Quantum state merging with bound entanglement

Quantum state merging is one of the most important protocols in quantum information theory. In this task two parties aim to merge their parts of a pure tripartite state by making use of additional singlets while preserving correlations with a third party. We study a variation of this scenario where the shared state is not necessarily pure, and the merging parties have free access to local operations, classical communication, and positive partial transpose (PPT) entangled states. We provide general conditions for a state to admit perfect merging, and present a family of fully separable states which cannot be perfectly merged if the merging parties have no access to additional singlets. We also show that free PPT entangled states do not give any advantage for merging of pure states, and the conditional entropy plays the same role as in standard quantum state merging quantifying the rate of additional singlets needed to perfectly merge the state.


Introduction
Quantum state merging can be understood as a game involving three players, which we will call Alice, Bob and Charlie in the following. Initially, they share a large number of copies of a joint pure state | | y y ñ = ñ ABC , and the aim of Bob and Charlie is to merge their parts of the state on Charlie's side while preserving correlations with Alice. For achieving this, Bob and Charlie have access to additional singlets and a classical communication channel. Taking into account that singlets are considered as an expensive resource in quantum information theory, the main question of quantum state merging can be formulated as follows: How many singlets are required for perfect asymptotic merging per copy of the state |yñ? The answer to this question was found in [1,2]: the minimal number of singlets per copy is given by the conditional entropy S(ρ BC )−S(ρ C ).
Noting that the conditional entropy can be positive or negative, it is surprising that it admits an operational interpretation in both cases. In particular, if the conditional entropy is positive, Bob and Charlie will require S(ρ BC )−S(ρ C ) singlets per copy for perfectly merging the total state |yñ in the asymptotic limit, and perfect merging cannot be accomplished if less singlets are available. On the other hand, if S(ρ BC )−S(ρ C ) is negative, Bob and Charlie can asymptotically merge the state |yñ without any additional singlets by only using local operations and classical communication (LOCC). Moreover, Bob and Charlie can gain additional singlets at rate S(ρ C )−S(ρ BC ), and store them for future use [1,2].
Another important concept in quantum information theory is the framework of entanglement distillation [3][4][5]. One of the most surprising features in this context is the phenomenon of bound entanglement: there exist entangled states from which no singlets can be distilled [6]. Moreover, it is known that all states with positive partial transpose (PPT) are nondistillable [6], while it is still an open question if there exist bound entangled states with nonpositive partial transpose (NPT) [7].
In this paper we introduce and study the task of PPT quantum state merging (PQSM). Similar to standard quantum state merging, PQSM can be considered as a game between three players who share a joint mixed state ρ=ρ ABC . The aim of the game for Bob and Charlie is to merge their parts of the state ρ on Charlie's side while preserving correlations with Alice. In contrast to standard quantum state merging, Bob and Charlie can use unlimited amount of PPT entangled states, see figure 1 for illustration. The situation where Bob and Charlie do not have access to PPT entangled states is known as LOCC quantum state merging (LQSM), and has been introduced in [8]. Any further distribution of this work must maintain attribution to the author(s) and the title of the work, journal citation and DOI.
Before we discuss the concept of PQSM and present our main results, we will introduce PPT assisted LOCC operations in the following.

PPT assisted LOCC
For a tripartite state ρ ABC shared between Alice, Bob and Charlie, a PPT assisted LOCC protocol performed by Charlie and Bob will be denoted by Λ PPT and has the following form: Here,˜m BC PPT is an arbitrary PPT state shared by Bob and Charlie, and Λ LOCC is an LOCC protocol between them, see figure 1.
We also introduce PPT distillable entanglement D PPT as the singlet rate which can be asymptotically obtained from a state via PPT assisted LOCC. This quantity is in full analogy to the standard distillable entanglement [3] that quantifies the singlet rate which can be obtained via LOCC only. We will denote the latter by D LOCC .
If we further introduce the PPT and LOCC entanglement cost C PPT and C LOCC as the entanglement cost for creating a state via the corresponding set of operations, we immediately obtain the following inequality: This relation follows by noting that PPT assisted LOCC is more general than LOCC only, and by the fact that the PPT entanglement cost cannot be below the PPT distillable entanglement. Since for all pure states D LOCC and C LOCC are equal to the von Neumann entropy of the reduced state [4], all quantities in equation (2) coincide for pure states. In the following, we will also use the logarithmic negativity [9, 10] is the trace norm of M. The logarithmic negativity is an upper bound on D LOCC [10].
We further note that PPT assisted LOCC is a subclass of general PPT preserving operations. It is however not clear whether or not these two classes coincide.

