so this formula is correct even in this case.) \ The
ground state of $H_0$ might be degenerate, however (because of
zero-modes). We {\it define} the density matrix for spin $\sigma$ in
the ground state to be $$\uprho_\sigma (x,y) = \sum \limits_{\lambda <
0} \phi^\lambda (x) \phi^\lambda (y)^* + \mfr1/2 \sum \limits_{\lambda
= 0} \phi^\lambda (x) \phi^\lambda (y)^*. \eqno(1)$$ We see that $\Tr
\ \uprho_\sigma = \sum \nolimits_x \uprho_\sigma (x,x) = \vert \Lambda
\vert /2$, as it should, and that $\uprho_\sigma$ agrees with the
$\beta \rightarrow \infty$ limit of the positive temperature density
matrix, defined in the {\it grand canonical} ensemble by
$$\uprho_{\beta \sigma} (x,y) = \sum \limits_\lambda \phi^\lambda (x)
\phi^\lambda (y)^* e^{-\beta \lambda}/(1 + e^{-\beta \lambda}).
\eqno(2)$$ Note that we have used zero chemical potential which, by
virtue of the $\lambda, - \lambda$ symmetry, always yields $\vert
\Lambda \vert$ as the average particle number. If there is more than
one zero-mode the ground state, and the $\uprho_\sigma$ in the ground
state, is not unique. Eq. (1) serves to fix $\uprho_\sigma$ for our
purposes. The Gibbs state is always unique for a finite volume.
Another Gibbs state with which we shall be concerned is the {\it
canonical} ensemble. The density matrix here will be denoted by
$\widetilde {\uprho}_{\beta \sigma} (x,y)$. Its definition is well
known and we shall not write it explicitly for $H_0$, but we note that
the $\beta \rightarrow \infty$ limit of $\widetilde{\uprho}_{\beta
\sigma}$ also equals $\uprho_\sigma$.
The following simple observation about (1) is the starting point of our
further analysis. If $x\in A$ and $y \in A$ then, using the fact that
$\phi^\lambda (x) = \phi^{-\lambda} (x)$, we have that $$\uprho_\sigma
(x,y) = \mfr1/2 \sum \limits_{{\rm all} \ \lambda} \phi^\lambda (x)
\phi^\lambda (y)^* = \mfr1/2 \delta_{x,y}$$ since the $\phi^\lambda$'s
form an orthonormal basis. A similar remark holds for $x,y \in B$.
Thus, $\uprho_\sigma (x,x) = \mfr1/2$ for all $x \in \Lambda$ and
$\uprho_\sigma (x,y) = 0$ for $x,y \in A$ or $x,y \in B$. (This result
is a slight improvement of an earlier theorem announced in [1] because
the zero-mode sum in eq. (1) is explicit here.) As we shall see from
the following general theorem (by specializing to zero interaction) the
same conclusion applies to $\uprho_{\beta \sigma} (x,y)$ and
$\widetilde{\uprho}_{\beta \sigma} (x,y)$.
The interacting system we shall be concerned with is the {\it
generalized} Hubbard model defined by the Hamiltonian (with spin
dependent hopping) $$H = \sum\limits_\sigma \sum \limits_{x,y \in
\Lambda} t^{\phantom{\dagger}}_{xy\sigma} c^\dagger_{x\sigma}
c^{\phantom{\dagger}}_{y \sigma} + \sum \limits_{\sigma,\tau} \sum
\limits_{x,y \in \Lambda} U_{xy\sigma\tau} (2n_{x \sigma} -1)
(2n_{y\tau} -1), \eqno(3)$$ where $n_{x \sigma} = c^\dagger_{x\sigma}
c^{\phantom{\dagger}}_{x \sigma}$ and with $U_{xy\sigma\tau}$ real (but
not necessarily of one sign). $T_\sigma = \{ t_{xy\sigma}\}$ is
hermitian and bipartite for each $\sigma = \uparrow$ or $\downarrow$.
