Monogamy of Entanglement Without Inequalities

Monogamy of entanglement without inequalities

Gilad Gour1 and Yu Guo2

1 Department of Mathematics and Statistics and Institute for Quantum Science and Technology, University of Calgary, Calgary, Alberta T2N 1N4, Canada 2Institute of Quantum Information Science, Shanxi Datong University, Datong, Shanxi 037009, China August 10, 2018

We provide a fine-grained definition of In such a preparation, any two coins are maxi- monogamous measure of entanglement that mally correlated. In contrast with the classical does not invoke any particular monogamy world, it is not possible to prepare three qubits relation. Our definition is given in terms of A, B, C in a way that any two qubits are maxi- an equality, as opposed to inequality, that mally entangled [1]. In fact, if qubit A is max- we call the “disentangling condition". We imally entangled with qubit B, then it must be relate our definition to the more traditional uncorrelated (not even classically) with qubit C. one, by showing that it generates standard This phenomenon of monogamy of entanglement monogamy relations. We then show that all was first quantified in a seminal paper by Coffman, quantum Markov states satisfy the disentan- Kundu, and Wootters (CKW) [1] for three qubits, gling condition for any entanglement mono- and later on studied intensively in more general set- tone. In addition, we demonstrate that tings [2,3,4,5, 14, 15,6, 16, 17, 18,7, 19, 20, 21, entanglement monotones that are given in 22, 23, 24, 25,8,9, 26, 27, 10, 11, 28, 29, 30, 12, terms of a convex roof extension are monog- 13, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40]. amous if they are monogamous on pure Qualitatively, monogamy of entanglement mea- states, and show that for any quantum state sures the shareability of entanglement in a com- that satisfies the disentangling condition, its posite quantum system, i.e., the more two subsys- entanglement of formation equals the entan- tems are entangled the less this pair has entangle- glement of assistance. We characterize all ment with the rest of the system. This feature of bipartite mixed states with this property, entanglement has found potential applications in and use it to show that the G-concurrence many quantum information tasks and other areas is monogamous. In the case of two qubits, of physics, such as quantum key distribution [41, we show that the equality between entan- 42, 43], classification of quantum states [44, 45, 46], glement of formation and assistance holds if condensed-matter physics [47, 48, 49], frustrated and only if the state is a rank 2 bipartite spin systems [50], statistical physics [51], and even state that can be expressed as the marginal black-hole physics [52, 53].

arXiv:1710.03295v2 [quant-ph] 9 Aug 2018 of a pure 3-qubit state in the W class. A monogamy relation is quantitatively displayed Monogamy of entanglement is one of the non- as an inequality of the following form: intuitive phenomena of quantum physics that dis- E(A|BC) ≥ E(A|B) + E(A|C) , (1) tinguish it from classical physics. Classically, three random bits can be maximally correlated. For ex- where E is a measure of bipartite entanglement and ample, three coins can be prepared in a state in A, B, C are three subsystems of a composite quan- which with 50% chance all three coins show “head", tum system. It states that the sum of the entan- and with the other 50% chance they all show “tail". glement between A and B, and between A and C, can not exceed the entanglement between A and Gilad Gour: [email protected] the joint system BC. While not all measures of Yu Guo: [email protected] entanglement satisfy this relation, some do. Con-

Accepted in Quantum 2018-08-05, click title to verify 1 sequently, any measure of entanglement that does ization for the disentangling condition in the form satisfy (1) was called in literature monogamous. of an equality between the entanglement of forma- Here we argue that this definition of monoga- tion (EoF) associated with the given entanglement mous measure of entanglement captures only par- measure (see Eq. (7) below) and the entanglement tially the property that entanglement is monoga- of assistance (EoA) [58], and discuss its relation mous. This is evident from the fact that many to quantum Markov chains [59]. In addition, we important measures of entanglement do not satisfy characterize all states for which EoF equals to EoA the relation (1). Some of these measures are not when the measure of entanglement is taken to be even additive under tensor product [54, 55, 56] (in the G-concurrence, and use that to show that the fact, some measures [57] are multiplicative under G-concurrence is monogamous. Finally, we show tensor product). Therefore, the summation in the that in the 2-dimensional case, the bipartite 2-qubit RHS of (1) is clearly only a convenient choice and mixed states that can be expressed as the marginal not a necessity. For example, it is well known that of the 3-qubit W -state, are the only 2-qubit entan- if E does not satisfy this relation, it is still possible gled states for which the EoF equals the EoA with α respect to any measure of entanglement. to find a positive exponent α ∈ +, such that E R A B AB satisfies the relation. This was already apparent in Let H ⊗ H ≡ H be a bipartite Hilbert AB AB the seminal work of [1] in which E was taken to space, and S(H ) ≡ S be the set of den- AB be the square of the concurrence and not the con- sity matrices acting on H . A function E : AB currence itself. More recently, it was shown that S → R+ is called a measure of entanglement if AB many other measures of entanglement satisfy the (1) E(σ ) = 0 for any separable density matrix AB AB monogamy relation (1) if E is replaced by Eα for σ ∈ S , and (2) E behaves monotonically un- some α > 1 [13,8, 12,5,7, 11,9,6, 10]. der local operations and classical communications (LOCC). That is, given an LOCC map Φ we have One attempt to address these issues with the cur- rent definition of a monogamous measure of entan- E Φ(ρAB) ≤ E ρAB , ∀ ρAB ∈ SAB . (2) glement, is to replace the relation (1) with a family of monogamy relations of the form, E(A|BC) ≥ The measure is said to be faithful if it is zero only f E(A|B),E(A|C) , where f is some function of on separable states. two variables that satisfy certain conditions [31]. The map Φ is completely positive and trace pre- However, such an approach is somewhat artificial serving (CPTP). In general, LOCC can be stochas- in the sense that the monogamy relations are not tic, in the sense that ρAB can be converted to AB derived from more basic principles. σj with some probability pj. In this case, the AB AB In this paper we take another approach to map from ρ to σj can not be described in monogamy of entanglement which is more “fine- general by a CPTP map. However, by introduc- 0 grained" in nature, and avoid the introduction of ing a ‘flag’ system A , we can view the ensemble AB A0AB such a function f. Therefore, our definition of {σj , pj} as a classical quantum state σ ≡ P A0 AB AB monogamous measure of entanglement (see Defi- j pj|jihj| ⊗ σj . Hence, if ρ can be con- AB nition 1 below) does not involve a monogamy rela- verted by LOCC to σj with probability pj, then tion, but instead a condition on the measure of en- there exists a CPTP LOCC map Φ such that tanglement that we call the disentangling condition Φ(ρAB) = σA0AB. Therefore, the definition above (following the terminology of [28]). Yet, we show of a measure of entanglement captures also proba- that our definition is consistent with the more tra- bilistic transformations. Particularly, E must sat- 0 ditional notion of monogamy of entanglement, by isfy E σA AB ≤ E ρAB . showing that E is monogamous according to our Almost all measures of entanglement studied in definition if and only if there exists an α > 0 such literature (although not all [60]) satisfy that Eα satisfies (1). Consequently, many more measures of entanglement are monogamous accord- A0AB X AB E σ = pjE(σj ) , (3) ing to our definition. We then provide a character- j