PPT quantum state merging
We are now in position to introduce the aforementioned task of PQSM. In this task, Bob and Charlie aim to merge their parts of the total state ρ=ρ ABC by using PPT assisted LOCC operations, see figure 1 for illustration. A natural figure of merit for this process is the fidelity of PQSM:  [ where ρ R is an arbitrary initial state of Charlie's register R. The supremum in equation (4) is taken over all PPT assisted LOCC operations Λ PPT between Bob's system B and Charlie's system CR, see also figure 1 for details. A state ρ admits perfect single-shot PQSM if and only if the corresponding fidelity is equal to one: otherwise. In the asymptotic scenario where a large number of copies of the state ρ is available, the figure of merit is the asymptotic fidelity of PQSM: This quantity can be regarded as a natural quantifier for asymptotic PQSM, since a state ρ admits perfect asymptotic PQSM if and only if

Perfect asymptotic PQSM
In the following we will focus on those states ρ=ρ ABC which admit perfect asymptotic PQSM: In particular, perfect asymptotic PQSM is always possible if the state ρ has nonpositive conditional entropy: This follows from the fact that in this situation Bob and Charlie can achieve perfect asymptotic merging for the purification of ρ just by using LOCC [1,2,8]. Moreover, equation (8) also implies that states satisfying equation (7) have nonzero measure in the set of all states, since this is evidently true for states satisfying equation (8).
At this point, we also note that perfect asymptotic PQSM is only possible if the state ρ satisfies the following condition: where X:Y denotes a bipartition between two (possibly multipartite) subsystems X and Y. To see this, consider the overall state ρ ABC ⊗ρ R , where R is a register in Charlie's hands. If this state allows for perfect asymptotic PQSM, there exists a PPT assisted LOCC protocol Λ PPT between Bob and Charlie such that Consider now the PPT distillable entanglement in the bipartition AB:CR. By its very definition, PPT distillable entanglement cannot grow under PPT assisted LOCC operations, and we obtain Noting that the states ρ ABC and ρ ACR differ only by relabeling B and R completes the proof of equation (9). For states which satisfy equation (9) but violate equation (8) no conclusive statement can be made in general. One important subclass of such states are fully separable states, and it is easy to provide examples for such states which violate equation (8), but still can be merged via LOCC even on the single-copy level. In the following we will show that the investigation of such states can be simplified significantly. This will also lead us to a new class of fully separable states which cannot be merged via asymptotic PQSM.

Single-shot versus asymptotic PQSM
In the following, we consider the situation where the total state ρ=ρ ABC is PPT with respect to the bipartition AB:C. The set of these states includes the aforementioned set of fully separable states. The following theorem shows that for all such states the single-copy fidelity is never smaller than for any number of copies.
Theorem 1. Given a tripartite state r r = ABC which is PPT with respect to AB:C, the following inequality holds for any  n 1: This also implies that in this case the single-shot fidelity cannot be smaller than the asymptotic fidelity: We refer to the appendix for the proof.Crucially, this result also means that perfect single-shot PQSM is fully equivalent to perfect asymptotic PQSM for all such states: The importance of this result lies in the fact that it remarkably simplifies the analysis, if one is interested in the question whether a state ρ admits perfect asymptotic PQSM or not. For all such states we only need to study the single-copy situation: if perfect PQSM is not possible in the single-copy case, it is also not possible asymptotically.
As an application of theorem 1, we will now present a general family of fully separable states which does not admit perfect asymptotic PQSM. These states are given by where all probabilities p i are nonzero, and the two-qubit states s i BC are all separable and chosen such that their generalized Bloch vectors are all linearly independent. For the proof that such states exist and that they indeed do not allow for perfect asymptotic PQSM we refer to the appendix.