If we take $T_\uparrow = T_\downarrow$ and $U_{xy\sigma \tau} = U
\delta_{xy}$ then $H$ is the usual Hubbard Hamiltonian (apart from a
trivial {\it additive} constant) with interaction $4U \sum
n_{x\uparrow} n_{x \downarrow}$. The noninteracting case, $H_0$,
corresponds to $U_{xy \sigma\tau} = 0$. In general, the total spin
angular momentum ($SU(2)$ symmetry) will be conserved if we require
$t_{xy\sigma}$ and $U_{xy\sigma\tau}$ to be independent of the spin
labels $\sigma$ and $\tau$; for our purposes we do not require this
$SU(2)$ invariance.
The positive temperature, grand canonical density matrix $\uprho_{\beta
\sigma}$ is defined to be $$\uprho_{\beta \sigma} (x,y) = \Tr
[c^\dagger_{x \sigma} c^{\phantom{\dagger}}_{y\sigma} e^{-\beta
H}]/\Tr[e^{-\beta H}]. \eqno(4)$$ Formula (4) reduces to (2) for the
noninteracting case. The trace is over the full Fock space containing
all particle numbers ranging from 0 to $2 \vert \Lambda \vert$. Again,
the zero chemical potential in (4) insures that $\Tr \ \uprho_{\beta
\sigma} = \vert \Lambda \vert /2$.
The canonical density matrix $\widetilde{\uprho}_{\beta \sigma} (x,y)$
for this model is also given by (4), but where the trace is only over
the $N$-particle sector (note that both $H$ and $c^\dagger_{x \sigma}
c^{\phantom{\dagger}}_{y \sigma}$ leave this sector invariant). The
half-filled band is defined by $N = \vert \Lambda \vert$. Since $(\Tr
ABC)^* = \Tr \ C^\dagger B^\dagger A^\dagger = \Tr \ B^\dagger
A^\dagger C^\dagger$ we see that $\uprho_{\beta \sigma}$ and
$\widetilde{\uprho}_{\beta \sigma}$ are hermitian matrices for each
$\sigma$.
{\bf THEOREM:} {\it The canonical and the grand canonical density
matrices satisfy: $$\eqalignno{\widetilde{\uprho}_{\beta\sigma} (x,x)
= \uprho_{\beta \sigma} (x,x) = 1/2 \quad &\hbox{for all} \ x \in
\Lambda \qquad&(5)\cr \widetilde{\uprho}_{\beta \sigma} (x,y) =
\uprho_{\beta \sigma} (x,y) = 0\phantom{/2} \quad &\hbox{if} \ x,y \in
A \ \hbox{or} \ x,y \in B. \qquad&(6)\cr}$$}
{\it Proof:} The proof for $\widetilde{\uprho}_{\beta\sigma}$ will be
the same as that for $\uprho_{\beta \sigma}$ so we shall only give the
proof for the latter.
First, we consider real $T$. The following version of the
hole-particle unitary transformation, $$c^{\phantom{\dagger}}_{x\sigma}
\leftrightarrow c^\dagger_{x \sigma} \ {\rm for} \ x \in A, \qquad
c^{\phantom{\dagger}}_{x \sigma} \leftrightarrow -c^\dagger_{x \sigma}
\ {\rm for} \ x \in B, \eqno(7)$$ evidently leaves the Hamiltonian $H$
and the relevant Hilbert spaces invariant. If this unitary
transformation is denoted by $W$ we have that $W^2 = 1$, $W =
W^\dagger$, $WHW = H$ and hence $\Tr [e^{-\beta H}]
\uprho_{\beta\sigma} (x,y) = \Tr [W c^\dagger_{x\sigma}
c^{\phantom{\dagger}}_{y\sigma} WW e^{-\beta H} W] = \Tr [W
c^\dagger_{x\sigma} W) (Wc_{y\sigma}W) e^{-\beta H}]$. If $x,y \in A$
we can use (7) and the fermion commutation rule to conclude that
$\uprho_{\beta \sigma} (x,y) = \delta_{x,y} - \uprho_{\beta \sigma}
(y,x)$. The same is true if $x,y \in B$. If $T$ is real,
$\uprho_{\beta\sigma}$ is evidently real; since $\uprho_{\beta \sigma}$
is also hermitian the theorem is proved in the real case.