Accepted in Quantum 2018-08-05, click title to verify 2 which is very intuitive since A0 is just a clas- to Definition1 if and only if there exists 0 < α < sical system encoding the value of j. In this ∞ such that A0AB AB case the condition E σ ≤ E ρ becomes Eα(ρA|BC ) ≥ Eα(ρAB) + Eα(ρAC ) , (5) P p E(σAB) ≤ E ρAB . That is, LOCC can not j j j for all ρABC ∈ SABC with fixed dim HABC = d < increase entanglement on average. Measures of en- ∞. tanglement that satisfy this property are called entanglement monotones, and they can also be shown We call the smallest possible value for α that sat- ABC to be convex [61]. In the following definition we isfies Eq. (5) in a given dimension d = dim(H ), denote density matrices acting on a finite dimen- the monogamy exponent associated with a measure sional tripartite Hilbert space HABC by ρA|BC , E, and denote it simply as α(E). In general, the where the vertical bar indicates the bipartite split monogamy exponent is hard to compute, and in across which we will measure the (bipartite) entan- the supplemental material we provide (along with glement. the proofs of the theorems in this paper) a com- prehensive list of known bounds for the monogamy Definition 1. Let E be a measure of entangle- exponent when E is one of the measures of entan- ment. E is said to be monogamous if for any glement that were studied extensively in literature. ρA|BC ∈ SABC that satisfies It is important to note that the relation given in (5) is not of the form E(A|BC) ≥ E(ρA|BC ) = E(ρAB) (4) f E(A|B),E(A|C) , where f is some function of two variables that satisfies certain conditions and AC we have that E(ρ ) = 0. is independent of d [31]. This is because the monogamy exponent in (5) depends on the dimen- The condition in (4) is a very strong one and sion d, whereas f as was used in [31] is universal in typically is not satisfied by most tripartite states the sense that it does not depend on the dimension. ρABC . Following the terminology of [28] we call Therefore, if a measure of entanglement such as the it here the “disentangling condition". We will see entanglement of formation is not monogamous ac- below that states that saturate the strong subaddi- cording to the class of relations given in [31], it tivity inequality for the von Neumann entropy (i.e. does not necessarily mean that it is not monoga- quantum Markov states) always satisfy this equal- mous according to our definition. Moreover, since ity, for any entanglement monotone E. our approach allows for dependence in the dimen- Note that the condition in (1) is stronger than sion, it avoids the issues raised in [31] that measures the one given in Definition1. Indeed, if E satis- of entanglement cannot be simultaneously monoga- A|BC fies (1), then any ρ that satisfies (4) must have mous and (geometrically) faithful (see [31] for their AC E(ρ ) = 0. At the same time, Definition1 cap- definition of faithfulness). tures the essence of monogamy: that is, if system In general, the class of all states ρABC that sat- A shares the maximum amount of entanglement isfy the disentangling condition (4) depends on the with subsystem B, it is left with no entanglement choice of the entanglement measure E. However, to share with C. there is a class of states that satisfy this condi- In Definition1 we do not invoke a particular tion for any choice of an entanglement monotone monogamy relation such as (1). Instead, we pro- E. These are precisely the states that saturate pose a minimalist approach which is not quanti- the strong subadditivity of the von-Neumann en- tative, in which we only require what is essential tropy [59]. For such states, the system B Hilbert from a measure of entanglement to be monoga- space HB must have a decomposition into a direct mous. Yet, this requirement is sufficient to generate BL BR sum of tensor products HB = L H j ⊗H j , such a more quantitative monogamy relation: j that the state ρABC has the form

ABC M ABL BRC Theorem 1. Let E be a continuous measure of ρ = qjρ j ⊗ ρ j , (6) entanglement. Then, E is monogamous according j

Accepted in Quantum 2018-08-05, click title to verify 3 where qj is a probability distribution [59]. pure states [28]. Similarly, we will use it to show that the G-concurrence is monogamous. However, Theorem 2. Let E be an entanglement mono- monogamy alone does not necessarily imply that tone. Then, E satisfies the disentangling condi- a tripartite state is a Markov state if it satisfies tion (4) for all Markov quantum states ρABC of the disentangling condition. In the following theo- the form (6). rem, we provide yet another property of a tripartite Note that a state of the form (6) has a separa- state ρABC that satisfies the disentangling condi- ble marginal state ρAC , and therefore the above tion. The property is that any LOCC protocol on theorem is consistent with Definition1. Conse- such a tripartite state can not increase on aver- quently, to check if an entanglement monotone is age the initial bipartite entanglement between A monogamous, one has to consider tripartite states and B. The maximum such average of bipartite that satisfy (4) but have a different form than (6). entanglement is known as entanglement of collabo- Perhaps such states do not exist for certain en- ration [62]. It is defined by: tanglement monotones. Indeed, partial results in n this direction were proved recently in [28]. Partic- AB|C ABC X AB Ec ρ = max pjE ρj , (8) ularly, it was shown that any pure tripartite en- j=1 ABC tangled state ψ with bipartite marginal state where E is a given measure of bipartite entangle- AB A|BC AB ρ , that satisfies N ψ = N ρ , where ment, and the maximum is taken over all tripartite N is the negativity measure, must have the form LOCC protocols yielding the bipartite state ρAB ABC ABL BRC j ψ = ψ ⊗ ψ . Note that this is precisely with probability pj. This measure of entanglement ABC the form (6) when ρ is pure. Like the Nega- is closely related to EoA, denoted here by Ea, in tivity, we will see later that also the G-concurrence which the optimization above is restricted to LOCC has this property for pure tripartite states. with one way classical communication from system Typically it is hard to check if a measure of AB|C C to systems A and B. Therefore, Ec ≥ Ea entanglement is monogamous since the conditions and in [62] it was shown that this inequality can be in (4) involves mixed tripartite states. However, as strict. we show below, it is sufficient to consider only pure tripartite states in (4), if the entanglement mea- Theorem 4. Let E be an entanglement monotone ABC sure E is defined on mixed states by a convex roof on bipartite mixed states, and let ρ be a (possi- extension; that is, bly mixed) tripartite state satisfying the disentangling condition (4). Then, n AB X AB Ef ρ ≡ min pjE |ψjihψj| , (7) AB AB|C ABC E ρ = Ec ρ . (9) j=1 where the minimum is taken over all pure state The condition in (9) is a very strong one. Par- AB Pn AB ticularly, it states that all measurements on system decompositions of ρ = j=1 pj|ψjihψj| . We C yield the same average of bipartite entanglement call Ef the entanglement of formation (EoF) associated with E. In general, it could be that E 6= E between A and B. Therefore, the entanglement of f AB on mixed states like the convex roof extended Neg- the marginal state ρ is resilient to any quantum ativity is different than the Negativity itself. process or measurement on system C. In the case ABC that ρ is pure, the EoA, Ea, depends only on Theorem 3. Let Ef be an entanglement monotone the marginal ρAB, and we get the following corol- as above. If Ef is monogamous (according to Def- lary: inition1) on pure tripartite states in HABC , then it is also monogamous on tripartite mixed states Corollary 5. Let E be an entanglement monotone ABC acting on HABC . on bipartite states, and let ρ be a pure tripartite state satisfying the disentangling condition (4). From the above theorem and Corollary5 be- Then, low it follows that the convex roof extended neg- AB AB AB ativity is monogamous, since it is monogamous for E ρ = Ef (ρ ) = Ea(ρ ) , (10)

