Any multipartite entangled state violating Mermin-Klyshko inequality can be distilled for almost all bipartite splits

Any multipartite entangled state violating Mermin-Klyshko inequality can be distilled for almost all bipartite splits

Soojoon Lee Department of Mathematics and Research Institute for Basic Sciences, Kyung Hee University, Seoul 130-701, Korea    Jinhyoung Lee Department of Physics, Hanyang University, Seoul 133-791, Korea    Jaewan Kim School of Computational Sciences, Korea Institute for Advanced Study, Seoul 130-722, Korea
July 5, 2019
Abstract

We study the explicit relation between violation of Bell inequalities and bipartite distillability of multi-qubit states. It has been shown that even though for there exist -qubit bound entangled states which violates a Bell inequality [Phys. Rev. Lett. 87, 230402 (2001)], for all the states violating the inequality there exists at least one splitting of the parties into two groups such that pure-state entanglement can be distilled [Phys. Rev. Lett. 88, 027901 (2002)]. We here prove that for all -qubit states violating the inequality the number of distillable bipartite splits increases exponentially with , and hence the probability that a randomly chosen bipartite split is distillable approaches one exponentially with , as tends to infinity. We also show that there exists at least one -qubit bound entangled state violating the inequality if and only if .

pacs:
03.67.Mn 03.65.Ud, 42.50.Dv

Entanglement has been considered as a key ingredient for quantum information science, and has brought a lot of its useful applications such as quantum key distribution and teleportation. Nevertheless, there still exist open problems related to entanglement, in particular, multipartite entanglement.

It is known that entanglement can be divided into two kinds of entanglement. One is called the distillable entanglement, from which some pure entanglement can be extracted by local quantum operations and classical communication, and the other is called the bound entanglement, which is not distillable. Since only pure entanglement is directly useful for quantum information processing, the bound entanglement seems to be useless. However, it has been recently shown that any bound entangled (BE) states are useful in quantum teleportation HHH (); Masanes1 (), all multipartite pure entangled states are interconvertible by stochastic local operations and classical communication with the assistance of BE states Ishizaka (), and there are several classes of BE states with a positive key rate in quantum key distribution HHHO1 (); HHHO2 (); HPHH (); CCKKL (); HA (). Thus, it is necessary to analyze BE states more profoundly.

If one of the two most significant features related to entanglement is distillability, then the other is nonlocality, which can be described as a physical property to explain that quantum correlation is quite different from classical correlations. Nonlocality can be seen from violation of some conditions, called Bell inequalities, that are satisfied by any local variable theory, and it is a well-known fact that any bipartite or multipartite pure state violates a Bell inequality if and only if the state is entangled Gisin91 (); PR92 (). However, for mixed states, there does not exist such a simple relation between nonlocality and entanglement. Since Werner Werner () found the existence of entangled mixed states described by a local hidden variable model, it has been known that some of these states can violate Bell inequalities after appropriately preprocessing the state Popescu (); Gisin96 ().

There is a simple relation between nonlocality and distillability in fewer-qubit systems: If any two-qubit HHH97 () or three-qubit LJK () (pure or mixed) state violates a specific form of the Bell inequality then it is distillable. However, Dür Dur () has shown that for there exist -qubit BE states which violate a Bell inequality. This result seems to show that nonlocality does not directly imply distillability in multipartite cases, even though it has been recently shown that asymptotic violation of a Bell inequality is equivalent to distillability in any multipartite quantum system Masanes ().

But, Acín Acin () has demonstrated that for all the states violating the inequality there exists at least one splitting of the parties into two groups such that pure-state entanglement can be distilled, and has more analyzed the relation of nonlocality to bipartite distillability in his subsequent works Acin_etal (). This does not only imply that there still exists a relation between nonlocality and distillability for a certain bipartite split, but also tells us that it is possible to make two-party quantum communications with respect to the bipartite split secure against eavesdropping. Then some questions naturally arise such as which bipartite split is distillable and how many splits are possible to be distillable if the Bell inequality is violated.

Assume that a multipartite entangled state violates the Bell inequality. If it could be distilled for almost all bipartite splits, then it would be possible for almost all two-party quantum communications over the multipartite state to be secure against eavesdropping, regardless of how it is divided into two parties. Thus, it would be important to answer the questions in quantum communication theory as well as in entanglement theory.

In this paper, we show that if any -qubit state violates the inequality then there exist much more than one distillable bipartite splits, to be exact, at least distillable bipartite splits. Hence, the distillation probability that a randomly chosen bipartite split is distillable approaches one exponentially with as tends to infinity. This means that if a given -qubit state violates the Bell inequality for sufficiently large then almost all bipartite splits are distillable. Furthermore, this result provides us with the following necessary and sufficient condition for the existence of -qubit BE states violating the inequality: At least one -qubit BE state violates the inequality if and only if .

Since it has been already known that there exists a four-qubit BE state, the so-called Smolin state Smolin (), violating some other Bell inequality AH (), our condition does not seem to be very strong. However, because our proof is based on the first main result counting distillable bipartite splits, this justifies some significance of considering the counting problem.

In order to introduce our main results, we first consider the family of -qubit states presented in DCT (); DC (),

(1)

where

(2)

and . We remark that any arbitrary -qubit state can be depolarized to a state in this family, and hence this family can be useful to find sufficient conditions for nonseparability and distillability in -qubit systems DCT (). Thus, this family may be regarded as a generalization of Werner states to multiqubit systems.