States with vanishing asymptotic fidelity
Taking into account the results discussed so far, it is natural to ask whether the asymptotic fidelity ¥  PPT can attain only one of two values, namely 0 or 1. We can neither prove nor disprove this at the moment. Nevertheless, we will provide strong evidence for this in the following, showing that a significant amount of quantum states has vanishing asymptotic fidelity: This happens for all states which are distillable between A and BC, and at the same time have PPT in the bipartition AB:C. These two conditions are summarized in the following inequality: : The proof of this statement can be found in the Appendix. At this point it is also interesting to note that the asymptotic fidelity ¥  PPT is not a continuous function of the state. This discontinuity is present even for pure states, and can be demonstrated on the following example: Note that this state admits perfect PQSM whenever |yñ is a product state, i.e. | | | y a b ñ = ñ Ä ñ A B . In this case, perfect merging can be accomplished without any communication if Charlie prepares his register R in the state |bñ R . Note however that the asymptotic fidelity ¥  PPT vanishes for any entangled state |yñ, as follows directly from the above discussion.
As is further shown in the appendix, the set of states having vanishing asymptotic fidelity has nonzero measure in the set of all states. Combining these results with our previous findings, namely that states satisfying ( ) r = ¥  1 PPT also have nonzero measure in the set of all states, this means that both of these sets have finite size. We hope that this result can serve as a starting point to prove that ¥  PPT can take as values only 0 or 1.

Absence of bound entanglement
The results presented in this work can also be applied to the scenario where Bob and Charlie do not have access to PPT entangled states. This task is known as LQSM, and has been presented in [8]. The figure of merit in this case will be denoted by  LOCC . Note that the quantities  LOCC and  PPT obey the following relation: Here, I A:BC is the mutual information between A and BC, and  is the concentrated information introduced in [8]. The concentrated information quantifies the maximal amount of mutual information between Alice and Charlie obtainable via LOCC operations performed by Charlie and Bob, and can be considered as a figure of merit for LQSM on its own right. The first inequality in (18) follows from the fact that PPT assisted LOCC operations are more general than LOCC operations alone. The second inequality in (18) crucially relies on results from [11][12][13], and the proof can be found in [8].
The second inequality in (18) further implies that  LOCC and  PPT are nonzero for any finite-dimensional state ρ. This follows directly by noting that the concentrated information  is non-negative, and that the mutual information I A:BC is finite. The first inequality in (18) implies that all states with vanishing asymptotic PQSM fidelity also have zero asymptotic LQSM fidelity: . This means that all states ρ which fulfill equation (16) also have vanishing asymptotic LQSM fidelity: This result can be slightly generalized by using the same arguments as in the proof of equation (15). In particular, all states ρ which are distillable between A and BC but nondistillable with respect to AB:C have vanishing asymptotic fidelity for LQSM, i.e. . For proving this, we can use the same proof as for equation (15), by noting that the final state shared by Alice and Charlie will never be distillable if the initial state ρ satisfies equation (19), and if Bob and Charlie use LOCC operations only.
At this point we also note that equation (19) does not guarantee vanishing PQSM fidelity. In particular, if there exist NPT bound entangled states-and it is strongly believed that this is indeed the case [7]-Bob and Charlie could use PPT entangled states to perfectly merge a state of the form where the particles B 1 and B 2 are in Bob's hands, | (| | ) f ñ = ñ + ñ + 00 11 2 is a maximally entangled two-qubit state, and ρ NPT is an NPT bound entangled state with the property that D LOCC (ρ NPT ⊗μ PPT )>1 for some PPT entangled state μ PPT . Note that states ρ NPT and μ PPT with the aforementioned properties exist if there are NPT bound entangled states [14]. Bob and Charlie can then use the state μ PPT to distill the state ρ NPT ⊗μ PPT , and by applying Schumacher compression [15] We also note that all states ρ which fulfill the condition(8) admit perfect asymptotic LQSM [8], which also implies that states with ( ) r = ¥  1 LOCC have nonzero measure in the set of all states. Moreover, a state ρ admits perfect asymptotic LQSM only if it satisfies the following condition: Similar to the condition (9) for perfect asymptotic PQSM, equation (21) follows from the fact that distillable entanglement cannot increase under LOCC operations.

