\

Valerie Coffman,$^{(1)}$ Joydip Kundu,$^{(2)}$

and William K. Wootters$^{(3)}$\

Abstract {#abstract .unnumbered}
--------

[Consider three qubits A, B, and C which may be entangled with each other. We show that there is a trade-off between A’s entanglement with B and its entanglement with C. This relation is expressed in terms of a measure of entanglement called the “tangle,” which is related to the entanglement of formation. Specifically, we show that the tangle between A and B, plus the tangle between A and C, cannot be greater than the tangle between A and the pair BC. This inequality is as strong as it could be, in the sense that for any values of the tangles satisfying the corresponding equality, one can find a quantum state consistent with those values. Further exploration of this result leads to a definition of the “three-way tangle” of the system, which is invariant under permutations of the qubits. ]{}

PACS numbers: 03.65.Bz, 89.70.+c

Quantum entanglement has rightly been the subject of much study in recent years as a potential resource for communication and information processing. As with other resources such as free energy and information, one would like to have a quantitative theory of entanglement giving specific rules about how it can and cannot be manipulated; indeed such a theory has begun to be developed. The first step in building the theory has been to quantify entanglement itself. In the last few years a number of entanglement measures for bipartite states have been introduced and analyzed \[1–7\], the one most relevant to the present work being the “entanglement of formation” [@entmeasures], which is intended to quantify the amount of quantum communication required to create a given state. In the present paper we draw on some of the earlier work on entanglement of formation [@Hill; @Wootters] in order to explore another basic quantitative question: To what extent does entanglement between two objects limit their entanglements with other objects?

Unlike classical correlations, quantum entanglement cannot be freely shared among many objects. For example, given a triple of spin-$\frac{1}{2}$ particles A, B and C, if particle A is fully entangled with particle B, [*e.g.*]{}, if they are in the singlet state $\frac{1}{\sqrt{2}}(|\hspace{-0.7mm}\uparrow\downarrow\rangle 
- |\hspace{-0.7mm}\downarrow\uparrow\rangle )$, then particle A cannot be simultaneously entangled with particle C. (If A were entangled with C, then the pair AB would also be entangled with C and would therefore have a mixed-state density matrix, whereas the singlet state is pure.) One expects that a less extreme form of this restriction should also hold: if A is [*partly*]{} entangled with B, then A can have only a limited entanglement with C. The first goal of this paper is to verify this intuition and express it quantitatively. We will see that the restriction on the sharing of entanglement takes a particularly elegant form in terms of a measure of entanglement called the “tangle,” which is closely related to the entanglement of formation. Further analysis of this result will lead us naturally to a quantity that measures a three-way entanglement of the system and is invariant under all permutations of the particles [@multi].

The present work is related to recent work on the characterization of multiparticle states in terms of invariants under local transformations \[9–12\]; indeed, both the tangle and our measure of three-way entanglement are invariants in this sense. Our work is also related to research exploring the connection between entanglement and cloning \[13–17\]. An example along these lines was studied by Bruß, who asked, in the case of a singlet pair AB, to what extent particle B’s entanglement with particle A can be shared symmetrically and isotropically with a third particle, for a purpose such as teleportation where isotropy is desired [@Bruss]. Our investigation is similar in spirit to that of Bruß but has a different focus in that we are looking for a general law governing the splitting or sharing of entanglement; thus, for example, we make no assumptions about the symmetry of the system. Some of the results presented here have been mentioned in a recent paper by one of us [@Royal], but the proofs and most of the details and observations have not been previously published. In this paper we confine our attention to binary quantum objects (qubits) such as spin-$\frac{1}{2}$ particles—we will use the generic basis labels $|0\rangle$ and $|1\rangle$ rather than $|\hspace{-0.1cm}\uparrow\rangle$ and $|\hspace{-0.1cm}\downarrow\rangle$—but the same questions could be raised for larger objects.