Accepted in Quantum 2018-08-05, click title to verify 4 where the EoF, Ef , is defined in (7), and EoA, (i.e. upper triangular matrices with zeros on the Ea, is also defined as in (7) but with a maximum diagonal) is a d(d − 1)/2-dimensional subspace in replacing the minimum. N . From Gerstenhaber’s theorem [63] it follows that the largest dimension of a subspace in N is The equality of E ρAB = E ρAB in corol- f a d(d − 1)/2, and if a subspace N0 ⊂ N has this lary5 implies that any pure state decomposition maximal dimension, then it must be similar to T of ρAB has the same average entanglement when (i.e. their exists an invertible matrix S such that AB −1 measured by E. Unless ρ is pure, it is almost N0 = ST S ). never satisfied. Nonetheless, there are non-trivial AB states (of measure zero) that do satisfy it. Such Theorem 7. Let ρ be a bipartite density matrix d d states have the following property: acting on C ⊗ C with rank r. Then, AB AB Theorem 6. Let E be a measure of bipartite G(ρ ) = Ga(ρ ) > 0 , (12) AB AB entanglement, and let ρ ∈ S be a bipar- if and only if r ≤ 1 + d(d − 1)/2, and there exists tite state with support subspace supp(ρAB). If AB d d a full Schmidt rank state |xi ∈ C ⊗ C , and a E (ρAB) = E (ρAB) then for any σAB ∈ SAB f a subspace N0 ⊂ N , with dim(N0) = r − 1, such that AB AB AB with supp(σ ) ⊆ supp(ρ ) we have Ef (σ ) = AB AB AB Ea(σ ). supp(ρ ) = {I ⊗ Y |xi Y ∈ K} , (13)

Remark 1. The theorem above follows from some of where K ≡ {I} ⊕ N0 ≡ {cI + N c ∈ C ; N ∈ N0}. the results presented in [33], and for completeness The direct sum above is consistent with the fact we provide its full proof in the appendix. that a nilpotent matrix has a zero trace, so that it Theorem6 demonstrates that the equality be- is orthogonal to the identity matrix in the Hilbert- tween the EoF and EoA corresponds to a property Schmidt inner product. The theorem above implies of the support of ρAB rather than ρAB itself. In the that the G-concurrence is monogamous on pure tri- following theorem, we characterize the precise form partite states: of the support of ρAB that yields such an equality. Corollary 8. Let |ψABC i ∈ d ⊗ d ⊗ n be a The entanglement monotone we are using is a gen- C C C pure tripartite state, with bipartite marginal ρAB. eralization of the concurrence measure known as If G(ψA|BC ) = G(ρAB) > 0 then |ψiABC = the G-concurrence [57]. |χiAB|ϕiC for some bipartite state |χiAB ∈ d⊗ d The G-concurrence is an entanglement mono- C C and a vector |ϕiC ∈ n. tone that on pure bipartite states is equal to the C geometric-mean of the Schmidt coefficients. On a The above result is somewhat surprising since the d d (possibly unnormalized) vector |xi ∈ C ⊗ C , it G-concurrence is not a faithful measure of entangle- can be expressed as: ment. Yet, it states that the disentangling condi- AC 2/d tion forces the marginal state ρ to be a product G(|xi) = |det(X)| ; |xi = X ⊗ I|φ+i , (11) state. This is much stronger than G(ρAC ) = 0 (which can even hold for some entangled ρAC ), where |φ i = Pd |iiA|iiB is the maximally en- + i=1 and in particular, it states that A and C can not tangled state, and X ∈ M ( ) is a d × d complex d C even share classical correlation. Combining the matrix. For mixed states the G-concurrence is de- above corollary with Theorem3 implies that the fined in terms of the convex roof extension; that is, G-concurrence is monogamous on any mixed state G(ρAB) := G (ρAB). f that is acting on d ⊗ d ⊗ n. Finally, in the qubit In the following theorem we denote by N ⊂ C C C case, Theorem7 takes the following form: Md(C), the nilpotent cone, consisting of all d × d nilpotent complex matrices (i.e. X ∈ N iff Xk = 0 Corollary 9. Let ρAB be an entangled two qubit for some 1 ≤ k ≤ d). While the set N is not a state with rank r > 1, and let E be any injective vector space, it contains subspaces. For example, (up to local unitaries) measure of pure two qubit the set T of all strictly upper triangular matrices entanglement. Then,

Accepted in Quantum 2018-08-05, click title to verify 5 AB AB 1. If Ef (ρ ) = Ea(ρ ) then r = 2, and No. 201608140008. Y.G thanks Professor C. Si- AB ABC ρ = TrC |W ihW | is the 2-qubit marginal mon and Professor G. Gour for their hospitality, of |W i = λ1|100i+λ2|010i+λ3|001i+λ4|000i, and thanks Sumit Goswami for helpful discussions. λi ∈ C. G.G research is supported by the Natural Sci- ences and Engineering Research Council of Canada 2. Conversely, if ρAB is a marginal of a state in (NSERC). Y.G is supported by the National Nat- AB AB the W-class then Cf (ρ ) = Ca(ρ ), where ural Science Foundation of China under Grant No. C is the concurrence. 11301312 and the Natural Science Foundation of Shanxi under Grant No. 201701D121001. Remark 2. The second part of the theorem implies AB AB AB that also Ef (ρ ) = Ea(ρ ), for any ρ that is a marginal of a W-state, and any measure E, References that can be expressed as a convex function of the concurrence C [64]. [1] V. Coffman, J. Kundu, and W. K. Wootters. Distributed entangle- In conclusions, we introduced a new definition ment. Phys. Rev. A, 61:052306, 2000. for a monogamous measure of entanglement. Our doi:10.1103/PhysRevA.61.052306. definition involves an equality (the disentangling [2] R. Horodecki, P. Horodecki, M. Horodecki, condition (4)) rather than the inequality (1). Yet, and K. Horodecki, Quantum entangle- we showed that our notion of monogamy can repro- ment. Rev. Mod. Phys., 81:865, 2009. duce monogamy relations like in (1) with a small doi:10.1103/RevModPhys.81.865. change that the measure E is replaced by Eα for some exponent α > 0. We then showed that convex [3] M. Koashi and A. Winter. Monogamy roof based entanglement monotones of the form (7) of quantum entanglement and other cor- are monogamous iff they are monogamous on pure relations. Phys. Rev. A, 69:022309, 2004. tripartite states, and showed further that the dis- doi:10.1103/PhysRevA.69.022309. entangling condition in (4) holds for any entangle- [4] G. Gour, D. A. Meyer, and B. C. ABC ment monotone if ρ is a quantum Markov state. Sanders. Deterministic entanglement While it is left open if the converse is also true (at of assistance and monogamy con- least for some entanglement monotones), we were straints. Phys. Rev. A, 72:042329, 2005. able to show that for the G-concurrence, the only doi:10.1103/PhysRevA.72.042329. pure tripartite states that satisfy the disentangling condition (4) are Markov states. In addition, we [5] T. J. Osborne and F. Verstraete. General related the disentangling condition to states that monogamy inequality for bipartite qubit en- have the same average entanglement for all convex tanglement. Phys. Rev. Lett., 96:220503, 2006. pure state decompositions, and found a character- doi:10.1103/PhysRevLett.96.220503. ization of such states in Theorem6 (for general [6] Y.-C. Ou and H. Fan, Monogamy in- measures of entanglement), and a complete char- equality in terms of negativity for three- acterization in Theorem7 (for the G-concurrence). qubit states. Phys. Rev. A, 75:062308, 2007. Clearly, much more is left to investigate along these doi:10.1103/PhysRevA.75.062308. lines. [7] J. S. Kim, A. Das, and B. C. Sanders. Entanglement monogamy of multipar- Acknowledgements tite higher-dimensional quantum systems using convex-roof extended nega- This work was completed while Y.G was visiting tivity. Phys. Rev. A, 79:012329, 2009. the Institute of Quantum Science and Technol- doi:10.1103/PhysRevA.79.012329. ogy of the University of Calgary under the sup- [8] X. N. Zhu and S. M. Fei. Entangle- port of the China Scholarship Council under Grant ment monogamy relations of qubit sys-

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A The Monogamy exponent

Theorem 1. Let E be a continuous measure of entanglement. Then, E is monogamous according to Deﬁnition1 if and only if there exists 0 < α < ∞ such that

Eα(ρA|BC ) ≥ Eα(ρAB) + Eα(ρAC ) , (14) for all ρABC ∈ SABC with ﬁxed dim HABC = d < ∞.

