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\begin{document}
\title[Entangled Chains]{Entangled Chains}
\author{William K. Wootters}
\address[William K. Wootters]{Department of Physics, Williams College \\
Williamstown, MA 01267, USA}
\email[William K. Wootters]{William.K.Wootters@williams.edu}
\urladdr{http://www.williams.edu:803/Physics/wwootters}
\date{July 27, 2001}
\subjclass[2000]{Primary 81Q99; Secondary 81P68, 81V70}
\keywords{Quantum mechanics, entanglement.}
\copyrightinfo{2002}% % copyright year
{American Mathematical Society}% copyright holder
\begin{abstract}
Consider an infinite collection of qubits arranged in a line, such that every
pair of nearest neighbors is entangled: an ``entangled chain.'' In this paper
we consider entangled chains with translational invariance and ask how large
one can make the nearest neighbor entanglement. We find that it is possible to
achieve an entanglement of formation equal to 0.285 ebits between each pair of
nearest neighbors, and that this is the best one can do under certain
assumptions about the state of the chain.
\end{abstract}
\maketitle
\section{Introduction: Example of an entangled chain}
Quantum entanglement has been studied for decades, first because of its
importance in the foundations of quantum mechanics, and more recently for its
potential technological applications as exemplified by a quantum computer
\cite{DiVincenzo}. The new focus has led to a quantitative theory of
entanglement \cite{BBPS,Vedral,BDSW} that, among other things, allows us to
express analytically the degree of entanglement between simple systems
\cite{HillWootters}. This development makes it possible to pose new
quantitative questions about entanglement that could not have been raised
before and that promise fresh perspectives on this remarkable phenomenon. In
this paper I would like to raise and partially answer such a question,
concerning the extent to which a collection of binary quantum objects (qubits)
can be linked to each other by entanglement.
Imagine an infinite string of qubits, such as two-level atoms or the spins of
spin-1/2 particles. Let us label the locations of the qubits with an integer
$j$ that runs from negative infinity to positive infinity. I wish to consider
special states of the string, satisfying the following two conditions: (i)
each qubit is entangled with its nearest neighbors; (ii) the state is
invariant under all translations, that is, under transformations that shift
each qubit from its original position $j$ to position $j+n$ for some integer
$n$. Let us call a string of qubits satisfying the first condition an
entangled chain, and if it also satisfies the second condition, a uniform
entangled chain. Note that each qubit need not be entangled with any qubits
other than its two nearest neighbors. In this respect an entangled chain is
like an ordinary chain, whose links are directly connected only to two
neighboring links. By virtue of the translational invariance, the degree of
entanglement between nearest neighbors in a uniform entangled chain must be
constant throughout the chain. The main question I wish to pose is this: How
large can the nearest-neighbor entanglement be in a uniform entangled chain?
This problem belongs to a more general line of inquiry about how entanglement
can be shared among more than two objects. Some work on this subject has been
done in the context of the cloning of entanglement [\textbf{6--11}], where one
finds limits on the extent to which entanglement can be copied. In a different
setting not particularly involving cloning, one finds an inequality bounding
the amount of entanglement that a single qubit can have with each of two other
qubits \cite{Coffman}. One can imagine more general ``laws of entanglement
sharing'' that apply to a broad range of configurations of quantum objects.
The present work provides further data that might be used to discover and
formulate such laws. The specific problem addressed in this paper could also
prove relevant for analyzing models of quantum computers in which qubits are
arranged along a line, as in an ion trap \cite{Zoller}. The infinite chain can
be thought of as an idealization of such a computer. Moreover, the analysis of
our question turns out to be interesting in its own right, being related, as
we will see, to a familiar problem in many-body physics.
To make the question precise we need a measure of entanglement between two
qubits. We will use a reasonably simple and well-justified measure called the
``concurrence,'' which is defined as follows \cite{HillWootters}.
Consider first the case of pure states. A general pure state of two qubits can
be written as
\begin{equation}
|\psi\rangle=\alpha|00\rangle+\beta|01\rangle+\gamma|10\rangle+\delta
|11\rangle.
\end{equation}
One can verify that such a state is factorizable into single-qubit
states---that is, it is unentangled---if and only if $\alpha\delta=\beta
\gamma$. The quantity $C=2\left| \alpha\delta-\beta\gamma\right| $, which
ranges from 0 to 1, is thus a plausible measure of the degree of entanglement.