We begin by defining the tangle. Let A and B be a pair of qubits, and let the density matrix of the pair be $\rho_{AB}$, which may be pure or mixed. We define the “spin-flipped” density matrix to be $$\tilde{\rho}_{AB} = 
(\sigma_y \otimes \sigma_y) \rho^*_{AB} (\sigma_y \otimes \sigma_y),$$ where the asterisk denotes complex conjugation in the standard basis $\{|00\rangle$, $|01\rangle$, $|10\rangle, |11\rangle\}$ and $\sigma_y$ expressed in the same basis is the matrix ${\left(\begin{array}{cc}0&{-i}\\i&0\end{array}\right)}$. As both $\rho_{AB}$ and $\tilde{\rho}_{AB}$ are positive operators, it follows that the product $\rho_{AB} \tilde{\rho}_{AB}$, though non-Hermitian, also has only real and non-negative eigenvalues. Let the square roots of these eigenvalues, in decreasing order, be $\lambda_1$, $\lambda_2$, $\lambda_3$, and $\lambda_4$. Then the tangle of the density matrix $\rho_{AB}$ is defined as $$\tau_{AB} = 
[{\rm max}\{\lambda_1 - \lambda_2 -\lambda_3 -\lambda_4,0\}]^2.
\label{tau}$$ For the special case in which the state of AB is pure, the matrix $\rho_{AB} \tilde{\rho}_{AB}$ has only one non-zero eigenvalue, and one can show that $\tau_{AB}=4\det \rho_{A}$, where $\rho_{A}$ is the density matrix of qubit A, that is, the trace of $\rho_{AB}$ over qubit B.

It is by no means obvious from the definition that the tangle is a measure of entanglement for mixed states. This interpretation comes from previous work, in which a specific connection is established between the tangle and the entanglement of formation of a pair of qubits [@Wootters]. For the purpose of this paper it is sufficient to note that $\tau = 0$ corresponds to an unentangled state, $\tau = 1$ corresponds to a completely entangled state, and the entanglement of formation is a monotonically increasing function of $\tau$.[^1] (The earlier work did not define the tangle [*per se*]{} but rather the “concurrence,” which is simply the square root of the tangle. It is with some hesitation that we introduce the new term “tangle” here instead of speaking of the square of the concurrence. However, for our present purpose the tangle does seem to be the more natural measure to use, and the discussion is significantly simplified by introducing this term.) At present, the tangle is defined only for a pair of qubits, not for higher-dimensional systems.

We now turn to the first problem of this paper: given a pure state of three qubits A, B, and C, how is the tangle between A and B related to the tangle between A and C? For this special case—a pure state of three qubits—the formula for the tangle simplifies: each pair of qubits, being entangled with only one other qubit in a joint pure state, is described by a density matrix having at most two non-zero eigenvalues. It follows that the product $\rho_{AB} \tilde{\rho}_{AB}$ also has only two non-zero eigenvalues. We can use this fact and Eq. (\[tau\]) to write the following inequality for the tangle $\tau_{AB}$ between A and B. $$\begin{aligned}
\tau_{AB} & = & (\lambda_1 - \lambda_2)^2
 = {\lambda_1}^2 + {\lambda_2}^2 - 2\lambda_1 \lambda_2  
 \nonumber\\
 & = & {\rm Tr} (\rho_{AB}\tilde{\rho}_{AB}) - 
2\lambda_1 \lambda_2
          \le {\rm Tr} (\rho_{AB}\tilde{\rho}_{AB}). 
  \label{2lambdas}\end{aligned}$$ Here $\rho_{AB}$ is the density matrix of the pair AB, obtained from the original pure state by tracing over qubit C. Eq. (\[2lambdas\]) and the analogous equation for $\tau_{AC}$ allow us to bound the sum $\tau_{AB} + \tau_{AC}$: $$\tau_{AB} + \tau_{AC} \le 
{\rm Tr} (\rho_{AB}\tilde{\rho}_{AB}) + 
{\rm Tr} (\rho_{AC}\tilde{\rho}_{AC}).   \label{tautau}$$ The next paragraph is devoted to evaluating the right-hand side of this inequality.