Proof. Let E be a monogamous measure of entanglement according to Def.1. Since E is a measure of entanglement, it is non-increasing under partial traces, and therefore E(ρA|BC ) ≥ max{E(ρAB),E(ρAC )} A|BC ABC A|BC AB A|BC for any state ρ ∈ S . We assume E(ρ ) > 0 and set x1 ≡ E(ρ )/E(ρ ) and x2 ≡ E(ρAC )/E(ρA|BC ). Clearly, there exists γ > 0 such that

γ γ x1 + x2 ≤ 1 , (15)

γ since either xj → 0 when γ increases, or if x1 = 1 then by assumption x2 = 0 and vise versa. We denote by f(ρABC ) the smallest value of γ that achieves equality in (15). Since E is continuous, so is f, and the compactness of SABC gives: α ≡ max f(ρABC ) < ∞ . (16) ρABC ∈SABC By deﬁnition, α satisﬁes the condition in (5).

As discussed in the paper, the expression for α in (16) is optimal in the sense that it provides the smallest possible value for α that satisﬁes Eq. (5). This monogamy exponent is a function of the measure E, and we denote it by α(E). It may depend also on the dimension d ≡ dim(HABC ) (see in Table1), and, in general, is hard to compute especially in higher dimensions [1,4,6,2,3,5]. By deﬁnition, α(E) can only increase with d (e.g. Table1). Table1 indicates that almost any entanglement measure is monogamous at least for multi-qubit systems. In addition, almost all the entanglement measures studied in the literature are continuous, and in particular C, N, Ncr, Ef , τ, Tq, Rα and Er are all continuous [27, 28, 29, 30].

B Quantum Markov States and Monogamy of Entanglement

Recall that quantum Markov states are states that saturate the strong subadditivity of the von-Neumann entropy. That is, they saturate the inequality:

S(ρAB) + S(ρBC ) ≥ S(ρABC ) + S(ρB) , (17) where S(ρ) = −Tr [ρ log ρ] is the von-Neumann entropy. In [31] it was shown that the inequality above is saturated if and only if the Hilbert space of system B, HB, can be decomposed into a direct sum of tensor products M BL BR HB = H j ⊗ H j (18) j

Accepted in Quantum 2018-08-05, click title to verify 1 Table 1: A comparison of the monogamy exponent of several entanglement measures. We denote the one-way distillable entanglement, concurrence, negativity, convex roof extended negativity, entanglement of formation (the original one deﬁned in [7]), tangle, squashed entanglement, Tsallis-q entanglement and Rényi-α entanglement by Ed, C, N,Ncr, Ef , τ, Esq, Tq and Rα, respectively.

E α(E) System Reference

Ed ≤ 1 any system [8] C 2 2⊗3 [9]a √ ≤ 2 2⊗n [10] ≤ 2 2 ⊗ 2 ⊗ 2m [11] ≤ 2 2⊗n [11] > 2 3⊗3 [12] ≤ 2 2 ⊗ 2 ⊗ 4 [13] > 3 3 ⊗ 2 ⊗ 2 [14] N ≤ 2 2⊗n [15, 16]b ≤ 2 2 ⊗ 2 ⊗ 2m [16]b ≤ 2 d ⊗ d ⊗ d,d = 2, 3, 4 [1] ⊗n Ncr ≤ 2 2 [17, 16, 18] E ≤ 2 2⊗n [19, 20] f √ ≤ 2 2⊗n [10] > 1 2 ⊗ 2 ⊗ 2 [20, 21] τ ≤ 1 2 ⊗ 2 ⊗ 4 [13]

Esq ≤ 1 any system [8] ⊗n Tq, 2 ≤ q ≤ 3 ≤ 1 2 [22] m c Tq ≤ 2 2 ⊗ 2 ⊗ 2 [23] R , α ≥ 2 ≤ 1 2⊗n [24, 25] α √ 7−1 ⊗n d Rα, α ≥ 2 ≤ 2 2 [26]

aα(C) ≤ 2 was shown in [9], and the equality α(C) = 2 follows from the saturation by W states (see, for example, Corollary 9). bFor pure states. √ √ c 5− 13 5+ 13 For mixed states, and q ∈ [ 2 , 2] ∪ [3, 2 ]. dFor mixed states.

Accepted in Quantum 2018-08-05, click title to verify 2 such that the state ρABC has the form

ABC M ABL BRC ρ = qjρ j ⊗ ρ j , (19) j where qj is a probability distribution.

Theorem 2. Let E be an entanglement monotone. Then, E satisﬁes the disentangling condition (4) for all entangled Markov quantum states ρABC of the form (19).

Proof. Since local ancillary systems are free in entanglement theory, one can append an ancillary system BL BR B0 that encodes the orthogonality of the subspaces H j ⊗ H j . This can be done with an isometry that BL BR L R 0 maps states in H j ⊗H j to states in HB ⊗HB ⊗|jihj|B , where systems BL and BR have dimensions BL BR maxj dim H j and maxj dim H j , respectively. Therefore, w.l.o.g. we can write the above state as

ABC X ABL B0 BRC ρ = qj ρj ⊗ |jihj| ⊗ ρj . (20) j Now, note that with any entanglement monotone E, the entanglement between A and BC is measured by: A|BC X A|BL BRC X ABL E ρ = qjE ρ j ⊗ ρ j = qjE ρ j , (21) j j where in the ﬁrst equality we used the property (3) of entanglement monotones. Similarly, the entanglement between A and B is measured by

AB X ABL BR X ABL E ρ = qjE ρ j ⊗ ρ j = qjE ρ j . (22) j j

A|BL We therefore obtain (4) as long as E ρ j > 0 for some j for which qj > 0. This completes the proof.

C Monogamy of entanglement: pure vs mixed tripartite states

As discussed in the paper, it is typically hard to check if a measure of entanglement is monogamous since the condition in (4) involves mixed tripartite states. On the other hand, it is significantly simpler to check the disentangling condition if ρABC that appears in the disentangling condition (4) is pure. We say that E is monogamous on pure states if for any pure tripartite state ρABC that satisfies (4), E(ρAC ) = 0. In the theorem below we shown that sometimes if E is monogamous on pure states it is also monogamous on mixed states (that is, it is monogamous according to Def.1). For any entanglement monotone E on the set of bipartite density matrices, SAB, we define a corresponding entanglement of formation measure, Ef which is defined by the following convex roof extension:

n AB X AB Ef ρ ≡ min pjE |ψjihψj| , (23) j=1

AB Pn AB where the minimum is taken over all pure state decompositions of ρ = j=1 pj|ψjihψj| . Clearly, E = Ef on pure bipartite states, but on mixed states they can be diﬀerent, like the convex roof extended Negativity is diﬀerent from the Negativity itself. Since we assume that E is entanglement monotone it AB AB AB AB is convex so E(ρ ) ≤ Ef (ρ ) for all ρ ∈ S . The corresponding entanglement of formation of

Accepted in Quantum 2018-08-05, click title to verify 3 a given entanglement monotone, is itself an entanglement monotone, and has the following remarkable property.