Comparison to standard quantum state merging
In the setting discussed so far we assumed that Bob and Charlie have free access to PPT entangled states together with LOCC. To compare our results to standard quantum state merging [1,2], we will now also allow Bob and Charlie to share singlets. The main question of this section can be formulated as follows: can shared PPT states reduce the singlet rate required for merging? As we will see in the following, the answer to this question is negative: also in the presence of PPT states the minimal singlet rate needed to achieve perfect merging of a tripartite pure state |yñ ABC corresponds to the conditional entropy S(ρ BC )−S(ρ C ). If Bob and Charlie have access to additional entangled states | ñ ¢ ¢ D i B C with initial distillable entanglement D i , perfect PQSM of the state | | y y ñ = ñ ABC can be seen as the following asymptotic transformation: Here, R is a register in Charlie's possession, and the state | ñ ¢ ¢ D f B C has final distillable entanglement D f . This condition means that by using additional singlets at rate D i , Bob and Charlie can perfectly merge the state |yñ in the asymptotic limit via PPT assisted LOCC, and will at the same time gain singlets at rate D f . The entanglement cost of the process is then given by D i −D f .
We will now show that the conditional entropy of the reduced state ρ BC is equal to the minimal entanglement cost of the above process. For this, we note that perfect merging is always possible at cost D i −D f =S(ρ BC )−S(ρ C ), since there exists an LOCC protocol accomplishing this task at this cost [1,2]. In the following, we will see that PPT assisted LOCC cannot lead to lower cost, i.e.
is true for any PPT assisted LOCC protocol achieving perfect merging as in equation (22). For proving this, we will introduce the initial state |Yñ i and the final state |Y ñ f . They correspond to the total state on the left-hand side and the right-hand side of equation (22), respectively. Using the fact that for pure states the PPT distillable entanglement D PPT is equal to the von Neumann entropy of the reduced state (see also equation (2) and discussion there), it is straightforward to verify the following equality: where the PPT distillable entanglement D PPT is considered with respect to the bipartition ABB′:CC′R. The desired inequality (23) follows by noting that D PPT cannot increase under PPT assisted LOCC, and thus

Conclusions
In this paper we introduced and studied the task of PQSM, where two parties-Bob and Charlie-aim to merge their shares of a tripartite mixed state by using PPT entanglement and classical communication, while preserving correlations with Alice. We considered the fidelity of this process, both in the single-copy and the asymptotic scenario, and showed that fully separable states can be perfectly merged asymptotically if and only if they can be perfectly merged on the single-copy level. We used this result to present a family of fully separable states which do not admit perfect asymptotic PQSM. We also identified very general conditions for a state to have vanishing fidelity of PQSM in the asymptotic limit. We showed that these conditions apply to a significant amount of quantum states having nonzero measure in the set of all states, thus proving that a large number of quantum states cannot be merged asymptotically with any nonzero precision. With respect to standard quantum state merging, our results imply that using additional PPT states does not change the entanglement cost of the process: the minimal singlet rate needed for perfectly merging a pure state in the asymptotic limit corresponds to the conditional entropy also in this extended setup.
We further note that the protocol considered here cannot be extended to the scenario where Bob and Charlie have access to arbitrary bound entangled states. In particular, if there exist NPT bound entangled states, the results presented in [14] immediately imply that Bob and Charlie also have access to an unlimited amount of singlets, and thus all states can be perfectly merged. On the other hand, if NPT bound entangled states do not exist, the scenario described here already represents the most general situation.
We expect that the tools presented here will find applications for other quantum communication protocols such as quantum state redistribution [16], also taking into account possible local constraints [17,18]. However, these questions are beyond the scope of this work.
where˜m BC PPT is a PPT state, Λ is an LOCC operation between Bob's total systemB B B 1 2 and Charlie's total system C C CR R 1 2 1 2 , and r R R 1 2 is an arbitrary initial state of Charlie's register. We will now prove equation (A1) by contradiction, assuming that it is violated for some state ρ which is PPT with respect to AB: C. In this case there must exist a PPT state μ PPT and an LOCC protocol Λ such that up to relabeling the parties B 1 and R 1 , and B 2 and R 2 .
We will now show that Bob and Charlie can 'simulate' such a two-copy protocol with just one copy of the state ρ, thus achieving a single-copy fidelity strictly above  PPT , which will be the desired contradiction. The basic idea of the proof is illustrated in figure 2. We assume that Alice, Bob and Charlie start with only one copy of the state ρ=ρ ABC , and that the state is PPT between AB and C. Since Bob and Charlie can prepare arbitrary PPT states, they can additionally prepare the state ρ A′ B′ C′ , which is equivalent to ρ ABC up to the fact that A′ and B′ are both in Bob's possession, see figure 2.
In the next step, Bob and Charly prepare a PPT state μ PPT and run the same LOCC protocol Λ which was leading to equation (A4). By following this strategy, they will end up with a final state s ¢ ¢ ¢ f AA CC RR having the property that Recalling that fidelity does not decrease under discarding subsystems, it follows that which is the desired contradiction. The proof for arbitrary n2 follows by applying the same arguments. Moreover, using the same ideas it is possible to show that the fidelity of LQSM  satisfies the inequality n for any n2 and any state ρ which is separable between AB and C.