Let us express the pure state $|\xi\rangle$ of the three-qubit system in the standard basis $\{|ijk\rangle \}$, where each index takes the values 0 and 1: $$|\xi\rangle = \sum_{ijk} a_{ijk}|ijk \rangle.$$ In terms of the coefficients $a_{ijk}$, we can write ${\rm Tr} (\rho_{AB}\tilde{\rho}_{AB})$ as $${\rm Tr} (\rho_{AB}\tilde{\rho}_{AB})
=\sum a_{ijk}a^*_{mnk}\epsilon_{mm^\prime}\epsilon_{nn^\prime}
a^*_{m^\prime n^\prime p}a_{i^\prime j^\prime p}
\epsilon_{i^\prime i}\epsilon_{j^\prime j},$$ where $\epsilon_{01}=-\epsilon_{10}=1$ and $\epsilon_{00}=\epsilon_{11}=0$ and the sum is over all the indices. We now replace the product $\epsilon_{nn^\prime}\epsilon_{j^\prime j}$ with the equivalent expression $\delta_{nj^\prime}\delta_{n^\prime j}-
\delta_{nj}\delta_{n^\prime j^\prime}$, and in the first of the two resulting terms (that is, the one associated with $\delta_{nj^\prime}\delta_{n^\prime j}$) we perform a similar substitution for $\epsilon_{mm^\prime}\epsilon_{i^\prime i}$. These substitutions directly give us $${\rm Tr} (\rho_{AB}\tilde{\rho}_{AB}) = 2\det {\rho_A}
-{\rm Tr} (\rho_B^2) + {\rm Tr} (\rho_C^2), \label{dettr}$$ where $\rho_A$, $\rho_B$, and $\rho_C$ are the 2x2 density matrices of the individual qubits. Because each of these matrices has unit trace, we can rewrite Eq. (\[dettr\]) as $${\rm Tr} (\rho_{AB}\tilde{\rho}_{AB}) = 
2(\det{\rho_A} + \det {\rho_B} - \det {\rho_C}).$$ By symmetry we must also have $${\rm Tr} (\rho_{AC}\tilde{\rho}_{AC}) = 
2(\det {\rho_A} + \det {\rho_C} - \det {\rho_B}).$$ Summing these last two equations, we finally get a simple expression for the right-hand side of Eq. (\[tautau\]), namely, $${\rm Tr} (\rho_{AB}\tilde{\rho}_{AB})
+ {\rm Tr} (\rho_{AC}\tilde{\rho}_{AC}) = 4\det {\rho_A}. \label{trdet}$$ Eqs. (\[tautau\]) and (\[trdet\]) give us our first main result: $$\tau_{AB}+\tau_{AC} \le 4\det\rho_A. \label{first}$$

We can interpret the right-hand side of Eq. (\[first\]) as follows. Thinking of the pair BC as a single object, it makes sense to speak of the tangle between qubit A and the pair BC, because, even though the state space of BC is four-dimensional, only two of those dimensions are necessary to express the pure state $|\xi\rangle$ of ABC. (The two necessary dimensions are those spanned by the two eigenstates of $\rho_{BC}$ that have non-zero eigenvalues. That there are only two such eigenvalues follows from the fact that A is only a qubit.) We may thus treat A and BC, at least for this purpose, as a pair of qubits, and so the tangle is well defined. In fact this tangle $\tau_{A(BC)}$ is simply $4\det{\rho_A}$ as we have mentioned before. We can therefore rewrite our result as $$\tau_{AB}+\tau_{AC} \le \tau_{A(BC)}.  \label{3taus}$$ Informally, Eq. (\[3taus\]) can be expressed as follows. Qubit A has a certain amount of entanglement with the pair BC. This amount bounds A’s entanglement with qubits B and C taken individually, and the part of the entanglement that is devoted to qubit B (as measured by the tangle) is not available to qubit C.

We will say more shortly about the case of three qubits in a pure state, but at this point it is worth mentioning a generalization to mixed states. If ABC is in a mixed state $\rho$, then $\tau_{A(BC)}$ is not defined, because all four dimensions of BC might be involved, but we can define a related quantity $\tau^{min}_{A(BC)}$ via the following prescription. Consider all possible pure-state decompositions of the state $\rho$, that is, all sets $\{(\psi_i,p_i)\}$ such that $\rho = \sum_i p_i 
|\psi_i\rangle\langle\psi_i|$. For each of these decompositions, one can compute the average value of $\tau_{A(BC)}$: $$\langle \tau_{A(BC)} \rangle = \sum_i p_i \tau_{A(BC)}(\psi_i).$$ The minimum of this average over all decompositions of $\rho$ is what we take to be $\tau^{min}_{A(BC)}(\rho)$. The following analogue of Eq. (\[3taus\]) then holds for mixed states: $$\tau_{AB}+\tau_{AC} \le \tau^{min}_{A(BC)}.  \label{mixed}$$ To prove this, consider the pure states $|\psi_i\rangle$ belonging to an optimal decomposition of $\rho$, that is, a decomposition that minimizes $\langle\tau_{A(BC)}\rangle$. We can write our basic inequality, Eq. (\[3taus\]), for each such pure state and then average both sides of the inequality over the whole decomposition. The right-hand side of the resulting inequality is $\tau^{min}_{A(BC)}(\rho)$, which is what we want on the right-hand side. On the left-hand side we have two terms: (i) the average of the tangle between A and B over a set of mixed states whose [*average*]{} is $\rho_{AB}$ ([*i.e.*]{}, ${\rm Tr}_C (\rho)$), and (ii) the average of the tangle between A and C over a set of mixed states whose average is $\rho_{AC}$. It is a fact that the tangle is a [*convex*]{} function on the set of density matrices.[^2] That is, the average of the tangles is greater than or equal to the tangle of the average. In this case the tangles of the averages are $\tau_{AB} = 
\tau(\rho_{AB})$ and $\tau_{AC}=\tau(\rho_{AC})$. The sum of these two tangles must thus be less than or equal to $\tau^{min}_{A(BC)}(\rho)$, which is what we wanted to prove.