Theorem 3. Let E be an entanglement monotone, and let Ef be its corresponding entanglement of ABC formation (23). If Ef is monogamous (according to Deﬁnition1) on pure tripartite states in H , then it is also monogamous on tripartite mixed states acting on HABC .

A|BC P ABC ABC ABC Proof. Let ρ = j pj|ψjihψj| be a tripartite state acting on H with {pj, |ψji } being the optimal decomposition such that

A|BC X A|BC Ef (ρ ) = pjEf |ψji . (24) j

A|BC AB AB We also assume w.l.o.g. that pj > 0. Now, suppose Ef (ρ ) = Ef (ρ ), and denote ρj ≡ ABC TrC |ψjihψj| . Since discarding a subsystem can only decrease the entanglement, we get

X A|BC X AB AB pjEf |ψji ≥ pjEf ρj ≥ Ef (ρ ) , (25) j j

AB P AB where the last inequality follows from the convexity of Ef and the fact that ρ = j pjρj . However, A|BC AB all the inequalities above must be equalities since Ef (ρ ) = Ef (ρ ). In particular, we get

X A|BC X AB pjEf |ψji = pjEf ρj . (26) j j

A|BC AB A|BC AB This in turn implies that Ef |ψji = Ef ρj for each j since Ef |ψji ≥ Ef ρj for each j (i.e. tracing out subsystem cannot increase entanglement). Since we assume that Ef is monogamous on AC AC ABC pure tripartite states, we conclude that for each j, Ef (ρj ) = 0, where ρj ≡ TrB|ψjihψj| . Hence, AC AC P AC Ef (ρ ) = 0 since ρ = j pjρj and Ef is convex.

D Entanglement of Collaboration and Monogamy of Entanglement

Monogamy of entanglement is closely related to entanglement of collaboration. Given a measure of bipar- AB|C tite entanglement E, its corresponding entanglement of collaboration, Ec , is a measure of entanglement on tripartite mixed states, ρABC , given by [32]:

n AB|C ABC X AB Ec ρ = max pjE ρj , (27) j=1

AB where the maximum is taken over all tripartite LOCC protocols yielding the bipartite state ρj with probability pj. The following theorem demonstrates the connection between the disentangling condition and entanglement of collaboration.

Theorem 4. Let E be an entanglement monotone on bipartite mixed states, and let ρABC be a (possibly mixed) tripartite state satisfying the disentangling condition (4). Then,

AB AB|C ABC E ρ = Ec ρ . (28)

Accepted in Quantum 2018-08-05, click title to verify 4 AB ABC Proof. Let {ρj , pj} be the optimal ensemble in (8) obtained by LOCC on ρ . Since E is a bipartite entanglement monotone, it does not increase on average:

A|BC X AB AB|C ABC E ρ ≥ pjE(ρj ) = Ec ρ . (29) j

AB AB|C ABC On the other hand, by deﬁnition E(ρ ) ≤ Ec ρ , so that together with (4) we get (9).

Entanglement of collaboration is different from entanglement of assistance, Ea, in which the optimization in (27) is restricted to LOCC of the following form: Charlie (system C) performs a measurement, and communicates the outcome j to Alice and Bob. In [32] the following LOCC protocol was considered: Alice performs a measurement, then sending the outcome to Charlie, and then Charlie performs his measurement, and sends back the result to Alice and Bob. It was shown that in such a scenario it is possible to increase the average entanglement between systems A and B to a value beyond the average AB|C entanglement that can be achieved if only Charlie performed a measurement. Therefore, Ec can be AB|C ABC strictly larger than Ea, and in general, Ec ≥ Ea. However, if ρ satisfies the disentangling condition AB|C ABC then we must have Ec = Ea. Indeed, if ρ satisfies (4) we get

ABC AB A|BC AB|C ABC Ea(ρ ) ≥ E(ρ ) = E ρ = Ec ρ . (30)

AB|C Therefore, one can replace Ec in (28) with Ea, which may be convenient since Ea is somewhat a AB|C AB|C A|BC AB|C ABC simpler measure than Ec . Note however that we left Ec in (28) since E ρ = Ec ρ A|BC ABC implies E ρ = Ea ρ but not vice versa.

D.1 When Entanglement of Formation equals Entanglement of Assistance? An immediate consequence of Theorem4 above is that if ρABC is a pure state that satisﬁes the disentangling condition then the entanglement of formation of ρAB must be equal to its entanglement of assistance.

Corollary 5. Let E be an entanglement monotone on bipartite states, and let ρABC be a pure tripartite state satisfying the disentangling condition (4). Then,

AB AB AB E ρ = Ef (ρ ) = Ea(ρ ) , (31) where the entanglement of formation, Ef , is deﬁned in (23), and the entanglement of assistance, Ea, is also deﬁned as in (23) but with a maximum replacing the minimum.

Proof. The proof follows straightforwardly from Theorem (4) recalling that

AB AB AB AB|C ABC A|BC E ρ ≤ Ef (ρ ) ≤ Ea(ρ ) ≤ Ec ρ = E ρ , (32) where the ﬁrst inequality follows from the fact that E is an entanglement monotone, and the last equality from Theorem (4). Therefore, all the inequalities above are equalities since we assume the disentangling condition E ρA|BC = E ρAB . This completes the proof.

In the following theorem we show that the equality between the entanglement of formation and entanglement of assistance is a property of the support space of the the bipartite state in question.

Theorem 6. Let E be a measure of bipartite entanglement, and let ρAB ∈ SAB be a bipartite state AB AB AB AB AB AB with support subspace supp(ρ ). If Ef (ρ ) = Ea(ρ ) then for any σ ∈ S with supp(σ ) ⊆ AB AB AB supp(ρ ) we have Ef (σ ) = Ea(σ ).

Accepted in Quantum 2018-08-05, click title to verify 5 AB P AB Proof. In the proof we use some of the ideas introduced in [33]. Suppose ρ = j pjρj , where {pj} AB AB AB AB are (non-zero) probabilities and ρj ∈ S . Then, the condition Ef (ρ ) = Ea(ρ ) together with the convexity (concavity) of Ef (Ea) gives

X AB AB AB X AB pjEf (ρj ) ≥ Ef (ρ ) = Ea(ρ ) ≥ pjEa(ρj ) . (33) j j

AB AB AB AB But since for all j we also have Ef (ρj ) ≤ Ea(ρj ), we get that Ef (ρj ) = Ea(ρj ) for all j. Let F(ρAB) be the set of all density matrices in SAB that appear in a convex decomposition of ρAB. In convex analysis, F(ρAB) is called a face of SAB. Now, from the argument above we have that if σAB ∈ AB AB AB AB AB F(ρ ) then Ef (σ ) = Ea(σ ). On the other hand, for any σ ∈ S with the property that supp(σAB) ⊂ supp(ρAB) there exists a small enough > 0 such that ρAB − σAB ≥ 0. Denoting by γAB ≡ ρAB − σAB /(1 − ) we get that γAB ∈ SAB and ρAB can be expressed as the convex AB AB AB AB AB combination ρ = σ + (1 − )γ . We therefore must have Ef (σ ) = Ea(σ ). This completes the proof.