We take this expression as the definition of concurrence for a pure state of
two qubits. For mixed states, we define the concurrence to be the greatest
convex function on the set of density matrices that gives the correct values
for pure states \cite{Uhlmann}.
Though this statement defines concurrence, it does not tell us how to compute
it for mixed states. Remarkably, there exists an explicit formula for the
concurrence of an arbitrary mixed state of two qubits \cite{HillWootters}: Let
$\rho$ be the density matrix of the mixed state, which we imagine expressed in
the standard basis $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$. Let
$\tilde{\rho}$, the ``spin-flipped'' density matrix, be
\hbox{$(\sigma_y\otimes\sigma_y)\rho^*
(\sigma_y\otimes \sigma_y)$} , where
the asterisk denotes complex conjugation in the standard basis and $\sigma
_{y}$ is the matrix $\left(
\begin{array}
[c]{cr}%
0 & -i\\
i & 0
\end{array}
\right) $. Finally, let $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}$ be
the square roots of the eigenvalues of $\rho\tilde{\rho}$ in descending
order---one can show that these eigenvalues are all real and non-negative.
Then the concurrence of $\rho$ is given by the formula
\begin{equation}
C(\rho)=\mathrm{max}\{\lambda_{1}-\lambda_{2}-\lambda_{3}-\lambda_{4},0\}.
\end{equation}
The best justification for using concurrence as a measure of entanglement
comes from a theorem \cite{HillWootters} showing that concurrence is a
monotonically increasing function of the ``entanglement of formation,'' which
quantifies the non-local resources needed to create the given state
\cite{BDSW}.\footnote{One can define the entanglement of formation as follows.
Let $\rho$ be a mixed state of a pair of quantum objects, to be shared between
two separated observers who can communicate with each other only via classical
signals. The entanglement of formation of $\rho$ is the asymptotic number of
singlet states the observers need, per pair, in order to create a large number
of pairs in \emph{pure} states whose average density matrix is $\rho$. (This
is conceptually different from the \emph{regularized} entanglement of
formation, which measures the cost of creating many copies of the \emph{mixed}
state $\rho$ \cite{Terhal}. However, it is conceivable that the two quantities
are identical.) Entanglement of formation is conventionally measured in
``ebits,'' and for a pair of binary quantum objects it takes values ranging
from 0 to 1 ebit.} As mentioned above, the values of $C$ range from zero to
one: an unentangled state has $C=0$, and a completely entangled state such as
the singlet state $\frac{1}{\sqrt{2}}(|01\rangle-|10\rangle)$ has $C=1$. Our
problem is to find the greatest possible nearest-neighbor concurrence of a
uniform entangled chain. At the end of the calculation we can easily
re-express our results in terms of entanglement of formation.
Another issue that needs to be addressed in formulating our question is the
meaning of the word ``state'' as applied to an infinite string of qubits; in
particular we need to discuss how such a state is to be normalized. Formally,
we can define a state of our system as follows. A state $w$ of the infinite
string is a function that assigns to every finite set $S$ of integers a
normalized (\emph{i.e.}, trace one) density matrix $w(S)$, which we interpret
to be the density matrix of the qubits specified by the set $S$; moreover the
function $w$ must be such that if $S_{2}$ is a subset of $S_{1}$, then
$w(S_{2})$ is obtained from $w(S_{1})$ by tracing over the qubits whose labels
are not in $S_{2}$. This formal definition is perfectly sensible but somewhat
bulky in practice. In what follows we will usually specify states of the
string more informally when it is clear from the informal specification how to
generate the density matrix of any finite subset of the string. We will also
usually use the symbol $\rho$ instead of $w(S)$ to denote the density matrix
of a pair of nearest neighbors.
It is not immediately obvious that there exists even a single example of an
entangled chain. Note, for example, that the limit of a Schr\"{o}dinger cat
state---an equal superposition of an infinite string of zeros with an infinite
string of ones---is not an entangled chain. In the cat state, the reduced
density matrix of a pair of neighboring qubits is an incoherent mixture of
$|00\rangle$ and $|11\rangle$, which exhibits a classical correlation but no
entanglement. (Note, by the way, that our informal statement ``an equal
\emph{superposition} of an infinite string of zeros with an infinite string of
ones,'' specifies exactly the same state as if we had taken an incoherent
\emph{mixture} of these two infinite strings: no finite set of qubits contains
information about the phase of the superposition.)