Returning to the case of pure states, one may wonder how tight the inequality (\[3taus\]) is. Could one find, for example, a more stringent bound of the same form, based on a different measure of entanglement? To address this question, consider the following pure state of ABC: $$|\phi\rangle = \alpha |100\rangle + \beta |010\rangle + \gamma 
|001\rangle,
\label{phi}$$ where the three positions in the kets refer to qubits A, B, and C in that order. For this state, one finds that $\tau_{AB}=4|\alpha|^2|\beta|^2$, $\tau_{AC}=4|\alpha|^2|\gamma|^2$, and $\tau_{A(BC)}=4|\alpha|^2(|\beta|^2+|\gamma|^2)$. Thus the inequality (\[3taus\]) becomes in this case an equality: $\tau_{AB}+\tau_{AC} = \tau_{A(BC)}$. This example shows that for any values of the tangles satisfying this equality, there is a quantum state that is consistent with those values.

Now let $T(\tau)$ be a monotonically increasing function of $\tau$ that we might propose as an alternative measure of entanglement. For simplicity let us assume that $T(0)=0$ and $T(1)=1$. Because of the above example, $T$ could satisfy the inequality $T_{AB} + T_{AC} \le T_{A(BC)}$ only if $T(x) + T(y) \le T(x+y)$ for all non-negative $x$ and $y$ such that $x+y \le 1$. Suppose $T$ has this property. Then could there exist some quantum state for which $T_{AB} + T_{AC} = T_{A(BC)}$ but $\tau_{AB}+\tau_{AC} < 
\tau_{A(BC)}$? That is, could $T$ yield an equality for some state for which $\tau$ gives only an inequality? The answer is no, because if $\tau_{AB}+\tau_{AC} < \tau_{A(BC)}$, then $T_{AB} + T_{AC}$ = $T(\tau_{AB}) + T(\tau_{AC}) \le 
T(\tau_{AB}+\tau_{AC}) < T(\tau_{A(BC)}) = T_{A(BC)}$. Moreover, the only way $T$ can match $\tau$ in those cases where $\tau$ gives an equality is for $T$ to be equal to $\tau$. In this sense, $\tau$ is an optimal measure of entanglement with respect to the inequality given in Eq. (\[3taus\]). Note, however, that the above argument applies only to functions of $\tau$. There could in principle be other measures of entanglement that are not functions of $\tau$ that could make an equal claim to optimality.

The entanglement of formation is a function of $\tau$, but it is a concave function and therefore does not satisfy an inequality of the form of Eq. (\[3taus\]). Consider, for example, the state $\frac{1}{\sqrt{2}}
|100\rangle + \frac{1}{2}|010\rangle + \frac{1}{2}|001\rangle$. One finds that the relevant entanglements of formation are $E_{AB} = 0.601$, $E_{AC} = 0.601$, and $E_{A(BC)} = 1$. Thus, contrary to what one might expect, the sum of the entanglements of formation between A and the separate qubits B and C is greater than 1 “ebit,” despite the fact that one can reasonably regard 1 ebit as the entanglement capacity of a single qubit. This is not a paradox; it simply shows us that entanglement of formation does not exhibit this particular kind of additivity. (This sense of “additivity” should not be confused with the additivity of entanglement when one combines pairs to make larger systems[@Wootters]. It is not known whether entanglement of formation satisfies the latter notion of additivity.)

We have just seen that there are some states for which the inequality (\[3taus\]) becomes an equality. Of course there are other states for which the inequality is strict. As we will see, it turns out to be very interesting to consider the [*difference*]{} between the two sides of Eq. (\[3taus\]). This difference can be thought of as the amount of entanglement between A and BC that [*cannot*]{} be accounted for by the entanglements of A with B and C separately. In the following paragraphs we refer to this quantity as the “residual tangle.”