In the proof above we called F(ρAB) a face. A face F of SAB is a convex subset of SAB that satisﬁes AB the following property: if tρ1 + (1 − t)ρ2 ∈ F for some ρ1, ρ2 ∈ S and 0 < t < 1, then ρ1, ρ2 ∈ F. To AB AB see that F(ρ ) is a face of S , we ﬁrst show that it is convex. Indeed, let σ ≡ tσ1 + (1 − t)σ2 for some AB AB AB t ∈ [0, 1] and σ1, σ2 ∈ F(ρ ). Since σ1, σ2 ∈ F(ρ ) there exists p, q ∈ (0, 1] and γ1, γ2 ∈ S such that

AB ρ = pσ1 + (1 − p)γ1 = qσ2 + (1 − q)γ2 . (34)

t AB t(1−p) 1−t AB The ﬁrst equality implies that p ρ = tσ1 + p γ1, and the second equality gives q ρ = (1 − t)σ2 + (1−t)(1−q) q γ2. Therefore,

t 1 − t t(1 − p) (1 − t)(1 − q) + ρAB = σAB + γ + γ . (35) p q p 1 q 2

t 1−t AB AB After dividing by p + q , we can see that σ appears in a convex combination of ρ . Therefore, AB AB F(ρ ) is convex. To complete the proof that F(ρ ) is a face, note that if τ ≡ tρ1 + (1 − t)ρ2 ∈ F for AB AB some ρ1, ρ2 ∈ S and 0 < t < 1 then clearly ρ1, ρ2 ∈ F(ρ ). Note that the condition supp(σAB) ⊆ supp(ρAB) is equivalent to σAB ∈ F(ρAB). The precise form of AB AB AB the support space of a bipartite state ρ that satisﬁes Ef (ρ ) = Ea(ρ ) depends on the measure of entanglement E. In the following sections we ﬁnd it precisely for the case where E is the G-concurrence, and we use it to show that the G-concurrence is monogamous.

D.2 Monogamy of the G-concurrence

d d Any pure bipartite state, |xi ∈ C ⊗ C , can be written as:

d X A B |xi = X ⊗ I|φ+i where |φ+i = |ii |ii , (36) i=1 and X is a d×d complex matrix. The relation above between a complex matrix X ∈ Md(C) and a bipartite d d d d vector |xi ∈ C ⊗C deﬁnes an isomorphism between Md(C) and C ⊗C . Using this isomorphism, in the remaining of this section we will view interchangeably the support of a density matrix both as a subspace d d of Md(C) or as a subspace of C ⊗ C , depending on the context. We will use capital letters X,Y,Z,W

Accepted in Quantum 2018-08-05, click title to verify 6 for matrices in Md(C), and use lower case letters |xi, |yi, |zi, |wi to denote their corresponding bipartite vectors. With these notations, the G-concurrence of |xi, which is the geometric mean of the Schmidt coeﬃcients of |xi, can be expressed as: G(|xi) = |det(X)|2/d , (37) AB AB and for mixed states it is deﬁned in terms of the convex roof extension; that is, G(ρ ) := Gf (ρ ) for any ρAB ∈ SAB. Note that the G-concurrence is homogeneous, and in particular, G(c|xi) = |c|2G(|xi). We start in proving the following Lemma: AB d d AB Lemma: Let ρ be a bipartite density matrix acting on C ⊗ C with G(ρ ) > 0. Then, there exists a pure state decomposition of ρAB with the following properties:

r AB X ρ = |wjihwj| ,G(|wji) = 0 , ∀ j = 2, ..., r, (38) j=1

AB d d where r is the rank of ρ and |wji are sub-normalized vectors in C ⊗ C (i.e. vectors with norm no greater than 1).

AB Pr AB Proof. Let ρ = j=1 |xjihxj| be the spectral decomposition of ρ with |xji being the sub-normalized AB eigenvectors of ρ . Clearly, G(|xji) > 0 for at least one j. Therefore, w.l.o.g. we assume G(|x1i) > 0. AB Let K2 ⊂ supp(ρ ) be the two dimensional subspace spanned by X1 and X2. We ﬁrst show that K2 contains a matrix with zero determinant. Indeed, if det(X2) = 0 we are done. Otherwise, for any λ ∈ C we get

2/d G(|x1i + λ|x2i) = |det(X1 + λX2)| 2/d 2/d −1 = |det(X2)| det(X1X2 + λI) . (39)

−1 Note that det(X1X2 + λI) is a polynomial of degree d in λ and must have at least one complex root. Therefore, there exists λ = λ0 6= 0 such that det(X1 + λ0X2) = 0. This completes the assertion that K2 contains a matrix with zero determinant. Next, denote a ≡ √ 1 and b ≡ √ λ0 , so that |a|2 +|b|2 = 1, and the vectors |w i ≡ a|x i+b|x i 2 2 2 1 2 1+|λ0| 1+|λ0| and |yi ≡ ¯b|x1i − a¯|x2i satisfy

|x1ihx1| + |x2ihx2| = |yihy| + |w2ihw2| (40) with G(|w2i) = 0 (by construction, W2 is proportional to X1 + λ0X2 and therefore has zero determinant). We can therefore write r AB X ρ = |w2ihw2| + |yihy| + |xjihxj| . (41) j=3 r Now, if G(|yi) = 0 then we denote it as |w3i and pick from {|xji}j=3 another state that does not have a vanishing G-concurrence. We therefore assume G(|yi) 6= 0 and from the same arguments as above we conclude that |yihy| + |x3ihx3| = |y˜ihy˜| + |w3ihw3| for some vectors |y˜i and |w3i, with G(|w3i) = 0. By replacing |yi and |x3i with |y˜i and |w3i, and repeating the process we arrive at the desired decomposition of ρAB.

In the following theorem we denote by N ⊂ Md(C), the nilpotent cone, consisting of all d × d nilpotent complex matrices (i.e. X ∈ N iﬀ Xk = 0 for some 1 ≤ k ≤ d). While the set N is not a vector space, it contains subspaces. For example, the set T of all strictly upper triangular matrices (i.e. upper triangular

Accepted in Quantum 2018-08-05, click title to verify 7 matrices with zeros on the diagonal) is a d(d − 1)/2-dimensional subspace in N . From Gerstenhaber’s theorem [34] it follows that the largest dimension of a subspace in N is d(d − 1)/2, and if a subspace N0 ⊂ N has this maximal dimension, then it must be similar to T (i.e. their exists an invertible matrix −1 S such that N0 = ST S ).

AB d d Theorem 7. Let ρ be a bipartite density matrix acting on C ⊗ C , and suppose it has rank r > 1. Then, AB AB G(ρ ) = Ga(ρ ) > 0 , (42) if and only if r ≤ 1 + d(d − 1)/2, and there exists a full rank matrix X ∈ Md(C) and a subspace N0 ⊂ N with dim(N0) = r − 1 such that

AB supp(ρ ) = XK ≡ {XY Y ∈ K} , (43) where K ≡ {I} ⊕ N0 ≡ {cI + N c ∈ C ; N ∈ N0}. AB d d Remark 3. In the statement of Theorem7 of the main text, we viewed supp(ρ ) as a subspace of C ⊗C . d d Here, for convenience of the proof, we use the isomorphism (36) between Md(C) and C ⊗ C , and view AB AB the supp(ρ ) as a subspace of Md(C). The matrix X in (43) is related to the bipartite state |xi in (13) via (36). Recall also that a nilpotent matrix has a zero trace, and is orthogonal to the identity matrix in the Hilbert-Schmidt inner product. This is consistent with the direct sum in the deﬁnition of K.