The complex case is a bit subtle. The Hamiltonian $H$ is no longer
invariant under the hole-particle transformation $W$, but it is
invariant under the {\it antiunitary} transformation $Y = JW$, in which
$J$ is complex conjugation. More precisely, any vector $\Psi$ in our
Hilbert space can be written as a linear combination, with complex
coefficients, of the basis vectors consisting of monomials in the
$c^\dagger_{x\sigma}$'s applied to the vacuum. The antiunitary map $J$
acts on $\Psi$ by replacing each coefficient by its complex conjugate.
We note that $JW = WJ$ and therefore $Y^2 = \1$. It is also easy to
see that $YHY = H$ and that $Y c_{x\sigma} Y = W c_{x\sigma} W$, which
is given by (7).
Now suppose that $K$ is an arbitrary linear operator, and consider $L
\equiv JKJ$. Although $J$ is nonlinear, it is easy to check that $L$
is linear. In fact the matrix elements of $L$ in the above mentioned
basis are simply the complex conjugates of the corresponding elements
of $K$. Therefore, even though $J^2 = 1$, it is {\it not} generally
true that $\Tr L = \Tr JKJ = \Tr KJ^2 = \Tr K$. What is true is that
$\Tr L = (\Tr K)^*$.
In our case we have, for $x,y \in A$ or $x,y \in B$, \ $\{ \Tr
e^{-\beta H} \} \uprho_{\beta\sigma} (x,y)^* = \Tr [JW
c^\dagger_{x\sigma} c^{\phantom{\dagger}}_{y\sigma} e^{-\beta H} WJ]
\hfill\break = \Tr [(Y c^\dagger_{x\sigma} Y)
(Yc^{\phantom{\dagger}}_{y\sigma}Y) (Ye^{-\beta H} Y)] = \Tr
[c^{\phantom{\dagger}}_{x\sigma} c^\dagger_{y\sigma} e^{-\beta H}] = \{
\Tr e^{-\beta H} \} \{\delta_{xy} - \uprho_{\beta\sigma} (y,x)\}$. The
hermiticity of $\uprho_{\beta\sigma}$ now implies the theorem. QED.
It is worth noting that only the invariance of $H$ under $Y$ and the
bipartite structure of the lattice have been used. Thus the theorem
(with some obvious modifications) applies to the case where the kinetic
energy has spin-flip terms and is given by $\sum
\limits_{x,y,\sigma,\tau} t^{\phantom{\dagger}}_{xy\sigma\tau}
c^\dagger_{x\sigma} c^{\phantom{\dagger}}_{y\tau}$ with
$t_{xy\sigma\tau} = \overline{t_{yx\tau\sigma}}$ and where, for fixed
$\sigma$ and $\tau$, $t_{xy\sigma\tau}$ is a hopping matrix on a
bipartite graph. $\uprho_{\beta\sigma} (x,y)$ is replaced by
$\uprho_{\beta \sigma \tau} (x,y) = \uprho_{\beta \tau \sigma}
(y,x)^*$, and the extended theorem states that it equals $\mfr1/2
\delta_{xy} \delta_{\sigma\tau}$ when $x,y$ are on the same
sublattice. In the interest of simplicity of notation and exposition
we have relegated this generalization --- relevant to some physical
models --- to the remark here.
We conclude with some additional remarks about some limiting cases of
the Theorem.
{\bf I. The Falicov-Kimball model [2]:} In this model there is one
species of mobile, spinless electrons and one species of arbitrarily
fixed particles. It is the same as the Hubbard model with the choice
$t_{xy\downarrow} \equiv 0$ and $t_{xy \uparrow} = t_{xy}$. The
theorem applies to this model as a special case. Note that it says
that {\it both} the mobile and immobile particles have density 1/2.