Let the system ABC be in a pure state $|\xi\rangle$, and as before, let the components of $|\xi\rangle$ in the standard basis be $a_{ijk}$: $$|\xi\rangle = \sum_{ijk}a_{ijk}|ijk\rangle.$$ According to Eqs. (\[2lambdas\]) and (\[trdet\]) and the discussion following Eq. (\[first\]), the residual tangle is equal to $$\tau_{A(BC)} - \tau_{AB} - \tau_{AC} = 
2(\lambda^{AB}_1\lambda^{AB}_2 + \lambda^{AC}_1\lambda^{AC}_2),$$ where $\lambda^{AB}_1$ and $\lambda^{AB}_2$ are the square roots of the two eigenvalues of $\rho_{AB}{\tilde{\rho}}_{AB}$, and $\lambda^{AC}_1$ and $\lambda^{AC}_2$ are defined similarly. We now derive an explicit expression for the residual tangle in terms of the coefficients $a_{ijk}$.

We focus first on the product $\lambda^{AB}_1\lambda^{AB}_2$. This product can almost be interpreted as the square root of the determinant of $\rho_{AB}{\tilde{\rho}}_{AB}$. But $\rho_{AB}{\tilde{\rho}}_{AB}$ is an operator acting on a four-dimensional space, and two of its eigenvalues are zero; so its determinant is also zero. However, if we consider the action of $\rho_{AB}{\tilde{\rho}}_{AB}$ only on its [*range*]{}, then $\lambda^{AB}_1\lambda^{AB}_2$ [*will be*]{} the square root of the determinant of this restricted transformation.

The range of $\rho_{AB}{\tilde{\rho}}_{AB}$ is spanned by the two vectors $|v_0\rangle = \sum_{ij}a_{ij0}|ij0\rangle$ and $|v_1\rangle = \sum_{ij}a_{ij1}|ij1\rangle$. (These vectors also span the range of $\rho_{AB}$.) To examine the action of $\rho_{AB}{\tilde{\rho}}_{AB}$ on this subspace, we consider its effect on vectors of the form $x|v_1\rangle
+y|v_2\rangle \equiv {\left(\begin{array}{c}x\\y\end{array}\right)}$. This effect is given by $${\left(\begin{array}{c}x^\prime \\ y^\prime\end{array}\right)} = R {\left(\begin{array}{c}x\\y\end{array}\right)},$$ where $R$ is a 2x2 matrix. The product $\lambda^{AB}_1\lambda^{AB}_2$ is the square root of the determinant of $R$. One finds that $$R_{ij} = \sum a_{klj}a^*_{mni}\sigma_{mp}
\sigma_{nq}a^*_{pqr}a_{str}\sigma_{sk}\sigma_{tl},$$ where the sum is over all repeated indices. (We have ordered the factors so as to suggest the expression $\rho_{AB}{\tilde{\rho}}_{AB}$ from which $R$ is derived.) Taking the determinant of $R$ involves somewhat tedious but straightforward algebra, with the following result: $$\lambda^{AB}_1\lambda^{AB}_2 = \sqrt{\det R} = |d_1 - 2d_2 + 4d_3|,$$ where $$\begin{aligned}
d_1 & = & a^2_{000}a^2_{111}+a^2_{001}a^2_{110}+
      a^2_{010}a^2_{101}+a^2_{100}a^2_{011}; 
      \nonumber \\
d_2 & = & a_{000}a_{111}a_{011}a_{100} + 
    a_{000}a_{111}a_{101}a_{010} + 
    a_{000}a_{111}a_{110}a_{001} \\ 
    & + & a_{011}a_{100}a_{101}a_{010} +
    a_{011}a_{100}a_{110}a_{001} + 
    a_{101}a_{010}a_{110}a_{001}; \nonumber \\
d_3 & = & a_{000}a_{110}a_{101}a_{011} + 
    a_{111}a_{001}a_{010}a_{100}.  \nonumber \end{aligned}$$ We can get a mental picture of this expression by imagining the eight coefficients $a_{ijk}$ attached to the corners of a cube. Then each term appearing in $d_1$, $d_2$, or $d_3$ is a product of four of the coefficients $a_{ijk}$ such that the “center of mass” of the four is at the center of the cube. Such configurations fall into three classes: those in which the four coefficients lie on a body diagonal and each one is used twice ($d_1$); those in which they lie on a diagonal plane ($d_2$), and those in which they lie on the vertices of a tetrahedron ($d_3$). Within each category, all the possible configurations are given the same weight.