AB Proof. Suppose there exist X and N0 as above such that supp(ρ ) ⊆ XK, and let r AB X ρ = |wjihwj| , (44) j=1 where |wji are the sub-normalized vectors given in (38); i.e. G(|wji) = 0 for j > 1. Since Wj ∈ XK, there exist constants cj ∈ C and matrices Zj ∈ N0 (corresponding to some sub-normalized vectors |zji) such that Wj = cjX + XZj . (45) For j > 1 we have

2/d 2/d 0 = G(|wji) = |det(cjX + XZj)| = G(|xi) |det(cjI + Zj)| (46)

d so that det(cjI + Zj) = 0 since G(|xi) > 0. But since Zj is a nilpotent matrix det(cjI + Zj) = (cj) . AB Hence, cj = 0 for j > 1 and we denote by c ≡ c1. Note that c 6= 0 since otherwise we will get G(ρ ) = 0. Therefore, the average G-concurrence of decomposition (44) is given by

r X 2/d 2 G(|wji) = G(|w1i) = |det(cX + XZ1)| = G(|xi) |c| . (47) j=1 Next, let m AB X ρ = |ykihyk| , (48) k=1 AB be another pure state decomposition of ρ with m ≥ r and {|yki} some sub-normalized states. Then, † there exists an m × r isometry U = (ukj), i.e. U U = Ir, such that r X |yki = ukj|wji . (49) j=1

Accepted in Quantum 2018-08-05, click title to verify 8 Using the form (45) with c1 ≡ c and c2 = c3 = ... = cr = 0, we get

r  r  X X Yk = ukjWj = X  ukj (cδ1jI + Zj) ≡ X (uk1cI + Nk) , (50) j=1 j=1

Pr where Nk ≡ j=1 ukjZj ∈ N0 are nilpotent matrices. Consequently,

2/d 2 2 G(|yki) = det X (uk1cI + Nk) = |uk1| |c| G(|xi) , (51)

d where we used the fact that det(uk1cI + Nk) = (uk1c) since Nk is nilpotent. Note that since the matrix Pm 2 U is an isometry, k=1 |uk1| = 1. Therefore, m r X 2 X G(|yki) = |c| G (|xi) = G(|wji) , (52) k=1 j=1 where the last equality follows from (47). Therefore, all pure states decomposition of ρAB have the same AB AB average G-concurrence so that G(ρ ) = Ga(ρ ). AB AB To prove the converse, suppose G(ρ ) = Ga(ρ ) > 0, and consider the decomposition (38) of AB AB AB ρ as in the Lemma above. Consider the unnormalized state σ ≡ ρ − |w1ihw1|. The state AB Pr σ = j=2 |wjihwj| has zero G concurrence since G(|wji) = 0 for all j = 2, ..., r. Consider another AB Pr AB AB decomposition of σ = j=2 |yjihyj|. Since G(ρ ) = Ga(ρ ) we must have that G(|yji) = 0 for all AB Pr j = 2, ..., r. Otherwise, the decomposition ρ = |w1ihw1| + k=2 |ykihyk| will have a higher average G-concurrence than the decomposition in (38). But since the {|yji} decomposition was arbitrary, we conclude that all the states in the subspace W ≡ span{|w2i, ..., |wri} have zero G-concurrence. −1 We now denote W1 ≡ X and for j = 2, ..., r we set Nj ≡ X Wj. Note that with these nota- AB tions, supp(ρ ) = span{X,XN2,XN3, ..., XNr}. Since X is invertible, and det(W ) = 0 for any matrix W in the span of W2, ..., Wr, we also have det(N) = 0 for any N in the span of N2, ..., Nr. Let N ∈ span{N2, ..., Nr} be a ﬁxed matrix, and consider the two dimensional subspace W2 ≡ {X,XN} ⊂ AB supp(ρ ). We ﬁrst show that W2 does not contain a matrix with zero determinant that is linearly independent of W ≡ XN. Indeed, if there exists a normalized matrix Z ∈ W such that det(Z) = det(W ) = 0 with W, Z being linearly independent, then W2 = span{W, Z}, and the rank 2 density matrix

σAB = |wihw| + |zihz| (53) must have zero G-concurrence (recall that G(|wi) = G(|zi) = 0). On the other hand, since |xi ∈ W2 = supp(σAB), the density matrix σAB must have a pure state decomposition containing |xi. Since G(|xi) > 0 AB AB this decomposition does not have a zero average G-concurrence. Therefore, 0 = G(σ ) < Ga(σ ). AB AB AB However, from Theorem6 any density matrix σ with a support supp(σ ) = W2 ⊂ supp(ρ ) has the same average G-concurrence for all pure state decompositions. We therefore get a contradiction with Theorem6, and thereby prove the assertion that W2 does not contain another matrix with zero determinant that is linearly independent of W . Since W = XN is the only matrix in W2 with zero determinant (up to multiplication by a constant), we must have that for any λ 6= 0 1 0 6= det(X + λW ) = det(X−1) det(I + λX−1W ) = λd det(X−1) det I + N . (54) λ

1 Setting t ≡ λ we conclude that the polynomial f(t) ≡ det (tI + N) (55)

Accepted in Quantum 2018-08-05, click title to verify 9 is never zero for t 6= 0. On the other hand, f(t) is a polynomial of degree d and the coeﬃcient of td is one (note that it is the characteristic polynomial of −N). But since t = 0 is the only root of f(t), we must have f(t) = td; that is, n det (tI + N) = t , ∀ t ∈ C . (56)

Therefore, N must be a nilpotent matrix. Since N was arbitrary, it follows that the subspace N0 ≡ span{N2, ..., Nr} is a subspace of nilpotent matrices. This completes the proof. The G-concurrence is deﬁned in (37) on pure bipartite states with the same local dimension. For a pure bipartite state |ψiAB ∈ HA ⊗ HB with dim(HA) < dim(HB) it is deﬁned by:

1/dA G(|ψiAB) = det(ρA) , (57)

A AB A where ρ ≡ TrB |ψihψ| is the reduced density matrix, and dA ≡ dim(H ). With this extended deﬁnition, we have the following result:

ABC d d n AB Corollary 8. Let |ψ i ∈ C ⊗ C ⊗ C be a pure tripartite state, with bipartite marginal ρ . If G(ψA|BC ) = G(ρAB) > 0 , (58)

ABC AB C AB d d C n then |ψi = |χi |ϕi for some bipartite state |χi ∈ C ⊗ C and a vector |ϕi ∈ C . Proof. Since |ψiABC satisﬁes the disentangling condition, we get from Corollary5 that G(ρAB) = AB Ga(ρ ) > 0. Therefore, from the theorem above there exists a pure state decomposition of the marginal AB state ρ consisting of sub-normalized bipartite states {|wji} as in (38), with the form

r A † † X † † ρ = X(cI + N1)(¯cI + N1 )X + X NjNj X . (61) j=2 Therefore,

1/d   r  1/d A|BC A 2/d 2 X 2 G(|ψ i) = det(ρ ) = | det(X)| det |cI + N1| + |Nj|  , (62) j=2 √ where we used the notation |A| ≡ AA† for any d × d matrix A. Hence, the condition G(|ψA|BC i) = G(ρAB) = | det(X)|2/d|c|2 gives

1/d   r  2 2 X 2 |c| = det |cI + N1| + |Nj|  . (63) j=2

Accepted in Quantum 2018-08-05, click title to verify 10 Since c 6= 0 the matrix cI + N1 is invertible. Denoting

 r  −1 X 2 −1 A ≡ |cI + N1| and B ≡ A  |Nj|  A , (64) j=2 we get that (63) can be expressed as

h i1/d |c|2 = [det (A(I + B)A)]1/d = det A2 [det(I + B)]1/d . (65)

But since N1 is nilpotent, we have

1/d h 2i 2/d 2/d 2 det A = [det (|cI + N1|)] = |det(cI + N1)| = |c| . (66)

We therefore conclude that det(I + B) = 1 . (67)

Pr 2 However, since B ≥ 0, Eq. (67) can hold only if B = 0. This in turn is possible only if j=2 |Nj| = 0; AB i.e. Nj = 0 for all j = 2, ..., d. We therefore conclude that ρ is a pure state which implies that ABC AB C C |ψi = |w1i |ϕi , where |ϕi is some pure state.