We can, however, construct a simple example of an entangled chain in the
following way. Let $w_{0}$ be the state such that for each \emph{even} integer
$j$, the qubits at sites $j$ and $j+1$ are entangled with each other in a
singlet state. We can write this state informally as\footnote{Alternatively,
we can characterize the state $w_{0}$ according to our formal definition by
specifying the density matrix of each finite collection of qubits: Let $S$
define such a collection. Then for each even integer $j$ such that both $j$
and $j+1$ are in $S$, the corresponding pair of qubits is in the singlet
state; all other qubits (\emph{i.e.}, the unpaired ones) are in the completely
mixed state $\left(
\begin{array}
[c]{cc}%
\frac{1}{2} & 0\\
0 & \frac{1}{2}%
\end{array}
\right) $, and the full density matrix $w(S)$ is obtained by taking the
tensor product of the pair states and single-qubit states.}
\begin{equation}
\cdots\otimes\left( \frac{|0\rangle_{-2}|1\rangle_{-1}-|1\rangle
_{-2}|0\rangle_{-1}}{\sqrt{2}}\right) \otimes\left( \frac{|0\rangle
_{0}|1\rangle_{1}-|1\rangle_{0}|0\rangle_{1}}{\sqrt{2}}\right) \otimes\left(
\frac{|0\rangle_{2}|1\rangle_{3}-|1\rangle_{2}|0\rangle_{3}}{\sqrt{2}}\right)
\otimes\cdots
\end{equation}
The state $w_{0}$ is not an entangled chain because the qubits are not
entangled with both of their nearest neighbors: qubits at even-numbered
locations are not entangled with their neighbors on the left. However, if we
let $w_{1}$ be the state obtained by translating $w_{0}$ one unit to the left
(or to the right---the result is the same), and let $w$ be an equal mixture of
$w_{0}$ and $w_{1}$---that is, $w=(w_{0}+w_{1})/2$---then $w$ is a uniform
entangled chain, as we now show.
That $w$ is translationally invariant follows from the fact that both $w_{0}$
and $w_{1}$ are invariant under \emph{even} displacements and that they
transform into each other under odd displacements. Thus we need only show that
neighboring states are entangled. For definiteness let us consider the qubits
in locations $j=1$ and $j=2$. In the state $w_{0}$, the density matrix for
these two qubits is
\begin{equation}
\rho^{(0)}=\left(
\begin{array}
[c]{cccc}%
\frac{1}{4} & 0 & 0 & 0\\
0 & \frac{1}{4} & 0 & 0\\
0 & 0 & \frac{1}{4} & 0\\
0 & 0 & 0 & \frac{1}{4}%
\end{array}
\right) ,
\end{equation}
that is, the completely mixed state. (The two qubits are from distinct singlet
pairs.) The density matrix of the same two qubits in the state $w_{1} $ is
\begin{equation}
\rho^{(1)}=\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & 0\\
0 & \frac{1}{2} & \frac{1}{2} & 0\\
0 & \frac{1}{2} & \frac{1}{2} & 0\\
0 & 0 & 0 & 0
\end{array}
\right) ,
\end{equation}
that is, the singlet state. In the state $w$, the qubits are in an equal
mixture of these two density matrices, which is
\begin{equation}
\rho=(\rho^{(0)}+\rho^{(1)})/2=\left(
\begin{array}
[c]{cccc}%
\frac{1}{8} & 0 & 0 & 0\\
0 & \frac{3}{8} & \frac{1}{4} & 0\\
0 & \frac{1}{4} & \frac{3}{8} & 0\\
0 & 0 & 0 & \frac{1}{8}%
\end{array}
\right) . \label{bike}%
\end{equation}
It is easy to compute the concurrence of this density matrix, because
$\tilde{\rho}$ is the same as $\rho$ itself. The values $\lambda_{i}$ in this
case are the eigenvalues of $\rho$, which are $\frac{5}{8},\frac{1}%
{8},\frac{1}{8},\frac{1}{8}$. The concurrence is therefore $C=\frac{5}%
{8}-\frac{1}{8}-\frac{1}{8}-\frac{1}{8}=\frac{1}{4}$. This same value of the
concurrence applies to any other pair of neighboring qubits in the string
because of the translational invariance. The fact that the concurrence is
non-zero implies that neighboring qubits are entangled, so that the state $w$
is indeed an entangled chain. For uniform entangled chains, we will call the
common value of $C$ for neighboring qubits the concurrence of the chain. Thus
in the above example the concurrence of the chain is $\frac{1}{4}$.
As we will see, it is possible to find uniform entangled chains with greater
concurrence. Let $C_{\mathrm{max}}$ be the least upper bound on the
concurrences of all uniform entangled chains. We would like to find this
number. We know that $C_{\mathrm{max}}$ is no larger than 1, since concurrence
never exceeds 1. In fact we can quickly get a somewhat better upper bound,
using the following fact: when a qubit is entangled with each of two other
qubits, the sum of the squares of the two concurrences is less than or equal
to one \cite{Coffman}. In a uniform entangled chain, each qubit must be
equally entangled with its two nearest neighbors; so the concurrence with each
of them cannot exceed $1/\sqrt{2}$. Thus, so far what we know about
$C_{\mathrm{max}}$ is this:
\begin{equation}
1/4 \leq C_{\mathrm{max}} \leq1/\sqrt{2}.
\end{equation}
This is still a wide range. Most of the rest of this paper is devoted to
getting a better fix on $C_{\mathrm{max}}$ by explicitly constructing
entangled chains.
\section{Building chains out of blocks}
Using the above example as a model, we will use the following construction to
generate other uniform entangled chains. (1) Break the string into blocks of
$n$ qubits, and define a state $w_{0}$ in which each block is in the same $n
$-qubit state $|\xi\rangle$; that is, $w_{0}$ is a tensor product of an
infinite number of copies of $|\xi\rangle$. (In the above example $n$ had the
value 2 and $|\xi\rangle$ was the singlet state.) (2) Define $w_{k}$,
$k=1,\ldots,n-1$, to be the state obtained by shifting $w_{0}$ to the left by
$k $ units. (3) Let the final state $w$ be the average $(w_{0}+\cdots
+w_{n-1})/n$. A state generated in this way will automatically be
translationally invariant. In order that the chain have a large concurrence,
we will need to choose the state $|\xi\rangle$ carefully. Finding an optimal
$|\xi\rangle$ and proving that it is optimal may turn out to be a difficult
problem. In this paper I will choose $|\xi\rangle$ according to a strategy
that makes sense and may well be optimal but is not proven to be so.
In the final state $w$, each pair of neighboring qubits has the same density
matrix because of the translational invariance. Our basic strategy for
choosing $|\xi\rangle$, described below, is designed to give this
neighboring-pair density matrix the following form:
\begin{equation}
\rho=\left(
\begin{array}
[c]{cccc}%
\rho_{11} & 0 & 0 & 0\\
0 & \rho_{22} & \rho_{23} & 0\\
0 & \rho_{32} & \rho_{33} & 0\\
0 & 0 & 0 & 0
\end{array}
\right) . \label{form}%
\end{equation}
(The ordering of the four basis states is the one given above: $|00\rangle
,|01\rangle,|10\rangle,|11\rangle$.) One can show that the concurrence of such
a density matrix is simply
\begin{equation}
C=2\big| \rho_{23}\big| . \label{simpleconc}%
\end{equation}
Besides making the concurrence easy to compute, the form (\ref{form}) seems a
reasonable goal because it picks out a specific kind of entanglement, namely,
a coherent superposition of $|01\rangle$ and $|10\rangle$, and limits the ways
in which this entanglement can be contaminated or diluted by being mixed with
other states. In particular, the form (\ref{form}) does not allow
contamination by an orthogonal entangled state of the form $\alpha
|00\rangle+\beta|11\rangle$---orthogonal entangled states when mixed together
tend to cancel each other's entanglement---or by the combination of the two
unentangled states $|00\rangle$ and $|11\rangle$. If the component $\rho_{44}$
were not equal to zero and the form were otherwise unchanged, the concurrence
would be $C=\mathrm{max}\{2(|\rho_{23}|-\sqrt{\rho_{11}\rho_{44}}),0\}$; so it
is good to make either $\rho_{11}$ or $\rho_{44}$ equal to zero if this can be
done without significantly reducing $\rho_{23}$. We have chosen to make
$\rho_{44}$ equal to zero.
As it happens, one can guarantee the form (\ref{form}) for the density matrix
of neighboring qubits by imposing the following three conditions on the
$n$-qubit state $|\xi\rangle$: (i) $|\xi\rangle$ is an eigenstate of the
operator that counts the number of qubits in the state $|1\rangle$. That is,
each basis state represented in $|\xi\rangle$ must have the same number $p$ of
qubits in the state $|1\rangle$. (ii) $|\xi\rangle$ has no component in which
two neighboring qubits are both in the state $|1\rangle$. (iii) The $n$th
qubit is in the state $|0\rangle$. (This last condition effectively extends
condition (ii) to the boundary between successive blocks.) Condition (i)
guarantees that the density matrix $\rho$ for a pair of nearest neighbors is
block diagonal, each block corresponding to a fixed number of 1's in the pair.
That is, there are two single-element blocks corresponding to $|00\rangle$ and
$|11\rangle$, and a 2x2 block corresponding to $|01\rangle$ and $|10\rangle$.
Conditions (ii) and (iii) guarantee that $\rho_{44}$ is zero. The conditions
thus give us the form (\ref{form}). We impose these three conditions because
they seem likely to give the best results; we do not prove that they are optimal.
To illustrate the three conditions and how they can be used, let us consider
in detail the case where the block size $n$ is 5 and the number $p$ of 1's in
each block is 2. (Our strategy does not specify the value of either $n$ or
$p$; these values will ultimately have to be determined by explicit
maximization.) In this case, the only basis states our conditions allow in the
construction of $|\xi\rangle$ are $|10100\rangle$, $|10010\rangle$, and
$|01010\rangle$. Any other basis state either would have a different number of
1's or would violate one of conditions (ii) and (iii). Thus we write
\begin{equation}
|\xi\rangle= a_{13}|10100\rangle+ a_{14}|10010\rangle+ a_{24}|01010\rangle.
\label{xi}%
\end{equation}
The subscripts in $a_{ij}$ indicate which qubits are in the state $|1\rangle$.
The state $w$ of the infinite string is derived from $|\xi\rangle$ as
described above. We now want to use Eq.~(\ref{xi}) to write the density matrix
$\rho$ of a pair of nearest neighbors when the infinite string is in the state
$w$. For definiteness let us take the two qubits of interest to be in
locations $j=1$ and $j=2$, and let us take the 5-qubit blocks in the state
$w_{0}$ to be given by $j=1,\ldots,5$, $j=6,\ldots,10$, and so on. Our final
density matrix $\rho$ will be an equal mixture of five density matrices,
corresponding to the five different displacements of $w_{0}$ (including the
null displacement).
For $w_{0}$ itself, the qubits at $j=1$ and $j=2$ are the first two qubits of
$|\xi\rangle$. The density matrix for these two qubits, obtained by tracing
out the other three qubits of the block, is
\begin{equation}
\rho^{(0)}=\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & 0\\
0 & \left| a_{24}\right| ^{2} & \overline{a}_{14}a_{24} & 0\\
0 & a_{14}\overline{a}_{24} & \left| a_{13}\right| ^{2}+\left|
a_{14}\right| ^{2} & 0\\
0 & 0 & 0 & 0
\end{array}
\right) .
\end{equation}
For $w_{1}$, the qubits at $j=1$ and $j=2$ are now the second and third qubits
of the block, since the block has been shifted to the left. Thus we trace over
the first, fourth, and fifth qubits to obtain
\begin{equation}
\rho^{(1)}=\left(
\begin{array}
[c]{cccc}%
\left| a_{14}\right| ^{2} & 0 & 0 & 0\\
0 & \left| a_{13}\right| ^{2} & 0 & 0\\
0 & 0 & \left| a_{24}\right| ^{2} & 0\\
0 & 0 & 0 & 0
\end{array}
\right) .
\end{equation}
In a similar way one can find $\rho^{(2)}$ and $\rho^{(3)}$:
\[
\rho^{(2)}=\left(
\begin{array}
[c]{cccc}%
0 & 0 & 0 & 0\\
0 & \left| a_{14}\right| ^{2}+\left| a_{24}\right| ^{2} & \overline
{a}_{13}a_{14} & 0\\
0 & a_{13}\overline{a}_{14} & \left| a_{13}\right| ^{2} & 0\\
0 & 0 & 0 & 0
\end{array}
\right) ;\rho^{(3)}=\left(
\begin{array}
[c]{cccc}%
\left| a_{13}\right| ^{2} & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & \left| a_{14}\right| ^{2}+\left| a_{24}\right| ^{2} & 0\\
0 & 0 & 0 & 0
\end{array}
\right) .
\]
The density matrix corresponding to $w_{4}$ is different in that the two
relevant qubits now come from different blocks: the qubit at $j=1$ is the last
qubit of one block and the qubit at $j=2$ is the first qubit of the next
block. The corresponding density matrix is thus the tensor product of two
single-qubit states:
\[
\rho^{(4)}=\left(
\begin{array}
[c]{cc}%
1 & 0\\
0 & 0
\end{array}
\right) \otimes\left(
\begin{array}
[c]{cc}%
\left| a_{24}\right| ^{2} & 0\\
0 & \left| a_{13}\right| ^{2}+\left| a_{14}\right| ^{2}%
\end{array}
\right) =\left(
\begin{array}
[c]{cccc}%
\left| a_{24}\right| ^{2} & 0 & 0 & 0\\
0 & \left| a_{13}\right| ^{2}+\left| a_{14}\right| ^{2} & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\right) .
\]
To get the neighboring-pair density matrix corresponding to our final state
$w$, we average the above five density matrices, with the following simple
result:
\begin{equation}
\rho=\frac{1}{5}\left(
\begin{array}
[c]{cccc}%
1 & 0 & 0 & 0\\
0 & 2 & x & 0\\
0 & \overline{x} & 2 & 0\\
0 & 0 & 0 & 0
\end{array}
\right) ,
\end{equation}
where
\begin{equation}
x=\overline{{a}}_{13}a_{14}+\overline{{a}}_{14}a_{24}. \label{examp}%
\end{equation}
According to Eq.~(\ref{simpleconc}), the concurrence of the pair is
\begin{equation}
C=\frac{2}{5}\big| \overline{{a}}_{13}a_{14}+\overline{{a}}_{14}a_{24}\big|
. \label{absoluteC}%
\end{equation}
Continuing with this example---$n=5$ and $p=2$---let us find out what values
we should choose for $a_{13}$, $a_{14}$, and $a_{24}$ in order to maximize $C
$. First, it is clear that we cannot go wrong by taking each $a_{ij}$ to be
real and non-negative---any complex phases could only reduce the absolute
value in Eq.~(\ref{absoluteC})---so let us restrict our attention to such
values. To take into account the normalization condition, we use a Lagrange
multiplier $\gamma/2$ and extremize the quantity
\begin{equation}
a_{13}a_{14}+a_{14}a_{24}-(\gamma/2)(a_{13}^{2}+a_{14}^{2}+a_{24}^{2}).
\end{equation}
Differentiating, we arrive at three linear equations expressed by the matrix
equation
\begin{equation}
\left(
\begin{array}
[c]{ccc}%
0 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 0
\end{array}
\right) \left(
\begin{array}
[c]{c}%
a_{13}\\
a_{14}\\
a_{24}%
\end{array}
\right) =\gamma\left(
\begin{array}
[c]{c}%
a_{13}\\
a_{14}\\
a_{24}%
\end{array}
\right) . \label{3x3}%
\end{equation}
Of the three eigenvalues, only one allows an eigenvector with non-negative
components, namely, $\gamma=\sqrt{2}$. The normalized eigenvector is
\begin{equation}
\left(
\begin{array}
[c]{c}%
a_{13}\\
a_{14}\\
a_{24}%
\end{array}
\right) =\left(
\begin{array}
[c]{c}%
\frac{1}{2}\\
\frac{1}{\sqrt{2}}\\
\frac{1}{2}%
\end{array}
\right) ,
\end{equation}
which gives $C=\sqrt{2}/5=0.283$. This is greater than the value 0.25 that we
obtained in our earlier example.
Before generalizing this calculation to arbitrary values of $n$ and $p$, we
adopt some terminology that will simplify the discussion. Let us think of the
qubits as ``sites,'' and let us call the two states of each qubit ``occupied''
($|1\rangle$) and ``unoccupied'' ($|0\rangle$). The states $|\xi\rangle$ that
we are considering have a fixed number $p$ of occupied sites in a string of
$n$ sites; so we can regard the system as a collection of $p$ ``particles'' in
a one-dimensional lattice of length $n$. Condition (ii) requires that two
particles never be in adjacent sites; it is as if each particle is an extended
object, taking up two lattice sites, and two particles cannot overlap. Thus
the number of particles is limited by the inequality $2p \le n$.
\section{Generalization to blocks of arbitrary size}
We now turn to the calculation of the optimal concurrence for general $n$ and
$p$ assuming our conditions are satisfied. It will turn out that this
calculation can be done exactly.
For any values of $n$ and $p$, the most general form of $|\xi\rangle$
consistent with condition (i) is
\begin{equation}
|\xi\rangle=\sum_{j_{1}<\cdots