We prove our first main result in the following way: (i) We assume that any -qubit state violates a specific form of Bell inequality. (ii) By some appropriate depolarizing process, the state can be transformed into , which also violates the same inequality. (iii) We show that the state violating the inequality has at least distillable bipartite splits. (iv) We conclude that the state also has at least distillable bipartite splits. In order to prove the main result, we need the following proposition and lemma.

For each ()-bit string , let be the bipartite split such that if and only if party belongs to the same set as the last party. Then the following proposition about bipartite distillability of the states has been known by Dür and Cirac DC ().

Proposition 1.

is distillable for the bipartite split if and only if .

We note that the quantity in Proposition 1 plays an important role in not only bipartite distillability but also a certain form of Bell inequality, which we will crucially use in this paper.

From Proposition 1, we can obtain the following key lemma for our first main result.

Lemma 2.

If

(3)

then there exist at least distillable bipartite splits in .

Proof.

Let be the number of distillable bipartite splits, . Suppose that . Then we readily obtain the following inequality:

It follows that

(5)

The inequality (5) leads to a contradiction. Therefore, we can conclude that . ∎

The Bell inequality that Dür and Acín have considered is called the Mermin-Klyshko (MK) inequality Mermin (); BK (), which generalizes the Clauser-Horne-Shimony-Holt inequality CHSH () into -qubit cases. Let be the Bell operator defined recursively as

(6)

where and are the two dichotomic observables measured on each particle , is obtained from by exchanging all the and , and . Then the MK inequality is as follows:

(7)

Choosing the same measurement directions in all locations, and for all , after local phase redefinition Acin (), can be written as

(8)

We note that, by the depolarizing process in DCT (), any -qubit state can be transformed into one in the family of with and . Thus, for the Bell operator in Eq. (8), we obtain the following equalities:

(9)

and hence we have the following theorem by Lemma 2.

Theorem 3.

For all the -qubit states violating the MK inequality with respect to the Bell operator (8), there exist at least distillable bipartite splits.

Let be the probability that a randomly chosen bipartite split on an -qubit state is distillable, when it violates the MK inequality with respect to the Bell operator (8). Then it follows from Theorem 3 that

(10)

This implies that approaches one exponentially with as tends to infinity as seen in FIG. 1.

Figure 1: The distillation probability that a randomly chosen bipartite split on an -qubit state is distillable, when it violates the MK inequality with respect to the Bell operator (8).

Interestingly, Theorem 3 provides us with a necessary and sufficient condition for the existence of -qubit BE states violating the MK inequality with respect to the Bell operator (8). In order to show the condition, we begin with reminding the following proposition about a relation between distillability and negative partial transposition (NPT), which has been shown by Dür and Cirac DCT ().

Proposition 4.

A maximally entangled pair between particles and can be distilled from if and only if all possible bipartite splits of where the particles and belong to different parties, have NPT.

By Theorem 3 and Proposition 4, we can prove the following theorem.

Theorem 5.

There exists at least one -qubit BE state violating the MK inequality with respect to the Bell operator (8) if and only if .

Proof.

We note that the number of total bipartite splits is , and that the number of all distillable bipartite splits is at least by Theorem 3.

We first assume that , that is, , , or .

(Case 1) ; It follows from Theorem 3 that all bipartite splits are distillable, and so have NPT. By Proposition 4, a maximally entangled state can be distilled between any particles and .

(Case 2) ; Since and , we obtain that all bipartite splits are distillable or there is only one non-distillable bipartite split. Hence, there is at least one pair and such that all bipartite splits whose two different parties contain the particles and respectively are distillable. As in the Case 1, since they have NPT, a maximally entangled pair can be distilled between the particles and .

(Case 3) ; Since and , we obtain that all bipartite splits are distillable, or there exist at most two non-distillable bipartite splits. Hence, there is at least one pair and between which a maximally entangled pair can be distilled by the same reason as the Case 2.

Conversely, if then there exists an -qubit BE state violating the MK inequality as follows: Take , , and if and otherwise. Under these conditions, the state becomes,

where . Then since if , the state violates the MK inequality with respect to the Bell operator (8).

Figure 2: Undistillable bipartite splits of in (LABEL:eq:BErho_N) when .

Furthermore, since if , by Proposition 1, the state is not distillable for the bipartite splits for .

As seen in FIG. 2, if two different particles and in the state are given then or is a bipartite split where the two particles belong to different parties, and neither nor is bipartite distillable, and hence a maximally entangled state between the particles and cannot be distilled. Since and are arbitrary, the state is not distillable, that is, it is BE since it is inseparable. Therefore, there exists an -qubit BE state violating the MK inequality if . ∎

As seen in Theorem 5, for , there exists no -qubit BE state that violates the inequality. Hence we can say that if then violation of the inequality implies distillability.

In conclusion, we have studied the explicit relation between violation of Bell inequalities and bipartite distillability of multi-qubit states, and have shown that if any -qubit state violates the MK inequality then there exist at least distillable bipartite splits. Hence, the probability that a randomly chosen bipartite split is distillable approaches one exponentially with as tends to infinity. We have also shown that an -qubit BE state violates the inequality if and only if .

It has been shown that while -qubit states in a class of BE states presented in Dur (); Acin () violate the MK inequality for , the states in the class violate different forms of Bell inequalities for in Ref. KKCO () and for in Ref. SSZ (). Furthermore, it has been also shown that there exists a four-qubit BE state which can maximally violate a certain form of Bell inequality AH (). Therefore, our results could be also improved by using Bell inequalities different from the MK inequality, and could be furthermore generalized to multipartite distillability.

S.L. was supported by the Korea Research Foundation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF-2007-331-C00049), and J.K. was partially supported by the IT R&D program of MKE/IITA (2005-Y-001-04, Development of Next Generation Security Technology).

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