This picture immediately yields an interesting fact: the quantity $\lambda^{AB}_1\lambda^{AB}_2$ is invariant under permutations of the qubits. (A permutation of qubits corresponds to a reflection or rotation of the cube, but each $d_i$ is invariant under such actions.) This means in particular that we need not carry out a separate calculation to find $\lambda^{AC}_1\lambda^{AC}_2$, since we know we will get the same result. We can now therefore write down an expression for the residual tangle: $$\tau_{A(BC)} - \tau_{AB} - \tau_{AC} = 4|d_1 - 2d_2 + 4d_3|.
\label{residual}$$ Note that the residual tangle does not depend on which qubit one takes as the “focus” of the construction. In our calculations we have focused on entanglements with qubit A, but if we had chosen qubit B instead, we would have found that $$\tau_{B(CA)} - \tau_{BC} - \tau_{BA} = 4|d_1 - 2d_2 + 4d_3|.$$ The residual tangle thus represents a collective property of the three qubits that is unchanged by permutations; it is really a kind of three-way tangle. If we call this quantity $\tau_{ABC}$, we can summarize the main results of this paper in the following equation. $$\tau_{A(BC)} = \tau_{AB} + \tau_{AC} + \tau_{ABC}. \label{final}$$ In words, the tangle of A with BC is the sum of its tangle with B, its tangle with C, and the essential three-way tangle of the triple. As an example, consider the Greenberger-Horne-Zeilinger state $\frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$ [@GHZ]. For this state the tangle of each qubit with the rest of the system is 1, the three-way tangle is also 1, and all the pairwise tangles are zero (the qubits in each pair are classically correlated but not entangled). Thus Eq. (\[final\]) in this case becomes $1 = 0 + 0 + 1$.

Finally, we note that the expression for $\tau_{ABC}$ in terms of $d_1$, $d_2$, and $d_3$ \[Eq. (\[residual\])\] can be rewritten, after a little more algebra, in a more standard form: $$\tau_{ABC} = 2\bigg|\sum a_{ijk}a_{i^\prime j^\prime m}
a_{npk^\prime}a_{n^\prime p^\prime m^\prime}
\epsilon_{ii^\prime}\epsilon_{jj^\prime}\epsilon_{kk^\prime}
\epsilon_{mm^\prime}\epsilon_{nn^\prime}\epsilon_{pp^\prime}
\bigg|,$$ where the sum is over all the indices. This form does not immediately reveal the invariance of $\tau_{ABC}$ under permutations of the qubits, but the invariance is there nonetheless.

It would be very interesting to know which of the results of this paper generalize to larger objects or to larger collections of objects. At this point it is not clear how one might begin to generalize this approach to qutrits or higher dimensional objects, because the spin-flip operation seems peculiar to qubits. On the other hand, it appears very likely that at least some of these results can be extended to larger collections of qubits. The one solid piece of evidence we can offer is the existence of a generalization of the state $|\phi\rangle$ of Eq. (\[phi\]) to $n$ qubits: $$|\phi\rangle = \alpha_1|100\ldots0\rangle+\alpha_2|010\ldots0\rangle+
\alpha_3|001\ldots0\rangle+\cdots+\alpha_n|000...1\rangle.$$ One can show that for this state, the following equality holds. $$\tau_{12}+\tau_{13}+\cdots+\tau_{1n} = \tau_{1(23\ldots n)},$$ where the qubits are now labeled by numbers rather than letters. We are willing to conjecture that the corresponding inequality, analogous to Eq. (\[3taus\]), is valid for all pure states of $n$ qubits.

This work was supported in part by the National Science Foundation’s Research Experiences for Undergraduates program. WKW is grateful for the support and hospitality of the Isaac Newton Institute. We would also like to thank Tony Sudbery and Noah Linden for discussions that helped streamline the derivation of Eq. (\[first\]).

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[^1]: The entanglement of formation is given by $E=h(\frac{1}{2} + \frac{1}{2}\sqrt{1-\tau})$, where $h$ is the binary entropy function $h(x)= -x\log x -(1-x)\log (1-x)$.

[^2]: It follows from Ref. [@Wootters] that the concurrence is a convex non-negative function on the set of density matrices for two qubits. The tangle, being the square of the concurrence, is therefore also convex.