Note that by combining the above corollary with Theorem3 we get that the G-concurrence is fully monogamous (even for mixed tripartite states). This is to our knowledge the ﬁst example of a monogamous measure of entanglement that is highly non-faithful. We conclude now with the 2-dimensional version of Theorem7. Recall that the concurrence is the two dimensional G-concurrence.

Corollary 9. Let ρAB be an entangled two qubit state with rank r > 1, and let E be any injective (up to local unitaries) measure of pure two qubit entanglement. Then,

AB AB 1. If Ef (ρ ) = Ea(ρ ) then r = 2, and

AB ABC ρ = TrC |W ihW |

is the 2-qubit marginal of the W-class state

|W i = λ1|100i + λ2|010i + λ3|001i + λ4|000i , (68)

P4 2 where λj ∈ C and j=1 |λj| = 1.

AB AB AB 2. Conversely, if ρ is a marginal of a state in the W-class then Cf (ρ ) = Ca(ρ ), where C is the concurrence. Proof. Part 1: In the two qubit case, we can also write E(|ψiAB) = g C(|ψiAB) , where C(|ψiAB) is the concurrence measure of entanglement, which is itself an injective measure. Moreover, since E is injective so is the function g. Now, in [35], Wootters showed that there always exists a pure state decomposition AB AB AB of ρ with each element in the decomposition being equal to Cf (ρ ). Denoting by {pj, |ψji } this decomposition, we have that

AB X AB AB Ef ρ ≤ pjg C(|ψji ) = g Cf (ρ ) . (69) j

Accepted in Quantum 2018-08-05, click title to verify 11 AB AB AB Similarly, there exists a pure state decomposition {qk, |φki } of ρ such that for each k, C |φki = AB Ca ρ [36]. Hence, AB X AB AB Ea ρ ≥ qkg C(|φki ) = g Ca(ρ ) . (70) k AB AB AB AB Therefore, the equality Ef (ρ ) = Ea(ρ ) implies that Cf (ρ ) = Ca(ρ ). so that we can ap- ply Theorem7. There is only one 1-dimensional subspace of N (up to similarity), given by N0 = ( ! ) 0 1 span S S−1 , where S is a 2 × 2 invertible matrix. Now, from (13) together with the decompo- 0 0 sition (38) we conclude that ρAB can be expressed as:

ρAB = |xihx|AB + |yihy|AB , (71)

AB B where |xi = X ⊗ I |φ+i for some invertible matrix X, and ! 0 1 −1 AB |yi ≡ I ⊗ S S xi . (72) 0 0

Therefore, ρAB is a marginal of the tripartite state

Multiplying the above 3-qubit state by the SLOCC element ST X−1 ⊗ S−1 ⊗ IC , and using the symmetry T −1 S ⊗ S |φ+i = |φ+i for any invertible matrix S, gives ! ! T −1 −1 C ABC T −1 AB C T 0 1 AB C S X ⊗ S ⊗ I |ψi = S ⊗ S |φ+i |0i + S ⊗ S|φ+i |1i 0 0 ! ! AB C A 0 1 AB C = |φ+i |0i + I ⊗ |φ+i |1i 0 0 = |000i + |110i + |101i . (74) note that the state above is the W state after a local flip of the first qubit. Therefore, |ψiABC is in the W class. q Proof of Part 2: We use for this part Wootters’ formula for the concurrence. Let R = ρ1/2ρρ˜ 1/2 AB ∗ be Wootters matrix [35], with ρ ≡ ρ and ρ˜ ≡ σy ⊗ σyρ σy ⊗ σy. Wootters showed in [35] that for AB entangled states Cf (ρ ) = λ1 − λ2 − λ3 − λ4, where λ1, ..., λ4 are the eigenvalues of R in decreasing AB order. Furthermore, in [37, 36] it was shown that Ca(ρ ) = Tr[R] = λ1 + λ2 + λ3 + λ4. Therefore, AB AB Cf (ρ ) = Ca(ρ ) if and only if R is a rank one matrix. A straightforward calculation shows that for the bipartite marginals, of the W-class states (68), Tr[R2] = (Tr[R])2 so that the rank of R is one (recall that R is positive semidefinite). This completes the proof.

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Accepted in Quantum 2018-08-05, click title to verify 13 [24] J. S. Kim and B. C. Sanders. Monogamy of multi-qubit entanglement using Rényi entropy. J. Phys. A, 43:445305, 2010. doi:10.1088/1751-8113/43/44/445305. [25] M. F. Cornelio and M. C. de Oliveira. Strong superadditivity and monogamy of the Rényi measure of entanglement. Phys. Rev. A, 81:032332, 2010. doi:10.1103/PhysRevA.81.032332. [26] Song et al. General monogamy relation of multiqubit systems in terms of squared Rényi-α entanglement. Phys. Rev. A, 93:022306, 2016. doi:10.1103/PhysRevE.93.022306. [27] Y. Guo, J. Hou, and Y. Wang. Concurrence for inﬁnite-dimensional quantum systems. Quant. Inf. Process., 12:2641-2653, 2013. doi:10.1007/s11128-013-0552-6. [28] Y. Guo and J. Hou. Entanglement detection beyond the CCNR criterion for inﬁnite-dimensions. Chin. Sci. Bull., 58(11):1250-1255, 2013. doi:10.1007/s11434-013-5738-x. [29] M. J. Donald and M. Horodecki. Continuity of relative entropy of entanglement. Phys. Lett. A, 264:257, 1999. doi:10.1016/S0375-9601(99)00813-0. [30] The continuty of the convex roof extended entanglement measure can be checked according to Propo- sition 2 in [27], the continuty of partial trace and partial transpose is proved in [28], the continuty of the realtive entropy entanglement is proved in [29]. [31] P. Hayden, R. Jozsa, D. Petz, and A. Winter. Structure of states which satisfy strong subadditivity of quantum entropy with equality. Commun. Math. Phys., 246(2):359-374, 2004. doi:10.1007/s00220- 004-1049-z. [32] G. Gour and R. W. Spekkens. Entanglement of assistance is not a bipartite measure nor a tripartite monotone. Phys. Rev. A, 73:062331, 2006. doi:10.1103/PhysRevA.73.062331. [33] A. Uhlmann. Roofs and Convexity. Entropy, 12:1799-1832, 2010. doi:10.3390/e12071799. [34] M. Gerstenhaber. On nilalgebras and linear varieties of nilpotent matrices (I). Amer. J. Math., 80:614- 622, 1958. doi:10.2307/2372773. [35] W. K. Wootters. Entanglement of formation of an arbitrary state of two qubits. Phys. Rev. Lett., 80:2245, 1998. doi:10.1103/PhysRevLett.80.2245. [36] G. Gour, D. A. Meyer, and B. C. Sanders. Deterministic entanglement of assistance and monogamy constraints. Phys. Rev. A, 72:042329, 2005. doi:10.1103/PhysRevA.72.042329. [37] T. Laustsen, F. Verstraete, and S. J. van Enk. Local vs. joint measurements for the entanglement of assistance. Quant. Inf. Comput., 3:64, 2003. arXiv:0206192.

Accepted in Quantum 2018-08-05, click title to verify 14