Appendix B. Fully separable states not admitting perfect asymptotic PQSM
Here we will present a family of fully separable tripartite states r ABC sep that cannot be merged via PPT assisted LOCC even in the asymptotic scenario. The desired family of states is given by Here, all states s i BC are separable two-qubit states and the particle A has dimension 15 (the reason for this will become clear below). The probabilities p i are strictly positive for all 0i14.
Note that any general d-dimensional Hilbert space has an associated Bloch vector space of dimension d 2 −1 [19]. In the case considered here, the particles B and C are qubits. Thus, the Bloch vector space associated with the Hilbert space of BC has dimension 15. Moreover, note that there exist 15 separable two-qubit states s i BC with the property that all their Bloch vectors are linearly independent. This follows from the fact that the set of separable states has finite size within the set of all states [20].
As we will see in the following, the state in equation (B1) cannot be merged via PPT assisted LOCC whenever the generalized Bloch vectors of the states σ i BC are linearly independent for all 0i14. Due to theorem 1 of the main text it is enough to focus on the single-shot scenario, since a fully separable state admits perfect asymptotic PQSM if and only if it admits perfect PQSM in the single-shot scenario.
Using the above result, we will now prove the desired statement by contradiction. Assume that the state r ABC sep with the above properties can be merged with some single-shot PPT assisted LOCC protocol Λ PPT between Bob and Charlie. It then immediately follows that this protocol must merge each of the states s i BC individually. Recall that the set of states of the form (B2) has finite size within all two-qubit states. By convexity, this implies that the protocol Λ PPT can be used for single-shot merging of any state shared by Bob and Charlie. In particular, this means that Λ PPT can merge both states | ñ 00 BC and |+ ñ 0 BC . The existence of such a protocol would thus imply that the states | ñ 0 and | (| | ) +ñ = ñ + ñ 0 1 2can be perfectly teleported with the aid of PPT states on the single-copy level. This is however impossible [21], which is the desired contradiction. This completes the proof that the aforementioned family of states does not admit perfect asymptotic PQSM.

Appendix C. States with vanishing asymptotic fidelity
Here we will show that all states satisfying the inequality : have zero fidelity in the asymptotic limit: PPT For this we note that for all such states the final state s s = f f ACR is PPT with respect to the bipartition A:CR, and thus is nondistillable with respect to this bipartition 1 . This means that for any number of copies n the fidelity of PQSM is bounded above as follows: , and the supremum in the last expression is taken over all states τ which are not distillable between Alice and Charlie.
In the next step, we introduce the geometric distillability Surprisingly, this is indeed the case for any distillable state ν, and the proof will be given in the following.

Appendix D. Asymptotic geometric distillability
In the following we consider the geometric distillability defined as As we will show in the following, both quantities D g and D t are discrete in the asymptotic limit: asymptotically they attain only the values 0 (if ρ is nondistillable) and 1 (if ρ is distillable). For nondistillable states ρ it is clear that D g and D t are both zero, and thus also zero asymptotically. We will now prove the following equality for any distillable state ρ: