%% This document created by Scientific Word (R) Version 3.5 \documentclass{conm-p-l}% \usepackage{amsmath} \usepackage{graphicx}% \usepackage{amsfonts}% \usepackage{amssymb} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{CSTFile=conm-p-l.cst} %TCIDATA{Created=Thursday, May 16, 2002 12:47:37} %TCIDATA{LastRevised=Wednesday, May 22, 2002 16:53:35} %TCIDATA{} %TCIDATA{} \theoremstyle{plain} \newtheorem{acknowledgement}{Acknowledgement} \newtheorem{algorithm}{Algorithm} \newtheorem{axiom}{Axiom} \newtheorem{case}{Case} \newtheorem{claim}{Claim} \newtheorem{conclusion}{Conclusion} \newtheorem{condition}{Condition} \newtheorem{conjecture}{Conjecture} \newtheorem{corollary}{Corollary} \newtheorem{criterion}{Criterion} \newtheorem{definition}{Definition} \newtheorem{example}{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}{Lemma} \newtheorem{notation}{Notation} \newtheorem{problem}{Problem} \newtheorem{proposition}{Proposition} \newtheorem{remark}{Remark} \newtheorem{solution}{Solution} \newtheorem{summary}{Summary} \newtheorem{theorem}{Theorem} \numberwithin{equation}{section} \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{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