{\bf II. Ground States:} If we define the ground state
$\uprho_\sigma$ as the limit $\beta \rightarrow \infty$ of the
canonical $\widetilde{\uprho}_{\beta\sigma}$, then (5) and (6) apply
there, too. (Note: We do not have a proof --- or even know if it is
true --- that the $\beta \rightarrow \infty$ limit of the canonical
ensemble is the same as for the grand canonical ensemble.) In general
the ground state is not unique, however, and there are other
possibilities for $\uprho_\sigma$, in which case (5) and (6) apply to
states that are invariant under $Y = WJ$. Not every ground state is
$Y$ invariant. In the case of the usual Hubbard model on a connected
lattice with $U > 0$ and real $Y$ it is known [3] that the ground state
has spin angular momentum $S = (\vert A \vert - \vert B \vert)/2$ and
it is unique apart from the $(2S+1)$-fold degeneracy associated with
$S^z = \mfr1/2 (N_\uparrow - N_\downarrow) \in \{ -S, -S+1, \dots ,
+S\}$. Ground states that are $Y$ invariant are mixtures of states
with $S^z$ and $-S^z$, i.e., $\mfr1/2 \vert S^z \rangle \langle S^z
\vert + \mfr1/2 \vert -\!S^z \rangle \langle -S^z \vert$ in Dirac's
notation. The reason we can be sure that $Y \vert S^z \rangle = \vert
-\!S^z \rangle$ is this: $Y \vert S^z \rangle$ is --- in any event ---
a state with $\mfr1/2 (N_\uparrow - N_\downarrow) = -S^z$. It is also
a ground state. By uniqueness, this state must be $\vert -\!S^z
\rangle$. In the general model (3), we cannot be sure that spin flip
$=$ hole-particle transformation.
{\bf III. Infinite Volume States:} These states can, of course, be
different from finite volume states. One way to define them is as
limits of finite volume states with specified boundary conditions that
need not respect $Y$ symmetry. One example concerns the
Falicov-Kimball model with $U > 0$ on a hypercubic lattice in $D$
dimensions: For $D \geq 2$ it has long-range order in the ground state
and at low temperatures, in which the up spins preferentially occupy
the $A$-sites and the down spins the $B$-sites (or vice versa) [4,5].
Similarly, the usual Hubbard model is expected to show the same
behavior when $D \geq 3$ if $U > 0$, at least if $U > 0$ is large
enough. This has not been proved, however. These examples violate
(5), but they do suggest that the charge density satisfies
$\uprho_{\beta \uparrow} (x,x) + \uprho_{\beta \downarrow} (x,x) = 1$
in these models. If we now introduce nearest neighbor repulsion (which
is allowed in our general model) then even this constancy of the charge
density might be violated, however.
What is significant about our finite volume theorem (5), (6) --- and
which does remain true in the infinite volume limit --- is that for
every state with non-constant (spin or charge) density there is another
equally good state with the complementary density. In other words, one
cannot invent a non-translationally invariant Hamiltonian (either by
altering the hopping matrix or the potential energy) with the property
that it {\it forces} the density to increase in some specified regions
and to decrease in others --- even though one might have thought {\it
a-priori} that the density can be controlled by the hopping or
potential energy. Any attempt to cause a non-constant density will
always result in the certainty that {\it exactly} the reverse of the
desired non-constancy will occur with the hole-particle reversed
boundary condition.
Partial support by U.S. National Science Foundation grants
PHY90-19433A01 (EHL) and DMS92-07703 (ML) is gratefully acknowledged,
as is a fruitful discussion with Daniel Fisher.
\bigskip\noindent
{\bf REFERENCES}
\item{[1]} E.H. Lieb, Helv. Phys. Acta {\bf 65}, 247 (1992). See Theorem
4.
\item{[2]} L.M. Falicov and J.C. Kimball, Phys. Rev. Lett. {\bf 22}, 997
(1969).
\item{[3]} E.H. Lieb, Phys. Rev. Lett. {\bf 62}, 1201 (1989). Errata {\bf
62}, 1927 (1989). For an extension to positive temperature see K. Kubo and
T. Kishi, Phys. Rev. {\bf 62}, 1201 (1989).
\item{[4]} T. Kennedy and E.H. Lieb, Physica {\bf A138}, 320 (1986).
\item{[5]} U. Brandt and R. Schmidt, Z. Phys. {\bf B67}, 43 (1986).
\bye